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Sometimes more complicated heteroatomic functional groups, containing bonds to more than one heteroatom, have slightly different spectra. Carboxylic acids feature a hydroxyl group bonded to a carbonyl. Hexanoic acid, a carboxylic acid in a six-atom chain, is one example. If you look at the IR spectrum of hexanoic acid: At first, the O-H peak appears to be absent. The C-H stretch appears to be very broad. The wide peak between 3000 and 2600 cm is really the usual C-H stretch with a broad O-H stretch superimposed on it. The low frequency vibration of this O-H bond is related to the partial dissociation of protons due to strong hydrogen bonding. ,

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Once we know what energy a given transition would have, we can ask, “Which transitions between energy levels or states are possible?” In answering this question, we also will learn why the longer cyanine dye molecules have stronger absorptions, or larger absorption coefficients. Clearly the transitions cannot violate the Pauli ; that is, they cannot produce an electron configuration with three electrons in the same orbital. Besides the Pauli Exclusion Principle, there are additional restrictions that result from the nature of the interaction between electromagnetic radiation and matter. These restrictions are summarized by spectroscopic selection rules. These rules tell whether or not a transition from one state to another state is possible. To obtain these selection rules, we consider light as consisting of perpendicular oscillating electric and magnetic fields. The magnetic field interacts with magnetic moments and causes transitions seen in electron spin resonance and nuclear magnetic resonance spectroscopies. The oscillating electric field interacts with electrical charges, i.e. the positive nuclei and negative electrons that comprise an atom or molecule, and cause the transitions seen in UV-Visible, atomic absorption, and fluorescence spectroscopies. The energy of interaction, \(E\), between a system of charged particles and an electric field \(\epsilon\) is given by the scalar product of the electric field and the dipole moment, \(μ\), for the system. Both of these quantities are vectors. \[E = - \mu \cdot \epsilon \label {4-20}\] The dipole moment is defined as the summation of the product of the charge \(q_j\) times the position vector \(r_j\) for all charged particles \(j\). \[ \mu = \sum _j q_j r_j \label {4-21}\] Calculate the dipole moment of HCl from the following information. The position vectors below use Cartesian coordinates (x, y, z), and the units are pm. What fraction of an electronic charge has been transferred from the chlorine atom to the hydrogen atom in this molecule? \(r_H = (124.0, 0, 0)\), \(r_{Cl} = (- 3.5, 0, 0)\), \(q_H = 2.70 x 10^{-20} C, q_{Cl} = - 2.70 x 10^{-20} C\). Sketch a diagram for Exercise \(\Page {1}\) showing the coordinate system, the \(\ce{HCl}\) molecule and the dipole moment. To calculate an expectation value for this interaction energy, we need to evaluate the expectation value integral. \[ \left \langle E \right \rangle = \int \psi ^*_n ( - \hat {\mu} \cdot \hat {\epsilon} ) \psi _n d \tau \label {4-22}\] The \(\int d \tau \) symbol simply means integrate over all coordinates. The operators \(\hat {\mu}\) and \(\hat {\epsilon}\) are vectors and are the same as the classical quantities, μ and \(\epsilon\). Usually the wavelength of light used in electronic spectroscopy is very long compared to the length of a molecule. For example, the wavelength of green light is 550 nm, which is much larger than molecules, which are closer to 1 nm in size. The magnitude of electric field then is essentially constant over the length of the molecule, and \(\epsilon\) can be removed from the integration since it is constant wherever ψ is not zero. In other words, ψ is finite only over the volume of the molecule, and the electric field is constant over the volume of the molecule. What remains for the integral is the expectation value for the permanent dipole moment of the molecule in state \(n\), namely, \[ \left \langle \mu \right \rangle= \int \psi ^*_n \hat {\mu} \psi _n d \tau \label {4-23}\] so \[ E = - \left \langle \mu \right \rangle \cdot \epsilon \label {4-24}\] Verify that the vectors in the scalar product in Equation \(\ref{4-24}\) commute by expanding \(\mu \cdot \epsilon\) and \(\epsilon \cdot \mu\). Use particle-in-a-box wave functions with HCl charges and coordinates. Equation \(\ref{4-24}\) shows that the strength or energy of the interaction between a charge distribution and an electric field depends on the dipole moment of the charge distribution. To obtain the strength of the interaction that causes transitions between states, the transition dipole moment is used rather than the dipole moment. The transition dipole moment integral is very similar to the dipole moment integral in Equation \(\ref{4-23}\) except the two wavefunctions are different, one for each of the states involved in the transition. Two different states are involved in the integral because the transition dipole moment integral has to do with the magnitude of the interaction with the electric field that causes a transition between the two states. For a transition where the state changes from ψi to ψf, the transition dipole moment integral is \[ \left \langle \mu \right \rangle _T = \int \psi ^*_f \hat {\mu} \psi _i d \tau = \mu _T \label {4-25}\] Just like the probability density is given by the absolute square of the wavefunction, the probability for a transition as measured by the absorption coefficient is proportional to the absolute square \(\mu ^*_T \mu _T\) of the transition dipole moment, which is calculated using Equation \(\ref{4-25}\). Since taking the absolute square always produces a positive quantity, it does not matter whether the transition moment itself is positive, negative, or imaginary. The transition dipole moment integral and its relationship to the absorption coefficient and transition probability can be derived from the time-dependent Schrödinger equation. Here we only want to introduce the concept of the transition dipole moment and use it to obtain selection rules and relative transition probabilities for the particle-in-a-box. Later it will be applied to other systems that we will be considering. If \(μ_T = 0\) then the interaction energy is zero and no transition occurs or is possible between states characterized by \(ψ_i\) and \(ψ_f\). Such a transition is said to be forbidden, or more precisely, electric-dipole forbidden. In fact, the electric-dipole electric-field interaction is only the leading term in a of the interaction energy, but the higher order terms in this expansion usually are not significant. If \(μ_T\) is large, then the probability for a transition and the absorption coefficient are large. It is very useful to be able to tell whether a transition is possible, \(μ_T ≠ 0\), or not possible, \(μ_T = 0\), without having to evaluate integrals. Properties of the wavefunctions such as symmetry or angular momentum can be used to determine the conditions that must exist for the transition dipole moment to be finite, i.e. not zero. Statements called spectroscopic selection rules summarize these conditions. Selection rules do not tell us how probable or intense a transition is. They only tell us whether a transition is possible or not possible. For the particle-in-a-box model, as applied to dye molecules and other appropriate molecular systems, we need to consider the transition moment integral for one electron. According to Equation \(\ref{4-21}\), the dipole moment operator for an electron in one dimension is –ex since the charge is –e and the electron is located at x. \[\mu _T = - e \int \limits _0^L \psi ^*_f (x) x \psi _i (x) dx \label {4-26}\] This is the integral that must be evaluated for various particle-in-a-box wavefunctions to see which transitions are allowed (i.e. \(μ_T ≠ 0\)) and forbidden (\(μ_T = 0\)) and to determine the relative strengths of the allowed transitions.

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Addition of an unsymmetrical substance such as \(\ce{HX}\) to an unsymmetrical alkene theoretically can give two products: and Both products are seldom formed in equal amounts; in fact, one isomer usually is formed to the exclusion of the other. For example, the hydration of propene gives 2-propanol (not 1-propanol), and hydrogen chloride adds to 2-methylpropene to give -butyl chloride (not isobutyl chloride): To understand the reason for the pronounced selectivity in the orientation of addition of electrophiles, it will help to consider one example, hydrogen bromide addition to 2-methylpropene. Two different carbocation intermediates could be formed by attachment of a proton to one or the other of the double bond carbons: Subsequent reactions of the cations with bromide ion give -butyl bromide and isobutyl bromide. In the usual way of running these additions, the product is very pure -butyl bromide. How could be have predicted which product would be favored? The first step is to decide whether the prediction is to be based on (1) which of the two products is the , or (2) which of the two products if formed . If we make a decision on the basis of product stabilities, we take into account \(\Delta H^0\) values, entropy effects, and so on, to estimate the equilibrium constants \(K_\text{eq}\) for the reactants and each product. When the ratio of the products is determined by the ratio of their equilibrium constants, we say the overall reaction is subject to (or ) . Equilibrium control requires that the reaction be . When a reaction is carried out under conditions in which it is , the ratio of the products is determined by the relative rates of formation of the various products. Such reactions are said to be under . The products obtained in a reaction subject to kinetic control are not necessarily the same as those obtained under equilibrium control. Indeed, the equilibrium constant for interconversion of -butyl bromide and isobutyl bromide at \(25^\text{o}\) is 4.5, and if the addition of hydrogen bromide to 2-methylpropene were under equilibrium control, the products would be formed in this ratio: \[K_\text{eq} = \dfrac{\left[ \text{tert-butyl bromide} \right]}{\left[ \text{isobutyl bromide} \right]} = 4.5\] But the addition product is \(99+\%\) -butyl bromide so the reaction clearly is kinetically controlled, -butyl being formed considerably faster than isobutyl bromide. . So to account for the formation of -butyl bromide we have to consider why the -butyl cation is formed more rapidly than the isobutyl cation: As we have seen in Section 8-7B, alkyl groups are more electron donating than hydrogen. This means that the more alkyl groups there are on the positive carbon of the cation, the more stable and the more easily formed the cation will be. The reason is that electron-donating groups can partially compensate for the electron deficiency of the positive carbon. As a result, we can predict that the -butyl cation with three alkyl groups attached to the positive center will be formed more readily than the primary isobutyl cation with one alkyl group attached to the positive center. Thus the problem of predicting which of the two possible products will be favored in the addition of unsymmetrical reagents to alkenes under kinetic control reduces to predicting which of two possible carbocation intermediates will be formed most readily. With simple alkenes, we shall expect the preference of formation of the carbocations to be in the order: > > . The reaction scheme can be represented conveniently in the form of an energy diagram (Figure 10-10). The activation energy, \(\Delta H^1_\text{tert}\) for the formation of the -butyl cation is less than \(\Delta H^1_\text{prim}\) for the formation of the isobutyl cation because the tertiary ion is much more stable (relative to the reactants) than the primary ion, and therefore is formed at the faster rate. The second step, to form the product from the intermediate cation, is very rapid and requires little activation energy. Provided that the reaction is , it will take the lowest-energy path and form exclusively -butyl bromide. However, if the reaction mixture is allowed to stand for a long time, isobutyl bromide begins to form. Over a long period, the products equilibrate and, at equilibrium, the product distribution reflects the relative stabilities of the rather than the stability of the transition states for formation of the intermediates. A rather simple rule, formulated in 1870 and known as , correlates the direction of additions of \(\ce{HX}\) to unsymmetrical alkenes. This rule, an important early generalization of organic reactions, may be stated as follows: . It should be clear that Markownikoff's rule predicts that addition of hydrogen bromide to 2-methylpropene will give -butyl bromide. We can extend Markownikoff's rule to cover additions of substances of the general type \(\ce{X-Y}\) to unsymmetrically substituted alkenes when a clear-cut decision is possible as to whether \(\ce{X}\) or \(\ce{Y}\) is the more electrophilic atom of \(\ce{X-Y}\). If the polarization of the \(\ce{X-Y}\) bond is such that \(\ce{X}\) is positive, \(^{\delta \oplus} \ce{X-Y} ^{\delta \ominus}\), then \(\ce{X}\) will be expected to add as \(\ce{X}^\oplus\) to the alkene to form the more stable carbocation. This step will determine the direction of addition. For example, if we know that the \(\ce{O-Br}\) bond of \(\ce{HOBr}\) is polarized as \(\overset{\delta \ominus}{\ce{HO}} - \overset{\delta \oplus}{\ce{Br}}\), then we can predict that addition of \(\ce{HOBr}\) to 2-methylpropene will give 1-bromo-2-methyl-2-propanol: Pauling's value for the electronegativity of carbon makes it slightly more electron-attracting than hydrogen. However, we expect that the electron-attracting power of a carbon atom (or of other elements) will depend also on the electronegativities of the groups to which it is attached. In fact, many experimental observations indicate that carbon in methyl or other alkyl groups is significantly electron-attracting than hydrogen. Conversely, the \(\ce{CF_3}-\) group is, as expected, far electron-attracting than hydrogen. The direction of polarization of bonds between various elements may be predicted from Figure 10-11. For example, an \(\ce{O-Cl}\) bond should be polarized so the oxygen is negative; a \(\ce{C-N}\) bond should be polarized so the nitrogen is negative: \[\overset{\delta \ominus}{\ce{O}}---\overset{\delta \oplus}{\ce{Cl}} \: \: \: \: \: \overset{\delta \oplus}{\ce{C}}---\overset{\delta \ominus}{\ce{N}}\] We then can predict that, in the addition of \(\ce{HOCl}\) to an alkene, the chlorine will add preferentially to form the more stable of two possible carbon cations. Generally, this means that chlorine will bond to the carbon carrying the greater number of hydrogens: A number of reagents that are useful sources of electrophilic halogen are included in Table 10-2. Some of these reagents, notably those with \(\ce{O}-\)halogen or \(\ce{N}-\)halogen bonds, actually are sources of hypohalous acids, \(\ce{HOX}\), and function to introduce halogen and hydroxyl groups at carbon. There are very few good fluorinating agents whereby the fluorine is added as \(\ce{F}^\oplus\). For alkenes that have halogen or similar substituents at the doubly bonded carbons, the same principles apply as with the simple alkenes. That is, under kinetic control the preferred product will be the one derived from the more stable of the two possible intermediate carbon cations. Consider a compound of the type \(\ce{Y-CH=CH_2}\). If \(\ce{Y}\) is electron-attracting than hydrogen, then hydrogen halide should add in such a way as to put the proton of \(\ce{HX}\) on the \(\ce{YCH=}\) end and \(\ce{X}\) on the \(\ce{=CH_2}\) end. The reason is that the positive carbon is expected to be more favorably located if it is not attached directly to an electron-attracting substituent: The addition goes as predicted, . For example, \[\ce{CF_3-CH=CH_2} + \ce{HCl} \rightarrow \ce{CF_3-CH_2-H_2-Cl}\] Such substituents are relatively uncommon, and most of the reported \(\ce{H-X}\) additions have been carried out with \(\ce{Y}\) groups having unshared electron pairs on an atom connected directly to a carbon of the double bond: These substituents usually are strongly electronegative relative to hydrogen, and this often causes diminished reactivity of the double bond toward electrophiles. Nonetheless, : The electron-attracting power of the substituent is more than counterbalanced by stabilization of the intermediate cation by the ability of the substituents to delocalize their electrons to the adjacent positive carbon (see ). and (1977)

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.11%3A_Pericyclic_Reactions

There are numerous reactions in organic chemistry that proceed through cyclic transition states. They may be classified generally as reactions. An important and familiar example is the Diels-Alder reaction, in which a conjugated diene cycloadds to an alkene or alkyne: This reaction has been described previously ( ) and is an example of a [4 + 2] cycloaddition. Such reactions occur thermally (by simply heating the reactants) and appear to be entirely concerted. By this we mean that the reactants are converted to products , without involving the formation of reaction intermediates. The principal evidence for the concertedness of [4 + 2] cycloadditions is the fact that they are highly stereospecific and involve suprafacial addition of both components. The configuration of substituents in the diene and the dienophile is retained in the adduct: In contrast to the [4 + 2] cycloaddition, thermal [2 + 2] cycloadditions seldom are observed, and when they are observed, they are not stereospecific and evidently are stepwise reactions (see ): Why are the [4 + 2] and [2 + 2] cycloadditions different? Simple molecular orbital theory provides an elegant explanation of this difference based on the \(4n + 2\) rule described in . To understand this, we need to look in more detail at how the \(p\) orbitals of the double bonds interact in concerted addition mechanisms by suprafacial overlap, as in \(36\) and \(37\): There is a way around the \(4n + 2\) rule that is not very important for substances analogous to benzene, but is quite important for cycloaddition reactions. Let us see how this works for a cyclic conjugated polyene. From the molecular-orbital diagrams of Figures 21-5, 21-7, 21-9, and 21-14, you will see that the -energy \(\pi\) molecular orbital has nodes (changes of phase). A model of such an orbital, which usually is called a , can be constructed by joining the ends of a ribbon or strip of parallel \(p\) orbitals, as represented on the left side of Figure 21-15. However, one could join the orbitals by making in the strip, which then would give a lowest-energy orbital with one node, as on the right side of Figure 21-15. A strip with one such twist is called a \(^8\) and has the topological property of having only one side. If we now calculate the orbital energies for the Möbius orbitals, as was down for the normal Hückel \(\pi\) orbitals in Figure 21-13, we get the results shown in Figure 21-16. From this, we see that the \(4n\) situation now is favored and \(4n + 2\) is unfavorable. Whereas the energies of the \(\pi\) molecular orbitals in the Hückel arrangement can be obtained by inscribing a polygon in a circle with a ( ), in the Möbius arrangement the orbital energies are obtained from the polygon inscribed with a . If you compare the orbital energies of the Hückel and Möbius cyclic \(\pi\) systems (Figures 21-13 and 21-16), you will see that the Hückel systems have only lowest-energy MO, whereas the Möbius systems have . Hückel systems have an odd number of bonding orbitals (which, when full, accommodate 2, 6, 10, 14, or \(4n + 2\) electrons) and the Möbius systems have an even number of bonding orbitals (which, when full, accommodate 4, 8, 12, or \(4n\) electrons). The Hückel molecular orbitals have or an of nodes (see, for example, the benzene MOs, Figure 21-5); the Möbius molecular orbitals are not shown, but they have or an of nodes. The relevance of all this may seem tenuous, especially because no example of a simple cyclic polyene with a Möbius \(\pi\) system is known. However, the Möbius arrangement is relevant to cycloaddition because we can conceive of alkenes, alkadienes, and so on approaching each other to produce Möbius transition states when \(4n\) electrons are involved. For example, consider two molecules of ethene, which we showed previously would violate the \(4n + 2\) rule by undergoing cycloaddition through a transition state represented by \(37\). There is an alternative transition state, \(38\), in which the four \(p\) orbitals come together in the Möbius arrangement (with one node for minimum energy). To achieve this arrangement the ethene molecules approach each other in roughly perpendicular planes so that the \(p\) orbitals overlap suprafacially in one ethene and antarafacially in the other, as shown in \(38\): This pathway is electronically favorable, but the steric interference between the groups attached to the double bond is likely to be severe. Such repulsions can be relieved if there are no groups sticking out sidewise at one end of the double bond, as with the central carbon of 1,2-propadiene, \(\ce{CH_2=C=CH_2}\), and ketene, \(\ce{CH_2=C=O}\). These substances often undergo [2 + 2] cycloadditions rather readily ( ), and it is likely that these are concerted additions occurring by the Möbius route. A much less strained Möbius [4 + 4] transition state can be formed from two molecules of 1,3-butadiene. When 1,3-butadiene is heated by itself, a few percent of 1,5-cyclooctadiene is formed, but it is not known for sure whether the mechanism is that shown: The principal reaction is a Diels-Alder [4 + 2] cycloaddition, with butadiene acting both as a diene and as a dienophile: Much of what we have said about the electronic factors controlling whether a cycloaddition reaction can be concerted or not originally was formulated by the American chemists R. B. Woodward and R. Hoffmann several years ago, in terms of what came to be called the principles, or the . Orbital symmetry arguments are too complicated for this book, and we shall, instead, use the \(4n + 2\) electron rule for normal Hückel arrangements of \(\pi\) systems and the \(4n\) electron rule for Möbius arrangements. This is a particularly simple approach among several available to account for the phenomena to which Woodward and Hoffmann drew special attention and explained by what they call "conservation of orbital symmetry". The cycloaddition reactions that we have discussed so far in this chapter ([2 + 2], [4 + 2], etc.) have involved ring formation by bringing two unsaturated molecules together. Thus [4 + 2] addition is represented by the Diels-Alder reaction of ethene and 1,3-butadiene: We can conceive of similar cyclizations involving only single molecules, that is, . Such reactions are called . Two examples follow to show cyclization of a diene and a triene: Cyclization of 1,3,5-hexatriene occurs only when the central double bond has the cis configuration. The reaction is reversible at elevated temperatures because of the gain in entropy on ring opening (see ). The cyclobutene-1,3-butadiene interconversion proceeds much less readily, even in the thermodynamically favorable direction of ring opening. However, substituted dienes and cyclobutenes often react more rapidly. A related group of reactions involves shifts of substituent groups from one atom to another; for example, with \(\ce{H}\), alkyl, or aryl groups as \(\ce{R}\): These reactions are called and, in general, they are subject to the \(4n + 2\) rule and the Möbius orbital modification of it. Potential sigmatropic rearrangements can be recognized by the fact that the single bond to the migrating group \(\left( \ce{R} \right)\) is "conjugated" with the \(\pi\) bonds, and the group moves from a saturated \(sp^3\) to an \(sp^2\) carbon at a different part of the \(\pi\) system. A striking feature of thermal electrocyclic reactions that proceed by concerted mechanisms is their high degree of stereospecificity. Thus when -3,4-dimethylcyclobutene is heated, it affords only one of the three possible cis-trans isomers of 2,4-hexadiene, namely, , -2,4-hexadiene: We can see how this can occur if, as the ring opens, the ends of the diene twist in the direction (\(\curvearrowright \curvearrowright\) or \(\curvearrowleft \curvearrowleft\), ) as indicated in the equation. You will notice that with this particular case, if conrotation occurs to the left, rather than the right, the same final product results: The conrotatory movement of groups is typical of thermal ring openings of cyclobutenes and other rings involving \(4n\) electrons. When a cyclobutene is so constituted that conrotation cannot occur for steric reasons, then the concerted reaction cannot occur easily. Substances that otherwise might be predicted to be highly unstable often turn out to be relatively stable. An example is bicyclo[2.1.0]-2-pentene, which at first sight might seem incapable of isolation because of the possibility of immediate arrangement to 1,3-cyclopentadiene. This rearrangement does occur, but not so fast as to preclude isolation of the substance: How can we explain the fact that this substance can be isolated? The explanation is that, if the reaction has to be conrotatory, then the product will not be ordinary 1,3-cyclopentadiene, but , -1,3-cyclopentadiene - surely a very highly strained substance. (Try to make a ball-and-stick model of it!) This means that the concerted mechanism is not favorable: It is of great interest and importance that, with systems of \(4n + 2\) electrons, the groups move in directions (\(\curvearrowleft \curvearrowright\) or \(\curvearrowright \curvearrowleft\), ). For example, In this case, the disrotation of the groups one another would lead to the cis,cis,cis product. Because this product is not formed, it seems likely that rotation of the methyl groups toward each other must be sterically unfavorable: How can we account for the stereoselectivity of thermal electrocyclic reactions? Our problem is to understand why it is that concerted \(4n\) electrocyclic rearrangements are conrotatory, whereas the corresponding \(4n + 2\) processes are disrotatory. From what has been said previously, we can expect that the conrotatory processes are related to the Möbius molecular orbitals and the disrotatory processes are related to Hückel molecular orbitals. Let us see why this is so. Consider the electrocyclic interconversion of a 1,3-diene and a cyclobutene. In this case, the Hückel transition state ( ) is formed by , but is unfavorable with four (that is, \(4n\)) electrons: In contrast, the Möbius transition state ( ) is formed by and is favorable with four \(\left( 4n \right)\) electrons: You will notice that the ring closure of a 1,3-diene through the favorable Möbius transition state may appear to be able to form only an arrangement of the overlapping \(\sigma\) orbitals, which would correspond to a high-energy cyclobutene. In fact, the normal cyclobutene would be formed, because on the way down from the transition state, the phases of the orbitals that will become the \(\sigma\) bond change to give the arrangement of the \(\sigma\) orbitals expected for the ground state. The reverse occurs in ring opening so that this reaction also can go through the favorable Möbius transition state. The same reasoning can be extended to electrocyclic reactions of 1,3,5-trienes and 1,3-cyclohexadienes, which involve \(4n + 2\) electrons and consequently favor Hückel transition states attained by . The three principal types of pericyclic reactions are , , and : The factors that control if and how these cyclization and rearrangement reactions occur in a concerted manner can be understood from the aromaticity or lack of aromaticity achieved in their transition states. For a concerted pericyclic reaction to be thermally favorable, the transition state must involve \(4n + 2\) participating electrons if it is a Hückel orbital system, or \(4n\) electrons if it is a Möbius orbital system. A Hückel transition state is one in which the cyclic array of participating orbitals has no nodes (or an even number) and a Möbius transition state has an odd number of nodes. We summarize here a procedure to predict the feasibility and the stereochemistry of reactions involving . The 1,2 rearrangement of carbocations will be used to illustrate the approach. This is a very important reaction of carbocations which we have discussed in other chapters. We use it here as an example to illustrate how qualitative MO theory can give insight into how and why reactions occur: The first step of the procedure is to draw the orbitals as they are expected to be involved in the transition state. There may be several possible arrangements. There are two such arrangements, \(41\) and \(42\), for the rearrangement of carbocations; the dotted lines show the regions of bond-making and bond-breaking (i.e., orbital overlap): The second step is to determine whether the transition states are Hückel or Möbius from the number of nodes. This is readily done by assigning signs to the lobes of the orbitals corresponding to their phases and counting the number of nodes that develop in the circle of overlapping orbitals. An odd number denotes a Möbius transition state, whereas an even number, including zero, denotes a Hückel transition state. There are alternative ways of node-counting for transition states \(41\) and \(42\). Diagrams \(43abc\) and \(44abc\) represent molecular orbitals of different energies - those with more nodes having the higher energies (cf. ).\(^9\) We show these diagrams with more than one node for the sake of completeness. It is not necessary to draw more than one such diagram to determine whether the transition state is Möbius or Hückel. Finally, we evaluate the transition states according to the \(4n\) or \(4n + 2\) rule. In the example here, because only two electrons occupy the molecular orbitals, the Hückel transition state (\(43a\)) is the favorable one. A bonus coming from these formulations is that the stereochemistry of the reaction can be predicted when we have predicted which transition state is the favored one. Thus the migrating group in 1,2-carbocation rearrangements should move with of configuration by a Hückel transition state - and this has been verified experimentally. The alternative Möbius transition state predicts of the configuration of the migrating group: You can use the procedures just outlined to determine whether any thermal reaction with a cyclic transition state is likely to be favorable. A good place to start is the Diels-Alder [4 + 2] cycloaddition, which proceeds thermally by a suprafacial (Hückel) transition state. We suggest that you apply the procedure to the Diels-Alder reaction of 1,3-butadiene and ethene, and following that, show the electrocyclic ring opening of a cyclobutene ring to be thermally favorable only by a conrotatory opening of the \(\ce{C-C}\) bond. Many pericyclic reactions take place photochemically, that is, by irradiation with ultraviolet light. One example is the conversion of norbornadiene to quadricyclene, described in . This reaction would have an unfavorable [2 + 2] mechanism if it were attempted by simple heating. Furthermore, the thermodynamics favor ring opening rather than ring closure. However, quadricyclene can be isolated, even if it is highly strained, because to reopen the ring thermally involves the reverse of some unfavorable [2 + 2] cycloaddition mechanism. Photochemical activation can be used to achieve forward or reverse cycloadditions and electrocyclic reactions that are thermodynamically unfavorable or have unfavorable concerted thermal mechanisms. Thus the thermodynamically unstable disrotatory [2 + 2] product can be obtained from 1,3-cyclopentadiene by irradiation with ultraviolet light: The stereochemical results of electrocyclic and cycloaddition reactions carried out photochemically often are opposite to what is observed for corresponding thermal reactions. However, exceptions are known and the degree of stereospecificity is not always as high as in the thermal reactions. Further examples of photochemical pericyclic reactions are given in . \(^8\)Named after the mathematician A. F. Möbius. \(^9\)The assignment of orbital phases must take appropriate account of molecular symmetry, and although this is easy for open-chain systems, it is much less straightforward for cyclic ones. You usually will be able to avoid this problem by always trying to set up the orbitals so that the transition state will have no nodes, or just one node at a point where a bond is being made or broken. and (1977)

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A hydrogen bond is an intermolecular force (IMF) that forms a special type of dipole-dipole attraction when a hydrogen atom bonded to a strongly electronegative atom exists in the vicinity of another electronegative atom with a lone pair of electrons. Intermolecular forces (IMFs) occur between molecules. Other examples include ordinary dipole-dipole interactions and dispersion forces. Hydrogen bonds are are generally stronger than ordinary dipole-dipole and dispersion forces, but weaker than true covalent and ionic bonds. Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the elements with hydrogen, you find that the boiling points increase as you go down the group. The increase in boiling point happens because the molecules are getting larger with more electrons, and so dispersion forces become greater. If you repeat this exercise with the compounds of the elements in , and with hydrogen, something odd happens. Although the same reasoning applies for group 4 of the periodic table, the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of \(NH_3\), \(H_2O\) and \(HF\) there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break the IMFs. These relatively powerful intermolecular forces are described as hydrogen bonds. The molecules capable of hydrogen bonding include the following: Notice that in each of these molecules: If you are not familiar with , you should follow this link before you go on. Consider two water molecules coming close together. The \(\delta^+\) hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a . It doesn't go that far, but the attraction is significantly stronger than an ordinary . Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status. Water is an ideal example of hydrogen bonding. Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules: two with the hydrogen atoms and two with the with the oxygen atoms. There are exactly the right numbers of \(\delta^+\) hydrogens and lone pairs for every one of them to be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there are not enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, two hydrogen bonds and two lone pairs allow formation of hydrogen bond interactions in a lattice of water molecules. Water is thus considered an ideal hydrogen bonded system. When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called . Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds. If you are interested in the bonding in hydrated positive ions, you could follow this link to . The diagram shows the potential hydrogen bonds formed with a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and would not normally be active enough to form hydrogen bonds, they are made more attractive by the full negative charge on the chlorine in this case. However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to. An alcohol is an organic molecule containing an -OH group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Hydrogen bonds also occur when hydrogen is bonded to fluorine, but the HF group does not appear in other molecules. Molecules with hydrogen bonds will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier," such that more heat (energy) is required to separate them. This phenomenon can be used to analyze boiling point of different molecules, defined as the temperature at which a phase change from liquid to gas occurs. Ethanol, \(\ce{CH3CH2-O-H}\), and methoxymethane, \(\ce{CH3-O-CH3}\), both have the same molecular formula, \(\ce{C2H6O}\). They have the same number of electrons, and a similar length. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be similar. However, ethanol has a hydrogen atom attached directly to an oxygen; here the oxygen still has two lone pairs like a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge. In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens are not sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules: The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two have similar chain lengths. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding. Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding; however, the values are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol. Hydrogen bonding also occurs in organic molecules containing N-H groups; recall the hydrogen bonds that occur with ammonia. Examples range from simple molecules like CH NH (methylamine) to large molecules like proteins and DNA. The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one. In order for a hydrogen bond to occur there must be both a hydrogen donor and an acceptor present. The donor in a hydrogen bond is usually a strongly electronegative atom such as N, O, or F that is covalently bonded to a hydrogen bond. The hydrogen acceptor is an electronegative atom of a neighboring molecule or ion that contains a lone pair that participates in the hydrogen bond. Since the hydrogen donor (N, O, or F) is strongly electronegative, it pulls the covalently bonded electron pair closer to its nucleus, and away from the hydrogen atom. The hydrogen atom is then left with a partial positive charge, creating a dipole-dipole attraction between the hydrogen atom bonded to the donor and the lone electron pair of the acceptor. This results in a hydrogen bond.(see Interactions Between Molecules With Permanent Dipoles) Although hydrogen bonds are well-known as a type of IMF, these bonds can also occur within a single molecule, between two identical molecules, or between two dissimilar molecules. Intramolecular hydrogen bonds are those which occur one single molecule. This occurs when two functional groups of a molecule can form hydrogen bonds with each other. In order for this to happen, both a hydrogen donor a hydrogen acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule. For example, intramolecular hydrogen bonding occurs in ethylene glycol (C H (OH) ) between its two hydroxyl groups due to the molecular geometry. Intermolecular hydrogen bonds occur separate molecules in a substance. They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present in positions where they can interact with one another. For example, intermolecular hydrogen bonds can occur between NH molecules alone, between H O molecules alone, or between NH and H O molecules. When we consider the points of molecules, we usually expect molecules with larger molar masses to have higher normal boiling points than molecules with smaller molar masses. This, without taking hydrogen bonds into account, is due to greater dispersion forces (see Interactions Between Nonpolar Molecules). Larger molecules have more space for electron distribution and thus more possibilities for an instantaneous dipole moment. However, when we consider the table below, we see that this is not always the case. We see that H O, HF, and NH each have higher boiling points than the same compound formed between hydrogen and the next element moving down its respective group, indicating that the former have greater intermolecular forces. This is because H O, HF, and NH all exhibit hydrogen bonding, whereas the others do not. Furthermore, \(H_2O\) has a smaller molar mass than HF but partakes in more hydrogen bonds per molecule, so its boiling point is higher. The same effect that is seen on boiling point as a result of hydrogen bonding can also be observed in the of certain substances. Substances capable of forming hydrogen bonds tend to have a higher viscosity than those that do not form hydrogen bonds. Generally, substances that have the possibility for multiple hydrogen bonds exhibit even higher viscosities. Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to. Thus, we see molecules such as PH , which do not participate in hydrogen bonding. PH exhibits a trigonal pyramidal molecular geometry like that of ammonia, but unlike NH it cannot hydrogen bond. This is due to the similarity in the electronegativities of phosphorous and hydrogen. Both atoms have an electronegativity of 2.1, and thus, there is no dipole moment. This prevents the hydrogen atom from acquiring the partial positive charge needed to hydrogen bond with the lone electron pair in another molecule. (see ) The size of donors and acceptors can also affect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction. Hydrogen bonding plays a crucial role in many biological processes and can account for many natural phenomena such as the . In addition to being present in water, hydrogen bonding is also important in the water transport system of plants, secondary and tertiary protein structure, and DNA base pairing. The cohesion-adhesion theory of transport in vascular plants uses hydrogen bonding to explain many key components of water movement through the plant's xylem and other vessels. Within a vessel, water molecules hydrogen bond not only to each other, but also to the cellulose chain that comprises the wall of plant cells. Since the vessel is relatively small, the attraction of the water to the cellulose wall creates a sort of capillary tube that allows for . This mechanism allows plants to pull water up into their roots. Furthermore, hydrogen bonding can create a long chain of water molecules, which can overcome the force of gravity and travel up to the high altitudes of leaves. Hydrogen bonding is present abundantly in the secondary structure of , and also sparingly in tertiary conformation. The secondary structure of a protein involves interactions (mainly hydrogen bonds) between neighboring polypeptide backbones which contain nitrogen-hydrogen bonded pairs and oxygen atoms. Since both N and O are strongly electronegative, the hydrogen atoms bonded to nitrogen in one polypeptide backbone can hydrogen bond to the oxygen atoms in another chain and vice-versa. Though they are relatively weak, these bonds offer substantial stability to secondary protein structure because they repeat many times and work collectively. In tertiary protein structure, interactions are primarily between functional R groups of a polypeptide chain; one such interaction is called a hydrophobic interaction. These interactions occur because of hydrogen bonding between water molecules around the hydrophobe that further reinforces protein conformation.

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The name borax (disodium tetraborate) generally describes a number of closely related compounds with different amounts of crystal water: Borax forms soft colorless/white crystals which dissolve easily in water and which effloresce in dry air. Wikimedia Commons, user Aramgutang When heated to temperatures beyond 350°C borax decahydrate loses its crystal water and forms anhydrous borax. Molten borax (m.p. 743°C) forms a glass-like bead which can readily dissolve metal oxides, developing a characteristic color which can be used in analytical chemistry for the detection of certain metals ("borax bead"). Borax is easily converted to boric acid by reaction with hydrochloric acid: Na B O 10H O + 2 HCl 4 H BO + 2 NaCl + 5 H O The "decahydrate" is sufficiently stable to find use as a primary standard for acid base titrimetry. Borax occurs naturally in evaporite deposits of seasonal lakes (California, Turkey, Chile, Tibet, Romania). The biggest borax producer is California. Most of the borax world production is used in the glass and ceramics industry (for ceramic glazes, optical glasses, and laboratory glassware).

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states that when a system at equilibrium undergoes a change in temperature, volume, or amount of moles present in a reactant or product, the system will respond in order to reach equilibrium. Think of a system at equilibrium as a balanced scale (equal weights on both sides) and when one side gains more weight, the scale will have to adjust the other side in order to reach equilibrium. In terms of volume changes within a system at equilibrium, the following applies: It's important to remember that these rules only apply to equations in which gases are involved. If only solids and aqueous solutions are present, volume changes will have no effect on the equilibrium. How could one increase the amount of reactants produced in the following equation: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\) a) increase volume b) decrease volume Explain what would happen in each of the following equations if there is a DECREASE in volume: \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\) \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\) \(AgCl(s) + \rightleftharpoons Ag^{+1}(aq) + Cl^-(aq)\) Le Chatlier's Principle applies to changes in the following: When any system at equilibrium is disturbed, it will attempt to reach a new equilibrium.

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Linkage isomerism occurs with ligands that are capable of coordinating in more than one way. The best known cases involve the monodentate ligands: \(SCN^- / NCS^-\) and \(NO_2^- / ONO^-\). The only difference is what . The ligand(s) must have more than one donor atom, but bind to ion in only one place. For example, the (NO ) ion is a ligand can bind to the central atom through the nitrogen or the oxygen atom, but cannot bind to the central atom with both oxygen and nitrogen at once, in which case it would be called a rather than an ligand. The names used to specify the changed ligands are changed as well. For example, the (NO ) ion is called when it binds with the N atom and is called when it binds with the O atom. The cationic cobalt complex [Co(NH ) (NO )]Cl exists in two separable linkage isomers of the complex ion: (NH ) (NO )] . When donation is from nitrogen to a metal center, the complex is known as a - complex and when donation is from one oxygen to a metal center, the complex is known as a - complex. An alternative formula structure to emphasize the different for the two isomers The formula of the complex is unchanged, but the properties of the complex may differ. Another example of an ambidentate ligans is thiocyanate, SCN , which can attach at either the sulfur atom or the nitrogen atom. Such compounds give rise to linkage isomerism. Polyfunctional ligands can bond to a metal center through different ligand atoms to form various isomers. Other ligands that give rise to linkage isomers include selenocyanate, SeCN – isoselenocyanate, NCSe and sulfite, SO . Are [FeCl (NO )] and [FeCl (ONO)] linkage isomers? Here, the difference is in how the ligand bonds to the metal. In the first isomer, the ligand bonds to the metal through an electron pair on the nitrogen. In the second isomer, the ligand bonds to the metal through an electron pair on one of the oxygen atoms. It's easier to see it:

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How does an electron get from one metal to another? This might be a more difficult task than it seems. In biochemistry, an electron may need to be transfered a considerable distance. Often, when the transfer occurs between two metals, the metal ions may be constrained in particular binding sites within a protein, or even in two different proteins. That means the electron must travel through space to reach its destination. Its ability to do so is generally limited to just a few Angstroms (remember, an Angstrom is roughly the distance of a bond). Still, it can react with something a few bond lengths away. Most things need to actually bump into a partner before they can react with it. This long distance hop is called an outer sphere electron transfer. The two metals react without ever contacting each other, without getting into each others' coordination spheres. Of course, there are limitations to the distance involved, and the further away the metals, the less likely the reaction. But an outer sphere electron transfer seems a little magical. So, what holds the electron back? What is the barrier to the reaction? at Caltech has developed a mathematical approach to understanding the kinetics of electron transfer, in work he did beginning in the late 1950's. We will take a very qualitative look at some of the ideas in what is referred to as "Marcus Theory". An electron is small and very fast. All those big, heavy atoms involved in the picture are lumbering and slow. The barrier to the reaction has little to do with the electron's ability to whiz around, although even that is limited by distance. Instead, it has everything to do with all of those things that are barely moving compared to the electron. Imagine an iron(II) ion is passing an electron to an iron(III) ion. After the electron transfer, they have switched identities; the first has become an iron(III) and the second has become an iron(II) ion. Nothing could be simpler. The trouble is, there are big differences between an iron(II) ion and an iron(III) ion. For example, in a coordination complex, they have very different bond distances. Why is that a problem? Because when the electron hops, the two iron atoms find themselves in sub-optimal coordination environments. Suppose an electron is transferred from an Fe(II) to a Cu(II) ion. Describe how the bond lengths might change in each case, and why. Don't worry about what the specific ligands are. In reality, a bond length is not static. If there is a little energy around, the bond can lengthen and shorten a little bit, or vibrate. A typical graph of molecular energy vs. bond length is shown below. The optimum C-O bond length in a carbon dioxide molecule is 1.116 Å. Draw a graph of what happens to internal energy when this bond length varies between 1.10 Å and 1.20 Å. Don't worry about quantitative labels on the energy axis. The optimum O-C-O bond angle in a carbon dioxide molecule is 180 °. Draw a graph of what happens to internal energy when this bond angle varies between 170 ° and 190 °. Don't worry about quantitative labels on the energy axis. The barrier to electron transfer has to do with reorganizations of all those big atoms before the electron makes the jump. In terms of the coordination sphere, those reorganizations involve bond vibrations, and bond vibrations cost energy. Outside the coordination sphere, solvent molecules have to reorganize, too. Remember, ion stability is highly influenced by the surrounding medium. Draw a Fe(II) ion and a Cu(II) ion with three water molecules located somewhere in between them. Don't worry about the ligands on the iron or copper. Show how the water molecules might change position or orientation if an electron is transferred from iron to copper. Thus, the energetic changes needed before electron transfer can occur involve a variety of changes, including bond lengths of several ligands, bond angles, solvent molecules, and so on. The whole system, involving both metals, has some optimum set of positions of minimum energy. Any deviations from those positions requires added energy. In the following energy diagram, the x axis no longer defines one particular parameter. Now it lumps all changes in the system onto one axis. This picture is a little more abstract than when we are just looking at one bond length or one bond angle, but the concept is similar: there is an optimum set of positions for the atoms in this system, and it would require an input of energy in order to move any of them move away from their optimum position. It is thought that these kinds of reorganizations -- involving solvent molecules, bond lengths, coordination geometry and so on -- actually occur prior to electron transfer. They happen via random motions of the molecules involved. However, once they have happened, there is nothing to hold the electron back. Its motion is so rapid that it can immediately find itself on the other atom before anything has a chance to move again. Consequently, the barrier to electron transfer is just the amount of energy needed for all of those heavy atoms to get to some set of coordinates that would be accessible in the first state, before the electron is transferred, but that would also be accessible in the second state, after the electron is transfered. Describe some of the changes that contribute to the barrier to electron transfer in the following case. In the drawing below, an electron is transferred from one metal to another metal of the same kind, so the two are just switching oxidation states. For example, it could be an iron(II) and an iron(III), as pictured in the problem above. In the blue state, one iron has the extra electron, and in the red state it is the other iron that has the extra electron. The energy of the two states are the same, and the reduction potential involved in this trasfer is zero. However, there would be some atomic reorganizations needed to get the coordination and solvation environments adjusted to the electron transfer. The ligand atoms and solvent molecules have shifted in the change from one state to another, and so our energy surfaces have shifted along the x axis to reflect that reorganization. That example isn't very interesting, because we don't form anything new on the product side. Instead, let's picture an electron transfer from one metal to a very different one. For example, maybe the electron is transferred from cytochrome c to the "copper A" center in cytochrome c oxidase, an important protein involved in respiratory electron transfer. In the drawing above, some water molecules are included between the two metal centres. The energy diagram for the case involving two different metals is very similar, except that now there is a difference in energy between the two states. The reduction potential is no longer zero. We'll assume the reduction potential is positive, and so the free energy change is negative. Energy goes down upon electron transfer. Compare this picture to the one for the degenerate case, when the electron is just transferred to a new metal of the same type. A positive reduction potential (or a negative free energy change) has the effect of sliding the energy surface for the red state downwards. As a result, the intersection point between the two surfaces also slides downwards. Since that is the point at which the electron can slide from one state to the other, the barrier to the reaction decreases. What would happen if the reduction potential were even more positive? Let's see in the picture below. The trend continues. According to this interpretation of the kinetics of electron transfer, the more exothermic the reaction, the lower its barrier will be. It isn't always the case that kinetics tracks along with thermodynamics, but this might be one of them. But is all of this really true? We should take a look at some experimental data and see whether it truly works this way. As the reduction potential becomes more positive, free energy gets more negative, and the rate of the reaction dramatically increases. So far, Marcus theory seems to get things right. When you look a little closer at Marcus theory, though, things get a little strange. Suppose we make one more change and see what happens when the reduction potential becomes positive. So, if Marcus is correct, at some point as the reduction potential continues to get more positive, reactions start to slow down again. They don't just reach a maximum rate and hold steady at that plateau; the barrier gets higher and higher and the reactions get slower and slower. If you feel a little skeptical about that, you're in good company. Marcus always maintained that this phenomenon was a valid aspect of the theory, and not just some aberration that should be ignored. The fact that nobody had ever actually observed such a trend didn't bother him. The reason we didn't see this kind of thing, he said, was that we just hadn't developed technology that was good enough to measure these kind of rates accurately. But technology did catch up. Just take a look at the following data (from Miller, , ). Don't worry that there are no metals involved anymore. An electron transfer is an electron transfer. Here, an electron is sent from the aromatic substructure on the right to the substructure on the left. By varying the part on the left, we can adjust the reduction potential (or the free energy change, as reported here. As the reaction becomes more exergonic, the rate increases, but then it hits a maximum and decreases again. Data like this means that the "Marcus Inverted Region" is a real phenomenon. Are you convinced? So were other people. In 1992, Marcus was awarded the Nobel Prize in Chemistry for this work. Take a look at the donor/acceptor molecule used in Williams' study, above. a) Why do you suppose the free energy change is pretty small for the first three compounds in the table? b) Why does the free energy change continue to get bigger over the last three compounds in the table? The rates of electron transfer between cobalt complexes of the bidentate bipyridyl ligand, Co(bipy) , are strongly dependent upon oxidation state in the redox pair. Electron transfer between Co(I)/Co(II) occurs with a rate constant of about 10 M s , whereas the reaction between Co(II)/Co(III) species proceeds with = 18 M s . ,

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An , being a proton donor, can only as an acid if there is a suitable present to accept the proton. What do we mean by "suitable'' in this context? Simply that a base, in order to accept a proton, must provide a lower-energy resting place for the proton. (We are actually referring to something called "free energy" here, but don't worry about that if you are not familiar with that term; just think of it as a form of potential energy.) -- the same as a book will fall (if you drop it) only downward, to a position of lower (gravitational) potential energy. Viewed in this way, , . The tendency for a proton to move from source to sink depends on how far the proton can fall in energy, and this in turn depends on the energy difference between the source and the sink. This is entirely analogous to measuring the tendency of water to flow down from a high elevation to a lower one; this tendency (which is related to the amount of energy that can be extracted in the form of electrical work if the water flows through a power station at the bottom of the dam) will be directly proportional to the difference in elevation (difference in potential energy) between the source (top of the dam) and the sink (bottom of the dam). Now look at the diagram at the right and study it carefully. In the center columns of the diagram, you see a list of acids and their conjugate bases. These acid-base pairs are plotted on an energy scale which is shown at the left side of the diagram. This scale measures the free energy released when one mole of protons is transferred from a given acid to H O. Thus if one mole of HCl is added to water, it dissociates completely and heat is released as the protons fall from the source (HCl) to the lower free energy that they possess in the H O ions that are formed when the protons combine with H O. Any acid shown on the left side of the vertical line running down the center of the diagram can donate protons to any base (on the right side of the line) that appears below it. The greater the vertical separation, the greater will be the fall in free energy of the proton, and the more complete will be the proton transfer at equilibrium. Notice the H O - H O pair shown at zero kJ on the free energy scale. This zero value of free energy corresponds to the proton transfer process \[\ce{H3O^{+ } + H_2O \rightarrow H2O + H3O^{+}}\] which is really no reaction at all, hence the zero fall in free energy of the proton. Since the proton is equally likely to attach itself to either of two identical H O molecules, the equilibrium constant is unity. Now look at the acid/base pairs shown at the top of the table, above the H O - H O line. All of these acids can act as proton sources to those sinks (bases) that appear below them. Since H O is a suitable sink for these acids, all such acids will lose protons to H O in aqueous solutions. These are therefore all that are 100% dissociated in aqueous solution; this total dissociation reflects the very large equilibrium constants that are associated with any reaction that undergoes a fall in free energy of more than a few kilojoules per mole. Because H O serves as a proton sink to any acid in which the proton free energy level is greater than zero, the strong acids such as HCl and H SO cannot "exist" (as acids) in aqueous solution; they exist as their conjugate bases instead, and the only proton donor present will be H O . This is the basis of the , which states that . Now consider a , such as HCN at about 50 kJ on the scale. This positive free energy means that in order for a mole of HCN to dissociate (transfer its proton to H O), the proton must gain 40 kJ of free energy per mole. In the absence of a source of energy, the reaction will simply "not go"; HCN dissociates only to a minute extent in water. Why is a weak acid such as HCN dissociated at all? The molecules in solution are continually being struck and bounced around by the thermal motions of neighboring molecules. Every once in a while, a series of fortuitous collisions will provide enough kinetic energy to a HCN molecule to knock off the proton, effectively boosting it to the level required to attach itself to water. This process is called , and its probability falls off very rapidly as the distance (in kJ ) that the proton must rise increases. The protons on a "stronger'' weak acid such as HSO or CH COOH will be thermally excited to the H O level much more frequently than will the protons on HCN or HCO , hence the difference in the dissociation constants of these acids. Although a weak acid such as HCN will not react with water to a significant extent, you are well aware that such an acid can still be titrated with strong base to yield a solution of NaCN at the equivalence point. To understand this process, find the H O-OH pair at about 80 kJ on the free energy scale. Because the OH ion can act as a proton sink to just about every acid shown on the diagram, the addition of strong base in the form of NaOH solution allows the protons at any acid above this level to fall to the OH level according to the reaction \[H_3O^+ + OH^– \rightleftharpoons 2 H_2O\] Titration, in other words, consists simply in introducing a low free energy sink that can drain off the protons from the acids initially present, converting them all into their conjugate base forms. There are two other aspects of the H O-H O pair that have great chemical significance. First, its location at 80 kJ/mol tells us that for a H O molecule to transfer its proton to another H O molecule (which then becomes a H O ion whose relative free energy is zero), a whopping 80 kJ/mol of free energy must be supplied by thermal excitation. This is so improbable that only one out of about 10 million H O molecules will have its proton elevated to the H O level at a given time; this corresponds to the small value of the ion product of water, about 10 . The other aspect of the H O-OH ; pair is that its location means the On our diagram only two stronger bases (lower proton free energy sinks) are shown: the amide ion NH , and the oxide ion O . What happens if you add a soluble oxide such as Na O to water? Since O is a proton sink to \water, it will react with the solvent, leaving OH as the strongest base present: O + H O OH + OH This again is the leveling effect; all bases stronger than OH appear equally strong in water, simply because they are all converted to OH . The pH of a solution is more than a means of expressing its hydrogen ion concentration on a convenient logarithmic scale. The concept of pH was suggested by the Swedish chemist Sørensen in 1909 as a means of compressing the wide range of [H ] values encountered in aqueous solutions into a convenient range. The modern definition of pH replaces [H ] with {H } in which the curly brackets signify the effective concentration of the hydrogen ion, which chemists refer to as the hydrogen ion \[pH = -\­log \{H^{+}\}\] The real physical meaning of pH is that it measures the of protons in the solution; that is, the ability of the solution to supply protons to a base such as H O. This is the same as the "hydrogen ion concentration" [H ] only in rather dilute solutions; at ionic concentrations greater than about 0.001 , electrostatic interactions between the ions cause the relation between the pH (as measured by direct independent means) and [H O ] to break down. Thus we would not expect the pH of a 0.100 M solution of HCl to be exactly 1.00. On the right side of the illustration is a pH scale. At the pH value corresponding to a given acid-base pair, the acid and base forms will be present at equal concentrations. For example, if you dissolve some solid sodium sulfate in pure water and then adjust the pH to 2.0, about half of the SO will be converted into HSO . Similarly, a solution of Na CO in water will not contain a very large fraction of CO unless the pH is kept above 10. Suppose we have a mixture of many different weak acid-base systems, such as exists in most biological fluids or natural waters, including the ocean. The available protons will fall to the lowest free energy levels possible, first filling the lowest-energy sink, then the next, and so on until there are no more proton-vacant bases below the highest proton-filled (acid) level. Some of the highest protonated species will donate protons to H O through thermal excitation, giving rise to a concentration of H O that will depend on the concentrations of the various species. The equilibrium pH of the solution is a measure of this concentration, but this in turn reflects the relative free energy of protons required to keep the highest protonated species in its acid form; it is in this sense that pH is a direct measure of proton free energy. In order to predict the actual pH of any given solution, we must of course know something about the nominal concentrations C of the various acid-base species, since this will strongly affect the distribution of protons. Thus if one proton-vacant level is present at twice the concentration of another, it will cause twice as many acid species from a higher level to become deprotonated. In spite of this limitation, the proton free energy diagram provides a clear picture of the relationships between the various acid and base species in a complex solution. This "proton-free energy" view of acid-base chemistry was first developed by Gurney in his classic "Ionic Processes in Solution" published in the early 1950's, but languished in the shadows until Stumm and Morgan revived it in their "Aquatic Chemistry" from the early 1970's. Why it still has not made it into the standard General Chmistry texts is something of a mystery.

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We view a chemical system as a collection of substances that occupies some volume. Let us consider a closed system whose volume is variable, and in which no work other than pressure–volume work is possible. If this system is undergoing a reversible change, it is at equilibrium, and it is in contact with its surroundings. Because the system is at equilibrium, all points inside the system have the same pressure and the same temperature. Since the change is reversible, the interior pressure is arbitrarily close to the pressure applied to the system by the surroundings. If the reversibly changing system can exchange heat with its surroundings, the temperature of the surroundings is arbitrarily close to the temperature of the system. (If a process takes place in a system that cannot exchange heat with its surroundings, we say that the process is .) We can measure the pressure, temperature, and volume of such a system without knowing anything about its composition. For a system composed of a known amount of a single phase of a pure substance, we know from experience that any cyclic change in pressure or temperature restores the initial volume. That is, for a pure phase, there is an equation of state that we can rearrange as \(V=V\left(P,T\right)\), meaning that specifying \(P\) and \(T\) is sufficient to specify \(V\) uniquely. For other reversible systems, the function \(V=V\left(P,T\right)\) may not exist. For example, consider a system that consists of a known amount of water at liquid–vapor equilibrium and whose pressure and temperature are known. For this system, the volume can have any value between that of the pure liquid and that of the pure gas. Specifying the pressure and temperature of this system is not sufficient to specify its state. However, if we specify the temperature of this system, the pressure is fixed by the equilibrium condition; and if we specify the volume of the system, we can find how much water is in each phase from the known molar volumes of the pure substances at the system pressure and temperature. For the water–water-vapor equilibrium system, we can write \(P=P\left(V,T\right)\). In each of these cases, we can view one of the variables as a function of the other two and represent it as a surface in a three dimensional space. The two independent variables define a plane. Projecting the system’s location in this independent-variable plane onto the surface establishes the value of the dependent variable. The two independent-variable values determine the point on the surface that specifies the state of the system. In the liquid–vapor equilibrium system, the pressure is a surface above the volume–temperature plane. A complete description of the state of the system must also include the number of moles of liquid and the number of mole of vapor present. Each of these quantities can also be described as a surface in a three dimensional space in which the other two dimensions are volume and temperature. asserts that these observations are special cases of a more general truth: For a closed, reversible system in which only pressure–volume work is possible, specifying how some pair of state functions changes is sufficient to specify how the state of the system changes. Duhem’s theorem asserts that two variables are sufficient to specify the state of the system in the following sense: Given the values of the system’s thermodynamic variables in some initial state, say \(\mathrm{\{}\)\(X_1\), \(Y_1\), \(Z_1\), \(W_1\),\(\dots\)\(\mathrm{\}}\), specifying the change in some pair of variables, say \(\Delta X\) and \(\Delta Y\), is sufficient to determine the change in the remaining variables, \(\Delta Z\), \(\Delta W\),\(\dots\) so that the system’s thermodynamic variables in the final state are \(\mathrm{\{}\)\(X_2\), \(Y_2\), \(Z_2\), \(W_2\),\(\dots\)\(\mathrm{\}}\), where \(W_2=W_1+\Delta W\), etc. The theorem does not specify which pair of variables is sufficient. In fact, from the discussion above of the variables that can be used to specify the state of a system containing only water, it is evident that a particular pair may not remain sufficient if there is a change in the number of phases present. In Chapter 10, we see that Duhem’s theorem follows from the first and second laws of thermodynamics, and we consider the particular pairs of variables that can be used. For now, let us consider a proof of Duhem’s theorem for a system in which the pressure, temperature, volume, and composition can vary. We consider systems in which only pressure–volume work is possible. Let the number of chemical species present be \(C^{'}\) and the number of phases be \(P\). (\(C\), the number of component in the phase rule, and \(C^{'}\) differ by the number of stoichiometric constraints that apply to the system: \(C\) is \(C^{'}\) less the number of stoichiometric constraints.) We want to know how many variables can be changed independently while the system remains at equilibrium. This is similar to the question we answered when we developed Gibbs’ phase rule. However, there are important differences. The phase rule is independent of the size of the system; it specifies the number of intensive variables required to prescribe an equilibrium state in which specified phases are present. The size of the system is not fixed; we can add or remove matter to change the size of any phase without changing the number of degrees of freedom. In the present problem, the system cannot exchange matter with its surroundings. Moreover, the number of phases present can change. We require only that any change be reversible, and a reversible process can change the number of phases. (For example, reversible vaporization can convert a two-phase system to a gaseous, one-phase system.) We want to impose a change on an initial state of a closed system. This initial state is an equilibrium state, and we want to impose a change that produces a new Gibbsian equilibrium state of the same system. This means that the change we impose can neither eliminate an existing chemical species nor introduce a new one. A given phase can appear or disappear, but a given chemical species cannot. We can find the number of independent variables for this system by an argument similar to the one we used to find the phase rule. To completely specify this system, we must specify the pressure, temperature, and volume of each phase. We must also specify the number of moles of each of \(C^{'}\) chemical species in each phase. This means that \(P\left(C^{'}+3\right)\) variables must be specified. Every relationship that exists among these variables decreases by one the number that are independent. The following relationships exist: Subtracting the number of constraints from the number of variables, we find that there are \[P\left(C^{'}+3\right)-\left(P-1\right)-\left(P-1\right)-P-C^{'}P=2\] independent variables for a reversible process in a closed system, if all work is pressure–volume work. The number of independent variables is constant; it is independent of the species that are present and the number of phases. It is important to appreciate that there is no conflict between Duhem’s theorem and the phase-rule conclusion that \(F\) degrees of freedom are required to specify an equilibrium state of a system containing specified phases. When we say that specifying some pair of variables is sufficient to specify the state of a particular closed system undergoing reversible change, we are describing a system that is continuously at equilibrium as it goes from a first equilibrium state to a second one. Because it is closed and continuously in an equilibrium state, the range of variation available to the system is circumscribed in such a way that specifying two variables is sufficient to specify its state. On the other hand, when we say that \(F\) degrees of freedom are required to specify an equilibrium state of a system containing specified phases, we mean that we must know the values of \(F\) intensive variables in order to establish that the state of the system is an equilibrium state. To illustrate the compatibility of these ideas and the distinction between them, let us consider a closed system that contains nitrogen, hydrogen, and ammonia gasammonia gases. In the presence of a catalyst, the reaction \[\ce{N_2 + 3H2 <=> 2NH3}\] occurs. For simplicity, let us assume that these gases behave ideally. (If the gases do not behave ideally, the argument remains the same, but more complex equations are required to express the equilibrium constant and the system pressure as functions of the molar composition.) This system has two components and three degrees of freedom. When we say that the system is closed, we mean that the total number of moles of the elements nitrogen and hydrogen are known and constant. Let these be \(n_N\) and \(n_H\), respectively. Letting the moles of ammonia present be \(n_{NH_3}=x\), the number of moles of dihydrogen and dinitrogen are \(n_{H_2}={\left(n_H-3x\right)}/{2}\) and \(n_{N_2}={\left(n_N-x\right)}/{2}\), respectively. If we know that this system is at equilibrium, we know that the equilibrium constant relationship is satisfied. We have \[K_P=\frac{P^2_{NH_3}}{P^3_{H_2}P_{N_2}}=\frac{n^2_{NH_3}}{n^3_{H_2}n_{N_2}}{\left(\frac{RT}{V}\right)}^{-2}=\frac{16x^2}{{\left(n_H-3x\right)}^3\left(n_N-x\right)}{\left(\frac{RT}{V}\right)}^{-2}\] where \(V\) is the volume of the system. The ideal-gas equilibrium constant is a function only of temperature. We assume that we know this function; therefore, if we know the temperature, we know the value of the equilibrium constant. The pressure of the system can also be expressed as a function of \(x\) and \(V\). We have \[P=P_{H_2}+P_{N_2}+P_{NH_3}=\left[\frac{\left(n_H+n_N\right)}{2}-x\right]\left(\frac{RT}{V}\right)\] If we know the system pressure and we know that the system is at equilibrium, we can solve the equations for \(K\) and \(P\) simultaneously to find the unknowns \(x\) and \(V\). From these, we can calculate the molar composition of the system and the partial pressure of each of the gases. (We discuss ideal-gas equilibrium calculations in detail in .) Thus, if we know that the system is at equilibrium, knowledge of the pressure and temperature is sufficient to determine its composition and all of its other properties. If we do not know that this system is at equilibrium, but instead want to collect sufficient experimental data to prove that it is, the phase rule asserts that we must find the values of some set of three intensive variables. Two are not sufficient. From the perspective provided by the equations developed above, we can no longer use the equilibrium constant relationship to find \(x\) and \(V\). Instead, our problem is to find the composition of the system by other means, so that we can test for equilibrium by comparing the value of the quantity \[\frac{P^2_{NH_3}}{P^3_{H_2}P_{N_2}}=\frac{16x^2}{{\left(n_H-3x\right)}^3\left(n_N-x\right)}{\left(\frac{RT}{V}\right)}^{-2}\] to the value of the equilibrium constant. We could accomplish this goal by measuring the values of several different combinations of three intensive variables. A convenient combination is pressure, temperature, and ammonia concentration, \({x}/{V}\). When we rearrange the equation for the system pressure to \[P=\left[\frac{\left(n_H+n_N\right)}{2V}-\left(\frac{x}{V}\right)\right]RT\] it is easy to see that knowing P, T, and \({x}/{V}\) enables us to find the volume of the system. Given the volume, we can find the molar composition of the system and the partial pressure of each of the gases. With these quantities in hand, we can determine whether the equilibrium condition is satisfied.

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It seems intuitive that a reaction goes faster as the temperature is raised, as more reactant molecules have the energy needed to overcome the activation barrier to the reaction. The Arrhenius equation relates the reaction rate constant (k) and temperature. One of the forms of the Arrhenius equation is given below: \[ \ln k = -{E_a \over RT} + \ln A \nonumber \] where E is the activation energy for the reaction, T is the absolute temperature (in Kelvin) at which a corresponding k is determined, R is the gas constant, and A is a pre-exponential factor. The activation energy may then be extracted from a plot of ln k vs. 1/T, which should be linear. This plot is called an "Arrhenius plot." Recall that y = mx + b. Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor. Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor. In practice, activation energies are not often cited in the current literature. Instead, a similar but more useful equation called the Eyring equation is used. The Eyring equation is: \[ \ln \left ( {k \over T} \right ) = - {\Delta H^{\ddagger} \over RT} + \ln \left ( {k_B \over h}\right ) + {\Delta S^{\ddagger} \over R} \nonumber \] where k, T and R are the same as in the Arrhenius equation, k is Boltzmann’s constant, h is Planck’s constant and ΔH and ΔS are the enthalpy and entropy of activation, respectively. Note that the activation parameters (ΔH and ΔS ) are not the same as the entropy and enthalpy of the reaction, which can usually be calculated from tables of values. Since they depend on how the reaction proceeds, not just the initial and final states of the reaction, they must be determined experimentally. Once that has been done, interpretation of the numerical values provides insight into the mechanism of the reaction.

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The heat capacity is a coefficient that gives the amount of energy to raise the temperature of a substance by one degree Celsius. The heat capacity can also be described as the temperature derivative of the average energy. The constant volume heat capacity is defined by using the notation that E = U - U(0) where U(0) is the energy at zero Kelvin. The molar internal energy of a monatomic ideal gas is E = 3/2RT. The heat capacity of a monatomic ideal gas is therefore C = 3/2R. For a monatomic gas there are three degrees of freedom per atom (these are the translations along the x, y, and z direction). Each of these translations corresponds to ½RT of energy. For an ideal diatomic gas some of the energy used to heat the gas may also go into rotational and vibrational degrees of freedom. For solids there is no translation or rotation and therefore the entire contribution to the heat capacity comes from vibrations. Given their extended nature the vibrations in solids are much lower in frequency than those of gases. Therefore, while vibrations in typical diatomic gases typically contribute little to the heat capacity, the vibrational contribution to the heat capacity of solids is the largest contribution. As the temperature is increased, there are more levels of the solid accessible by thermal energy and therefore Q increases. This also means that U increases and finally that C increases. In the high temperature limit in an ideal solid there are 3N vibrational modes that are accessible giving rise to a contribution to the molar heat capacity of 3R.

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These bonds are pretty polar, so they show up strongly in IR spectroscopy. IR spectroscopy is therefore a good way to determine what heteroatom-containing functional groups are present in a molecule. Oxygen forms two bonds. An oxygen atom could be found in between two carbons, as in dibutyl ether, or between a carbon and a hydrogen as in 1-butanol. Dibutyl ether is an example of an ether and 1-butanol is an example of an alcohol. If you look at an IR spectrum of dibutyl ether, you will see: Japan, 14 July 2008) If you look at an IR spectrum of 1-butanol, you will see: Japan, 14 July 2008) Peak shapes are sometimes very useful in recognizing what kind of bond is present. The rounded shape of most O-H stretching modes occurs because of hydrogen bonding between different hydroxy groups. Because protons are shared to varying extent with neighboring oxygens, the covalent O-H bonds in a sample of alcohol all vibrate at slightly different frequencies and show up at slightly different positions in the IR spectrum. Instead of seeing one sharp peak, you see a whole lot of them all smeared out into one broad blob. Since C-H bonds don't hydrogen bond very well, you don't see that phenomenon in an ether, and an O-H peak is very easy to distinguish in the IR spectrum. Even though there are only two C-O bonds in dibutyl ether, the C-O stretching mode is even stronger than the peak at 2900 cm arising from 10 different C-H bonds. Explain why. The IR spectrum of methyl phenyl ether (aka anisole) has strong peaks at 1050 and 1250 cm . Japan, 14 July 2008) ,

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This page starts by looking at the extraction of aluminum from its ore, bauxite, including some economic and environmental issues. It finishes by looking at some uses of aluminum. Aluminum is too high in the (reactivity series) to extract it from its ore using carbon reduction. The temperatures needed are too high to be economic. Instead, it is extracted by electrolysis. The ore is first converted into pure aluminum oxide by the Bayer Process, and this is then electrolyzed in solution in molten cryolite - another aluminum compound. The aluminum oxide has too high a melting point to electrolyse on its own. The usual aluminum ore is bauxite. Bauxite is essentially an impure aluminum oxide. The major impurities include iron oxides, silicon dioxide and titanium dioxide. Crushed bauxite is treated with moderately concentrated sodium hydroxide solution. The concentration, temperature and pressure used depend on the source of the bauxite and exactly what form of aluminum oxide it contains. Temperatures are typically from 140°C to 240°C; pressures can be up to about 35 atmospheres. High pressures are necessary to keep the water in the sodium hydroxide solution liquid at temperatures above 100°C. The higher the temperature, the higher the pressure needed. With hot concentrated sodium hydroxide solution, aluminum oxide reacts to give a solution of sodium tetrahydroxoaluminate. \[ Al_2O_3 + 2NaOH + 3H_2O \longrightarrow 2NaAl(OH)_4\] The impurities in the bauxite remain as solids. For example, the other metal oxides present tend not to react with the sodium hydroxide solution and so remain unchanged. Some of the silicon dioxide reacts, but goes on to form a sodium aluminosilicate which precipitates out. All of these solids are separated from the sodium tetrahydroxoaluminate solution by filtration. They form a "red mud" which is just stored in huge lagoons. The sodium tetrahydroxoaluminate solution is cooled, and "seeded" with some previously produced aluminum hydroxide. This provides something for the new aluminum hydroxide to precipitate around. \[ NaAl(OH)_4 \longrightarrow Al(OH)_3 + NaOH \] Aluminum oxide (sometimes known as alumina) is made by heating the aluminum hydroxide to a temperature of about 1100 - 1200°C. \[ 2Al(OH)_3 \longrightarrow Al_2O_3 + 3H_2O \] The aluminum oxide is electrolyzed in solution in molten cryolite, Na AlF . Cryolite is another aluminum ore, but is rare and expensive, and most is now made chemically. The diagram shows a very simplified version of an electrolysis cell. Although the carbon lining of the cell is labelled as the cathode, the effective cathode is mainly the molten aluminum that forms on the bottom of the cell. Molten aluminum is syphoned out of the cell from time to time, and new aluminum oxide added at the top. The cell operates at a low voltage of about 5 - 6 volts, but at huge currents of 100,000 amps or more. The heating effect of these large currents keeps the cell at a temperature of about 1000°C. These are very complicated - in fact one source I've looked at says that they aren't fully understood. For chemistry purposes at this level, they are always simplified (to the point of being wrong! - see comment below). This is the simplification: Aluminum is released at the cathode. Aluminum ions are reduced by gaining 3 electrons. \[ Al^3+ + 3e^- \longrightarrow Al\] Oxygen is produced initially at the anode. \[ 2O^{2-} \longrightarrow O_2 + 4e^-\] However, at the temperature of the cell, the carbon anodes burn in this oxygen to give carbon dioxide and carbon monoxide. Continual replacement of the anodes is a major expense. This section is designed to give you a brief idea of the sort of economic and environmental issues involved with the extraction of aluminum. I wouldn't claim that it covers everything! Think about: Think about: Think about: Think about: Aluminum is usually alloyed with other elements such as silicon, copper or magnesium. Pure aluminum isn't very strong, and alloying it adds to it strength. Aluminum is especially useful because it Anodizing essentially involves etching the aluminum with sodium hydroxide solution to remove the existing oxide layer, and then making the aluminum article the anode in an electrolysis of dilute sulphuric acid. The oxygen given of at the anode reacts with the aluminum surface, to build up a film of oxide up to about 0.02 mm thick. As well as increasing the corrosion resistance of the aluminum, this film is porous at this stage and will also take up dyes. (It is further treated to make it completely non-porous afterwards.) That means that you can make aluminum articles with the colour built into the surface. Some uses include: Jim Clark ( )

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The term refers to the tendency of molecules to generate induced electric dipole moments when subjected to an electric field. It originates from the fact that nuclei and electrons are generally not fixed in space and that when molecules are subject to an electric field, the negatively charged electrons and positively charged atomic nuclei are subject to opposite forces and undergo charge separation. Polarizability allows us to better understand the interactions between nonpolar atoms and molecules and other electrically charged species, such as ions or polar molecules with dipole moments. Neutral nonpolar species have spherically symmetric arrangements of electrons in their electron clouds. When in the presence of an electric field, their electron clouds can be distorted (Figure \(\Page {1}\)). The ease of this distortion is the polarizability of the atom or molecule. The created distortion of the electron cloud causes the originally nonpolar molecule or atom to acquire a dipole moment. This induced dipole moment is related to the polarizability of the molecule or atom and the strength of the electric field by the following equation: \[μ_{ind} = \alpha E \label{1}\] where \(E\) denotes the strength of the electric field and \(\alpha\) is the polarizability of the atom or molecule with units of C m V . In general, polarizability correlates with the interaction between electrons and the nucleus. The amount of electrons in a molecule affects how tight the nuclear charge can control the overall charge distribution. Atoms with fewer electrons will have smaller, denser electron clouds, as there is a strong interaction between the few electrons in the atoms’ orbitals and the positively charged nucleus. There is also less shielding in atoms with fewer electrons contributing to the stronger interaction of the outer electrons and the nucleus. With the electrons held tightly in place in these smaller atoms, these atoms are typically not easily polarized by external electric fields. In contrast, large atoms with many electrons, such as negative ions with excess electrons, are easily polarized. These atoms typically have very diffuse electron clouds and large atomic radii that limit the interaction of their external electrons and the nucleus. The relationship between polarizability and the factors of electron density, atomic radii, and molecular orientation is as follows: The dispersion force is the weakest intermolecular force. It is an attractive force that arises from surrounding temporary dipole moments in nonpolar molecules or species. These temporary dipole moments arise when there are instantaneous deviations in the electron clouds of the nonpolar species. Surrounding molecules are influenced by these temporary dipole moments and a sort of chain reaction results in which subsequent weak, dipole-induced dipole interactions are created (Figure \(\Page {2}\)). These cumulative dipole-induced dipole interactions create attractive dispersion forces. Dispersion forces are the forces that make nonpolar substances condense to liquids and freeze into solids when the temperature is low enough. Polarizability affects dispersion forces in the following ways: The influence of polarizability on the strength of the dispersion forces is also modulated by the 3-D structure of the molecules. For example, neo-pentane and n-pentane (Figure \(\Page {3}\)) exhibit comparable polarizabilities, but have different boiling points. The lack of polar bonding in both indicates that dispersion is the dominant interaction responsible in both substances.  The higher boiling point of n-pentane indicates there are stronger intermolecular interactions. Both neopentane and n-pentane would be expected to exhibits comparable dispersion forces since they have comparable polarizabilities. However, the structure of the n-pentane molecule facilitates greater dispersion forces due to more contacts between n-pentane molecules. In contrast, neopentane has a smaller dispersion force since the interactions between neopentane are reduced since neopentane is more compact and symmetrical. The relationship between polarizability and dispersion forces can be seen in the following equation, which can be used to quantify the interaction between two like nonpolar atoms or molecules (e.g., \(\ce{A}\) with \(\ce{A}\)): \[ V(r) = \dfrac{-3}{4} \dfrac{\alpha^2 I}{r^6} \label{2}\] where \(r\) is the distance between the atoms or molecules, \(I\) is the of the atom or molecule, and \(\alpha\) is the polarizability constant expressed in units of m . This expression of \(\alpha\) is related to \(\alpha'\) by the following equation: \[\alpha' = \dfrac{\alpha}{4 \pi \epsilon_o} \label{3}\] To quantify the interaction between unlike atoms or molecules (\(\ce{A}\) and \(\ce{B}\)), Equation \(\ref{2}\) becomes: \[ V(r) = \dfrac{-3}{2}\dfrac{I_AI_B}{I_A+I_B} \dfrac{\alpha_A \alpha_B}{r^6} \label{4}\]

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A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by (in some cases, the bonding is actually more complicated than that. The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions. What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. All ligands are lone pair donors. In other words, all ligands function as . We are going to look in detail at the bonding in the complex ion formed when water molecules attach themselves to an aluminum ion to give Al(H O) . Start by thinking about the structure of a naked aluminum ion before the water molecules bond to it. Aluminum has the electronic structure 1s 2s 2p 3s 3p When it forms an Al ion it loses the 3-level electrons to leave 1s 2s 2p That means that all the 3-level orbitals are now empty. The aluminium uses all six of these empty 3-level orbitals to accept lone pairs from six water molecules. It re-organizes (hybridizes) the 3s, the three 3p, and two of the 3d orbitals to produce six new orbitals all with the same energy. You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules that will fit around an aluminum ion (and most other metal ions) due to steric constraints. By making the maximum number of bonds, it releases most energy and so becomes most energetically stable. Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminum and so isn't involved in the bonding. The resulting ion looks like this: Because of the movement of electrons towards the center of the ion, the 3+ charge is no longer located entirely on the aluminum, but is now spread over the whole of the ion. Because the aluminum is forming 6 bonds, the co-ordination number of the aluminum is said to be 6. The co-ordination number of a complex ion counts the number of co-ordinate bonds being formed by the metal ion at its center. In a simple case like this, that obviously also counts the number of ligands - but that is not necessarily so, as you will see later. Some ligands can form more than one co-ordinate bond with the metal ion. Iron has the electronic structure 1s 2s 2p 3s 3p 3d 4s When it forms an Fe ion it loses the 4s electrons and one of the 3d electrons to leave 1s 2s 2p 3s 3p 3d Looking at this as electrons-in-boxes, at the bonding level: The single electrons in the 3d level are involved in the bonding in any way. Instead, the ion uses 6 orbitals from the 4s, 4p and 4d levels to accept lone pairs from the water molecules. Before they are used, the orbitals are re-organized (hybridized) to produce 6 orbitals of equal energy. Once the co-ordinate bonds have been formed, the ion looks exactly the same as the equivalent aluminium ion. Because the iron is forming 6 bonds, the co-ordination number of the iron is 6. This is a simple example of the formation of a complex ion with a negative charge. Copper has the electronic structure 1s 2s 2p 3s 3p 3d 4s When it forms a Cu ion it loses the 4s electron and one of the 3d electrons to leave 1s 2s 2p 3s 3p 3d To bond the four chloride ions as ligands, the empty 4s and 4p orbitals are used (in a hybridized form) to accept a lone pair of electrons from each chloride ion. Because chloride ions are bigger than water molecules, you can't fit 6 of them around the central ion - that's why you only use 4. Only one of the 4 lone pairs on each chloride ion is shown. The other three are pointing away from the copper ion, and aren't involved in the bonding. That gives you the complex ion: The ion carries 2 negative charges overall. That comes from a combination of the 2 positive charges on the copper ion and the 4 negative charges from the 4 chloride ions. In this case, the co-ordination number of the copper is 4. Jim Clark ( )

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The largest class of oxygen-containing molecules is , which contain C=O bonds. A C=O stretch is normally easy to find in an IR spectrum, because it is very strong and shows up in a part of the spectrum that is not cluttered with other peaks. Examples of carbonyl compounds include 2-octanone, a ketone, and butanal, an aldehyde. In an aldehyde, the carbonyl is at the end of a chain, with a hydrogen attached to the carbonyl carbon. If you look at the IR spectrum of 2-octanone: Even though there is just one C=O bond, the carbonyl stretch is often the strongest peak in the spectrum. That makes carbonyl compounds easy to identify by IR spectroscopy. If you look at the IR spectrum of butanal: The aldehyde C-H bond absorbs at two frequencies because it can vibrate in phase with the C=O bond (a symmetric stretch) and out of phase with the C=O bond (an asymmetric stretch), and these vibrations are of different energies. The probability of the symmetric stretch and the asymmetric stretch are about equal, so the two peaks are always about the same size. This unusual C-H peak can often be used to distinguish between an aldehyde and a ketone. ,

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The intermolecular forces have important contribution in helping us to understand the interaction between atoms in same or different kind of molecules. The interaction can involve polar or non polar molecules and ions. There are three common and this module will focus more in-depth the interaction involving polar molecules and interaction between polar molecules and ions. Dipole moment (\(\mu\)) is the measure of net molecular polarity, which is the magnitude of the charge \(Q\) at either end of the molecular dipole times the distance \(r\) between the charges. \[\mu = Q \times r\] Dipole moments tell us about the charge separation in a molecule. The larger the difference in electronegativities of bonded atoms, the larger the dipole moment. For example, NaCl has the highest dipole moment because it has an ionic bond (i.e. highest charge separation). In the Chloromethane molecule (CH Cl), chlorine is more electronegative than carbon, thus attracting the electrons in the C—Cl bond toward itself (Figure 1). As a result, chlorine is slightly negative and carbon is slightly positive in C—Cl bond. Since one end of C-Cl is positive and the other end is negative, it is described as a polar bond. To indicate the increased in electron density, the dipole is represented by an arrow with a cross at one end. The cross end of the arrow represents the positive end and the point of the arrow represents the negative end of the dipole. The vector will point from plus to minus charge and run parallel with the bond between 2 atoms. The symbol δ indicates the partial charge of an individual atom. In addition, the direction of vector implies the physical movement of electrons to an atom that has more electronegativity when 2 atoms are separated by a distance of r. In other words, the electrons will spend more time around atom that has larger electronegativity. A polar molecule is a molecule where one end has a positive electrical charge and the other end has a negative charge due to the arrangement or geometry of its atoms. Because polar molecules have a positive and negative charge ends, the positive charge end of a molecule will attract to the negative end of adjacent molecule with the same or different kind of molecule. The attraction beween two polar molecules is called . The attraction between two dipoles create a very strong intermolecular force, which have great influence in the evaporation of liquid and condensation of gas. Since water are polar molecules, the interaction between water molecules are so strong that it takes a lot of energy to break the bond between the water molecules. Therefore, the boiling point of polar substances are higher than those of nonpolar substance due to stronger intermolecular force among polar molecules. When a polar molecule is mixed with ion, the positive charge end of the polar molecule will be attracted to the negative charge called anion on the ion. Also, the positive charge called cation on the ion will be attracted to the negative charge end of the polar molecule. This is stronger than the dipole-dipole interaction between polar molecules, but is weaker than the ion-ion interaction. Since ions and polar molecule have positive and negative charge, we can use to evaluate the force between them \[ F = k_e \dfrac{q_1q_2}{r^2}\] with The strength of ion-dipole interaction is based on the distance between ion and polar molecule, the charge of the ion, and dipole magnitude. The closer ion and polar molecule are, the stronger the intermolecular force is between polar molecule and ion. An ion with higher charge will make the attraction stronger. Last, a greater magnitude of dipole will cause stronger attraction.

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Boranes are compounds consisting of boron and hydrogen. They were investigated systematically by the german scientist Alfred Stock at the beginning of the 19th century. The most basic example is diborane (\(\ce{B2H6}\)), all boranes are electron-deficient compounds. For \(\ce{B2H6}\) usually 14 electrons are needed to form 2c,2e-bonds, but only 12 valence electrons are present. Because of this there are two B-H-B bonds, which have three centers, but only two electrons (3c, 2e bond). This can be interpreted as a molecular orbital that is formed by combining the contributed atomic orbitals of the three atoms. In more complicated boranes not only B-H-B bonds but also B-B-B 3c, 2e-bonds occur. In such a bond the three B-atoms lie at the corners of an equilateral triangle with their sp hybrid orbitals overlapping at its center. One of the common properties of boranes is, that they are flammable or react spontaneously with air. They burn with a characteristic green flame. And they are colorless, diamagnetic substances. In neutral boranes the number of boron atoms is given by a prefix and the number of Hydrogen-atoms is given in parentheses behind the name. example: \(\ce{B5H11}\) -> pentaborane(11), \(\ce{B4H10}\) -> tetraborane(10) For ions primarily the number of hydrogen-atoms and than the number of boron-atoms is given, behind the name the charge is given in parentheses. example: \(\ce{[B6H6]^{2-}}\) -> hexahydrohexaborat(2-) Wades rule helps to predict the general shape of a borane from its formula. The polyhedra are always made up of triangular faces, so they are called . Usually there are three possible structure types: There exist also other structures like the hypho-boranes, but they are less important. Diborane can be synthesized by an exchange reaction (metathesis) of a boron halide with \(\ce{LiAlH4}\) or \(\ce{LiBH4}\) in ether, for example: \[\ce{3 LiAlH4 + 4 BF3 -> 2 B2H6 + 3 LiAlF4}\] The reaction has to be done under vacuum or with exclusion of air, because diborane burns in contact with air. Higher boranes are obtained by controlled pyrolysis of diborane in the gas phase. example: \[\ce{H2B6 (g) -> 2BH3 (g)}\] \[\ce{B2H6 (g) + BH3(g) -> B3H7(g) + H2(g)}\] \[\ce{BH3 (g) + B3H7(g) -> B4H10(g)}\]

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The carbonyl bond is both a strong bond and a reactive bond. The bond energy varies widely with structure, as we can see from the carbonyl bond energies in Table 16-1. Methanal has the weakest bond \(\left( 166 \: \text{kcal} \right)\) and carbon monoxide the strongest \(\left( 237.3 \: \text{kcal} \right)\). Irrespective of these variations, the carbonyl bond not only is significantly but also is than a carbon-carbon double bond. A typical difference in stability and reactivity is seen in hydration: The equilibrium constant for ethene hydration is considerably greater than for methanal hydration, largely because the carbon-carbon double bond is . Even so, methanal adds water rapidly and reversibly at room temperature without need for a catalyst. The corresponding addition of water to ethene occurs only in the presence of strongly acidic catalysis (Section 10-3E, Table 15-2). The reactivity of the carbonyl bond is primarily due to the difference in electronegativity between carbon and oxygen, which leads to a considerable contribution of the dipolar resonance form with oxygen negative and carbon positive: The polarity of the carbonyl bond facilitates addition of water and other polar reagents relative to addition of the same reagent to alkene double bonds. This we have seen previously in the addition of organometallic compounds \(\overset{\delta \ominus}{\ce{R}} \overset{\delta \oplus}{\ce{-MgX}}\) and \(\overset{\delta \ominus}{\ce{R}} \overset{\delta \oplus}{\ce{-Li}}\) to carbonyl compounds (Section 14-12A). Alkene double bonds are normally untouched by these reagents: Likewise, alcohols add readily to carbonyl compounds, as described in Section 15-4E. However, we must keep in mind the possibility that, whereas additions to carbonyl groups may be rapid, the equilibrium constants may be small because of the strength of the carbonyl bond. The important reactions of carbonyl groups characteristically involve addition at one step or another. For the reactions of organometallic reagents and alcohols with carbonyl compounds (Chapters 14 and 15), you may recall that steric hindrance plays an important role in determining the ratio between addition and other, competing reactions. Similar effects are observed in a wide variety of other reactions. We expect the reactivity of carbonyl groups in addition processes to be influenced by the size of the substituents thereon, because when addition occurs the substituent groups are pushed back closer to one another. In fact, reactivity and equilibrium constant decrease with increasing bulkiness of substituents, as in the following series (also see Table 15-3): Strain effects also contribute to reactivity of cyclic carbonyl compounds. The normal bond angles around a carbonyl group are about \(120^\text{o}\): Consequently if the carbonyl group is on a small carbocyclic ring, there will be substantial angle strain and this will amount to about \(120^\text{o} - 60^\text{o} = 60^\text{o}\) of strain for cyclopropanone, and \(120^\text{o} - 90^\text{o} = 30^\text{o}\) of strain for cyclobutanone (both values being for the \(\angle \ce{C-C-C}\) at the carbonyl group). Addition of a nucleophile such as \(\ce{CH_3OH}\) (cf. Section 15-4E) to these carbonyl bonds creates a tetrahedral center with less strain in the ring bonds to \(\ce{C_1}\): Thus the hemiketal from cyclopropanone will have \(109.5^\text{o} = 60^\text{o} = 49.5^\text{o}\), and that from cyclobutanone \(109.5^\text{o} - 90^\text{o} = 19.5^\text{o}\) of strain at \(\ce{C_1}\). This change in the angle strain means that a sizable enhancement of the reactivity and equilibrium constant for addition is expected. In practice, the strain effect is so large that cyclopropanone reacts rapidly with methanol to give a stable hemiketal from which the ketone cannot be recovered. Cyclobutanone is less reactive than cyclopropanone, but more reactive than cyclohexanone or cyclopentanone. Electrical effects also are important in influencing the ease of addition to carbonyl groups. Electron-attracting groups facilitate the addition of nucleophilic reagents to carbon by increasing its positive character: Thus compounds such as the following add nucleophilic reagents readily: \(^1\)An electrical dipole results when unlike charges are separated. The magnitude of the dipole, its , is given by \(e \times r\), where \(e\) is the magnitude of the charges and \(r\) is the distance the charges are separated. Molecular dipole moments are measured in debye units \(\left( \text{D} \right)\). A pair of ions, \(\overset{\oplus}{\ce{C}}\) and \(\overset{\ominus}{\ce{O}}\), as at the \(\ce{C=O}\) distance of \(1.22 \: Å\), would have a dipole moment of \(5.9 \: \text{D}\). Thus, if the dipole moment of a carbonyl compound is \(2.7 \: \text{D}\), we can estimate the "\(\%\) ionic character" of the bond to be \(\left( 2.7/5.9 \right) \times 100 = 46\%\). The analysis is oversimplified in that the charges on the atom are not point charges and we have assumed that all of the ionic character of the molecule is associated with the \(\ce{C=O}\) bond. One should be cautious in interpreting dipole moments in terms of the ionic character of bonds. Carbon dioxide has dipole moment, but certainly has polar \(\ce{C=O}\) bonds. The problem is that the dipoles associated with the \(\ce{C=O}\) bonds of \(\ce{CO_2}\) are equal and to each other and, as a result, cancel. Thus, \(\overset{\delta \ominus}{\ce{O}} \overset{\delta \oplus}{\ce{-C}} \overset{\delta \ominus}{\ce{-O}}\) has no dipole moment, even though it has highly polar bonds. and (1977)

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This page discusses a few compounds of the Group 1 elements (lithium, sodium, potassium, rubidium and cesium), including some information about the nitrates, carbonates, hydrogen carbonates and hydrides of the metals. Group 1 compounds are more resistant to heat than the corresponding compounds in Group 2. Lithium compounds often behave similarly to Group 2 compounds, but the rest of Group 1 act differently in various ways. Most nitrates tend to decompose on heating to the metal oxide, brown fumes of nitrogen dioxide, and oxygen. For example, a typical Group 2 nitrate like magnesium nitrate decomposes this way: \[ 2Mg(NO_3)_2 (s) \rightarrow 2MgO(s) + 4NO_2 (g) + O_2 (g)\] In Group 1, lithium nitrate behaves in the same way, producing lithium oxide, nitrogen dioxide, and oxygen as shown: \[ 4LiNO_3 (s) \rightarrow 2Li_2O (s) + 4NO_2 (g) + O_2 (g)\] The other Group 1 nitrates, however, do not decompose completely at regular laboratory temperatures. They produce the metal nitrite and oxygen, but no nitrogen dioxide: \[ 2XNO_3 (s) \rightarrow 2XNO_2(s) + O_2 (g)\] Each of the nitrates from sodium to cesium decomposes in this way; the only difference is in the temperature required for the reaction to proceed. For larger metals, the decomposition is more difficult and requires higher temperatures. Most carbonates decompose on heating to the metal oxide and carbon dioxide. For example, a typical Group 2 carbonate like calcium carbonate decomposes like this: \[ CaCO_3 (s) \rightarrow CaO(s) + CO_2 \] In Group 1, lithium carbonate behaves in the same way, producing lithium oxide and carbon dioxide: \[ Li_2CO_3 (s) \rightarrow Li_2O(s) + CO_2 \] The rest of the Group 1 carbonates do not decompose at laboratory temperatures, although at higher temperatures this becomes possible. The decomposition temperatures again increase down the Group. The Group 2 hydrogen carbonates such as calcium hydrogen carbonate are so unstable to heat that they only exist in solution. Any attempt to extract them from solution causes them to decompose to the carbonate, carbon dioxide and water as shown: \[ Ca(HCO_3)_2 (aq) \rightarrow CaCO_3 (s) + CO_2 (g) + H_2O (l)\] By contrast, the Group 1 hydrogen carbonates are stable enough to exist as solids, although they do decompose easily on heating. For example, this is the reaction for sodium hydrogen carbonate: \[ 2NaHCO_3 (s) \rightarrow Na_2CO_3 (s) + CO_2 (g) + H_2O (l)\] Detailed explanations are given for the carbonates because the diagrams are easier to draw. Exactly the same arguments apply to the nitrates or hydrogen carbonates. There are two ways of explaining the increase in thermal stability down the Group. The hard way is in terms of the energetics of the process; the simple way is in terms of the polarizing ability of the positive ions. A small positive ion has a large amount of charge packed into a small volume of space; this is especially true if it has a charge greater than +1. An ion with a high charge density has a marked distorting effect on any negative ions which happen to be nearby. A larger positive ion has the same charge spread over a larger volume of space. Its charge density is therefore lower, and it causes less distortion to nearby negative ions. The molecular structure of carbonate is given below: This figure shows two carbon-oxygen oxygen atoms delocalized This is a more complicated version of the bonding in benzene or in ions like ethanoate. The next diagram shows the delocalized electrons. The shading shows electron density, implying a greater chance of finding electrons around the oxygen atoms than near the carbon. Imagine that this ion is placed next to a positive ion. The positive ion attracts the delocalized electrons in the carbonate ion towards itself. The carbonate ion becomes polarized. The diagram shows what happens with an ion from Group 2, carrying two positive charges: If this system is heated, the carbon dioxide breaks free, leaving a metal oxide. The amount of heat required depends on how polarized the ion was. If it is highly polarized, less heat is required than if it is only slightly polarized. If the positive ion only has one positive charge, the polarizing effect is lessened. This is why the Group 1 compounds are more thermally stable than those in Group 2. The Group 1 compound must be heated more because the carbonate ion is less polarized by a singly-charged positive ion. The smaller the positive ion, the higher the charge density, and the greater the effect on the carbonate ion. As the positive ions get bigger down the group, they have less effect on the carbonate ions near them. To compensate, the compound must be heated more in order to force the carbon dioxide to break off and leave the metal oxide. In other words, carbonates become more thermally stable down the group. The polarization argument exactly the same for these compounds. The small positive ions at the top of the Group polarize the nitrate or hydrogen carbonate ions to a greater extent than the larger positive ions at the bottom. Again, the Group 1 compounds need more heat than those in Group 2 because the Group 1 ions are less polarizing. Group 1 compounds are more soluble than the corresponding ones in Group 2. Group 2 carbonates are virtually insoluble in water. Magnesium carbonate (the most soluble Group 2 carbonate) has a solubility By contrast, the least soluble Group 1 carbonate is lithium carbonate. A saturated solution has a concentration of about 1.3 g per 100 g of water at 20°C. The other carbonates in the group are very soluble, with solubilities increasing to an astonishing 261.5 g per 100 g of water at this temperature for cesium carbonate. Solubility of the carbonates increases down Group 1. The least soluble hydroxide in Group 1 is lithium hydroxide, but it is still possible to make a solution with a concentration of 12.8 g per 100 g of water at 20°C. The other hydroxides in the group are even more soluble. Solubility of the hydroxides increases down Group 1. In Group 2, the most soluble is barium hydroxide—it is only possible to make a solution of concentration around 3.9 g per 100 g of water at the same temperature of . It is difficult to explain the trends in solubility. The discussion on Group 2 of the periodic table explains why the usual explanations for these trends are not accurate. Group 1 metal hydrides are white crystalline solids; each contains the metal ion and a hydride ion, H . They have the same crystal structure as sodium chloride, which is why they are called saline or salt-like hydrides. Because they can react violently with water or moist air, they are normally supplied as suspensions in mineral oil. Group 1 hydrides are made by passing hydrogen gas over the heated metal. For example, for lithium hydride: \[2Li + H_2 \rightarrow 2LiH\] Two of the most common reactions include electrolysis and reactions with water. On heating, most of these hydrides decompose into the metal and hydrogen before they melt. It is, however, possible to melt lithium hydride and to electrolyze the melt. The metal is deposited at the cathode as expected. Hydrogen is given off at the anode (the positive electrode); this is convincing evidence for the presence of the negative hydride ion in lithium hydride. The anode equation is: \[ 2H^- \rightarrow H_2 + 2e^-\] The other Group 1 hydrides can be electrolyzed in solution in various molten mixtures such as a mixture of lithium chloride and potassium chloride. These mixtures melt at lower temperatures than the pure chlorides. Group 1 hydrides react violently with water releasing hydrogen gas and producing aqueous metal hydroxide. For example, sodium hydride reacts with water to produce sodium hydroxide and hydrogen gas: \[ NaH + H_2O \rightarrow NaOH + H_2\] Jim Clark ( )

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This page explains what happens when organic compounds absorb UV or visible light, and why the wavelength of light absorbed varies from compound to compound. When we were talking about the various sorts of orbitals present in organic compounds on the introductory page (see above), you will have come across this diagram showing their relative energies: Remember that the diagram isn't intended to be to scale - it just shows the relative placing of the different orbitals. When light passes through the compound, energy from the light is used to promote an electron from a bonding or non-bonding orbital into one of the empty anti-bonding orbitals. The possible electron jumps that light might cause are: In each possible case, an electron is excited from a full orbital into an empty anti-bonding orbital. Each jump takes energy from the light, and a big jump obviously needs more energy than a small one. Each wavelength of light has a particular energy associated with it. If that particular amount of energy is just right for making one of these energy jumps, then that wavelength will be absorbed - its energy will have been used in promoting an electron. We need to work out what the relationship is between the energy gap and the wavelength absorbed. Does, for example, a bigger energy gap mean that light of a lower wavelength will be absorbed - or what? It is easier to start with the relationship between the frequency of light absorbed and its energy: You can see that if you want a high energy jump, you will have to absorb light of a higher frequency. The greater the frequency, the greater the energy. That's easy - but unfortunately UV-visible absorption spectra are always given using wavelengths of light rather than frequency. That means that you need to know the relationship between wavelength and frequency. You can see from this that the higher the frequency is, the lower the wavelength is. So, if you have a bigger energy jump, you will absorb light with a higher frequency - which is the same as saying that you will absorb light with a lower wavelength. Important summary: The larger the energy jump, the lower the wavelength of the light absorbed. An absorption spectrometer works in a range from about 200 nm (in the near ultra-violet) to about 800 nm (in the very near infra-red). Only a limited number of the possible electron jumps absorb light in that region. Look again at the possible jumps. This time, the important jumps are shown in black, and a less important one in grey. The grey dotted arrows show jumps which absorb light outside the region of the spectrum we are working in. Remember that bigger jumps need more energy and so absorb light with a shorter wavelength. The jumps shown with grey dotted arrows absorb UV light of wavelength less that 200 nm. The important jumps are: That means that in order to absorb light in the region from 200 - 800 nm (which is where the spectra are measured), the molecule must contain either pi bonds or atoms with non-bonding orbitals. Remember that a non-bonding orbital is a lone pair on, say, oxygen, nitrogen or a halogen. Groups in a molecule which absorb light are known as chromophores. The diagram below shows a simple UV-visible absorption spectrum for buta-1,3-diene - a molecule we will talk more about later. Absorbance (on the vertical axis) is just a measure of the amount of light absorbed. The higher the value, the more of a particular wavelength is being absorbed. You will see that absorption peaks at a value of 217 nm. This is in the ultra-violet and so there would be no visible sign of any light being absorbed - buta-1,3-diene is colorless. You read the symbol on the graph as "lambda-max". In buta-1,3-diene, CH =CH-CH=CH , there are no non-bonding electrons. That means that the only electron jumps taking place (within the range that the spectrometer can measure) are from pi bonding to pi anti-bonding orbitals. A chromophore such as the carbon-oxygen double bond in ethanal, for example, obviously has pi electrons as a part of the double bond, but also has lone pairs on the oxygen atom. That means that both of the important absorptions from the last energy diagram are possible. You can get an electron excited from a pi bonding to a pi anti-bonding orbital, or you can get one excited from an oxygen lone pair (a non-bonding orbital) into a pi anti-bonding orbital. The non-bonding orbital has a higher energy than a pi bonding orbital. That means that the jump from an oxygen lone pair into a pi anti-bonding orbital needs less energy. That means it absorbs light of a lower frequency and therefore a higher wavelength. Ethanal can therefore absorb light of two different wavelengths: Both of these absorptions are in the ultra-violet, but most spectrometers won't pick up the one at 180 nm because they work in the range from 200 - 800 nm. Consider these three molecules: Ethene contains a simple isolated carbon-carbon double bond, but the other two have conjugated double bonds. In these cases, there is delocalization of the pi bonding orbitals over the whole molecule. Now look at the wavelengths of the light which each of these molecules absorbs. All of the molecules give similar UV-visible absorption spectra - the only difference being that the absorptions move to longer and longer wavelengths as the amount of delocalization in the molecule increases. Why is this? You can actually work out what must be happening. . . . and that's what is happening. Compare ethene with buta-1,3-diene. In ethene, there is one pi bonding orbital and one pi anti-bonding orbital. In buta-1,3-diene, there are two pi bonding orbitals and two pi anti-bonding orbitals. This is all discussed in detail on the introductory page that you should have read. The highest occupied molecular orbital is often referred to as the HOMO - in these cases, it is a pi bonding orbital. The lowest unoccupied molecular orbital (the LUMO) is a pi anti-bonding orbital. Notice that the gap between these has fallen. It takes less energy to excite an electron in the buta-1,3-diene case than with ethene. In the hexa-1,3,5-triene case, it is less still. If you extend this to compounds with really massive delocalisation, the wavelength absorbed will eventually be high enough to be in the visible region of the spectrum, and the compound will then be seen as colored. A good example of this is the orange plant pigment, beta-carotene - present in carrots, for example. Beta-carotene has the sort of delocalization that we've just been looking at, but on a much greater scale with 11 carbon-carbon double bonds conjugated together. The diagram shows the structure of beta-carotene with the alternating double and single bonds shown in red. The more delocalization there is, the smaller the gap between the highest energy pi bonding orbital and the lowest energy pi anti-bonding orbital. To promote an electron therefore takes less energy in beta-carotene than in the cases we've looked at so far - because the gap between the levels is less. Remember that less energy means a lower frequency of light gets absorbed - and that's equivalent to a longer wavelength. Beta-carotene absorbs throughout the ultra-violet region into the violet - but particularly strongly in the visible region between about 400 and 500 nm with a peak about 470 nm. If you have read the page in this section about electromagnetic radiation, you might remember that the wavelengths associated with the various colors are approximately: So if the absorption is strongest in the violet to cyan region, what color will you actually see? It is tempting to think that you can work it out from the colors that are left - and in this particular case, you wouldn't be far wrong. Unfortunately, it isn't as simple as that! Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors. If you arrange some colors in a circle, you get a "color wheel". The diagram shows one possible version of this. An internet search will throw up many different versions! colors directly opposite each other on the color wheel are said to be complementary colors. Blue and yellow are complementary colors; red and cyan are complementary; and so are green and magenta. Mixing together two complementary colors of light will give you white light. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. In the beta-carotene case, the situation is more confused because you are absorbing such a range of wavelengths. However, if you think of the peak absorption running from the blue into the cyan, it would be reasonable to think of the color you would see as being opposite that where yellow runs into red - in other words, orange. You have probably used phenolphthalein as an acid-base indicator, and will know that it is colorless in acidic conditions and magenta (bright pink) in an alkaline solution. How is this color change related to changes in the molecule? The structures of the two differently colored forms are: Both of these absorb light in the ultra-violet, but the one on the right also absorbs in the visible with a peak at 553 nm. The molecule in acid solution is colorless because our eyes can't detect the fact that some light is being absorbed in the ultra-violet. However, our eyes do detect the absorption at 553 nm produced by the form in alkaline solution. 553 nm is in the green region of the spectrum. If you look back at the color wheel, you will find that the complementary color of green is magenta - and that's the color you see. So why does the color change as the structure changes? What we have is a shift to absorption at a higher wavelength in alkaline solution. As we've already seen, a shift to higher wavelength is associated with a greater degree of delocalisation. Here is a modified diagram of the structure of the form in acidic solution - the colorless form. The extent of the delocalization is shown in red. Notice that there is delocalization over each of the three rings - extending out over the carbon-oxygen double bond, and to the various oxygen atoms because of their lone pairs. But the delocalization doesn't extend over the whole molecule. The carbon atom in the centre with its four single bonds prevents the three delocalized regions interacting with each other. Now compare that with the magenta form: The rearrangement now lets the delocalization extend over the entire ion. This greater delocalization lowers the energy gap between the highest occupied molecular orbital and the lowest unoccupied pi anti-bonding orbital. It needs less energy to make the jump and so a longer wavelength of light is absorbed. Increasing the amount of delocalization shifts the absorption peak to a higher wavelength. You will know that methyl orange is yellow in alkaline solutions and red in acidic ones. The structure in alkaline solution is: In acid solution, a hydrogen ion is (perhaps unexpectedly) picked up on one of the nitrogens in the nitrogen-nitrogen double bond. This now gets a lot more complicated! The positive charge on the nitrogen is delocalized (spread around over the structure) - especially out towards the right-hand end of the molecule as we've written it. The normally drawn structure for the red form of methyl orange is . . . But this can be seriously misleading as regards the amount of delocalization in the structure for reasons discussed below (after the red warning box) if you are interested. Let's work backwards from the absorption spectra to see if that helps. The yellow form has an absorption peak at about 440 nm. That's in the blue region of the spectrum, and the complementary color of blue is yellow. That's exactly what you would expect. The red form has an absorption peak at about 520 nm. That's at the edge of the cyan region of the spectrum, and the complementary color of cyan is red. Again, there's nothing unexpected here. Notice that the change from the yellow form to the red form has produced an increase in the wavelength absorbed. An increase in wavelength suggests an increase in delocalisation. That means that there must be more delocalization in the red form than in the yellow one. Here again is the structure of the yellow form: delocalization will extend over most of the structure - out as far as the lone pair on the right-hand nitrogen atom. If you use the normally written structure for the red form, the delocalization seems to be broken in the middle - the pattern of alternating single and double bonds seems to be lost. But that is to misunderstand what this last structure represents. If you draw the two possible Kekulé structures for benzene, you will know that the real structure of benzene isn't like either of them. The real structure is somewhere between the two - all the bonds are identical and somewhere between single and double in character. That's because of the delocalization in benzene. The two structures are known as canonical forms, and they can each be thought of as adding some knowledge to the real structure. For example, the bond drawn at the top right of the molecule is neither truly single or double, but somewhere in between. Similarly with all the other bonds. The two structures we've previously drawn for the red form of methyl orange are also canonical forms - two out of lots of forms that could be drawn for this structure. We could represent the delocalized structure by: These two forms can be thought of as the result of electron movements in the structure, and curly arrows are often used to show how one structure can lead to the other. In reality, the electrons haven't shifted fully either one way or the other. Just as in the benzene case, the actual structure lies somewhere in between these. You must also realize that drawing canonical forms has no effect on the underlying geometry of the structure. Bond types or lengths or angles don't change in the real structure. For example, the lone pairs on the nitrogen atoms shown in the last diagram are both involved with the delocalisation. For this to happen all the bonds around these nitrogens must be in the same plane, with the lone pair sticking up so that it can overlap sideways with orbitals on the next-door atoms. The fact that in each of the two canonical forms one of these nitrogens is shown as if it had an ammonia-like arrangement of the bonds is potentially misleading - and makes it look as if the delocalization is broken. The problem is that there is no easy way of representing a complex delocalized structure in simple structural diagrams. It is bad enough with benzene - with something as complicated as methyl orange any method just leads to possible confusion if you aren't used to working with canonical forms. It gets even more complicated! If you were doing this properly there would be a host of other canonical forms with different arrangements of double and single bonds and with the positive charge located at various places around the rings and on the other nitrogen atom. The real structure can't be represented properly by any one of this multitude of canonical forms, but each gives a hint of how the delocalization works. If we take the two forms we have written as perhaps the two most important ones, it suggests that there is delocalization of the electrons over the whole structure, but that electron density is a bit low around the two nitrogens carrying the positive charge on one canonical form or the other. Finally, we get around to an attempt at an explanation as to why the delocalization is greater in the red form of methyl orange in acid solution than in the yellow one in alkaline solution. The answer may lie in the fact that the lone pair on the nitrogen at the right-hand end of the structure as we've drawn it is more fully involved in the delocalization in the red form. The canonical form with the positive charge on that nitrogen suggests a significant movement of that lone pair towards the rest of the molecule. Doesn't the same thing happen to the lone pair on the same nitrogen in the yellow form of methyl orange? Not to the same extent. Any canonical form that you draw in which that happens produces another negatively charged atom somewhere in the rest of the structure. Separating negative and positive charges like this is energetically unfavourable. In the red form, we aren't producing a new separation of charge - just shifting a positive charge around the structure.

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The production of iron from its ore involves an carried out in a blast furnace. Iron ore is usually a mixture of iron and vast quantities of impurities such as sand and clay referred to as gangue. The iron found in iron ores are found in the form of iron oxides. As a result of these impurities, iron must be first separated from the gangue and then converted to pure iron. This is accomplished by the method of , a high temperature process. The high temperatures are needed for the reduction of iron and the oxidation of the limestone which will be seen below. The production of iron from its ore involves a redox reaction carried out in a blast furnace. The furnace is filled at the top with the iron ore oxide most commonly hematite (\(Fe_2O_3\)) but can also magnetite (\(Fe_3O_4\)), carbon called coke and limestone (\(CaCO_3\)). For the purpose of this discussion the iron ore oxide hematite (\(Fe_2O_3\)) will be shown. On a side note, Hematite gets its name from the Greek word meaning blood like because of the color of one form of its powder. The Ancient Greeks believed that large deposits of hematite were formed from battles that were fought and the blood from these battles flowed into the ground. To begin the process a blast of hot air is forced in at the bottom of the furnace that helps create a large temperature variation with the bottom being 2273 K and the top 473 K. The amount of oxygen is strictly controlled so that carbon monoxide is the main product as shown: \[2C (s) + O_2 \; (g) \longrightarrow 2CO (g) + \rm{heat}\] Similarly carbon and carbon monoxide both contribute in the reduction of the iron (III) oxide to give the impure metal as shown: \[Fe_2O_3 \; (s) + 3C \, (s) \longrightarrow 2Fe (l) + 3CO_2 (g)\] \[Fe_2O_3 \; (s) + 3CO_2 (g) \longrightarrow 2Fe (l) + 3CO_2 \; (g)\] One of the most interesting part of this redox reaction is that the majority of the carbon dioxide formed is itself reduced when it comes to contact with the unburned coke and produce more reducing agent. As the process continue the molten iron flow down through the furnace and collects at the bottom, where it is removed through an opening in the side. When it cools the impure iron is brittle and some cases soft due to the presence of the small impurities, such as sulfur and phosphorus. Thus the impure iron coming from the bottom of the furnace is further purified. The most common method is the basic oxygen furnace. In the furnace, oxygen is blown into the impure iron. This is vital because the oxygen oxidizes the phosphorus and sulfur shown in the following redox reactions: \[P_4 (s) + 5O_2 \;(g) \longrightarrow P_4O_{10} \; (g)\] \[S_8 (s) + 8O_2 \; (g) \longrightarrow 8SO_2 \; (g)\] The oxides either escapes as gases or react with basic oxides that are added or used to line the furnace. This final purification step removes much of the impurities and the result is ordinary carbon steel. Thus iron is obtained through the process of oxidation-reduction.

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There are a variety of reactions whereby rings are formed through addition to double or triple bonds. An especially simple example is the addition of ethene to 1,3-butadiene to give cyclohexene: This is the prototype reaction, which has proved so valuable in synthesis that it won its discoverers, O. Diels and K. Alder, the Nobel Prize in chemistry in 1950. The Diels-Alder reaction is both a 1,4 addition or ethene to 1,3-butadiene and a 1,2 addition of butadiene to ethene. It can be called a and as such results in the formation of a six-membered ring. Many other cycloadditions are known, such as [2 + 2], other types of [4 + 2], and [2 + 2 + 2], which give different size of rings. Some specific examples follow: The synthetic importance of these reactions is very great and, because many of them often involve dienes, we will discuss their general characteristics in this chapter. The most valuable cycloaddition reaction almost certainly is the [4 + 2], or Diels-Alder, reaction and will be discussed in detail. There is one very important point you should remember about the Diels-Alder reaction: The reaction usually occurs well only when the [2] component is substituted with electron-attracting groups and the [4] component is substituted with electron-donating groups, or the reverse. The most common arrangement is to have the alkene (usually referred to as the ) substituted with electron-attracting groups such as \(\ce{-CO_2H}\), \(\ce{-COR}\), or \(\ce{-C \equiv N}\). For example, A list of the more reactive dienophiles carrying electron-attracting groups is given in Table 13-1. Ethene and other simple alkenes generally are poor dienophiles and react with 1,3-butadiene only under rather extreme conditions and in low yield. However, when the diene is substituted with several electron-attracting groups such as chlorine or bromine, electron-donating groups on the facilitate the reaction. Many substances, such as 2-methylpropene, that act as dienophiles with hexachlorocyclopentadiene simply will not undergo [4 + 2] addition with cyclopentadiene itself: The Diels-Alder reaction is . The diene reacts in an unfavorable conformation in which its double bonds lie in a plane on the same side (cis) of the single bond connecting them. This (or ) conformation is required to give a stable product with a cis double bond. Addition of ethene to the alternate and more stable ( ) conformation would give an impossibly strained -cyclohexene ring. Possible transition states for reaction in each conformation follow, and it will be seen that enormous molecular distortion would have to take place to allow addition of ethene to the transoid conformation: Cyclic dienes usually react more readily than open-chain dienes, probably because they have their double bonds fixed in the proper conformation for [4 + 2] cycloaddition, consequently the price in energy of achieving the configuration already has been paid: Further evidence of stereospecificity in [4 + 2] additions is that the configurations of the diene and the dienophile are in the adduct. This means that the reactants (or addends) come together to give addition. Two examples follow, which are drawn to emphasize how suprafacial addition occurs. In the first example, dimethyl -butadioate adds to 1,3-butadiene to give a cis-substituted cyclohexene: In the second example, suprafacial approach of a dienophile to the 2,5 carbons of -2,4-hexadiene is seen to lead to a product with two methyl groups on the same side of the cyclohexene ring: (The use of models will help you visualize these reactions and their stereochemistry.) There is a further feature of the Diels-Alder reaction that concerns the stereochemical orientation of the addends. In the addition of -butanedioic anhydride (maleic anhydride) to cyclopentadiene there are two possible ways that the diene and the dienophile could come together to produce different products. These are shown in Equations 13-3 and 13-4: In practice, the adduct with the endo\(^2\) configuration usually is the major product. As a general rule, Diels-Alder additions tend to proceed to favor that orientation that corresponds to having the diene double bonds and the unsaturated substituents of the dienophile closest to one another. This means that addition by Equation 13-3 is more favorable than by Equation 13-4, but the degree of endo-exo stereospecificity is not as high as the degree of stereospecificity of suprafacial addition to the diene and dienophile. There are exceptions to favored endo stereochemistry of Diels-Alder additions. Some of these exceptions arise because the , dissociation being particularly important at high temperature. The exo configuration is generally more stable than the endo and, given time to reach equilibrium (cf. ), the exo isomer may be the major adduct. Thus endo stereospecificity can be expected only when the additions are subject to . The reactivities of dienes in the Diels-Alder reaction depend on the number and kind of substituents they possess. The larger the substituents are, or the more of them, at the ends of the conjugated system, the slower the reaction is likely to be. There also is a marked difference in reactivity with diene configuration. Thus -1,3-pentadiene is substantially less reactive toward a given dienophile (such as maleic anhydride) than is -1,3-pentadiene. In fact, a mixture of the cis and trans isomers can be separated by taking advantage of the difference in their reactivities on cycloaddition: There is little evidence to support simple radical or polar mechanisms (such as we have discussed previously) for the Diels-Alder reaction. As the result of many studies the reaction seems best formulated as a process in which the bonds between the diene and the dienophile are formed essentially simultaneously: We already have discussed a few addition reactions that appear to occur in a concerted manner. These include the addition of diimide, ozone, and boron hydrides to alkenes ( , , and ). Concerted reactions that have cyclic transition states often are called . Other examples will be considered in later chapters. We indicated previously that sulfur dioxide \(\left( \ce{SO_2} \right)\) and 1,3-butadiene form a [4 + 1] cycloaddition product: This reaction is more readily reversible than most Diels-Alder reactions, and the product largely dissociates to the starting materials on heating to \(120^\text{o}\). The cycloadduct is an unsaturated cyclic , which can be hydrogenated to give the saturated cyclic sulfone known as "sulfolane": This compound is used extensively in the petrochemical industry as a selective solvent. The reversibility of the diene-\(\ce{SO_2}\) cycloaddition makes it useful in the purification of reactive dienes. 2-Methyl-1,3-butadiene (isoprene) is purified commercially in this manner prior to being polymerized to rubber ( ): Neither 1,3-cyclopentadiene nor 1,3-cyclohexadiene react with sulfur dioxide, probably because the adducts would be too highly strained: Many naturally occurring organic compounds contain six-membered carbon rings, but there are relatively few with four-membered carbon rings. After encountering the considerable ease with which six-membered rings are formed by [4 + 2] cycloaddition, we might expect that the simpler [2 + 2] cycloadditions to give four-membered rings also should go well, provided that strain is not too severe in the products. In fact, the dimerization of ethene is thermodynamically favorable: Nonetheless, this and many other [2 + 2] cycloaddition reactions do not occur on simple heating. However, there are a few exceptions. One is the dimerization of tetrafluoroethene, which perhaps is not surprising, considering the favorable thermodynamic parameters: What is surprising is that addition of \(\ce{CF_2=CF_2}\) to 1,3-butadiene gives a cyclobutane and a cyclohexane, although the [2 + 2] product probably is about \(25 \: \text{kcal mol}^{-1}\) less stable than the [4 + 2] product: Such [2 + 2] additions generally are limited to polyhaloethenes and a few substances with cumulated double bonds, such as 1,2-propadiene \(\left( \ce{CH_2=C=CH_2} \right)\) and ketenes \(\left( \ce{R_2C=C=O} \right)\). Some examples follow: Many [2 + 2] cycloadditions that do not occur by simply heating the possible reactants can be achieved by with ultraviolet light. The following example, [2 + 2] addition of 2-cyclopentenone to cyclopentene, occurs photochemically but not thermally: In all such photochemical cycloadditions the energy required to achieve a transition state, which can amount to \(100 \: \text{kcal mol}^{-1}\) or more, is acquired by absorption of light. . A striking example is the photochemical conversion of norbornadiene to quadricyclene. The reverse of this reaction can occur with almost explosive violence in the presence of appropriate metal catalysts or on simple heating: Why do some [2 + 2] cycloadditions occur thermally and others photochemically? What is special about fluoroalkenes and cumulated dienes? The answers are complex, but it appears that most [2 + 2] cycloadditions, unlike the Diels-Alder [4 + 2] cycloadditions, go by routes (see ). Why the two types of thermal cycloaddition have different mechanisms will be discussed in and . \(^2\)In general, the designation or refers to configuration in bridged or polycyclic ring systems such as those shown in Equations 13-3 and 13-4. With reference to the bridge atoms, a substituent is exo if it is on the same side as the bridge, or endo if it is on the opposite side. Further examples are In drawing endo and exo isomers, it is best to represent the actual spatial relationships of the atoms as closely as possible. The cyclohexane ring is shown here in the boat form ( ) because it is held in this configuration by the methylene group that bridges the 1,4 positions. If you do not see this, we strongly advise that you construct models. and (1977)

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This page describes reactions of the halogens that do not fall under the other categories in other pages in this section. All the reactions described here are redox reactions. The following examples illustrate the decrease in reactivity of the halogens down Group 7. Fluorine combines explosively with hydrogen even under cold, dark conditions, evolving hydrogen fluoride gas. A mixture of chlorine and hydrogen explodes if exposed to sunlight or a flame, producing hydrogen chloride gas. This reaction can be controlled by lighting a jet of hydrogen and then lowering it into a gas jar of chlorine. The hydrogen burns at a slower, constant rate, and hydrogen chloride gas is formed as before. Bromine vapor and hydrogen combine with a mild explosion when ignited. Hydrogen bromide gas is formed. Iodine and hydrogen combine only partially even on constant heating. An equilibrium exists between the hydrogen and the iodine and hydrogen iodide gas. Each of these reactions has an equation of the form: \[ H_2 + X_2 \rightarrow 2HX\] A minor exception is made for iodine: the single arrow is replaced with a reversible sign. \[ P_4 + 6Br_2 \rightarrow 4PBr_3\] \[ 2P + 3Br_2 \rightarrow 2PBr_3\] \[ P_4 + 10Cl_2 \rightarrow 4PCl_5\] \[ PCl_3 + Cl_2 \rightleftharpoons PCl_5\] \[ 2Na + Cl_2 \rightarrow 2NaCl\] \[ 2Fe + 3F_2 \rightarrow 2FeF_3\] Chlorine gas in contact with hot iron forms iron(III) chloride. Anhydrous iron(III) chloride forms black crystals; any trace of water present in the apparatus or in the chlorine reacts with the crystals, turning them reddish-brown. The equation for this reaction is given below: \[ 2Fe + 3Cl_2 \rightarrow 2FeCl_3\] The iron is again oxidized from a state of zero to +3. Bromine vapor passed over hot iron triggers a similar, slightly less vigorous reaction, shown below; iron(III) bromide is produced. Anhydrous iron(III) bromide usually appears as a reddish-brown solid. \[2Fe + 3Br_2 \rightarrow 2FeBr_3\] In this reaction the iron is again oxidized to a +3 state. The reaction between hot iron and iodine vapor produces gray iron(II) iodide, and is much less vigorous. This reaction, the equation for which is given below, is difficult to carry out because he product is always contaminated with iodine. \[ Fe + 2I_2 \rightarrow FeI_2\] Iodine is only capable of oxidizing iron to the +2 oxidation state. \[ 2Fe_3+ + 2I^- \rightarrow 2Fe^{2+} + I_2\] \[ 2NaOH + Cl_2 \rightarrow NaCl + NaClO + H_2O\] \[ 6NaOH + 3Cl_2 \rightarrow 5NaCl + NaClO_3 + 3H_2O\] \[ NaOH + Cl_2 \rightarrow NaCl + NaClO_3 + ? \] \[ NaOH + Cl_2 \rightarrow 5NaCl + NaClO_3 + ? \] Jim Clark ( )

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A metal (from Greek μέταλλον métallon, "mine, quarry, metal") is a material (an element, compound, or alloy) that is typically hard, opaque, shiny, and has good electrical and thermal conductivity. Metals are generally - that is, they can be hammered or pressed permanently out of shape without breaking or cracking - as well as (able to be fused or melted) and (able to be drawn out into a thin wire). About 91 of the 118 elements in the periodic table are metals (some elements appear in both metallic and non-metallic forms). Atoms of metals readily lose their outer shell electrons, resulting in a free flowing cloud of electrons within their otherwise solid arrangement. This provides the ability of metallic substances to easily transmit heat and electricity. While this flow of electrons occurs, the solid characteristic of the metal is produced by electrostatic interactions between each atom and the electron cloud. This type of bond is called a metallic bond. Crystalline solids consist of repeating patterns of its components in three dimensions (a crystal lattice) and can be represented by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a . Many metals adopt close packed structures i.e. cubic close packed (face centred cubic) and hexagonal close packed structures. A simple model for both of these is to assume that the metal atoms are spherical and are packed together in the most efficient way (close packing or closest packing). For closest packing, every atom has 12 equidistant nearest neighbours, and therefore a coordination number of 12. If the close packed structures are considered as being built of layers of spheres then the difference between hexagonal close packing and cubic close packed is how each layer is positioned relative to others. It can be envisaged that for a regular buildup of layers: This is not a close packed structure. Here each metal atom is at the centre of a cube with 8 nearest neighbors, however the 6 atoms at the centres of the adjacent cubes are only approximately 15% further away so the coordination number can therefore be considered to be 14 when these are included. Note that if the body centered cubic unit cell is compressed along one 4 fold axis the structure becomes cubic close packed (face centred cubic). Melting points are chosen as a simple measure of the stability or strength of the metallic lattice. Some simple trends can be noted. The transition metals have generally higher melting points than the others. In the alkali metals (Group 1) and alkaline earth metals (Group 2) the melting point decreases as atomic number increases, but in transition metal groups with incomplete d-orbital subshells, the heavier elements have higher melting points. For a given period, the melting points reach a maximum at around Group 6 and then fall with increasing atomic number. Mercury, caesium and gallium have melting points below 30 °C whereas all the other metals have sufficiently high melting points to be solids at "room temperature". The structures of the metals can be summarised by the table below which shows that most metals crystallise in roughly equal amounts of bcc, hcp and ccp lattices. The alkali metals have their outermost electron in an s-orbital and this electronic configuration results in their characteristic properties. The alkali metals provide the best example of group trends in properties in the periodic table, with elements exhibiting well-characterized homologous behaviour. The alkali metals have very similar properties: they are all shiny, soft, highly reactive metals at standard temperature and pressure and readily lose their outermost electron to form cations with charge +1. They can all be cut easily with a knife due to their softness, exposing a shiny surface that tarnishes rapidly in air due to oxidation by atmospheric moisture and oxygen. Because of their high reactivity, they must be stored under oil to prevent reaction with air, and are found naturally only in salts and never as the free element. In the modern IUPAC nomenclature, the alkali metals comprise the group 1 elements, excluding hydrogen (H), which is only nominally considered a group 1 element. Much of the information in these course notes has been sourced from under the Creative Commons . 'Inorganic Chemistry' - C. Housecroft and A.G. Sharpe, Prentice Hall, 4th Ed., 2012, ISBN13: 978-0273742753, pps 24-27, 43-50, 172-176, 552-558, 299-301, 207-212 'Basic Inorganic Chemistry' - F.A. Cotton, G. Wilkinson and P.L. Gaus, John Wiley and Sons, Inc. 3rd Ed., 1994. 'Introduction to Modern Inorganic Chemistry' - K.M. Mackay, R.A. Mackay and W. Henderson, International Textbook Company, 5th Ed., 1996.

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In , we derive Boyle’s law from Newton’s laws using the assumption that all gas molecules move at the same speed at a given temperature. This is a poor assumption. Individual gas molecules actually have a wide range of velocities. In Chapter 4, we derive the Maxwell–Boltzmann distribution law for the distribution of molecular velocities. This law gives the fraction of gas molecules having velocities in any range of velocities. Before developing the Maxwell–Boltzmann distribution law, we need to develop some ideas about distribution functions. Most of these ideas are mathematical. We discuss them in a non-rigorous way, focusing on understanding what they mean rather than on proving them. The overriding idea is that we have a real-world source of data. We call this source of data the . We can collect data from this source to whatever extent we please. The datum that we collect is called the distribution’s . We call each possible value of the random variable an . The process of gathering a set of particular values of the random variable from a distribution is often called or . The set of values that is collected is called . The set of values that comprise the sample is often called “the data.” In scientific applications, the random variable is usually a number that results from making a measurement on a physical system. Calling this process “drawing a sample” can be inappropriate. Often we call the process of getting a value for the random variable “doing an experiment”, “doing a test”, or “making a trial”. As we collect increasing amounts of data, the accumulation quickly becomes unwieldy unless we can reduce it to a mathematical model. We call the mathematical model we develop a , because it is a function that expresses what we are able to learn about the data source—the distribution. A distribution function is an equation that summarizes the results of many measurements; it is a mathematical model for a real-world source of data. Specifically, it models the of an event with which we obtain a particular outcome. We usually believe that we can make our mathematical model behave as much like the real-world data source as we want if we use enough experimental data in developing it. Often we talk about . By a statistic, we mean any mathematical entity that we can calculate from data. Broadly speaking a distribution function is a statistic, because it is obtained by fitting a mathematical function to data that we collect. Two other statistics are often used to characterize experimental data: the and the . The mean and variance are defined for any distribution. We want to see how to estimate the mean and variance from a set of experimental data collected from a particular distribution. We distinguish between discrete and continuous distributions. A is a real-world source of data that can produce only particular data values. A coin toss is a good example. It can produce only two outcomes—heads or tails. A is a real-world source of data that can produce data values in a continuous range. The speed of an automobile is a good example. An automobile can have any speed within a rather wide range of speeds. For this distribution, the random variable is automobile speed. Of course we can generate a discrete distribution by aggregating the results of sampling a continuous distribution; if we lump all automobile speeds between 20 mph and 30 mph together, we lose the detailed information about the speed of each automobile and retain only the total number of automobiles with speeds in this interval.

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Geometric Isomers are isomers that differ in the arrangement of the ligands around the metal or the central atom. In other words, these isomers differ from each other based on where the ligands are placed in the coordinate compound. This will be much easier to understand as examples will be considered. There are 2 main types of geometric isomers: Cis-Trans Isomers are isomers that differ in the arrangement of two ligands in square planar and octahedral geometry. Cis isomers are isomers where the two ligands are 90 degrees apart from one another in relation to the central molecule. This is because Cis isomers have a bond angle of 90 , between two same atoms. Trans isomers, on the other hand, are isomers where the two ligands are on opposite sides in a molecule because trans isomers have a bond angle of 180 , between the two same atoms. When naming cis or trans isomers, the name begins either with cis or trans, whichever applies, followed by a hyphen and then the name of a molecule. For example a cis isomer of CoCl F would be called cis-CoCl F . Finally, the last thing to keep in mind when examining cis and trans isomers is that geometries can have cis or trans isomers. Examples of both isomers are provided below. CoCl F : (Color scheme: pink=cobalt, blue=fluorine, green=chlorine) This above is an example of the molecule cis-CoCl F or cis-dichlorodifluorocobalt (IV). The molecule pictured above is a cis isomer because both fluorine and chlorine ligands, respectively, are on the same side of the molecule. Additionally, one can approximate that the bond angle between each of the chlorine atoms and between each of the fluorine atoms is 90 . CoCl F : (Color scheme: pink=cobalt, blue=fluorine, green=chlorine) (Note, the differences in the length of the bond between the two pictures are not intentional and have nothing to do with cis-trans isomerism) This above is an example of the molecule trans-CoCl F or trans-dichlorodifluorocobalt (IV). We know the above molecule is a trans isomer because the two same chlorine atoms and the two same fluorine atoms are opposite each other. Furthermore, the bond angle between the two chlorine atoms and between the two fluorine atoms is 180 . The above examples were all for square planar geometry but as the examples below illustrate, cis-trans isomerism can also occur in octahedral geometry. Both the molecules below are isomers of the molecule SCl F (color scheme: yellow=sulfur, blue=fluorine, green=chlorine). SCl F : We know this isomer above is a cis isomer because both the chlorine ligands are on the same side and the bond angle between the chlorine atoms appears to be 90 . SCl F : The isomer above is a trans isomer because the chlorine ligands are on opposite sides and the bond angle between the chlorine atoms is 180 . All other isomers are essentially just rotations of these two isomers. Once again when trying to find cis and trans isomers look at the arrangement of the ligands. If two same ligands are on the same side, it is a cis isomer and if the ligands are on opposite sides, it is a trans isomer. Another way to tell the isomers apart is the bond angles: cis isomers have a 90 bond angle whereas trans isomers have a 180 bond angle. Mer-Fac isomers are easier to notice than cis-trans isomers in the sense that they only exist in octahedral geometry. Just like cis-trans isomers, mer-fac isomers are determined based on whether or not the ligands exist on the same side. Instead of dealing with 2 ligands, mer-fac isomers deal with 3 ligands. If the 3 ligands are all on the same side, the isomer is called a fac-isomer. Another way to identify fac isomers is to look at the bond angle between the ligands because fac isomers have a 90 bond angle between each of the 3 atoms. The mer isomer on the other hand is where only 2 of the 3 ligands are on the same side. In mer isomers, there exists a 90 -90 -180 bond angle between the 3 same ligands. In terms of nomenclature, mer-fac isomers follow the same rule as cis-trans isomers where you put the isomer type, followed by a hyphen, followed by the molecular formula. Examples have been provided below. Below is an example of the fac isomer, fac-CoCl F : (note the color scheme: pink=cobalt, green=chlorine, blue=fluorine) Through the 2d version, it is easier to see how the ligands are all on the same side. Nonetheless, in the 3D version, one can observe that the bond angle between the 3 same ligands is 90 , thus making this isomer a fac-isomer. Below is an example of the mer isomer, mer-CoCl F : (note the color scheme: pink=cobalt, green=chlorine, blue=fluorine) ligands

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This page looks at the assumptions which are made in the Kinetic Theory about ideal gases, and takes an introductory look at the Ideal Gas Law: pV = nRT. There is no such thing as an ideal gas, of course, but many gases behave approximately as if they were ideal at ordinary working temperatures and pressures. Real gases are dealt with in more detail on another page. The assumptions are: And then two absolutely key assumptions, because these are the two most important ways in which real gases differ from ideal gases: The ideal gas equation is: \[ pV = nRT \] On the whole, this is an easy equation to remember and use. The problems lie almost entirely in the units, which should be in strict SI units Pressure is measured in Pascals, Pa - sometimes expressed as newtons per square meter, N m . These mean exactly the same thing. Be careful if you are given pressures in kPa (kilopascals). For example, 150 kPa is 150,000 Pa. You must make that conversion before you use the ideal gas equation. Should you want to convert from other pressure measurements: This is the most likely place for you to go wrong when you use this equation. That's because the SI unit of volume is the cubic metre, m - not cm or dm with 1 m = 1000 dm = 1,000,000 cm . So if you are inserting values of volume into the equation, you first have to convert them into cubic metres. You would have to divide a volume in dm by 1000, or in cm by a million. Similarly, if you are working out a volume using the equation, remember to covert the answer in cubic metres into dm or cm if you need to - this time by multiplying by a 1000 or a million. If you get this wrong, you are going to end up with a silly answer, out by a factor of a thousand or a million. So it is usually fairly obvious if you have done something wrong, and you can check back again. This is easy, of course - it is just a number. You already know that you work it out by dividing the mass in grams by the mass of one mole in grams. You will most often use the ideal gas equation by first making the substitution to give: A value for R will be given you if you need it, or you can look it up in a data source. The SI value for R is 8.31441 J K mol . The temperature has to be in Kelvin. Don't forget to add 273 if you are given a temperature in degrees Celsius. Calculations using the ideal gas equation are included in my calculations book (see the link at the very bottom of the page), and I can't repeat them here. There are, however, a couple of calculations that I haven't done in the book which give a reasonable idea of how the ideal gas equation works. If you have done simple calculations from equations, you have probably used the molar volume of a gas. 1 mole of any gas occupies 22.4 dm at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm at room temperature and pressure (taken as about 20°C and 1 atmosphere). These figures are actually only true for an ideal gas, and we'll have a look at where they come from. We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure. First, we have to get the units right. 0°C is 273 K. T = 273 K 1 atmosphere = 101325 Pa. p = 101325 Pa We know that n = 1, because we are trying to calculate the volume of 1 mole of gas. And, finally, R = 8.31441 J K mol . Slotting all of this into the ideal gas equation and then rearranging it gives: And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres. The molar volume of an ideal gas is therefore 22.4 dm at stp. And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure. This is about as tricky as it gets using the ideal gas equation. The density of ethane is 1.264 g dm at 20°C and 1 atmosphere. Calculate the relative formula mass of ethane. The density value means that 1 dm of ethane weighs 1.264 g. Again, before we do anything else, get the awkward units sorted out. A pressure of 1 atmosphere is 101325 Pa. The volume of 1 dm has to be converted to cubic metres, by dividing by 1000. We have a volume of 0.001 m . The temperature is 293 K. Now put all the numbers into the form of the ideal gas equation which lets you work with masses, and rearrange it to work out the mass of 1 mole. The mass of 1 mole of anything is simply the relative formula mass in grams.So the relative formula mass of ethane is 30.4, to 3 sig figs. Now, if you add up the relative formula mass of ethane, C H using accurate values of relative atomic masses, you get an answer of 30.07 to 4 significant figures. Which is different from our answer - so what's wrong? There are two possibilities. If you need to know about real gases, now is a good time to read about them. Jim Clark ( )

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An reaction occurs when the temperature of a system increases due to the evolution of heat. This heat is released into the surroundings, resulting in an overall negative quantity for the heat of reaction (\(q_{rxn} < 0\)). An reaction occurs when the temperature of an isolated system decreases while the surroundings of a non-isolated system gains heat. Endothermic reactions result in an overall positive heat of reaction (\(q_{rxn} > 0\)). Exothermic and endothermic reactions cause energy level differences and therefore differences in enthalpy (\(ΔH\)), the sum of all potential and kinetic energies. ΔH is determined by the system, not the surrounding environment in a reaction. A system that heat to the surroundings, an exothermic reaction, has a ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system. \[ \ce{C(s) + O2(g) -> CO2 (g)} \tag{ΔH = –393.5 kJ} \] \[\ce{ H2 (g) + 1/2 O2 (g) -> H2O(l)} \tag{ΔH = –285.8 kJ} \] The enthalpies of these reactions are than zero, and are therefore exothermic reactions. A system of reactants that heat from the surroundings in an endothermic reaction has a \(ΔH\), because the enthalpy of the products is higher than the enthalpy of the reactants of the system. \[ \ce{N2(g) + O2(g) -> 2NO(g)} \tag{ΔH = +180.5 kJ > 0}\] \[ \ce{ C(s) + 2S(s) -> CS2(l)} \tag{ΔH = +92.0 kJ > 0}\] Because the enthalpies of these reactions are than zero, they are endothermic reactions. The equilibrium constant (\(K_c\)) defines the relationship among the concentrations of chemical substances involved in a reaction at equilibrium. The states that if a stress, such as changing temperature, pressure, or concentration, is inflicted on an equilibrium reaction, the reaction will shift to restore the equilibrium. For exothermic and endothermic reactions, this added stress is a change in temperature. The equilibrium constant shows how far the reaction will progress at a specific temperature by determining the ratio of products to reactions using equilibrium concentrations. The equilibrium expression for the following equation \[aA + bB \rightleftharpoons cC + dD \] is given below: \[ K_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{Equation:Kc}\] where If the products dominate in a reaction, the value for K is greater than 1. The larger the K value, the more the reaction will tend toward the right and thus to completion. Suppose that the following reaction is at equilibrium and that the concentration of N is 2 M, the concentration of H is 4 M, and the concentration of NH is 3 M. What is the value of K ? \[\ce{ N2 + 3H2 <=> 2NH3} \nonumber\] The coefficients and the concentrations are plugged into the \(K_c\) expression (Equation \ref{Equation:Kc}) to calculate its value. \[\begin{align*} K_c &= \dfrac{[NH_3]^2}{[N_2]^1[H_2]^3} \\[4pt] &= \dfrac{[3]^2}{[2]^1[4]^3} \\[4pt] &= \dfrac{9}{128} \\[4pt] &= 0.07 \end{align*}\] Determine \(K_c\) for the following chemical reaction at equilibrium if the molar concentrations of the molecules are: \[\ce{2H2 (g) + 2NO (g) <=> 2H2O (g) + N2 (g)} \nonumber\] Using the \(K_c\) expression (Equation \ref{Equation:Kc}) and plugging in the concentration values of each molecule: \[ \begin{align*} K_c &= \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \\[4pt] &= \dfrac{[\ce{H2O}]^2[\ce{N2}]^1}{[\ce{H2}]^2[\ce{NO}]^2} \\[4pt] &= \dfrac{0.20^2\, 0.1}{0.20^2 \, 0.10 ^2 }\\[4pt] &= 10 \end{align*}\] For the previous equation, does the equilibrium favor the products or the reactants? Because \(K_c = 10 > 1\), the reaction favors the products. In the following reaction, the temperature is increased and the \(K_ c\) value decreases from 0.75 to 0.55. Is this an exothermic or endothermic reaction? \[\ce{N_2 (g) + 3H_2 <=>2NH_3 (g) } \nonumber\] Because the K value decreases with an increase in temperature, the reaction is an exothermic reaction. In the following reaction, in which direction will the equilibrium shift if there is an increase in temperature and the enthalpy of reaction is given such that \(ΔH\) is -92.5 kJ? \[\ce{PCl3(g) + Cl2(g) <=> PCl_5(g)} \nonumber\] In the initial reaction, the energy given off is negative and thus the reaction is exothermic. However, an increase in temperature allows the system to absorb energy and thus favor an endothermic reaction; the equilibrium will shift to the left.

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The search for less problematic hydrogen-atom transfers for use in the Barton-McCom­bie reaction has led to com­pounds with phos­phorus–hydrogen bonds. These include dialkyl­phos­phine oxides ( ), alkyl phosphites ( ), hypo­phos­phorous acid ( ), and salts of hypo­phos­phorous acid ( ) (Figure 2). All of these compounds can function as inexpensive, nontoxic hydrogen-atom transfers that form the chain-carrying radicals needed for reaction and do not produce by­pro­ducts difficult to remove. An example of a reaction in which hydrogen donation is from a P–H bond is shown in eq 13. Alkyl phosphites ( ) are excellent hydrogen-atom transfers, but reactions involv­ing these com­pounds have the disadvantage of not being able to be initi­ated by 2,2'-azobis(isobutyronitrile); ben­zoyl peroxide usually is the initi­ator. Reac­tions in which the hydrogen-atom transfer is a dialkyl­phosphine oxide ( , hypophosphorous acid ( ), or a salt of hypo­phos­phorous acid ( ) can be initiated by AIBN. Because it is difficult to completely remove water from hypo­phos­phorous acid and its salts, these donors are less attrac­tive choices when moisture sen­si­tive compounds are reacting.

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Transition metals consist of elements from the d-block found between the group IIa and the group IIb elements of the . A transition metal is an element that forms one or more stable ions which have incompletely filled d sub-orbitals. While these transition metals contain at most two electrons in their outer shell, they are called d-block elements because their next highest energy shell has incompletely filled d sub-orbitals. The d sub-orbitals in transition metals are filled up, progressively going across the periodic table from left to right. These d sub-orbitals of transition metals are readily available for metallic bonding. They give way to general properties of transition metals such as good electrical and heat conductivity and high melting points. Since both the outer shell and the d sub-orbital shells can be used for bonding, it allows transition metals to bond with a variety of elements in many shapes. It should be noted, however, that not all d-block elements are transition metals. Some of the elements, such as and Zinc, have either empty or completely filled d sub-orbitals and do not meet the criteria to be considered transition metals. The of transition metal elements are characterized as having full outer sub-orbitals and the second outermost d sub-orbitals incompletely filled, with the exception of Copper which loses one 4s orbital electron to the 3d sub-orbital for increased stability. The electron configuration for copper is 1s 2s 2p 3s 3p 4s 3d or given in terms of a noble gas configuration which is [Ar]4s 3d and not [Ar]4s 3d , as would be expected from the Aufbau principle. The loss of the electron from the 4s orbital completes the 3d sub-orbital and leaves it in a more stable, lower energy state. This lowest energy electron configuration is determined by the electron-pairing energy and the exchange energy. The two electrons in the 4s orbital will repel each other slightly, and this will cause a minor increase in the energy of the atom. The removal of one of the electrons from the 4s shell will remove this energy increase which is caused by the pairing energy. When electrons can exchange between orbitals, the energy of the atom is lowered. This electron exchange can occur when there are degenerate electron configurations, and the electrons of the same spin can exchange between orbitals. This exchange energy is defined by the spin-Hamiltonian: \[ H_o = -2S_1S_2 \] where J is the set of many electron exchange parameters, and S and S are the spin operators of the two electrons. Manganese, on the other hand, has an electron configuration of 1s 2s 2p 3s 3p 4s 3d and a noble gas configuration of [Ar]4s 3d , resulting in one unpaired electron in each 3d sub-orbital. Similar to copper, the exchange energy is maximized since the 3d sub-orbital is exactly half-filled with 5 electrons without the need to steal electrons from the 4s orbital. The electrons in the d sub-orbitals of both copper and manganese are subject to which says that no two fermions in an atom can have the same quantum number. Copper is found in two common ionic forms, Cuprous Cu(I) with a noble gas electron configuration of [Ar]4s 3d and Cupric Cu(II) with [Ar]4s 3d . Cu(III) and even Cu(IV) ionic forms of copper have been formed. \[ 2Cu(I) \rightarrow Cu(0) + Cu(II)\] Manganese has seven ionic forms from Mn(I) to Mn(VII). The two most common forms are Mn(II), with a noble gas electronic configuration of [Ar]4s 3d and Mn(VII), with a configuration of [Ar]4s 3d and a formal loss of all seven electrons from the 3d and 4s orbitals. Crystal field splitting diagrams can be used to determine the way strong and weak ligands will affect the magnitude of ligand field, splitting parameters and determining the number of unpaired electrons are in each transition metal. Using H O as a weak field ligand, complexed with Mn(II) in the form of [Mn(H O) ] , the low oxidation state of manganese will cause a high spin with five unpaired electrons. The crystal field splitting of a ligand in the case with water, has an important effect on the d-orbital energy levels. Since the ligand electrons have repulsions with the d sub-orbital electrons, the d-orbital energy levels are raised. However, they are not raised equally. In the case of this octahedral complex ion, ligands approach along the x-, y- and z-axis. When the six ?-donor water molecules are coordinated with the manganese ion, the orbitals (\(d_{z^2}\) and \(d_{x^2-y^2}\)) are head-to-head with the ligand and have a greater repulsion and thus, a higher energy. The energy levels of the (\(d_{xy}\), \(d_{xz}\), \(d_{yz}\)) orbitals are also increased, due to this repulsion but to a lesser extent. Using NH as a strong field splitting ligand coordinated to copper, such as [Cu(NH ) ] , which has an ionic form of copper Cu(II). The Cu(II) has nine d sub-orbital electrons, and since there are only five orbitals, there can only be one configuration of the electrons which results in one unpaired electron. The complex is paramagnetic, but the structure cannot be determined from magnetic properties. With a Mn(II) ion with a weak field ligand, the energy splitting is low and the cost of placing an electron into a higher energy d orbital is lower than the repulsion energy of pairing two electrons together. Thus, the Mn(II) ion in a weak field ligand setting, produces a high spin as all five electrons are unpaired and the ion is paramagnetic. One electron is placed into each d orbital according to , and the “high spin” octahedral complex has all five orbitals singly occupied. The magnetic susceptibility is determined to be 5.92, using the formula: \[ \chi= [n(n+2)]^{1/2} \] where n is the number of unpaired electrons. In the case of Cu(II) in a strong ligand field such as NH , the energy required to pair the electrons is not relevant since there are nine d sub-orbital electrons that need to be placed in five d orbital shells. This results in a single unpaired electron which means that the complex is paramagnetic, but the high spin vs low spin associated energy splittings are not relevant. The ground state electronic term symbols represent a shorthand notation which is predicted, using Hund’s rule for the angular momentum quantum numbers of an atom with multiple electrons. This is related to the energy level of the electronic configurations of the atoms. The term symbol has the form: \[ \large ^{2S+1}L_J \] where S is the spin quantum number; L is the orbital momentum quantum number in spectroscopic notation given in L= 0,1,2,3,4,5 are S,P,D,F,G,H, respectively; and finally, J is the angular momentum quantum number. This symbol assumes that all spins are combined to produce S; all orbital angular momenta are combined to produce L; and all the spin and orbital angular momenta are combined to produce J. The multiplicity of a term is the value of 2S+1, provided that L ? S, and is the number of levels of the term. The values for S, L, and J are constructed by the application of Clebsch-Gordan series. To construct a term symbol arising from an electronic configuration, first determine the possible values of L, and identify the spectroscopic letters for the terms. The possible values of S are determined similarly, using a Clebsch-Gordan series, and the multiplicities are calculated. Finally, the values for J are constructed again, using a Clebsch-Gordan series by combining S and L. \[ j = j_1+j_2, j_1+j_{2-1},…,|j_1-j_2| \] Using the ground state electronic configuration, the electrons are distributed according to Pauli’s exclusion principle to the available orbitals. The first orbitals filled are those with the highest ml values; S is calculated by added all the ms values; L is calculated by added together the m values for each electron; and J is then determined, using a set of rules. If the subshell is half-filled, then the L=0 and J will be equal to S. If the subshell is more than half-filled, then J=L+S; if the subshell is less than half-filled, then J=|L-S|. The ground electronic term symbol for copper is thus determined to be S . The term symbol for Cu(II), with an electronic configuration [Ar]4s 3d is calculated to be D . The ground electronic term symbol for manganese is thus determined to be S . The term symbol for Mn(II) is also S , since the orbital with the highest ml value still has 5 electrons in it and thus, L=0. .  .  .  .  . 

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In spontaneous processes for an isolated system, there is a competition between minimizing the energy of the system and increasing the dispersal of energy within the system. If energy is constant, then the system will evolve to maximize energy dispersal. If energy dispersal is not a factor, the system will evolve to minimize its energy. We already have a quantitative basis for the energy of a system and we need to do the same for energy dispersal. Suppose we have a small reversible change \(dU\) in the energy of an ideal gas. We know that \(U\) only depends on temperature: \[dU = C_vdT \nonumber \] We also know that any reversible work would be volume work. \[δw_{rev} = -PdV \nonumber \] This means that we can write: \[\begin{align*} δq_{rev} &= dU - δw_{rev} \\[4pt] &= C_vdT + PdV \end{align*} \nonumber \] Let us examine if this represents an . If \(\delta q\) were an exact differential, we could write the total differential: \[\delta q_{rev} =\left(\frac{\partial q}{dT}\right)_V dT + \left(\frac{\partial q}{dV}\right)_T dV \nonumber \] And the following would be true: \[\frac{\partial^2 q_{rev}}{\partial T\partial V}=\frac{\partial^2 q_{rev}}{\partial V\partial T} \nonumber \] From our equation above, we know that: \[\frac{\partial q_{rev}}{\partial T}=C_V \nonumber \] \[\frac{\partial q_{rev}}{\partial V}=P \nonumber \] Therefore, the following should be true: \[\frac{\partial C_V}{\partial V}=\frac{\partial P}{\partial T} \nonumber \] However, \[\dfrac{\partial C_v}{\partial V}=0 \nonumber \] Because \(C_v\) does not depend on volume (only \(T\), just like \(U\): it is its derivative). And: \[\dfrac{\partial P}{\partial T} = \dfrac{\partial nRT}{\partial T} = \dfrac{nR}{V} \nonumber \] Which is not zero!! Clearly, \(δq_{rev}\) is , but look what happens if we multiply everything with an 'integration factor' \(1/T\): \[\dfrac{δq_{rev}}{T} = \dfrac{C_v}{T}dT + \dfrac{P}{T}dV \nonumber \] \[\dfrac{\partial C_v/T}{\partial V} = 0 \nonumber \] Because \(\dfrac{C_v}{T}\) does not depend on volume. However, \[\dfrac{\partial (P/T)}{\partial T} = \dfrac{\partial (nR/V)}{\partial T} = 0 \nonumber \] , the quantity \(dS = \dfrac{δq_{rev}}{T}\) is an exact differential, so \(S\) is a state function and it is called . Entropy is the dispersal of energy, or equivalently, the measure of the number of possible microstates of a system.

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The reactions of alkanes discussed in are processes, which means that the bonds are made and broken through radical or atomic intermediates. In contrast, the \(S_\text{N}\) and \(E\) reactions of alkyl halides, considered in Chapter 8, involve heterolytic bond cleavage and ionic reagents or products. An especially important factor contributing to the differences between the reactions of the alkanes and alkyl halides is the slight ionic character of \(\ce{C-H}\) compared to \(\ce{C}\)-halide bonds (see ). The alkenes are like the alkanes in being nonpolar compounds ( ) and it may come as a surprise that many important reactions of alkenes are heterolytic reactions. Why should this be so? No doubt because the electrons in the alkene double bonds are more exposed and accessible than the electrons in an alkane \(\ce{C-C}\) bond. This is evident from the atomic-orbital models of ethene described in . The electrons of the double bond are pushed outward by their mutual repulsions, and their average positions are considerably farther from the bond axis than the electron positions of a single bond (Figure 10-6). In such circumstances, electrophilic reagents, which act to acquire electrons in chemical reactions ( ), are expected to be particularly reactive. This is actually the case. Furthermore, reagents that are primarily nucleophilic (electron-donating) are notoriously poor for initiating reactions at carbon-carbon double bonds. Exceptions occur when the double bonds carry substituents with a sufficiently high degree of electron-attracting power to reduce the electron density in the double bond enough to permit attack by a nucleophilic agent. Example of electrophilic reagents that normally add to carbon-carbon double bonds of alkenes to give saturated compounds include halogens (\(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\)), hydrogen halides (\(\ce{HCl}\) and \(\ce{HBr}\)), hypohalous acids (\(\ce{HOCl}\) and \(\ce{HOBr}\)), water, and sulfuric acid: The mechanisms of these reactions have much in common and have been studied extensively from this point of view. They also have very considerable synthetic utility. The addition of water to alkenes (hydration) is particularly important for the preparation of a number of commercially important alcohols. Thus ethanol and 2-propanol (isopropyl alcohol) are made on a very large scale by the hydration of the corresponding alkenes (ethene and propene) using sulfuric or phosphoric acids as catalysts. The nature of this type of reaction will be described later. We shall give particular attention here to the addition of bromine to alkenes because this reaction is carried out very conveniently in the laboratory and illustrates a number of important points about electrophilic addition reactions. Much of what follows applies to addition of the other halogens, except fluorine. A significant observation concerning bromine addition is that it and many of the other reactions listed above proceed in the dark and are influenced by radical inhibitors. This is evidence against a radical-chain mechanism of the type involved in the halogenation of alkanes ( ). However, it does not preclude the operation of radical-addition reactions under other conditions, and, as we shall see later in this chapter, bromine, chlorine, and many other reagents that commonly add to alkenes by ionic mechanisms also can add by radical mechanisms. One alternative to a radical-chain reaction for bromine addition to an alkene would be the simple four-center, one-step process shown in Figure 10-7. The mechanism of Figure 10-7 cannot be correct for bromine addition to alkenes in solution for two important reasons. First, notice that this mechanism requires that the two \(\ce{C-Br}\) bonds be formed on the side of the double bond, and hence produce . However, there is much evidence to show that bromine and many other reagents add to alkenes to form products (Figure 10-8). Cyclohexene adds bromine to give -1,2-dibromocyclohexane: The cis isomer is not formed at all. To give the trans isomer, the two new \(\ce{C-Br}\) bonds have to be formed on of the double bond by antarafacial addition. But this is impossible by a one-step mechanism because the \(\ce{Br-Br}\) bond would have to stretch too far to permit the formation of both \(\ce{C-Br}\) bonds at the same time. The second piece of evidence against the mechanism of Figure 10-7 is that bromine addition reactions carried out in the presence of more than one nucleophilic reagent usually give mixtures of products. Thus the addition of bromine to an alkene in methanol solution containing lithium chloride leads not only to the expected dibromoalkane, but also to products resulting from attack by chloride ions and by the solvent: The intervention of extraneous nucleophiles suggests a mechanism in which the nucleophiles compete for a reactive intermediate formed in one of the steps. A somewhat oversimplified two-step mechanism that accounts for most of the foregoing facts is illustrated for the addition of bromine to ethene. [In the formation shown below, the curved arrows are not considered to have real mechanistic significance, but are used primarily to show which atoms can be regarded as nucleophilic (donate electrons) and which as electrophilic (accept electrons). The arrowheads always should be drawn to point to the atoms that are formulated as accepting a pair of electrons.] The first step (which involves electrophilic attack by bromine on the double bond) produces a bromide ion and a carbocation, as shown in Equation 10-1.\(^1\) As we know from our study of \(S_\text{N}1\) reactions ( ), carbocations react readily with nucleophilic reagents. Therefore in the second step of the bromine-addition mechanism, shown in Equation 10-2, the bromoethyl cation is expected to combine rapidly with bromide ion to give the dibromo compound. However, if other nucleophiles, such as \(\ce{Cl}^\ominus\) or \(\ce{CH_3OH}\), are present in solution, they should be able to compete with bromide ion for the cation, as in Equations 10-3 and 10-4, and mixtures of products will result: To account for the observation that all of these reactions result in antarafacial addition, we must conclude that the first and second steps take place from . The simple carbocation intermediate of Equation 10-1 does not account for formation of the antarafacial-addition product. The results with \(S_\text{N}1\) reactions (Section 8-6) and the atomic-orbital representation (see ) predict that the bonds to the positively charged carbon atom of a carbocation should lie in a plane. Therefore, in the second step of addition of bromine to cycloalkenes, bromide ion could attack either side of the planar positive carbon to give a mixture of - and -1,2-dibromocyclohexanes. Nonetheless, antarafacial addition occurs exclusively: To account for the stereospecificity of bromine addition to alkenes, it has been suggested that in the initial electrophilic attack of bromine a cyclic intermediate is formed that has bromine bonded to carbons of the double bond. Such a "bridged" ion is called a because the bromine formally carries the positive charge: An \(S_\text{N}2\)-type of attack of bromide ion, or other nucleophile, at carbon on the side to the bridging group then results in formation of the antarafacial-addition product: We may seem to have contradicted ourselves because Equation 10-1 shows a carbocation to be formed in bromine addition, but Equation 10-5 suggests a bromonium ion. Actually, the formulation of intermediates in alkene addition reactions as "open" ions or as cyclic ions is a controversial matter, even after many years of study. Unfortunately, it is not possible to determine the structure of the intermediate ions by any direct physical method because, under the conditions of the reaction, the ions are so reactive that they form products more rapidly than they can be observed. However, it is possible to generate stable bromonium ions, as well as the corresponding chloronium and iodonium ions. The technique is to use low temperatures in the absence of any strong nucleophiles and to start with a 1,2-dihaloalkane and antimony pentafluoride in liquid sulfur dioxide: The \(\ce{C_2H_4Br}^\oplus\) ions produced in this way are relatively stable and have been shown by nmr to have the cyclic halonium ion structure. There is a further aspect of polar additions to alkenes that we should consider, namely, that electrophilic reagents form loose complexes with the \(\pi\) electrons of the double bonds of alkenes to reaction by addition. Complexes of this type are called (or ). Formation of a complex between iodine and cyclohexene is demonstrated by the fact that iodine dissolves in cyclohexene to give a solution, whereas its solutions in cyclohexane are . The brown solution of iodine in cyclohexene slowly fades as addition occurs to give colorless -1,2-diiodocyclohexane. Precise Lewis structures cannot be written for charge-transfer complexes, but they commonly are represented as with the arrow denoting that electrons of the double bond are associated with the electrophile. These complexes probably represent the first stage in the formation of addition products by a sequence such as the following for bromine addition: We have seen that electrophiles can react with alkenes to form carbon-halogen bonds by donating positive halogen, \(\ce{Br}^\oplus\), \(\ce{Cl}^\oplus\), or \(\ce{I}^\oplus\). Likewise, carbon-hydrogen bonds can be formed by appropriately strong proton donors, which, of course, are typically strong proton acids. These acids are more effective in the absence of large amounts of water because water can compete with the alkene as a proton acceptor (also see ). Hydrogen chloride addition to ethene occurs by way of a proton-transfer step to give the ethyl cation and a chloride ion (Equation 10-6) followed by a step in which the nucleophilic chloride ion combines with the ethyl cation (Equation 10-7): All of the hydrogen halides \(\ce{HF}\), \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\)) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of \(70\%\) anhydrous hydrogen fluoride and \(30\%\) of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent \(\ce{I_2}\) addition products resulting from iodine formed by oxidation reactions such as \[4 \ce{HI} + \ce{O_2} \rightarrow 2 \ce{I_2} + 2 \ce{H_2O}\] With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled carefully (this will be discussed in ). The stereochemistry of addition depends largely on the structure of the alkene, but for simple alkenes and cycloalkenes, addition occurs predominantly in an antarafacial manner. For example, hydrogen bromide reacts with 1,2-dimethylcyclohexene to give the antarafacial addition product: We mentioned previously that the hydration of alkenes required a strong acid as a catalyst, because water itself is too weak an acid to initiate the proton-transfer step. However, if a small amount of a strong acid such as sulfuric acid is present, hydronium ions, \(\ce{H_3O}^\oplus\), are formed in sufficient amount to protonate reasonably reactive alkenes, although by no means as effectively as does concentrated sulfuric acid. The carbocation formed then is attacked rapidly by a nucleophilic water molecule to give the alcohol as its conjugate acid,\(^2\) which regenerates hydronium ion by transferring a proton to water. The reaction sequence follows for 2-methylpropene: In this sequence, the acid acts as a because the hydronium ion used in the proton addition step is regenerated in the final step. Sulfuric acid (or phosphoric acid) is preferred as an acid catalyst for addition of water to alkenes because the conjugate base, \(\ce{HSO_4-}\) (or \(\ce{H_2PO_4-}\)), is a poor nucleophile and does not interfere in the reaction. However, if the water concentration is kept low by using concentrated acid, addition occurs to give sulfate (or phosphate) esters. The esters formed with sulfuric acid are either alkyl acid sulfates \(\ce{R-OSO_3H}\) or dialkyl sulfates \(\ce{(RO)_2SO_2}\). In fact, this is one of the major routes used in the commercial production of ethanol and 2-propanol. Ethen and sulfuric acid give ethyl hydrogen sulfate, which reacts readily with water in a second step to give ethanol: One of the more confusing features of organic chemistry is the multitude of conditions that are used to carry out a given kind of reaction, such as the electrophilic addition of proton acids to different alkenes. Strong acids, weak acids, water, no water - Why can't there be a standard procedure? The problem is that alkenes have very different tendencies to accept protons. In the vapor phase, \(\Delta H^0\) for addition of a proton to ethene is about \(35 \: \text{kcal}\) more positive than for 2-methylpropene, and although the difference should be smaller in solution, it still would be large. Therefore we can anticipate (and we find) that a much more powerful proton donor is needed to initiate addition of an acid to ethene than to 2-methylpropene. But why not use in all cases a strong enough acid to protonate alkene one might want to have a proton acid add to? Two reasons: First, strong acids can induce undesirable side reactions, so that one usually will try not to use a stronger acid than necessary; second, very strong acid may even prevent the desired reaction from occurring! In elementary chemistry, we usually deal with acids in more or less dilute aqueous solution and we think of sulfuric, hydrochloric, and nitric acids as being similarly strong because each is essentially completely disassociated in dilute water solution: \[\ce{HCl} + \ce{H_2O} \overset{\longrightarrow}{\leftarrow} \ce{H_3O}^\oplus + \ce{Cl}^\ominus\] This does not mean they actually are equally strong acids. It means only that each of the acids is sufficiently strong to donate all of its protons to water. We can say that water has a "leveling effect" on acid strengths because as long as an acid can donate its protons to water, the solution has but one acid "strength" that is determined by the \(\ce{H_3O}^\oplus\) concentration, because \(\ce{H_3O}^\oplus\) is where the protons are. Now, if we use poorer proton acceptors as solvent we find the proton-donating powers of various "strong" acids begin to spread out immensely. Furthermore, new things begin to happen. For example, ethene is not hydrated appreciably by dilute aqueous acid; it just is too hard to transfer a proton from hydronium ion to ethene. So we use concentrated sulfuric acid, which is strong enough to add a proton to ethene. But now we don't get hydration, because any water that is present in concentrated sulfuric acid is virtually all converted to \(\ce{H_3O}^\oplus\), which is non-nucleophilic! \[\ce{H_2SO_4} + \ce{H_2O} \rightarrow \ce{H_3O}^\oplus + \ce{HSO_4-}\] However, formation of \(\ce{H_3O}^\oplus\) leads to formation of \(\ce{HSO_4-}\), which has enough nucleophilic character to react with the \(\ce{CH_3CH_2+}\) to give ethyl hydrogen sulfate and this is formed instead of the conjugate acid of ethanol ( ). The epitome of the use of stronger acid and weaker nucleophile is with liquid \(\ce{SO_2}\) (bp \(\sim 10^\text{o}\)) as the solvent and \(\ce{HBF_6}\) as the acid. This solvent is a very poor proton acceptor (which means that its conjugate acid is a very good proton donor) and \(\ce{SbF_6-}\) is an extremely poor nucleophile. If we add ethene to such a solution, a stable solution of \(\ce{CH_3CH_2+} \ce{SbF_6-}\) is formed. The reason is that there is no better proton acceptor present than \(\ce{CH_2=CH_2}\) and no nucleophile good enough to combine with the cation. The conversion of fumaric acid to malic acid is an important biological hydration reaction. It is one of a cycle of reactions (Krebs citric acid cycle) involved in the metabolic combustion of fuels (amino acids and carbohydrates) to \(\ce{CO_2}\) and \(\ce{H_2O}\) in a living cell. \(^1\)An alternative to Equation 10-1 would be to have \(\ce{Br_2}\) ionize to \(\ce{Br}^\oplus\) and \(\ce{Br}^\ominus\), with a subsequent attack of \(\ce{Br}^\oplus\) on the double bond to produce the carbocation. The fact is that energy required for such an ionization of \(\ce{Br_2}\) is prohibitively large even in water solution \(\left( \Delta H^0 \geq 80 \: \text{kcal} \right)\). One might well wonder why Equation 10-1 could possibly be more favorable. The calculated \(\Delta H^0\) for \(\ce{CH_2=CH_2} + \ce{Br_2} \rightarrow \cdot \ce{CH_2-CH_2Br} + \ce{Br} \cdot\) is \(+41 \: \text{kcal}\), which is only slightly more favorable than the \(\Delta H^0\) for \(\ce{Br_2} \rightarrow 2 \ce{Br} \cdot\) of \(46.4 \: \text{kcal}\). However, available thermochemical data suggest that the ease of transferring an electron from \(\cdot \ce{CH_2CH_2Br}\) to \(\ce{Br} \cdot\) to give \(^\oplus \ce{CH_2CH_2Br} + \ce{Br}^\ominus\) is about \(80 \: \text{kcal}\) more favorable than \(2 \ce{Br} \cdot \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). Thus the overall \(\Delta H^0\) of Equation 10-1 is likely to be about \(85 \: \text{kcal}\) more favorable than \(\ce{Br_2} \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). \(^2\)The terms and are very convenient to designate substances that are difficult to name simply as acids, bases, or salts. The conjugate acid of a compound \(\ce{X}\) is \(\ce{XH}^\oplus\) and the conjugate base of \(\ce{HY}\) is \(\ce{Y}^\ominus\). Thus \(\ce{H_3O}^\oplus\) is the conjugate acid of water, while \(\ce{OH}^\ominus\) is its conjugate base. Water itself is then both the conjugate base of \(\ce{H_3O}^\oplus\) and the conjugate acid of \(\ce{OH}^\ominus\) and (1977)

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In much the same fashion as the partial molar volume is defined, the is defined for compound \(i\) in a mixture: \[ \mu_i = \left( \dfrac{\partial G}{\partial n_i} \right) _{P,T,n_j\neq i} \label{eq1}\] The partial molar function is of particular importance and is called the . The chemical potential tells how the Gibbs function will change as the composition of the mixture changes. Since systems tend to seek a minimum aggregate Gibbs function, the chemical potential will point to the direction the system can move in order to reduce the total Gibbs function and reach equilibrium. In general, the total change in the Gibbs function (\(dG\)) can be calculated from: \[dG = \left( \dfrac{\partial G}{\partial P} \right) _{T,n_i} dP + \left( \dfrac{\partial G}{\partial T} \right) _{P, n_i }dT + \sum_i \left( \dfrac{\partial G}{\partial n_i} \right) _{T,n_j\neq i} dn_i\] Or, by substituting the definition for the chemical potential, and evaluating the pressure and temperature derivatives: \[dG = VdP - SdT + \sum_i \mu_i dn_i\] But as it turns out, the chemical potential can be defined as the partial molar quantity of any of the four major thermodynamic functions \(U\), \(H\), \(A\), or \(G\): The last definition, in which the chemical potential is defined as the partial molar Gibbs function, is the most commonly used, and perhaps the most useful (Equation \ref{eq1}). As the partial most Gibbs function, it is easy to show that: \[d\mu = VdP - SdT\] where \(V\) is the molar volume, and \(S\) is the molar entropy. Using this expression, it is easy to show that: \[\left( \dfrac{\partial \mu}{\partial P} \right) _{T} = V\] and so at constant temperature: \[ \int_{\mu^o}^{\mu} d\mu = \int_{P^o}^{P} V\,dP \label{eq5}\] So that for a substance for which the molar volume is fairly independent of pressure at constant temperature (i. e., \(\kappa_T\) is very small), therefore Equation \ref{eq5} becomes: \[ \int_{\mu^o}^{\mu} d\mu = V \int_{P^o}^{P} dP\] \[ \mu - \mu^o = V(P-P^o)\] or: \[ \mu = \mu^o + V(P-P^o)\] Where \(P^o\) is the standard state pressure (1 bar) and \(\mu^o\) is the chemical potential at the standard pressure. If the substance is highly compressible (such as a gas) the pressure dependence of the molar volume is needed to complete the integral. If the substance is an ideal gas: \[V =\dfrac{RT}{P}\] So at constant temperature, Equation \ref{eq5} then becomes: \[ \int_{\mu^o}^{\mu} d\mu = RT int_{P^o}^{P} \dfrac{dP}{P} \label{eq5b}\] or: \[ \mu = \mu^o + RT \ln \left(\dfrac{P}{P^o} \right) \] A lot of chemistry takes place in solution and therefore this topic is of prime interest for chemistry. The Gibbs free energy of an ideal gas depends logarithmically on pressure: \[G = G^o + RT \ln \dfrac{P}{P^o} \nonumber \] P is and we write: \[G = G^o + RT \ln P \nonumber \] Notice however that although \(P\) and \(P/P^o\) have the same numerical value, the dimensions are different. \(P\) usually has dimensions of bar, but \(P/P^o\) is . If we have a gas mixture we can hold the same logarithmic argument for each partial pressure as the gases do not notice each other. We do need to take into account the number of moles of each and work with (partial) molar values, i.e. the thermodynamic potential: \[μ_j = μ_j^o + RT \ln \dfrac{P_j}{P^o} \label{B} \] If we are dealing with an equilibrium over an ideal liquid solution the situation in the gas phase gives us a probe for the situation in the liquid. The equilibrium must hold for each of components j (say two in binary mixture). That means that for each of them the thermodynamic potential in the liquid and in the gas must be equal: \[μ_j^{sln} = μ_j^{gas} \nonumber \] for all \(j\). Consider what happens to a pure component, e.g. \(j=1\) in equilibrium with its vapor. We can write: \[μ_1^{pure \,liq}= μ_1^{pure\, vapor}=μ_1^o + RT \ln \dfrac{P^*_1}{P^o} \nonumber \] The asterisk in \(P^*_1\) denotes the equilibrium vapor pressure of pure component 1 and we will use that to indicate the thermodynamic potential of pure compounds too: \[μ_1^{*liq}= μ_1^o + RT \ln \dfrac{P^*_1}{P^o} \label{A} \] Combining Equations \(\ref{A}\) and \(\ref{B}\) we find a relationship between the solution and the pure liquid: \[μ_j^{sln}=μ^*_j + RT \ln \dfrac{P_j}{P^*_j} \nonumber \] Notice that the and its pressure is used to link the mixture and the pure compound.

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Wien's displacement law states that the blackbody radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature. The shift of that peak is a direct consequence of the Planck radiation law which describes the spectral brightness of black body radiation as a function of wavelength at any given temperature. However it had been discovered by Wilhelm Wien several years before Max Planck developed that more general equation, and describes the entire shift of the spectrum of black body radiation toward shorter wavelengths as temperature increases. Derive from Planck's law. Proceed as follows: \[\rho (\nu, T) = \dfrac {2 h \nu^3 }{c^3\left(e^{\frac {h\nu}{k_B T}}-1\right)} \label{Planck2}\] We need to evaluate the derivative of Equation \ref{Planck2} with respect to \(\nu\) and set it equal to zero to find the peak wavelength. \[\dfrac{d}{d\nu} \left \{ \rho (\nu, T) \right \} = \dfrac{d}{d\nu} \left \{ \dfrac {2 h \nu^3 }{c^3\left(e^{\frac {h\nu}{k_B T}}-1\right)} \right \} =0 \label{eq2}\] This can be solved via the quotient rule or product rule for differentiation. Selecting the latter for convenience requires rewriting Equation \ref{eq2} as a product: \[ \dfrac{d}{d\nu} \left \{ \rho (\nu, T) \right \} = \dfrac{2h}{c^3} \dfrac{d}{d\nu} \left \{ ( \nu^3) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1} \right \} = 0\] applying the product rule (and power rule and chain rule) \[ = \dfrac{2h}{c^3} \left [ (3 \nu^2) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1} - ( \nu^3) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-2} \left(\dfrac{h}{k_BT}\right) e^{\frac {h\nu}{k_B T}} \right] = 0\] so this expression is zero when \[ (3 {\nu^2}) {\left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-1}} = ( \nu^{3}) \left( e^{\frac {h\nu}{k_B T}}-1 \right)^{-2} \left(\dfrac{h}{k_BT}\right) e^{\frac {h\nu}{k_B T}} \] or when simplified \[ 3 \left( e^{\frac {h\nu}{k_B T}}-1 \right) - \left(\dfrac{hv}{k_BT}\right) e^{\frac {h\nu}{k_B T}} =0 \label{eq10}\] We can do a substitution \(u=\frac {h\nu}{k_B T} \) and Equation \ref{eq10} becomes \[3 (e^u - 1) - u e^u = 0\] Finding the solutions to this equation requires using Lambert's W-functions and results numerically in \[ u = 3 +W(-3e^{-3}) \approx 2.8214\] so unsubstituting the \(u\) variable \[ u = \dfrac {h\nu}{k_B T} \approx 2.8214 \label{eq20}\] or \[ \begin{align} \nu &\approx \dfrac{2.8214\, k_B}{h} T \\[4pt] &\approx \dfrac{(2.8214 )(1.38 \times 10^{-23} J/K) }{6.63 \times 10^{-34} J\,s} T \\[4pt] &\approx (5.8 \times 10^{10} Hz/K)\, T \end{align}\] The consequence is that the shape of the blackbody radiation function would shift proportionally in frequency with temperature. When Max Planck later formulated the correct blackbody radiation function it did not include Wien's constant explicitly. Rather, Planck's constant was created and introduced into his new formula. From Planck's constant and the Boltzmann constant , Wien's constant (Equation \ref{eq20}) can be obtained.

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Let's go back to that first ion pair which was formed when the positive ion and the negative ion came together. If the electronegativities of the elements are sufficiently different (like an alkali metal and a halide), the charges on the paired ions will not change appreciably - there will be a full electron charge on the blue ion and a full positive charge on the red ion. The bond formed by the attraction of these opposite charges is called an ionic bond. If the difference in electronegativity is not so great, however, there will be some degree of sharing of the electrons between the two atoms. The result is the same whether two ions come together or two atoms come together: The combination of atoms or ions is no longer a pair of ions, but rather a polar molecule which has a measureable dipole moment. The dipole moment (D) is defined as if there were a positive (+q) and a negative (-q) charge separated by a distance (r): \[D = qr\] If there is no difference in electronegativity between the atoms (as in a diatomic molecule such as \(O_2\) or \(F_2\)) there is no difference in charge and no dipole moment. The bond is called a covalent bond, the molecule has no dipole moment, and the molecule is said to be non-polar. Bonds between different atoms have different degrees of ionicity depending on the difference in the electronegativities of the atoms. The degree of ionicity may range from zero (for a covalent bond between two atoms with the same electronegativity) to one (for an ionic bond in which one atom has the full charge of an electron and the other atom has the opposite charge). In some cases, two or more partially ionic bonds arranged symmetrically around a central atom may mutually cancel each other's polarity, resulting in a non-polar molecule. An example of this is seen in the carbon tetrachloride (\(CCl_4\)) molecule. There is a substantial difference between the electronegativities of carbon (2.55) and chlorine (3.16), but the four chlorine atoms are arranged symmetrically about the carbon atom in a tetrahedral configuration, and the molecule has zero dipole moment. Saturated hydrocarbons (\(C_nH_{n+2}\)) are non-polar molecules because of the small difference in the electronegativities of carbon and hydrogen plus the near symmetry about each carbon atom.

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The hydrogen bond is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. The hydrogen then has the partial positive charge. To recognize the possibility of hydrogen bonding, examine the Lewis structure of the molecule. The electronegative atom must have one or more unshared electron pairs as in the case of oxygen and nitrogen, and has a negative partial charge. The hydrogen, which has a partial positive charge tries to find another atom of oxygen or nitrogen with excess electrons to share and is attracted to the partial negative charge. This forms the basis for the hydrogen bond. In other words - In the graphic on the left, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations which are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon. Link to animation of Hydrogen Bonding in Water - Northland Community and Technical College Hydrogen bonding is usually stronger than normal dipole forces between molecules. Of course hydrogen bonding is not nearly as strong as normal covalent bonds within a molecule - it is only about 1/10 as strong. This is still strong enough to have many important ramifications on the properties of water. The graphic on the left shows a cluster of water molecules in the liquid state. Water is a polar molecule, with the oxygen (red) being the negative area and the hydrogen (white) being the more positive area. Opposite charges attract. The bond lengths give some indication of the bond strength. A normal covalent bond is 0.96 Angstroms, while the hydrogen bond length is is 1.97 A. The molecular electrostatic potential is the potential energy of a proton at a particular location near a molecule. The polarity of the water molecule with the attraction of the positive and negative partial charges is the basis for the hydrogen bonding.

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Two compounds that have the same formula and the same connectivity do not always have the same shape. There are two reasons why this may happen. In one case, the molecule may be flexible, so that it can twist into different shapes via rotation around individual sigma bonds. This phenomenon is called conformation, and it is covered in a different chapter. The second case occurs when two molecules appear to be connected the same way on paper, but are connected in two different ways in three dimensional space. These two, different molecules are called . One simple example of stereoisomers from inorganic chemistry is diammine platinum dichloride, (NH ) PtCl . This important compound is sometimes called "platin" for short. As the formula implies, it contains a platinum ion that is coordinated to two ammonia ligands and two chloride ligands (remember, a ligand in inorganic chemistry is an electron donor that is attached to a metal atom, donating a pair of electrons to form a bond). Platin is an example of a coordination compound. The way the different pieces of bond together is discussed in the chapter of . For reasons arising from molecular orbital interactions, platin has a square planar geometry at the platinum atom. That arrangement results in two possible ways the ligands could be connected. The two sets of like ligands could be connected on the same side of the square or on opposite corners. These two arrangements result in two different compounds; they are isomers that differ only in three-dimensional space. Although these two compounds are very similar, they have slightly different physical properties. Both are yellow compounds that decompose when heated to 270 degrees C, but trans-platin forms pale yellow crystals and is more soluble than cis-platin in water. Cis-platin has clinical importance in the treatment of ovarian and testicular cancers. The biological mechanism of the drug's action was long suspected to involve binding of the platinum by . Further details were worked out by chemist Steve Lippard and graduate student Amy Rosenzweig in the 1990's. Inside the cell nucleus, the two ammines in cis-platin can be replaced by nitrogen donors from a DNA strand. To donate to the Lewis acidic platinum, the DNA molecule must bend slightly. Normally that bend is detected and repaired by proteins in the cell. However, ovarian and testicular cells happen to contain a protein that is just the right shape to fit around this slightly bent DNA strand. The DNA strand becomes lodged in the protein and can't be displaced, and so it is unable to bind with other proteins used in DNA replication. The cell becomes unable to replicate, and so cancerous growth is stopped. Draw the cis and trans isomers of the following compounds: Only one isomer of (tmeda)PtCl is possible [tmeda = (CH ) NCH CH N(CH ) ; both nitrogens connect to the platinum]. Draw this isomer and explain why the other isomer is not possible. The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called . Because isomers usually have different physical and chemical properties, it is important to know which isomer we are dealing with if more than one isomer is possible. Recall that in many cases more than one structure is possible for organic compounds with the same molecular formula; examples discussed previously include n-butane versus isobutane and cis-2-butene versus trans-2-butene. As we will see, coordination compounds exhibit the same types of isomers as organic compounds, as well as several kinds of isomers that are unique. Metal complexes that differ only in which ligands are adjacent to one another ( ) or directly across from one another ( ) in the coordination sphere of the metal are called . They are most important for square planar and octahedral complexes. Because all vertices of a square are equivalent, it does not matter which vertex is occupied by the ligand B in a square planar MA B complex; hence only a single geometrical isomer is possible in this case (and in the analogous MAB case). All four structures shown here are chemically identical because they can be superimposed simply by rotating the complex in space: For an MA B complex, there are two possible isomers: either the A ligands can be adjacent to one another (cis), in which case the B ligands must also be cis, or the A ligands can be across from one another (trans), in which case the B ligands must also be trans. Even though it is possible to draw the cis isomer in four different ways and the trans isomer in two different ways, all members of each set are chemically equivalent: The anticancer drug cisplatin and its inactive trans isomer. Cisplatin is especially effective against tumors of the reproductive organs, which primarily affect individuals in their 20s and were notoriously difficult to cure. For example, after being diagnosed with metastasized testicular cancer in 1991 and given only a 50% chance of survival, Lance Armstrong was cured by treatment with cisplatin. Square planar complexes that contain symmetrical bidentate ligands, such as [Pt(en) ] , have only one possible structure, in which curved lines linking the two N atoms indicate the ethylenediamine ligands: Octahedral complexes also exhibit cis and trans isomers. Like square planar complexes, only one structure is possible for octahedral complexes in which only one ligand is different from the other five (MA B). Even though we usually draw an octahedron in a way that suggests that the four “in-plane” ligands are different from the two “axial” ligands, in fact all six vertices of an octahedron are equivalent. Consequently, no matter how we draw an MA B structure, it can be superimposed on any other representation simply by rotating the molecule in space. Two of the many possible orientations of an MA B structure are as follows: If two ligands in an octahedral complex are different from the other four, giving an MA B complex, two isomers are possible. The two B ligands can be cis or trans. Cis- and trans-[Co(NH ) Cl ]Cl are examples of this type of system: Replacing another A ligand by B gives an MA B complex for which there are also two possible isomers. In one, the three ligands of each kind occupy opposite triangular faces of the octahedron; this is called the fac isomer (for facial). In the other, the three ligands of each kind lie on what would be the meridian if the complex were viewed as a sphere; this is called the mer isomer (for meridional): Draw all the possible geometrical isomers for the complex [Co(H O) (ox)BrCl] , where ox is O CCO , which stands for oxalate. : formula of complex : structures of geometrical isomers This complex contains one bidentate ligand (oxalate), which can occupy only adjacent (cis) positions, and four monodentate ligands, two of which are identical (H O). The easiest way to attack the problem is to go through the various combinations of ligands systematically to determine which ligands can be trans. Thus either the water ligands can be trans to one another or the two halide ligands can be trans to one another, giving the two geometrical isomers shown here: In addition, two structures are possible in which one of the halides is trans to a water ligand. In the first, the chloride ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens. Exchanging the chloride and bromide ligands gives the other, in which the bromide ligand is in the same plane as the oxalate ligand and trans to one of the oxalate oxygens: This complex can therefore exist as four different geometrical isomers. Draw all the possible geometrical isomers for the complex [Cr(en) (CN) ] . Two geometrical isomers are possible: trans and cis. Many metal complexes form isomers, which are two or more compounds with the same formula but different arrangements of atoms. Structural isomers differ in which atoms are bonded to one another, while geometrical isomers differ only in the arrangement of ligands around the metal ion. Ligands adjacent to one another are cis, while ligands across from one another are trans. ,

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Magnesium is a group two element and is the eighth most common element in the earth's crust. Magnesium is light, silvery-white, and tough. Like aluminum, it forms a thin layer around itself to help prevent itself from rusting when exposed to air. Fine particles of magnesium can also catch on fire when exposed to air. Magnesium is essential in nutrition for animals and plants. It is also used as an alloy to combine with other metals to make them lighter and easier to weld, for purposes in the aerospace industry along with other industries. It is also used in medicine, in the forms of magnesium hydroxides, sulfates, chlorides, and citrates. General Information Magnesium takes it name from magnesite ore, named for the district Magnesia in Thessaly, Greece. Magnesium is a strong metal that is light and silvery-white. Recognized as a element as far back as 1775, it was first isolated in pure form by Davy in 1805. Magnesium has the ability to tarnish, which creates an oxide layer around itself to prevent it from rusting. It also has the ability to react with water at room temperature. When exposed to water, bubbles form around the metal. Increasing the temperature speeds up this reaction. One property of magnesium is high flammability. Like many other things, magnesium is more flammable when it has a higher surface area to volume ratio. An example of surface area to volume ratio is seen in the lighting of fire wood. It is easier to light kindling and smaller branches than a whole log. This property of magnesium is used in war, photography, and in light bulbs. Magnesium is used in war for incendiary bombs, flares, and tracer bullets. When these weapons are used, they ignite immediately and cause fires. The only way to extinguish a magnesium fire is to cover it with sand. Water does not extinguish the fire as water reacts with the hot magnesium and releases even more hydrogen. Magnesium is one of the lightest metals, and when used as an alloy, it is commonly used in the automotive and aeronautical industries. The use of magnesium has increased and peaked in 1943. One reason the use of magnesium has increased is that it is useful in alloys. Alloys with magnesium are able to be welded better and are lighter, which is ideal for metals used in the production of planes and other military goods. Another characteristic of magnesium is that it aids in the digestive process. Magnesium is commonly used in milk of magnesia and Epsom salts. These forms of magnesium can range from magnesium hydroxide, magnesium sulfate, magnesium chloride, and magnesium citrate. Magnesium not only aids in humans and animals, but also in plants. It is used to convert the sun's lights into energy for the plant in a process known as photosynthesis. The main component of this process is chlorophyll. This is a pigment molecule that is composed of magnesium. Without magnesium, photosynthesis as we know it would not be possible. Magnesium has three stable isotopes, Mg-24, Mg-25, Mg-26. The most common isotope is Mg-24, which is 79% of all Mg found on Earth. Mg and Mg are used to study the absorption and metabolism of magnesium in the human body. They are also used to study heart disease. Magnesium not only has stable isotopes, but also has radioactive isotopes, which are isotopes that have an unstable nuclei. These isotopes are Mg--22, Mg23, Mg-27, Mg-28, and Mg-29. Mg-28 was commonly used in nuclear sites for scientific experiments from the 1950s to 1970s. When exposed to steam, magnesium changes from magnesium to magnesium oxide and hydrogen. \[Mg(s) +H_2O(g) \rightarrow MgO(s) + H_2(g) \] When exposed to cold water, the reaction is a bit different. The reaction does not stop because the magnesium hydroxide gets insoluble in water. \[Mg(s) +2H_2O(g) \rightarrow Mg(OH)_2(s) + H_2(g)\] When exposed to oxygen, magnesium turns into magnesium oxide. \[2Mg(s) +O_2(g) \rightarrow 2MgO(s)\] When exposed to hydrogen, magnesium turns into magnesium hydride. \[Mg(s) + H_2(g) \rightarrow MgH_2(s)\] When reacted with nitrogen, magnesium turns into magnesium nitride. \[3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)\] When reacted with a halogen, magnesium is very reactive. An example will be with chloride. When reacted with chloride, the product is magnesium(II) chloride. \[Mg(s) + Cl_2(g) \rightarrow MgCl_2(s)\] When reacted with acids, magnesium dissolves and forms solutions that have both the Mg(II) ion and hydrogen gas. \[Mg(s) + 2HCl(aq) \rightarrow Mg^{2+}(aq) + 2Cl^-(aq) + H_2(g)\] When reacted with bases, magnesium react.

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Dioxygen has a rich redox chemistry that is not explicitly exploited in the oxygen carriers, but which is central to enzymes that coordinate and activate dioxygen for subsequent reaction with a substrate. On reduction of dioxygen by one electron, the superoxide anion radical O is formed. Concomitant with a reduction in bond order from 2.0 to 1.5 is an increase in bond length from 1.21 to 1. 30 Å. A second reduction step produces the peroxide anion 0 ; the bond order is one, and the O—O separation is 1.49 Å. Each of these reduced species, O and O , has a characteristic O—O stretching vibration in the infrared region. The free-energy changes and electrochemical potentials for the reduction of dioxygen at unit activity, pH = 1 (E°), are different from those at pH 7.0 (E°'), as shown in Figure 4.11. The values at pH 7.0 are more relevant to physiological conditions. Note that the superoxide anion may function as either an oxidant or a reductant. In coordinating to metals, dioxygen shows a great variety of geometries and two formal oxidation states. Many complexes have (O—O) values in the range 740 to 930 cm , and, where known, an O—O separation in the range 1.40 to 1.50 Å. By analogy with the peroxide anion, these species are designated peroxo, O . Similarly, the designation superoxo O is applied to those complexes where (O—O) values are in the range 1075 to 1200 cm , and the O—O separation is around 1.30 Å. Although such O—O separations and vibrations are consistent with coordinated peroxide or superoxide moieties, the net amount of charge transferred onto the dioxygen ligand from the metal and its other ligands is difficult to measure experimentally and is probably variable. Thus the oxidation state of the dioxygen ligand and that of the metal are best considered in a formal sense rather than literally—hence the use of the terminology O to indicate oxidation state I- for the O moiety as a unit (not each O atom). Because of the high degree of covalency in the M—O bond, a more sensible comparison, at least for the peroxo class of compounds, is with organic peroxides, ROOH or ROOR. The clear separation of coordinated dioxygen into either the superoxo or the peroxo class is shown in Figure 4.12. those compounds for which both stretching frequencies ( (O—O)) and O—O separations ( (O—O)) are available are shown; for the purpose of the plot, non-coordinated anions and cations, replacement of ethylenendiamine by two ammonia ligands, and replacement of triphenylphosphine by alkylphenylphosphines are assumed not to perturb significantly (O—O) or (O—O). At least seven different geometries have been observed for the coordination of dioxygen (Figure 4.13), only three or four of which are currently known to be biologically relevant—the superoxo (for oxyhemoglobin), the peroxo or (for oxyhemocyanin), and the hydroperoxo (for oxyhemerythrin). The geometry is a function of the metal, its oxidation state, and its ligands. For the late transition metals of the cobalt and nickel triads, with soft \(\pi\)-acid ligands, such as phosphines and carbonyls, and with an initially low oxidation state of the metal, triangular coordination of a peroxo species with covalent M-O bonds is common. Concomitant with the formal reduction of dioxygen, the metal center undergoes a formal two-electron oxidation: \(\tag{4.26}\) In this example, where the metal has undergone, at least formally, a two-electron oxidation, the UV-visible properties of the metal-dioxygen complex tend to resemble those of bona fide M rather than M species. Early transition metals (Ti, V, Cr triads) often coordinate several peroxo species, leaving the metal in formally a very high oxidation state (e.g., Cr(O ) , a Cr ion). The M—O links have more ionic character, with the \(\eta\)-peroxo groups acting as bidentate ligands. Titanium and molybdenum(II) porphyrins bind, respectively, one and two dioxygen molecules in this manner. With harder \(\sigma\)-donor ligand systems, such as those containing nitrogen and oxygen donors, and the metal center in a normal oxidation state, a formal one-electron reduction to an end-on coordinated superoxo species occurs with a bent bond. Metal-dioxygen species can also be formed by adding the superoxide anion to the oxidized species: \(\tag{4.27}\) In the absence of steric constraints, dimerization to a (bridging) \(\mu\)-peroxo species frequently occurs, especially for cobalt-dioxygen complexes: These dicobalt species (right-hand side of Equation 4.28) may be oxidized by one electron to give a \(\mu\)-superoxo moiety. A clear shortening of the O—O bond and concomitant increase in the value of (O—O) are observed in several superoxo-peroxo pairs. These and other modes of O attachment are illustrated in Figure 4.13. Some geometries are represented by only one or two examples, and some geometries, for example, a linear M—O—O species, have never been observed. In binding to metals, O effectively functions both as a \(\pi\) acid, accepting into its \(\pi\)* orbitals electron density from the filled d orbitals of the metal, and as a \(\sigma\) donor, donating electron density into an empty metal d orbital. Thus other \(\sigma\) donor or \(\pi\) acceptor ligands, such as nitric oxide (NO), alkyl isocyanides (R—NC), alkyl nitroso (R—NO), and carbon monoxide (CO), are often observed to bind to the same metal complexes that bind O . The nature of the metal-dioxygen linkage in biological oxygen carriers and their models will be examined in more detail later.

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The reactions of 1,3-butadiene are reasonably typical of conjugated dienes. The compound undergoes the usual reactions of alkenes, such as catalytic hydrogenation or radical and polar additions, but it does so than most alkenes or dienes that have isolated double bonds. Furthermore, the products frequently are those of 1,2 Formation of both 1,2- and 1,4-addition products occurs not only with halogens, but also with other electrophiles such as the hydrogen halides. The mechanistic course of the reaction of 1,3-butadiene with hydrogen chloride is shown in Equation 13-1. The first step, as with alkenes, is formation of a carbocation. However, with 1,3-butadiene, if the proton is added to \(\ce{C_1}\) (but not \(\ce{C_2}\)), the resulting cation has a substantial , with the charge distributed over two carbons (review and if this is not clear to you). Attack of \(\ce{Cl}^\ominus\) as a nucleophile at one or the other of the positive carbons yields the 1,2- or the 1,4- addition product: An important feature of reactions in which 1,2 and 1,4 additions occur in competition with one another is that the ratio of the products can depend on the temperature, the solvent, and also on the . The reason for the dependence on the reaction time is that the formation of the carbocation is , and the ratio of products at equilibrium need not be the same as the ratio of the rates of attack of \(\ce{Cl}^\ominus\) at \(\ce{C_1}\) and \(\ce{C_3}\) of the carbocation. This is another example of a difference in product ratios resulting from kinetic control equilibrium control. The fact is that at low temperatures the 1,2 product predominates because it is formed more rapidly, and the back reactions, corresponding to \(k_{-1}\) or \(k_{-3}\), are slow (Equation 13-2). However, at equilibrium\(^1\) the 1,4 product is favored because it is more stable, not because it is formed more rapidly. Conjugated dienes also undergo addition reactions by radical-chain mechanisms. Here, the addition product almost always is the 1,4 adduct. Thus radical addition of hydrogen bromide to 1,3-butadiene gives l-bromo-2-butene, presumably by the following mechanism: \(^1\)The equilibrium ratio is obtained as follows. At equilibrium \(k_1/k_{-1} = \left[ \ce{CH_3CH=CHCH_2Cl} \right]/\left[ \ce{R}^\oplus \right] \left[ \ce{Cl}^\ominus \right]\) and \(k_{-3}/k_3 = \left[ \ce{R}^\oplus \right] \left[ \ce{Cl}^\ominus \right]/\left[ \ce{CH_3CHClCH=CH_2} \right]\), in which \(\ce{R}^\oplus\) is the concentration of delocalized carbocation. Multiplication of these equations gives \(k_1k_{-3}/k_{-1}k_3 = \left[ \ce{CH_3CH=CHCH_2Cl} \right]/\left[ \ce{CH_3CHClCH=CH_2} \right] = K_\text{eq}\). and (1977)

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Strontium is a group 2 element that does not occur as a free element due to its extreme reactivity with oxygen and water. It occurs naturally only in compounds with other elements such as strontianite. It is softer than calcium and decomposes water more vigorously. It has a silver appearance but then turns yellow with the formation of oxide. The element strontium is named for a Scottish town, Strontian. It was isolated in 1808 by Davy and is a silvery and malleable metal that reacts vigorously with water to produce hydrogen gas. It has the same relative abundance as carbon and sulfur but does not occur in pure form. Strontium is softer than calcium and decomposes vigorously in water. It is a silvery color but rapidly oxidizes to yellow due to the formation of strontium oxide. Because of its propensity for oxidation and ignition, strontium is stored typically under kerosene. Finely powdered strontium metal is sufficiently reactive to ignite spontaneously in air. It reacts with water quickly (but not violently like group 1 elements) to produce strontium hydroxide and hydrogen gas. Strontium and its compounds burn with a crimson flame and are used in fireworks. Strontium compounds are useful in pyrotechnic devices and signal flares because of the bright crimson coloring they give to flames. Strontium is used for producing glass for color televisions. It is also used in producing ferrite ceramic magnets and in refining zinc. Strontium atoms helped develop the world's most accurate atomic clock, which is accurate to one second in 200 million years. Toothpaste for sensitive teeth uses strontium chloride, and strontium oxide is used to improve the quality of pottery glazes. The isotope Sr is one of the world's best long-lived, high energy beta emitters known. It is also used in cancer therapy. Strontium-90, a radioactive isotope of the metal produced by fission reactions is a dangerous environmental menace because its chemistry is similar to calcium and it may take its place in bones. The strong radiation emitted by the isotope interferes with the production of new blood cells and can cause death.

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Discovered by von Reichenstein in 1782, tellurium is a brittle metalloid that is relatively rare. It is named from the Latin tellus for "earth". Tellurium can be alloyed with some metals to increase their machinability and is a basic ingredient in the manufacture of blasting caps. Elemental tellurium is occasionally found in nature but is more often recovered from various gold ores, all containing \(AuTe_2\). The ore was known as "Faczebajer weißes blättriges Golderz" (white leafy gold ore from Faczebaja) or antimonalischer Goldkies (antimonic gold pyrite). In 1782, while serving as the Austrian chief inspector of mines in Transylvania Franz-Joseph Müller von Reichenstein concluded that ore did not contain antimony, but that it contained bismuth sulfide. However the following year, he reported that this was erroneous and that the ore contained mostly gold and an unknown metal very similar to antimony. After 3 years of testing Müller determined the specific gravity of the mineral and noted the radish-like odor of the white smoke, which passed off, when the new metal was heated. In 1789, another Hungarian scientist, Pál Kitaibel, also discovered the element independently in an ore from Deutsch-Pilsen which had been regarded as argentiferous molybdenite, but later he gave the credit to Müller. In 1798, it was named by Martin Heinrich Klaproth who earlier isolated it from the mineral calaverite. Tellurium is a semimetallic, lustrous, crystalline, brittle, silver-white element. It is usually available as a dark grey powder and has metal and non-metal properties. Te forms many compounds corresponding to those of sulfur and selenium. When burned in the air tellurium has a greenish-blue flame and forms tellurium dioxide as a result. Tellurium is unaffected by water or hydrochloric acid, but dissolves in nitric acid. It as an atomic mass of 127.6 g/mol and a density of 6.24 g-cm . It's boiling point is at 450 degrees Celsius and its melting point is at 1390 °C. There are eight naturally occurring isotopes of Tellurium, of which three are radioactive. Tellurium is among the rarest stable solid elements in the Earth's crust. At 0.005 ppm, it is comparable to in abundance. However, tellurium is far more abundant in the wider universe. Tellurium was originally and is most commonly found in gold tellurides. However, the largest sources for modern production of tellurium is as a byproduct of blister copper refinement. The treatment of 500 tons of copper ore results in 0.45 kg of tellurium. Tellurium can also be found in lead deposits. Other tellurium sources, known as subeconomic deposits because the cost of abstraction outweighs the yield in tellurium, are lower-grade copper and some coal. Originally, the copper tellurium ore is treated with sodium bicarbonate and elemental oxygen to produce a tellurium oxide salt, copper oxide, and carbon dioxide: \[Cu_2Te + Na_2CO_3 + 2O_2 \rightarrow 2CuO + Na_2TeO_3 + CO_2\] Then, the sodium tellurium oxide is treated with sulfuric acid to precipitate out tellurium dioxide which can be treated with aqueous sodium hydroxide to reduce to pure tellurium and oxygen gas: \[TeO_2 + 2NaOH \rightarrow Na_2TeO_3 + H_2O \rightarrow Te + 2NaOH + O_2\] Tellurium has many unique industrial and commercial uses that improve product quality and quality-of-life. Many of these technologies that utilize tellurium have important uses for the energy industry, the military, and health industries. Tellurium is used to color glass and ceramics and can improve the machining quality of metal products. When added to copper alloys, tellurium makes the alloy more ductile, whereas it can prevent corrosion in lead products. Tellurium is an important component of infrared detectors used by the military as well as x-ray detectors used by a variety of fields including medicine, science, and security. In addition, tellurium-based catalysts are used to produce higher-quality rubber. CdTe films are one of the highest efficiency photovoltaics, metals that convert sunlight directly into electrical power, at 11-13% efficiency and are, therefore, widely used in solar panels. CdTe is a thin-film semiconductor that absorbs sunlight. Tellurium can be replaced by other elements in some of its uses. For many metallurgical uses, selenium , bismuth , or lead are effective substitutes. Both selenium and sulfur can replace tellurium in rubber production. Technologies based on tellurium have global impacts. As a photovoltaic, CdTe is the second most utilized solar cell in the world, soon said to surpass crystalline silicon and become the first. According to the US military, the tellurium-based infrared detectors are the reason that the military has such an advantage at night, an advantage which, in turn, has an effect on global and domestic politics. Tellurium extraction, as a byproduct of copper refinement, shares environmental impacts associated with copper mining and extraction. While a generally safe process, the removal of copper from other impurities in the ore is can lead to leaching of various hazardous sediments. In addition, the mining of copper tends to lead to reduced water flow and quality, disruption of soils and erosion of riverbanks, and reduction of air quality. About 215-220 tons of tellurium are mined across the globe every year. In 2006, the US produced 40% of production, Peru produced 30%, Japan produced 20%, and Canada produced 10% of the world's tellurium supply (since the chart can't be any bigger). The leading countries in production are the United States with 50 tons per year, Japan with 40 tons per year, Canada with 16 tons per year, and Peru with 7 tons per year (year 2009). When pure, tellurium costs $24 per 100 grams. Because tellurium is about as rare as platinum on earth, the United States Department of Energy expects a supply shortfall by the year 2025, despite the always improving extraction methods. As demand increases to provide the tellurium needed for solar panels and other such things, supply will continue to decrease and thus the price will skyrocket. This will cause waves in the sustainable energy movement as well as military practices and modern medicine.

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If two operators commute then both quantities can be measured at the same time, if not then there is a tradeoff in the accuracy in the measurement for one quantity vs. the other. Operators are commonly used to perform a specific mathematical operation on another function. The operation can be to take the derivative or integrate with respect to a particular term, or to multiply, divide, add or subtract a number or term with regards to the initial function. Operators are commonly used in physics, mathematics and chemistry, often to simplifiy complicated equations such as the Hamiltonian operator, used to solve the Schrödinger equation. Operators are generally characterized by a hat. Thus they generally appear like the following equation with \(\hat{E}\) being the operator operating on \(f(x)\) \(\hat{E} f(x) = g(x)\) For example if \(\hat{E} = d/dx\) then: \(g(x) = f'(x)\) One property of operators is that the order of operation matters. Thus: \(\hat{A}\hat{E}f(x) \not= \hat{E}\hat{A}f(x)\) unless the two operators commute. Two operators commute if the following equation is true: \(\left[\hat{A},\hat{E}\right] = \hat{A}\hat{E} - \hat{E}\hat{A} = 0\) To determine whether two operators commute first operate \(\hat{A}\hat{E}\) on a function \(f(x)\). Then operate\(\hat{E}\hat{A}\) the same function \(f(x)\). If the same answer is obtained subtracting the two functions will equal zero and the two operators will commute.on

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Previously, we introduced Serpentinization, the geothermal processes that heat water up to 91°C (196°F) in the geothermal vents in the Atlantic Massif . A "Fly-in video" puts the formation into perspective.   The Atlantis Massif rises ~14,000 feet above the surrounding seafloor and is formed by long-lived faulting The actively venting 'IMAX' flange protrudes from the side of the massive Poseidon structure, photographed by the Hercules submersible . By now chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Now we'll see how the enthalpy of the serpentinization reaction, and many variations on it that might never get tabulated, can be easily calculated. For example, if we're interested in one of the main serpentinization reactions which provides heat to the thermal vents: Forsterite + aqueous silica → serpentine (crysotile) 3 Mg SiO ( ) + SiO ( ) + 4 H O( ) → 2 Mg Si O ( )(OH) ( ) (1) We can use Hess' law to calculate the Δ from a single list of Δ , for all compounds. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C). For example, if we know that Δ [H O( )] = –285.8 kJ mol , we can immediately write the thermochemical equation H ( ) + ½O ( ) → H O( ) Δ = –285.8 kJ mol (2) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off of H O( ) formed. Equation (1) must specify formation of mol H O( ), and so the coefficient of O must be ½. Calculate the Enthalpy change for Equation (1) from the heats of formation of the reactants and products. values from SUPCRT We can imagine that the reaction takes place in four steps, each of which involves only a standard enthalpy of formation. First, the reactants will be decomposed into their elements, then the elements will be recombined into the product. In the first step H O( ) (water) is decomposed to its elements: 4H O( ) → 2 O ( ) + 4H ( ) Δ = (3) Since this is the of formation of mol H O( ) from its elements, the enthalpy change is Δ = 4 mol × {–Δ [H O( )]} = 4 mol × [– (–285.8 kJ mol )] = +1143.2 kJ In the second step the SiO ( )similarly decomposes into its elements, with an enthalpy change equal to the negative of its heat of formation: SiO ( ) → Si( ) + O ( ) Δ (4) Δ = 1 mol × - Δ [SiO ( )] = 1 mol × -(– 876.9 kJ mol ) = + 876.9 kJ The final reactant, 2 Mg SiO decomposes into elements as follows, and the enthalpy change will be twice the negative Δ 3 Mg SiO → 6 O + 6 Mg ( ) + 2 Si( )Δ (5) Δ = 3 mol × {–Δ [Mg SiO ]} = 3 mol × [– (–2173.6 kJ mol )] = +6520.8 kJ Finally, we write the reaction for the formation of the product from elements: 6 Mg ( ) + 4 Si( ) + 9 O + 2 H O → 2 Mg Si O (OH) ( ) Δ (6) Δ = 2 mol × {Δ [Mg Si O (OH) ]} = 2 mol × [(–4360.3 kJ mol )] = -8720.6 kJ You can easily verify that the sum of Equations (3)-(6) is Equation (1). Therefore Δ = Δ + Δ + Δ + Δ = +1143.2 kJ + 876.9 kJ +6520.8 kJ -8720.6 kJ = – 179.7 kJ mol Note carefully how Example 1 was solved. The compounds Mg SiO ( ) SiO ( ) and H O( ) were hypothetically decomposed to its elements. These equations were the reverse of formation of the compounds, and so Δ was opposite in sign from Δ . Step 1 also involved 4 mol H O( ) and so the enthalpy change had to be multiplied by 4. In step 2, we had the hypothetical decomposition of SiO ( ), with an enthalpy change which is the negative of Δ ; finally, the hypothetical decomposition of 3 mol of Mg SiO contributing 3 x -Δ , and finally, we had the hypothetical formation of the Mg Si O (OH) ( ) from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same. Any chemical reaction can be approached similarly. To calculate Δ we all the Δ values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of Δ for the reactants had to be reversed in step 1, we them, again multiplying by appropriate coefficients. This can he summarized by the equation Δ = ∑ Δ (products) – ∑ Δ (reactants) (7) The symbol Σ means “the sum of.” Since Δ values are given of compound, you must be sure to multiply each Δ by an appropriate coefficient derived from the equation for which Δ is being calculated. Applying this equation to the Example we've just completed, Δ = [2 mol x -4360.6 kJ/mol] - [3 mol x 2173.6 kJ/mol) + (1 mol x -876.9 kJ/mol) + (4 mol x -285.8 kJ/mol)] = -179.7 kJ. This is the enthalpy change for the , forming 2 mol of product. In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine Δ . Quite often, however, elements do not react directly with each other to form the desired compound, and Δ must be calculated by combining the enthalpy changes for other reactions. In the case of "aqueous silica", SiO (aq), the species is not actually formed from Si and O , yet its enthalpy of formation can be calculated from other known enthalpy changes, and used legitimately as long as the species is specified and well defined. When silicates dissolve, they form H SiO (silicic acid, sometimes written Si(OH) ). We know the heat of formation of these silica solutions: And we know the enthalpy change for the reaction So combining the two, and viewing Si(OH) ( ) as SiO ( )•2 H O, we get a valid heat of formation for : for the fictional species SiO (aq) (dissolved SiO with no Si-O-H bonds), and we can replace SiO ( ) + 2 H O( ) (Δ = -910.9 + 571.6 = 1457.3 kJ) with SiO (aq) (Δ = -885.7 kJ) in geothermal equations. In other words, we could have written Equation (1) as 3 Mg SiO ( ) + Si(OH) ( ) + 2 H O( ) → 2 Mg Si O (OH) { ) Δ = -179.7 kJ mol (25°C, 1 atm pressure) (1b) One further point arises from the definition of Δ . . If we form oxygen from its elements, for example, we are talking about the reaction O ( ) → O ( ) Since the oxygen is unchanged, there can be no enthalpy change, and Δ = 0 kJ mol . There are many sources of standard enthalpies of formation for geologically important species. A article has references to several of them. Several print compilations are available, including those cited in this exemplar Many include software that adjusts the values to the high pressures (hundreds of bars (or atmospheres) and temperatures required by geologists. For example, or SUPCRT92 . Standard enthalpies of formation for some common compounds are given in the table below. These values may be used to calculate Δ for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example. Use the table of standard enthalpies of formation at 25°C below to calculate Δ for the reaction Forsterite + water → serpentine (chrysotile) + brucite 3 Mg SiO ( ) +3 H O( ) → 2 Mg Si O (OH) ( ) + Mg(OH) Δ =? (25°C, 1 atm pressure) (3) Using Eq. (4), we have Δ = ∑ Δ (products) – ∑ Δ (reactants) Reactions like this supply the large amounts of heat necessary to drive the thermal vents of the "Lost City". Use the table of standard enthalpies of formation at 25°C to calculate Δ for the reaction 4NH ( ) + 5O ( ) → 6H O( ) + 4NO( ) Using Eq. (4), we have Δ = ∑ Δ (products) – ∑ Δ (reactants) = [6 Δ (H O) + 4 Δ (NO)] – [4 Δ (NH ) + 5 Δ (O )] = 6(–241.8) kJ mol + 4(90.3) kJ mol – 4(–46.1 kJ mol ) – 5 × 0 = –1450.8 kJ mol + 361.2 kJ mol + 184.4 kJ mol = –905.2 kJ mol Note that we were careful to use Δ [H O( )] not Δ [H O( )]. Even though water vapor is not the most stable form of water at 25°C, we can still use its Δ value. Also the standard enthalpy of formation of the element O ( ) is zero . Obviously it would be a waste of space to include it in the table above.

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This page discusses the tests for halide ions (fluoride, chloride, bromide and iodide) using silver nitrate and ammonia. This test is carried out in a solution of halide ions. The solution is acidified by adding dilute nitric acid. The nitric acid reacts with, and removes, other ions that might also form precipitates with silver nitrate. Silver nitrate solution is then added, and the halide can be identified from the following products: The chloride, bromide and iodide precipitates are shown in the photograph: The chloride precipitate is easily identified, but the other two are quite similar to each other. They can only be differentiated in a side-by-side comparison. All the precipitates change color if they are exposed to light, taking on gray or purple tints. The absence of a precipitate with fluoride ions is unhelpful unless it is known that a halogen is present; otherwise, it indicates that there is no chloride, bromide, or iodide. The precipitates are insoluble silver halides: silver chloride, silver bromide or silver iodide. The formation of these is illustrated in the following equations: \[ Ag^+_{aq} + Cl^-_{(aq)} \rightarrow AgCl_{(s)}\] \[ Ag^+_{aq} + Br^-_{(aq)} \rightarrow AgBr_{(s)}\] \[ Ag^+_{aq} + I^-_{(aq)} \rightarrow AgI_{(s)}\] Silver fluoride is soluble, so no precipitate is formed. \[ Ag^+_{aq} + F^-_{(aq)} \rightarrow Ag^+_{aq} + F^-_{(aq)} \] Ammonia solution is added to the precipitates. There are no absolutely insoluble ionic compounds. A precipitate forms if the concentrations of the ions in solution in water exceed a certain value, unique to every compound. This value is known as the . For the silver halides, the solubility product is given by the expression: \[ K_{sp} = [Ag^+,X^-]\] The square brackets indicate molar concentrations, with units of mol L . Essentially, the product of the ionic concentrations is never be greater than the solubility product value. Enough solid is always precipitated to lower the ionic product to the solubility product. The table below lists solubility products from silver chloride to silver iodide (a solubility product for silver fluoride cannot be reported because it is too soluble). The compounds are all quite insoluble, but become even less so down the group. The ammonia combines with silver ions to produce a complex ion called the diamminesilver(I) ion, [Ag(NH ) ] . This is a reversible reaction, but the complex is very stable, and the position of equilibrium lies well to the right. The equation for this reaction is given below: A solution in contact with one of the silver halide precipitates contains a very small concentration of dissolved silver ions. The effect of adding the ammonia is to lower this concentration still further. If the adjusted silver ion concentration multiplied by the halide ion concentration is less than the solubility product, some precipitate dissolves to restore equilibrium. This occurs with silver chloride, and with silver bromide if the ammonia is concentrated. The more concentrated ammonia pushes the equilibrium even further to the right, lowering the silver ion concentration even more. The silver iodide is so insoluble that ammonia cannot lower the silver ion concentration enough for the precipitate to dissolve. Adding concentrated sulfuric acid to a solid sample of one of the halides gives the following results: The only possible confusion is between a fluoride and a chloride—they behave identically under these conditions. They can be distinguished by dissolving the original solid in water and then testing with silver nitrate solution. The chloride gives a white precipitate; the fluoride produces none. Jim Clark ( )

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The Presidential Green Chemistry Challenge Awards began in 1995 as an effort to recognize individuals and businesses for innovations in green chemistry. IWC replica Réplica de reloj Typically five awards are given each year, one in each of five categories: Academic, Small Business, Greener Synthetic Pathways, Greener Reaction Conditions, and Designing Greener Chemicals. A committee of the American Chemical Society selects the awardees. Through 2006, a total of 57 technologies have been recognized for the award, and over 1000 nominations have been submitted. Propylene glycol is a potential replacement for toxic ethylene glycol in antifreeze used in cars, solar heating systems, and deicing airplane wings. It is also used as a solvent in pharmaceuticals, or as a moisturizer in cosmetics, as well as the main ingredient in deodorant sticks, in fog machines, and a multitude of other uses. It's an important enough industrial chemical that the 2006 Presidential Green Chemistry Award was given to Professor Galen J. Suppes of the University of Missouri/Columbia for an improved synthesis. Propylene glycol is synthesized in the laboratory by oxidation of propylene (CH CHCH , see Figure above) with potassium permanganate: \[3 \ce{CH_2CHCH_3} + 2 \ce{KMnO_4} + 4 \ce{H_2O} \rightarrow 3 \ce{HO-CH_2-CHOH-CH_3} + 2 \ce{MnO_2} + 2 \ce{KOH} \label{1} \] This is considered a "good" synthesis, because it converts nearly all of the starting material to the desired product. But environmentally (and economically) it is extremely undesirable, for several reasons. It uses a non-renewable petrochemical as a reactant, uses an expensive oxidizing agent (KMnO ), produces a strong caustic (KOH) as a product and an undesirable byproduct (MnO ). It does have one advantage: It proceeds at low temperatures. Let's abbreviate Equation \(\ref{1}\) by symbolizing propylene as P and propylene glycol as PG to make the discussion easier: \[\ce{3 P + 2 KMnO_4 + 4 H_2O \rightarrow 3 PG + 2 MnO_2 + 2 KOH} \label{1a} \] A balanced chemical equation such as Equation (1) above not only tells how many molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved. Equation (1) says that 3 propylene (P) react with 2 KMnO and 4 H O to give 3 propylene glycol (PG) , 2 MnO and 2 KOH . Here we use the term "formula unit" in cases where the substances are not necessarily molecules, but ionic or other species. The equation also says that 3 of P reacts with 2 of KMnO and 4 of water yielding 2 of PG, 2 MnO and 2 of KOH. The balanced equation does more than this, though. It also tells us that 2 × 3 = = 6 mol P will react with 2 x 2 = 4 mol KMnO and 2 x 4 = 8 mol water, and that ½ × 3 = 1.5 mol P requires only ½ × 2 =1 mol KMnO and ½ × 4 = 2 mol of H O. In other words, the equation indicates that exactly 4 mol H O must react 3 mol of P and 2 mol of KMnO consumed. For the purpose of calculating how much KMnO is required to react with a certain amount of P therefore, the significant information contained in Equation \(\ref{1}\) is the \[\dfrac{\text{2 mol KMnO}_{\text{4}}}{\text {3 mol P}} \nonumber \] We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Equation \(\ref{1}\), \[\text{S}\left( \frac{\text{KMnO}_{\text{4}}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{\text{4}}}{\text{3 mol P}} \label{2} \] The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. Derive all possible stoichiometric ratios from Equation \(\ref{1}\). Any ratio of amounts of substance given by coefficients in the equation may be used: \[\text{S}\left( \frac{\text{KMnO}_{4}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{4 mol H}_{\text{2}}\text{O}} \nonumber \] \[\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{P}} \right)=\frac{\text{4 mol H}_{\text{2}}\text{O}}{\text{2 mol P}}\nonumber \] \[\text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{3 mol P}}\nonumber \] \[\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{MnO}_{\text{2}}} \right)=\frac{\text{4 mol H}_{\text{2}}\text{O}}{\text{2 mol MnO}_{\text{2}}}\nonumber \] \[\text{S}\left( \frac{\text{KMnO}_{4}}{\text{MnO}_{\text{2}}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{2 mol MnO}_{\text{2}}}\nonumber \] \[\text{S}\left( \frac{\text{P}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{3 mol P}}{\text{4 mol H}_{\text{2}}\text{O}}\nonumber \] There are several more stoichiometric ratios, because they can link any reactant to any other reactant, any reactant to any product, or any product to any other product. . Using Equation \ref{1} as an example, this means that the ratio of the amount of KMnO consumed to the amount of propylene (P) consumed must be the stoichiometric ratio S(KMnO /P): \[\frac{n_{\text{KMnO}_{\text{4}}\text{ consumed}}}{n_{\text{P consumed}}} = \text{S}\left( \frac{\text{KMnO}_{\text{4}}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{\text{4}}}{\text{3 mol P}} \nonumber \] Similarly, the ratio of the amount of MnO produced to the amount of P consumed must be S(MnO /P): \[\frac{n_{\text{MnO}_{\text{2}}\text{ produced}}}{n_{\text{P consumed}}} = \text{S}\left( \frac{\text{MnO}_{\text{2}}}{\text{P}} \right) = \frac{\text{2 mol MnO}_{\text{2}}}{\text{3 mol P}} \nonumber \] In general we can say that \(\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)} \nonumber \] or, in symbols, \[\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)} \nonumber \] Note that in the word Equation \ref{3a} and the symbolic Equation \ref{3b}, X and Y may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. If 3.68 mol propylene (P) is consumed according to Equation \ref{1}, find the amounts of (a) potassium permanganate (KMnO ) required, and the amounts of a. The amount of KMnO required must be in the stoichiometric ratio S(KMnO / P) to the amount of propylene consumed: \[\text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right) = \frac{n_{\text{KMnO}_{\text{4 consumed}}}}{n_{\text{P consumed}}} \nonumber \] Multiplying both sides , by we have \(n_{\text{KMnO}_{\text{4 consumed}}} = n_{\text{P consumed}}\times \text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{2 mol KMnO}_{4}}{\text{3 mol P}}=\text{2}\text{.45 mol KMnO}_{4}\) b. The amount of propylene glycol produced must be in the stoichiometric ratio S(PG/P) to the amount of propylene consumed: \[\text{S}\left( \frac{\text{PG}}{\text{P}} \right)=\frac{n_{\text{PG produced}}}{n_{\text{P consumed}}} \nonumber \] Multiplying both sides , by we have \[n_{\text{PG produced}}=n_{\text{P consumed}}\times \text{S}\left( \frac{\text{PG}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{3 mol PG}}{\text{3 mol P}}=\text{3}\text{.68 mol PG} \nonumber \] c. Similarly, the amount of MnO produced must be in the stoichiometric ratio S(MnO / P) to the amount of propylene consumed: \[\text{S}\left( \frac{\text{MnO}_{2}}{\text{P}} \right) = \frac{n_{\text{MnO}_{\text{2 produced}}}}{n_{\text{P consumed}}} \nonumber \] Multiplying both sides , by we have \[n_{\text{MnO}_{\text{2 produced}}} = n_{\text{P consumed}}\times \text{S}\left( \frac{\text{MnO}_{2}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{2 mol MnO}_{2}}{\text{3 mol P}}=\text{2}\text{.45 mol MnO}_{2} \nonumber \] These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations. This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to , where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Equation \ref{3} when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form \[\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber \] or symbolically. \[n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber \] When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol KMnO cancels 1 mol KMnO but does not cancel 1 mol H O. We can also calculate the masses of reactants and products. The molar masses can be determined from the formulas, and the masses are calculated as follows: m (g)= n (mol) x M (g/mol) For KMnO , m = n x M m = 2.45 mol x 158.03 g/mol = 387.17 g This result is added to the table below, and you may want to see if you can verify the other results in the table. Notice that there seems to be an aweful lot of extraneous stuff in this synthesis. It requires 387 g of KMnO , but that results only in the addition of two -OH groups to each propylene molecule (and they may come,in part, from H O). As a measure of this inefficiency, Barry trost developed the concept of , and for this work he received the Presidential Green Chemistry Challenge Award in 1998 . The percentage atom economy is defined as \[\text{% atom economy} = \dfrac{\text{mass of atoms utilized}}{\text{mass of all reactant atoms}} \times  100\% \label{atomecon}\] So we see that the mass of reactants that actually end up in products is 155 g (3.68 mol) of propylene and 125.2 g of OH groups (2 x 3.68 mol, or 7.36 mol, with a molar mass of 17.01 g/mol OH), for a total of 280 g. The mass of all reactants is 155 g + 387 g + 88.5 g = 630 g, so the percent atom efficiency is (via Equation \ref{atomecon}) \[\text{% atom economy} = \dfrac{280 g}{630 g} x 100% = 44.4% \nonumber\] We see that a lot of atoms are wasted in this synthesis. For comparison, let's look at several industrial processes that have been developed. Professor Suppes’s system that won the Green Chemistry Award couples a new copper-chromite catalyst to convert glycerin to propylene glycol. First, it utilizes a byproduct of biodiesel synthesis, glycerine (rather than the propylene used above, which comes from petroleum). Second, it uses a catalyst, which is not consumed in a chemical reaction and thus does not reduce the atom economy. Finally, this process uses a lower temperature and lower pressure than do previous systems (428 °F versus 500 °F and <145 psi versus >2,170 psi), and produces less byproduct than do similar catalysts . The synthesis of propylene glycol from glycerine The reaction involves dehydration (removal of -H and -OH, or H O) of glycerine, (yielding 2,3-dihydroxypropene or 3-hydroxypropanal) followed by hydrogenation of the resulting double bond to give the final product. The overall equation for the reaction is \[\ce{H2 + HO-CH2CH(OH)CH2OH → HO-CH2-CHOH-CH3 + H2O } \label{4} \] Calculate the mass of hydrogen that's required to react with 100 g of glycerine, and the masses of propylene glycol and water that result in the Suppes Green Synthesis of propylene glycol. First, we'll start the same kind of table that we used above, but we'll start with the information we have: the mass of glycerine, and the molar masses that we can calculate from formulas: We know that the stoichiometry of the reaction involves the amounts of reactants and products, not their masses, so we'll convert each mass to an amount. For example, the amount of glycerine is: \[\begin{align*} n(mol) &= \dfrac{m(g)}{M\, (g/mol)} \\[4pt] &= \dfrac{100\, g}{92.1\, g/mol} \\[4pt] &= 1.09\, mol \end{align*} \] We can easily calculate the other amounts, because the stoichiometric ratios are all 1. So for the amount of hydrogen consumed: \[\text{n}_{\text{H}_{2}} = \text{n}_{\text{HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} ~x~\frac{\text{1 mol H}_{2}}{\text{1 mol HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} \nonumber \] \[\text{n}_{\text{H}_{2}} = \text{1.09 mol} ~x~\frac{\text{1 mol H}_{2}}{\text{1 mol HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} = 1.09 mol \nonumber \] Now we can use the amounts to calculate the masses of the other reactants and products, remembering that m(g) = n(mol) x M(g/mol). So for hydrogen, m = 1.09 mol x 2.02 g/mol = 2.20 g. Verify the other table entries: Now we can calculate the percent atom economy for this process: % atom economy = \(\frac{mass ~of ~atoms ~utilized}{mass ~of~all ~reactant ~atoms} x 100%\) All of the atoms in the reactants except one oxygen atom which is removed from the glycerine. Since we have 1.09 mol of glycerine, the mass of oxygen not incorporated into product is 1.09 mol O x 15.999 g/mol O = 17.44 g O. The mass of reactants utilized is therefore 100 g - 17.44 g = 82.56 g, and the mass of product formed is 83.0 g. \(\frac{82.56}{83.0} ~x~ 100% = 99.5%\) Not only that, but the product, water, is innocuous, and it uses an overproduced reactant! The main cost is heating and pressurizing the reactor. What mass of oxygen would be consumed when 3.3 × 10 g, 3.3 Pg (petagrams), of octane (C H ) is burned to produce CO and H O? First, write a balanced equation \[\ce{2C8H18 + 25O2 → 16CO2 + 18H2O} \nonumber \] The problem gives the mass of C H burned and asks for the mass of O required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O consumed. Finally, the molar mass of O permits calculation of the mass of O . Symbolically \[m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}} \nonumber \] \[m_{\text{O}_{\text{2}}}=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g } \nonumber \] Thus 12 Pg (petagrams) of O would be needed. The large mass of oxygen obtained in this example is an estimate of how much O is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 10 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O . Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.

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The alkynes behave in many ways as if they were doubly unsaturated alkenes. For example, bromine adds to ethyne in two stages - first to give -1,2-dibromoethene by antarafacial addition, and finally to give 1,1,2,2-tetrabromoethane: Likewise, anhydrous hydrogen fluoride adds first to give fluoroethene and ultimately to give 1,1-difluoroethane: However, there is an interesting contrast in reactivity. Alkynes are substantially reactive than corresponding alkenes toward many electrophiles. This is perhaps surprising because the electrons of a triple bond, like those of a double bond, are highly exposed, which suggests that the reactivity (nucleophilicity) of a triple bond should be high. Evidently this is not the case. A simple but reasonable explanation is that the carbocation formed from the alkyne is less stable than that from the alkene because it cannot achieve the \(sp^2\) hybrid-orbital configuration expected to be the most stable arrangement for a carbocation (see ): Alkynes, unlike alkenes, are not hydrated readily in aqueous acid unless a mercuric salt is present as a catalyst. Also, the products that are isolated are either aldehydes or ketones instead of alcohols. Even though the addition of one molecule of water to ethyne probably gives ethenol (vinyl alcohol) initially, this compound is unstable relative to its structural isomer (ethanal) and rapidly rearranges: Similarly, addition of water to propyne leads to 2-propanone by way of its unstable isomer, 2-propenol: In general, the position of equilibrium for interconversion of a carbonyl compound with the corresponding alkenol (or ), having the hydroxyl group attached to the double bond, lies far on the side of the carbonyl compound: Because mercuric salts catalyze the hydration of alkynes, they probably are acting as electrophiles. Mercuric salts are known to add to both alkenes and alkynes, and if the reaction mixture is acidic, the carbon-mercury bond is cleaved to form a carbon-hydrogen bond. The overall sequence in propyne hydration may be written as follows: and (1977)

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Write the Schrödinger equation for a particle in a two dimensional box with infinite potential barriers and adjacent sides of unequal length (a rectangle). Solve the equation by separating variables with a product function X(x)Y(y) to obtain the wavefunctions X(x) and Y(y) and energy eigenvalues. How many different sets of quantum numbers are needed for this case? Sketch an energy level diagram to illustrate the energy level structure. What happens to the energy levels when the box is a square? When two or more states have the same energy, the states and the energy level are said to be degenerate. What is the zero point energy for an electron in a square box of length 0.05 nm? A materials scientist is trying to fabricate a novel electronic device by constructing a two dimensional array of small squares of silver atoms. She thinks she has managed to produce an array with each square consisting of a monolayer of 25 atoms. You are an optical spectroscopist and want to test this conclusion. Use the particle-in-a-box model to predict the wavelength of the lowest energy electronic transition for these quantum dots. Which electrons do you want to describe by the particle-in-a-box model, or do you think you can apply this model to all the electrons in silver and get a reasonable prediction? In which spectral region does this transition lie? What instrumentation would you need to observe this transition? Model the pi electrons of benzene by adapting the electron in a box model. Consider benzene to be a ring of radius r and circumference 2πr. You can find r by using the bond length of benzene (0.139 nm) and some trigonometry. Show how the electron on a ring is analogous to the electron in a linear box. Derive this analogy by thinking, not by copying from some book. What is the boundary condition for the case of the particle on a ring? Find mathematical expressions for the energy and the wavefunctions. Draw an energy level diagram. What is the physical reason that the energy levels are degenerate for this situation? Predict the wavelength of the lowest energy electronic transition for benzene. Compare your prediction with the experimental value (256 nm). What insight do you gain from this comparison? Explain how and why the following two sets of selection rules for the particle-in-a-box are related to each other: (1) If Δn is even, the transition is forbidden; if Δn is odd, the transition is allowed. (2) If the transition is g to g or u to u, it is forbidden; if the transition is g to u or u to g, it is allowed. The factor \(fi/(f^2-i^2)^2\) in Equation (4-32) determines the relative intensity of transitions in the particle-in-a-box model. Make plots of [\(fi/(f^2-i^2)^2\)] vs f for several values of i with f starting at i+1 and increasing. What conclusions can you make about particle-in-a-box spectra from your plots? Starting with the mathematical definition of uncertainty as the standard or root mean square deviation σ from the average, show by evaluating the appropriate expectation value integrals that \[ \sigma _x = \frac {L}{2\pi n } \left ( \frac {\pi ^2 n^2}{3} -2 \right ) \text {and} \sigma _p = \frac {n \pi \hbar }{L} \] for a particle in a one-dimensional box of length L as given in the chapter. Then show that the product \(\sigma _x \sigma _p \ge \frac {\hbar}{2}\). Use the symbolic processor in Mathcad to help you carry out the steps leading from Equation (4-27) to Equation (4-31). See Activity 4.3 for an introduction to the symbolic processor. An electron is confined to a one-dimensional space with infinite potential barriers at x = 0 and x = L and a constant potential energy between 0 and L. The electron is described by the wavefunction \(ψ(x) = N (Lx - x^2)\) In responding to the following questions (a through g), do not leave your answers in the form of integrals, i.e. do the integrals. Note: \[ \text {Note} : \int x^n dx = \frac {1}{n + 1} x ^{n + 1} + C \text {for} x \ne 0 \] How does choosing the potential energy inside the box to be –100 eV rather than 0 modify the description of the particle-in-a-box? ")

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Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces. The amplitude of interaction is important in the interpretation of the various properties listed above. There are four types of intermolecular forces. Most of the intermolecular forces are identical to bonding between atoms in a single molecule. Intermolecular forces just extend the thinking to forces between molecules and follows the patterns already set by the bonding within molecules. The forces holding ions together in ionic solids are electrostatic forces. Opposite charges attract each other. These are the strongest intermolecular forces. Ionic forces hold many ions in a crystal lattice structure. According to Coulomb’s law: \[ V=-\dfrac{q_1 q_2}{4 \pi \epsilon r} \tag{1}\] Based on Coulomb’s law, we can find the potential energy between different types of molecules. Polar covalent molecules are sometimes described as "dipoles", meaning that the molecule has two "poles". One end (pole) of the molecule has a partial positive charge while the other end has a partial negative charge. The molecules will orientate themselves so that the opposite charges attract principle operates effectively. In the figure below, hydrochloric acid is a polar molecule with the partial positive charge on the hydrogen and the partial negative charge on the chlorine. A network of partial + and - charges attract molecules to each other. \[ V=-\dfrac{3}{2}\dfrac{I_A I_B}{I_A+I_B}\dfrac{\alpha_A \alpha_B}{r^6} \tag{3}\] When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of the other polar molecule. Many molecules have dipoles, and their interaction occur by dipole-dipole interaction. For example: SO ↔SO . (approximate energy: 15 kJ/mol). Polar molecules have permanent dipole moments, so in this case, we consider the electrostatic interaction between the two dipoles: \[ V=-\dfrac{2}{3}\dfrac{\mu_1^2 \mu_2^2}{(4\pi\epsilon_0)^2 r^6} \tag{4}\] µ is the permanent of the molecule 1 and 2. Ion-Dipole interaction is the interaction between an ion and polar molecules. For example, the sodium ion/water cluster interaction is approximately 50 KJ/mol. \[ Na^+ ↔ (OH_2)_n \tag{5}\] Because the interaction involves in the charge of the ion and the dipole moment of the polar molecules, we can calculate the potential energy of interaction between them using the following formula: \[ V=-\dfrac{q\mu}{(4\pi\epsilon_0) r^2} \tag{6}\] The is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine. In other words - The hydrogen on one molecule attached to O or N that is attracted to an O or N of a different molecule. In the graphic below, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen or nitrogen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations that are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon. Forces between essentially non-polar molecules are the weakest of all intermolecular forces. "Temporary dipoles" are formed by the shifting of electron clouds within molecules. These temporary dipoles attract or repel the electron clouds of nearby non-polar molecules. The temporary dipoles may exist for only a fraction of a second but a force of attraction also exist for that fraction of time. The strength of induced dipole forces depends on how easily electron clouds can be distorted. Large atoms or molecules with many electrons far removed from the nucleus are more easily distorted. London Dispersion Force, also called induced dipole-induced dipole, is the weakest intermolecular force. It is the interaction between two nonpolar molecules. For example: interaction between methyl and methyl group: (-CH ↔ CH -). (approximate energy: 5 KJ/mol) When two non polar molecules approach each other, attractive and repulsive forces between their electrons and nuclei will lead to distortions in these electron clouds, which mean they are induced by the neighboring molecules, and this leads to the interaction. We can calculate the potential energy between the two identical nonpolar molecule as followed formula: \[ V=-\dfrac{3}{4} \dfrac{\alpha^2 I}{r^6} \tag{2}\] Negative sign indicates the attractive interaction. For two nonidentical nonpolar molecules A and B, we have the formula:

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Selenium was discovered by Berzelius in 1818. It is named for the Greek for "moon", selene. Selenium can exist in multiple allotropes that are essentially different molecular forms of an element with varying physical properties. For example, one allotrope of selenium can be seen as an amphorous (“without crystalline shape”) red powder. Selenium also takes a crystalline hexagonal structure, forming a metallic gray allotrope which is known to be stable. The most thermodynamically stable allotrope of selenium is trigonal selenium and also appears as a gray. Most selenium is recovered from the electrolytic copper refining process. This is usually in the form of the red allotrope. Selenium is mostly noted for its important chemical properties, especially those dealing with electricity. Unlike sulfur, selenium is a semiconductor, meaning that it conducts some electricity, but not as well as conductors. Selenium is a photoconductor, which means it has the ability to change light energy into electrical energy. Not only is selenium able to convert light energy into electrical energy, but it also displays the property of photoconductivity. Photoconductivity is the idea that the electrical conductivity of selenium increases due to the presence of light or in other words, it becomes a better photoconductor as light intensity increases. Isotopes of an element are atoms that have the same atomic numbers but a different number of neutrons (different mass numbers) in their nuclei. Selenium is known to have over 20 different isotopes; however, only 5 of them are stable. The five stable isotopes of selenium are Se, Se, Se, Se, Se. Due to selenium’s property of photoconductivity, it is known to be used in photocells, exposure meters in photography, and also in solar cells. Selenium can also be seen in its production in plain-paper photocopiers, laser printers and photographic toners. Besides its uses in the electronic industry, selenium is also popular in the glass-making industry. When selenium is added to glass, it is able to negate the color of other elements found in the glass and essentially decolorizes it. Selenium is also able to create a ruby-red colored glass when added. The element can also be used in the production of alloys and is an additive to stainless steel. Selenium, a trace element, is important in the diet and health of both plants and animals, but can be only taken in very small amounts. Exposure to an excess amount of selenium is known to be toxic and cause health problems. With a tolerable upper intake level of 400 micrograms per day, too much selenium can lead to selenosis and may result in health problems and even death. Compounds of selenium are also known to be carcinogenic. Selenium forms hydrogen selenide, H Se, a colorless flammable gas when reacted with hydrogen. Selenium burns in air displaying a blue flame and forms solid selenium dioxide. \[Se_{8(s)} + 8O_{2(g)} \rightarrow 8SeO_{2(s)}\] Selenium is also known to form selenium trioxide, SeO . Selenium reacts with fluorine, F , and burns to form the selenium hexafluoride. \[Se_{8(s)} + 24F_{2(g)} \rightarrow 8SeF_{6(l)}\] Selenium also reacts with chlorine and bromine to form diselenium dichloride, \(Se_2Cl_2\) and diselenium dibromide, \(Se_2Br_2\). \[Se_8 + 4Cl_2 \rightarrow 4Se_2Cl_{2(l)}\] \[Se_8 + 4Br_2 \rightarrow 4Se_2Br_{2(l)}\] Selenium also forms \(SeF_4\), \(SeCl_2\) and \(SeCl_4\), Selenium reacts with metals to form selenides. Example: Aluminum selenide \[3 Se_8 + 16 Al \rightarrow 8 Al_2Se_3\] Selenium reacts to form salts called selenites, e.g., silver selenite (Ag SeO ) and sodium selenite (Na SeO )

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Another important feature of the The Lanthanide Contraction refers to the fact that the 5s and 5p orbitals penetrate the 4f sub-shell so the 4f orbital is not shielded from the increasing nuclear change, which causes the atomic radius of the atom to decrease. This decrease in size continues throughout the series. The Lanthanide Contraction applies to all 14 elements included in series. This series includes Cerium(Ce), Praseodymium(Pr), Neodymium(Nd), Promethium(Pm), Samarium(Sm), Europium(Eu), Gadolinium(Gd), Terbium(Tb), Dysprosium(Dy), Holmium(Ho), Erbium(Er), Thulium(Tm), Ytterbium(Yb), and Lutetium(Lu). The atomic radius, as according to the Lanthanide Contraction, of these elements decreases as the atomic number increases. We can compare the elements Ce and Nd by looking at a . Ce has an atomic number of 58 and Nd has an atomic number of 60. Which one will have a smaller atomic radius? Nd will because of its larger atomic number. The Lanthanide Contraction is the result of a poor shielding effect of the 4f electrons. The shielding effect is described as the phenomenon by which the inner-shell electrons shield the outer-shell electrons so they are not effected by nuclear charge. So when the shielding is not as good, this would mean that the positively charged nucleus has a greater attraction to the electrons, thus decreasing the atomic radius as the atomic number increases. The s orbital has the greatest shielding while f has the least and p and d in between the two with p being greater than d. The Lanthanide Contraction can be seen by comparing the elements with f electrons and those without f electrons in the d block orbital. Pd and Pt are such elements. Pd has 4d electrons while Pt has 5d and 4f electrons. These 2 elements have roughly the same atomic radius. This is due to Lanthanide Contraction and shielding. While we would expect Pt to have a significantly larger radius because more electrons and protons are added, it does not because the 4f electrons are poor at shielding. When the shielding is not good there will be a greater nuclear charge, thus pulling the electrons in closer, resulting in a smaller than expected radius. The graphs depict the atomic radii of the first three rows of transition metals. We can apply the same principle as applied with the elements Pd and Pt to whole rows and columns. As we can see by comparing Row 1 with Row 2, the atomic radii differ greatly between the elements, but if we compare Row 2 with Row 3, the atomic radii do not have much difference. Elements with atomic number 23 and 41 lie in the same column of the periodic table and have a significantly large difference in atomic radii (atomic radii increases from Row 1 to Row 2), but elements 41 and 73, also in the same column, only differ slightly. This is the cause of introducing 4f electrons in Row 3. In Row 3, we would expect the elements to carry on the same trend as was witnessed between Rows 1 and 2 (large increase in atomic radii) but we do not. This is because the 4f orbitals are not doing a great job of shielding. The d block contraction, also known as the Scandide Contraction, describes the atomic radius trend that the d block elements (Transition metals) experience. Normally the trend for atomic radius, moving across the periodic table is that the atomic radius decreases significantly. In the transition metals with D electrons as we move from left to right across the periodic table, the element’s atomic radius only decreases slightly. This is because they have the same amount of s electrons, but are only differing in d electrons. These d electrons are in an inner shell (penultimate shell) and electrons are getting added to this shell, another shell is not created. The d electrons are not good at shielding the nuclear charge, so the atomic radius does not change much as electrons are added. Almost like disregarding the D electrons being added. As the proton number increases and the atomic radius decreases, the ionization energy increases. This is due to a more positively charged nucleus and a greater pull on the electrons by the nucleus. A greater pull is the result of an increased effective nuclear charge. Effective nuclear charge is caused by the nucleus having a more positive charge than the negative charge on the electron (net positive charge). The density, melting point, and hardness increase from left to right throughout the Lanthanide Series. The Lanthanide Contraction makes chemical separation of the Lanthanides easier. The Lanthanide Contraction, while making the chemical separation of Lanthanides easier, it makes the separation of elements following the series a bit more difficult. Answers:

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Nuclear chemistry is the study of changes associated with the nuclei of atoms. Positron emission tomography (PET) is one of the beneficial real-life uses of nuclear chemistry. Simply, it is a handy instrument that physicians use to take images of an individual's body to determine if a person is at risk for a certain disease or carries one. This module will focus on the procedures of a patient receiving a PET scan, the nuclear reactions associated with the images that are produced, commonly used tracer molecules, and useful applications of PET scans in clinical diagnosis. The The research that went into creating the first PET scanner dates several decades back to the early 1950s. It was suggested by William Sweet, Chief at the Massachusetts General Hospital, that the radiation from positron emission has the potential to increase the resolution of nuclear images taken in detecting any brain damages. Soon after this proposal, many patients under the suspicion of brain tumors were imaged with a simple positron scanner and the image results were much clearer, as was predicted by Sweet. The simple positron imaging device uses two-dimensional arrays and only a few radiation sensors to trace the radioactive substances, so the images were very low quality. In 1952, the first clinical positron scanner device was invented. This device was essentially the same as the first simple positron scanner from 1950 but with a few upgrades. It not only produced coincidence scans, but unbalance scans as well. The unbalance scan was a huge refinement because it produced a lower resolution image that allowed for easier determination of a brain tumor to the right or left of the midline of a patient's brain. Then in 1962, the first multiple detector positron imaging device was invented. A few unique features on this new scanner include higher sensitivity as well as being able to obtain another three-dimensional image by focusing on different planes of the two detector arrays. This allowed for the imaging of lesions in two dimensions while the exact position of the tumor growth can be determined in three dimension. Refinement of the PET scanner focuses on increasing the sensitivity and adding features to find more ways to have the scanners produce images in higher definition. In 1970, the PET scanner was formally introduced for use in the medical field. A few years following the introduction of the PET scanner, the radiopharmaceutical fluorine-18-2-fluoro-2-deoxyglucose (FDG) was made by chemists as a tracer so studying of the images were easier in locating brain tumors. Technological advances in the 1980s allowed for the invention of PET scanners with higher resolution than ever before that produced very precise, clear images. The procedure of PET scans are made possible by nuclear reactions involving the emission of the positive beta particle known as the positron. The positron is a mass-less particle but has a positive charge of one. In other words, it is very similar to an electron (they have the same mass), but instead of a negative charge, it carries a positive charge. In nuclear reactions, it is commonly represented as any of the symbols \(\ce{^0_{+1}e}\), \(\ce{e^+}\), or \(\ce{_{+1}\beta}\). The emission of a positron is represented by: \[\ce{^1_1p \rightarrow ^1_0n + ^0_{+1}e}\] This shows that the positron (represented here by \(\ce{^0_{+1}e}\)) speeds out of the nucleus while the neutron stays inside the nucleus. Consider the following nuclear reaction that is common in PET scans of the brain where carbon-11 is used as the tracer molecule. \[\ce{^{11}_6C \rightarrow ^{11}_{5}B + ^0_{+1}e}\] Notice that in this example of positron emission, the nuclide changes into a different element and as it gives off a positron particle, the atomic number is lowered by one, but the mass of the new element stays the same as the carbon that has decayed. The first step of the process is to inject the patient with a solution commonly used by the body to produce energy, such as glucose. It must be a positron-emitting substance because when the scanner with its array of detectors detect the collisions of the positrons with electrons, the two species disappear, and in the process, create two gamma photons to move apart in opposite directions. The scanner's electronics record these detected gamma rays and map an image of the area where the active glucose is located. PET allows the physician to locate the exact areas where metabolic activities are occurring. For example, the PET scan shows exactly where the injected glucose is being used in the brain, the heart muscle, or a growing tumor. In a cancer patient, the cancer cells show up as denser areas on a PET scan due to the cells' high metabolic rates in comparison with normal cells. Computed Axial Tomography (CAT) scanners are very similar to PET scanners. The major difference is that patients are scanned with x-rays in a CAT scan so they are at risk of developing cancer if exposure to the ionizing radiation is too long or too often. The Magnetic Resonance Imaging (MRI), which was invented after the CAT scanner, is an expensive test and uses radio waves and magnets that show information on a computer screen. Patients are not at risk with MRIs because the test does not use ionizing radiation of any kind. The reason CAT scans are still used is most likely because it is much more affordable than MRIs. Compared to CAT and MRI scanners, PET is much more efficient in detecting the differences between benign and malignant tumors because it focuses on the metabolic activity of radiopharmaceutical chemicals (glucose). Radionuclides used in PET scanning are typically isotopes with short half-lives that stay active just long enough to trace its activities in the body effectively and produce corresponding images. Below is a list of some of the more common isotopes used as tracers in the body and what each of their primary uses are: PET scanning is useful in the medical field as well as a research tool. In the medical field, it is heavily used in the clinical diagnosis of certain diseases and disorders because it is effective in targeting radionuclides (tracers) used in particular bodily functions. It is also useful in researching certain brain diseases and checking whether cardiovascular functions in an individual are normal because the produced images map this out very clearly as shown in the graphic representation below. PET scanning using the tracer radionuclide fluorine-18 (commonly known as FDG) is used because glucose-using cells take this molecule up and the images show where the metabolism of glucose is abnormal. The FDG molecule is trapped inside of the cell that takes it up until it decays completely, so a relatively short half-life is important. The use of FDG in oncology PET scans make up the majority of all PET scans in current practice. PET scanning is especially effective in the detection of cancer because the activity of the injected glucose is especially active and is easily traced in the produced images. Listed below are a few reasons why PET scanning is effective in detecting cancer: PET imaging of the brain is based on the assumption that brain activity is associated with high radioactivity. The tracer nuclide that is used is oxygen-15. Its short half-life of two minutes makes it so that PET scans of the brain must be processed immediately after the oxygen-15 is made. The brain’s rapid use of glucose produces very vividly colored images that indicate where the most activities in the brain are. This makes it relatively easy to narrow down where abnormal activity is occurring in a patient with a brain disease because the color on the image will be different in the areas where a disease has settled in. Though relatively new, detecting patients at risk of stroke may have been suggested to be the next focus of the process of PET scanning. The tracer nuclides that are used include krypton-79 and nitrogen-13. Carbon-11 and fluorine-18 are two radionuclide tracers that selectively bind to receptors in the brain, thus are a target for further research in the biological aspects of psychiatry. The states of dopamine, serotonin, and opioid receptors have been studied in humans because of their affect in patients with psychiatric conditions such as schizophrenia and mood disorders. \[\ce{^{13}N \rightarrow \underline{\;\;\;} + _{+1}\beta}\] \[\ce{^{15}O \rightarrow \underline{\;\;\;} + _{+1}\beta}\]

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Polonium was discovered in 1898 by Marie Curie and named for her native country of Poland. The discovery was made by extraction of the remaining radioactive components of pitchblende following the removal of uranium. There is only about 10 g per ton of ore! Current production for research purposes involves the synthesis of the element in the lab rather than its recovery from minerals. This is accomplished by producing Bi-210 from the abundant Bi-209. The new isotope of bismuth is then allowed to decay naturally into Po-210. The sample pictured above is actually a thin film of polonium on stainless steel. Although radioactive, polonium has a few commercial uses. You can buy your own sample of polonium at a photography store. It is part of the special anti-static brushes for dusting off negatives and prints.

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Hydrogen chloride (HCl) is a colorless gas which forms white fumes of hydrochloric acid when brought into contact with atmospheric humidity. Inhalation of the gas can cause severe burns of the nose, throat, and upper respiratory tract (which may lead to death in severe cases). Hydrogen chloride may also result in severe burns of the eyes. The hydrogen and the chlorine atom are connected by a covalent single bond, which is highly polar, since the chlorine atom is much more electronegative than the hydrogen atom. Thus the molecule shows a large dipole moment with the negative charge at the chlorine atom. When dissolved in water, the HCl gas dissociates and forms hydronium and chloride ions: HCl + H O H O + Cl This solution is called hydrochloric acid , which is a strong acid with a high acid dissociation constant.

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Hydrogen peroxide (H O ) is a viscous liquid (mp.: -0.89°C, bp.: 152.1°C, density: 1.448 g/cm at 20°C) that has strong oxidizing properties. It is commonly used (in concentrations typically around 5%) to bleach human hair, hence the phrases "peroxide blonde" and "bottle blonde". It burns the skin upon contact in sufficient concentration. In lower concentrations (3%), it is used medically for cleaning wounds and removing dead tissue. However, recent studies have indicated that hydrogen peroxide is toxic to new cells and is therefore not recommended for wound care. The same solution is often used by medical professionals to clean blood from cloth and equipment. H O is produced by a combination of electrolysis of sulfuric acid and subsequent hydrolysis of the resulting peroxo-disulfuric acid: 2 H SO H S O + H H S O + 2 H O H O 2 H SO 2 H O H O + H Hydrogen peroxide decomposes exothermically into water and oxygen gas (46.87 kcal/mol). However, at room temperature the rate of decomposition is very low, so that pure H O is stable (metastable). Catalysts (like pulverized silver, gold, platinum, manganese dioxed, iron and copper salts, alkali salts, dust, activated carbon, etc.) dramatically increase the rate of decomposition. High concentrations of H O may lead to explosions when it comes into contact with catalysts. H O may be stabilized by adding phosphoric acid, sodium diphosphate, uric acid, or barbituric acid. Pure H O without any stabilizers can only be stored in glass bottles which are coated with paraffin way, or in pure aluminum (better than 99.9 %) bottles.

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This page explains what a transition metal is in terms of its electronic structure, and then goes on to look at the general features of transition metal chemistry. These include variable oxidation state (oxidation number), complex ion formation, colored ions, and catalytic activity. The terms transition metal (or element) and d block element are sometimes used as if they mean the same thing. They don't - there's a subtle difference between the two terms. We'll explore d block elements first: You will remember that when you are building the and working out where to put the electrons using the , something odd happens after argon. At argon, the 3s and 3p levels are full, but rather than fill up the 3d levels next, the 4s level fills instead to give potassium and then calcium. Only after that do the 3d levels fill. The elements in the Periodic Table which correspond to the d levels filling are called d block elements. The first row of these is shown in the shortened form of the Periodic Table below. The electronic structures of the d block elements shown are: You will notice that the pattern of filling is not entirely tidy! It is broken at both chromium and copper.Transition metals Not all d block elements count as transition metals! A transition metal is one that forms one or more stable ions which have filled d orbitals. On the basis of this definition, scandium and zinc count as transition metals - even though they are members of the d block. By contrast, copper, [Ar] 3d 4s , forms two ions. In the Cu ion the electronic structure is [Ar] 3d . However, the more common Cu ion has the structure [Ar] 3d . Copper is definitely a transition metal because the Cu ion has an incomplete d level. Here you are faced with one of the most irritating facts in chemistry at this level! When you work out the electronic structures of the first transition series (from scandium to zinc) using the , you do it on the basis that the 3d orbitals have higher energies than the 4s orbitals. That means that you work on the assumption that the 3d electrons are added after the 4s ones. However, in all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first. The 2+ ion is formed by the loss of the two 4s electrons. The 4s electrons are lost first followed by one of the 3d electrons. One of the key features of transition metal chemistry is the wide range of (oxidation numbers) that the metals can show. It would be wrong, though, to give the impression that only transition metals can have variable oxidation states. For example, elements like Sulfur or nitrogen or chlorine have a very wide range of oxidation states in their compounds - and these obviously aren't transition metals. However, this variability is less common in metals apart from the transition elements. Of the familiar metals from the main groups of the Periodic Table, only lead and tin show variable oxidation state to any extent. You will find the above examples and others looked at in detail if you explore the chemistry of individual metals from the transition metal menu. There is a link to this menu at the bottom of the page. We'll look at the formation of simple ions like Fe and Fe . When a metal forms an ionic compound, the formula of the compound produced depends on the energetics of the process. On the whole, the compound formed is the one in which most energy is released. The more energy released, the more stable the compound. There are several energy terms to think about, but the key ones are: The more highly charged the ion, the more electrons you have to remove and the more ionization energy you will have to provide. But off-setting this, the more highly charged the ion, the more energy is released either as lattice enthalpy or the hydration enthalpy of the metal ion. The formula for Calcium chloride is CaCl . Why is that? If you tried to make CaCl, (containing a Ca ion), the overall process is slightly exothermic. By making a Ca ion instead, you have to supply more ionization energy, but you get out lots more lattice energy. There is much more attraction between chloride ions and Ca ions than there is if you only have a 1+ ion. The overall process is very exothermic. Because the formation of CaCl releases much more energy than making CaCl, then CaCl is more stable - and so forms instead. What about CaCl ? This time you have to remove yet another electron from calcium. The first two come from the 4s level. The third one comes from the 3p. That is much closer to the nucleus and therefore much more difficult to remove. There is a large jump in ionization energy between the second and third electron removed. Although there will be a gain in lattice enthalpy, it is not anything like enough to compensate for the extra ionization energy, and the overall process is very endothermic. It definitely is not energetically sensible to make CaCl ! Here are the changes in the electronic structure of iron to make the 2+ or the 3+ ion. The 4s orbital and the 3d orbitals have very similar energies. There is not a huge jump in the amount of energy you need to remove the third electron compared with the first and second. The figures for the first three ionization energies (in kJ mol ) for iron compared with those of calcium are: There is an increase in ionization energy as you take more electrons off an atom because you have the same number of protons attracting fewer electrons. However, there is much less increase when you take the third electron from iron than from calcium. In the iron case, the extra ionization energy is compensated more or less by the extra lattice enthalpy or hydration enthalpy evolved when the 3+ compound is made. The net effect of all this is that the overall enthalpy change is not vastly different whether you make, say, FeCl or FeCl . That means that it is not too difficult to convert between the two compounds. A complex ion has a metal ion at its center with a number of other molecules or ions surrounding it. These can be considered to be attached to the central ion by coordinate ( ) bonds (in some cases, the bonding is actually more complicated). The molecules or ions surrounding the central metal ion are called ligands. Simple ligands include water, ammonia and chloride ions. What all these have got in common is active lone pairs of electrons in the outer energy level. These are used to form co-ordinate bonds with the metal ion. [Fe(H O) ] [Co(NH ) ] [Cr(OH) ] [CuCl ] Other metals also form complex ions - it is not something that only transition metals do. Transition metals do, however, form a very wide range of complex ions. The diagrams show approximate colors for some common transition metal complex ions. You will find these and others discussed if you follow links to individual metals from the transition metal menu (link at the bottom of the page). Alternatively, you could explore the complex ions menu (follow the link in the help box which has just disappeared off the top of the screen). When white light passes through a solution of one of these ions, or is reflected off it, some colors in the light are absorbed. The color you see is how your eye perceives what is left. Attaching ligands to a metal ion has an effect on the energies of the d orbitals. Light is absorbed as electrons move between one d orbital and another. This is explained in detail on another page. Transition metals and their compounds are often good catalysts. A few of the more obvious cases are mentioned below, but you will find catalysis explored in detail elsewhere on the site (follow the link after the examples). Transition metals and their compounds function as catalysts either because of their ability to change oxidation state or, in the case of the metals, to adsorb other substances on to their surface and activate them in the process. All this is explored in the main catalysis section. The combines hydrogen and nitrogen to make ammonia using an iron catalyst. This reaction is at the heart of the manufacture of from vegetable oils. However, the simplest example is the reaction between ethene and hydrogen in the presence of a nickel catalyst. At the heart of the Contact Process is a reaction which converts Sulfur dioxide into Sulfur trioxide. Sulfur dioxide gas is passed together with air (as a source of oxygen) over a solid vanadium(V) oxide catalyst. Persulphate ions (peroxodisulphate ions), S O , are very powerful oxidizing agents. Iodide ions are very easily oxidized to iodine. And yet the reaction between them in solution in water is very slow. The reaction is catalyzed by the presence of either iron(II) or iron(III) ions. \[ S_2O_8^{2-} +2I^- \rightarrow 2SO_4^{2-} + I_2\] Jim Clark ( )

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In this section we examine how the results of the various approximation methods considered in this chapter can be used to understand and predict the physical properties of multi-electron atoms. Our results include total electronic energies, orbital energies and single-electron wavefunctions that describe the spatial distribution of electron density. Physical properties that can be used to describe multi-electron atoms include total energies, atomic sizes and electron density distributions, ionization energies and electron affinities. Trends in these properties as Z increases form the basis of the periodic table and, as we see in Chapter 10, control chemical reactivity. Spectroscopic properties are considered in a link that includes a development of term symbols for multi-electron systems.

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There are a variety of reactions whereby rings are formed through addition to double or triple bonds. An especially simple example is the addition of ethene to 1,3-butadiene to give cyclohexene: This is the prototype reaction, which has proved so valuable in synthesis that it won its discoverers, O. Diels and K. Alder, the Nobel Prize in chemistry in 1950. The Diels-Alder reaction is both a 1,4 addition or ethene to 1,3-butadiene and a 1,2 addition of butadiene to ethene. It can be called a and as such results in the formation of a six-membered ring. Many other cycloadditions are known, such as [2 + 2], other types of [4 + 2], and [2 + 2 + 2], which give different size of rings. Some specific examples follow: The synthetic importance of these reactions is very great and, because many of them often involve dienes, we will discuss their general characteristics in this chapter. The most valuable cycloaddition reaction almost certainly is the [4 + 2], or Diels-Alder, reaction and will be discussed in detail. There is one very important point you should remember about the Diels-Alder reaction: The reaction usually occurs well only when the [2] component is substituted with electron-attracting groups and the [4] component is substituted with electron-donating groups, or the reverse. The most common arrangement is to have the alkene (usually referred to as the ) substituted with electron-attracting groups such as \(\ce{-CO_2H}\), \(\ce{-COR}\), or \(\ce{-C \equiv N}\). For example, A list of the more reactive dienophiles carrying electron-attracting groups is given in Table 13-1. Ethene and other simple alkenes generally are poor dienophiles and react with 1,3-butadiene only under rather extreme conditions and in low yield. However, when the diene is substituted with several electron-attracting groups such as chlorine or bromine, electron-donating groups on the facilitate the reaction. Many substances, such as 2-methylpropene, that act as dienophiles with hexachlorocyclopentadiene simply will not undergo [4 + 2] addition with cyclopentadiene itself: The Diels-Alder reaction is . The diene reacts in an unfavorable conformation in which its double bonds lie in a plane on the same side (cis) of the single bond connecting them. This (or ) conformation is required to give a stable product with a cis double bond. Addition of ethene to the alternate and more stable ( ) conformation would give an impossibly strained -cyclohexene ring. Possible transition states for reaction in each conformation follow, and it will be seen that enormous molecular distortion would have to take place to allow addition of ethene to the transoid conformation: Cyclic dienes usually react more readily than open-chain dienes, probably because they have their double bonds fixed in the proper conformation for [4 + 2] cycloaddition, consequently the price in energy of achieving the configuration already has been paid: Further evidence of stereospecificity in [4 + 2] additions is that the configurations of the diene and the dienophile are in the adduct. This means that the reactants (or addends) come together to give addition. Two examples follow, which are drawn to emphasize how suprafacial addition occurs. In the first example, dimethyl -butadioate adds to 1,3-butadiene to give a cis-substituted cyclohexene: In the second example, suprafacial approach of a dienophile to the 2,5 carbons of -2,4-hexadiene is seen to lead to a product with two methyl groups on the same side of the cyclohexene ring: (The use of models will help you visualize these reactions and their stereochemistry.) There is a further feature of the Diels-Alder reaction that concerns the stereochemical orientation of the addends. In the addition of -butanedioic anhydride (maleic anhydride) to cyclopentadiene there are two possible ways that the diene and the dienophile could come together to produce different products. These are shown in Equations 13-3 and 13-4: In practice, the adduct with the endo\(^2\) configuration usually is the major product. As a general rule, Diels-Alder additions tend to proceed to favor that orientation that corresponds to having the diene double bonds and the unsaturated substituents of the dienophile closest to one another. This means that addition by Equation 13-3 is more favorable than by Equation 13-4, but the degree of endo-exo stereospecificity is not as high as the degree of stereospecificity of suprafacial addition to the diene and dienophile. There are exceptions to favored endo stereochemistry of Diels-Alder additions. Some of these exceptions arise because the , dissociation being particularly important at high temperature. The exo configuration is generally more stable than the endo and, given time to reach equilibrium (cf. ), the exo isomer may be the major adduct. Thus endo stereospecificity can be expected only when the additions are subject to . The reactivities of dienes in the Diels-Alder reaction depend on the number and kind of substituents they possess. The larger the substituents are, or the more of them, at the ends of the conjugated system, the slower the reaction is likely to be. There also is a marked difference in reactivity with diene configuration. Thus -1,3-pentadiene is substantially less reactive toward a given dienophile (such as maleic anhydride) than is -1,3-pentadiene. In fact, a mixture of the cis and trans isomers can be separated by taking advantage of the difference in their reactivities on cycloaddition: There is little evidence to support simple radical or polar mechanisms (such as we have discussed previously) for the Diels-Alder reaction. As the result of many studies the reaction seems best formulated as a process in which the bonds between the diene and the dienophile are formed essentially simultaneously: We already have discussed a few addition reactions that appear to occur in a concerted manner. These include the addition of diimide, ozone, and boron hydrides to alkenes ( , , and ). Concerted reactions that have cyclic transition states often are called . Other examples will be considered in later chapters. We indicated previously that sulfur dioxide \(\left( \ce{SO_2} \right)\) and 1,3-butadiene form a [4 + 1] cycloaddition product: This reaction is more readily reversible than most Diels-Alder reactions, and the product largely dissociates to the starting materials on heating to \(120^\text{o}\). The cycloadduct is an unsaturated cyclic , which can be hydrogenated to give the saturated cyclic sulfone known as "sulfolane": This compound is used extensively in the petrochemical industry as a selective solvent. The reversibility of the diene-\(\ce{SO_2}\) cycloaddition makes it useful in the purification of reactive dienes. 2-Methyl-1,3-butadiene (isoprene) is purified commercially in this manner prior to being polymerized to rubber ( ): Neither 1,3-cyclopentadiene nor 1,3-cyclohexadiene react with sulfur dioxide, probably because the adducts would be too highly strained: Many naturally occurring organic compounds contain six-membered carbon rings, but there are relatively few with four-membered carbon rings. After encountering the considerable ease with which six-membered rings are formed by [4 + 2] cycloaddition, we might expect that the simpler [2 + 2] cycloadditions to give four-membered rings also should go well, provided that strain is not too severe in the products. In fact, the dimerization of ethene is thermodynamically favorable: Nonetheless, this and many other [2 + 2] cycloaddition reactions do not occur on simple heating. However, there are a few exceptions. One is the dimerization of tetrafluoroethene, which perhaps is not surprising, considering the favorable thermodynamic parameters: What is surprising is that addition of \(\ce{CF_2=CF_2}\) to 1,3-butadiene gives a cyclobutane and a cyclohexane, although the [2 + 2] product probably is about \(25 \: \text{kcal mol}^{-1}\) less stable than the [4 + 2] product: Such [2 + 2] additions generally are limited to polyhaloethenes and a few substances with cumulated double bonds, such as 1,2-propadiene \(\left( \ce{CH_2=C=CH_2} \right)\) and ketenes \(\left( \ce{R_2C=C=O} \right)\). Some examples follow: Many [2 + 2] cycloadditions that do not occur by simply heating the possible reactants can be achieved by with ultraviolet light. The following example, [2 + 2] addition of 2-cyclopentenone to cyclopentene, occurs photochemically but not thermally: In all such photochemical cycloadditions the energy required to achieve a transition state, which can amount to \(100 \: \text{kcal mol}^{-1}\) or more, is acquired by absorption of light. . A striking example is the photochemical conversion of norbornadiene to quadricyclene. The reverse of this reaction can occur with almost explosive violence in the presence of appropriate metal catalysts or on simple heating: Why do some [2 + 2] cycloadditions occur thermally and others photochemically? What is special about fluoroalkenes and cumulated dienes? The answers are complex, but it appears that most [2 + 2] cycloadditions, unlike the Diels-Alder [4 + 2] cycloadditions, go by routes (see ). Why the two types of thermal cycloaddition have different mechanisms will be discussed in and . \(^2\)In general, the designation or refers to configuration in bridged or polycyclic ring systems such as those shown in Equations 13-3 and 13-4. With reference to the bridge atoms, a substituent is exo if it is on the same side as the bridge, or endo if it is on the opposite side. Further examples are In drawing endo and exo isomers, it is best to represent the actual spatial relationships of the atoms as closely as possible. The cyclohexane ring is shown here in the boat form ( ) because it is held in this configuration by the methylene group that bridges the 1,4 positions. If you do not see this, we strongly advise that you construct models. and (1977)

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The reactions of alkanes discussed in are processes, which means that the bonds are made and broken through radical or atomic intermediates. In contrast, the \(S_\text{N}\) and \(E\) reactions of alkyl halides, considered in Chapter 8, involve heterolytic bond cleavage and ionic reagents or products. An especially important factor contributing to the differences between the reactions of the alkanes and alkyl halides is the slight ionic character of \(\ce{C-H}\) compared to \(\ce{C}\)-halide bonds (see ). The alkenes are like the alkanes in being nonpolar compounds ( ) and it may come as a surprise that many important reactions of alkenes are heterolytic reactions. Why should this be so? No doubt because the electrons in the alkene double bonds are more exposed and accessible than the electrons in an alkane \(\ce{C-C}\) bond. This is evident from the atomic-orbital models of ethene described in . The electrons of the double bond are pushed outward by their mutual repulsions, and their average positions are considerably farther from the bond axis than the electron positions of a single bond (Figure 10-6). In such circumstances, electrophilic reagents, which act to acquire electrons in chemical reactions ( ), are expected to be particularly reactive. This is actually the case. Furthermore, reagents that are primarily nucleophilic (electron-donating) are notoriously poor for initiating reactions at carbon-carbon double bonds. Exceptions occur when the double bonds carry substituents with a sufficiently high degree of electron-attracting power to reduce the electron density in the double bond enough to permit attack by a nucleophilic agent. Example of electrophilic reagents that normally add to carbon-carbon double bonds of alkenes to give saturated compounds include halogens (\(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\)), hydrogen halides (\(\ce{HCl}\) and \(\ce{HBr}\)), hypohalous acids (\(\ce{HOCl}\) and \(\ce{HOBr}\)), water, and sulfuric acid: The mechanisms of these reactions have much in common and have been studied extensively from this point of view. They also have very considerable synthetic utility. The addition of water to alkenes (hydration) is particularly important for the preparation of a number of commercially important alcohols. Thus ethanol and 2-propanol (isopropyl alcohol) are made on a very large scale by the hydration of the corresponding alkenes (ethene and propene) using sulfuric or phosphoric acids as catalysts. The nature of this type of reaction will be described later. We shall give particular attention here to the addition of bromine to alkenes because this reaction is carried out very conveniently in the laboratory and illustrates a number of important points about electrophilic addition reactions. Much of what follows applies to addition of the other halogens, except fluorine. A significant observation concerning bromine addition is that it and many of the other reactions listed above proceed in the dark and are influenced by radical inhibitors. This is evidence against a radical-chain mechanism of the type involved in the halogenation of alkanes ( ). However, it does not preclude the operation of radical-addition reactions under other conditions, and, as we shall see later in this chapter, bromine, chlorine, and many other reagents that commonly add to alkenes by ionic mechanisms also can add by radical mechanisms. One alternative to a radical-chain reaction for bromine addition to an alkene would be the simple four-center, one-step process shown in Figure 10-7. The mechanism of Figure 10-7 cannot be correct for bromine addition to alkenes in solution for two important reasons. First, notice that this mechanism requires that the two \(\ce{C-Br}\) bonds be formed on the side of the double bond, and hence produce . However, there is much evidence to show that bromine and many other reagents add to alkenes to form products (Figure 10-8). Cyclohexene adds bromine to give -1,2-dibromocyclohexane: The cis isomer is not formed at all. To give the trans isomer, the two new \(\ce{C-Br}\) bonds have to be formed on of the double bond by antarafacial addition. But this is impossible by a one-step mechanism because the \(\ce{Br-Br}\) bond would have to stretch too far to permit the formation of both \(\ce{C-Br}\) bonds at the same time. The second piece of evidence against the mechanism of Figure 10-7 is that bromine addition reactions carried out in the presence of more than one nucleophilic reagent usually give mixtures of products. Thus the addition of bromine to an alkene in methanol solution containing lithium chloride leads not only to the expected dibromoalkane, but also to products resulting from attack by chloride ions and by the solvent: The intervention of extraneous nucleophiles suggests a mechanism in which the nucleophiles compete for a reactive intermediate formed in one of the steps. A somewhat oversimplified two-step mechanism that accounts for most of the foregoing facts is illustrated for the addition of bromine to ethene. [In the formation shown below, the curved arrows are not considered to have real mechanistic significance, but are used primarily to show which atoms can be regarded as nucleophilic (donate electrons) and which as electrophilic (accept electrons). The arrowheads always should be drawn to point to the atoms that are formulated as accepting a pair of electrons.] The first step (which involves electrophilic attack by bromine on the double bond) produces a bromide ion and a carbocation, as shown in Equation 10-1.\(^1\) As we know from our study of \(S_\text{N}1\) reactions ( ), carbocations react readily with nucleophilic reagents. Therefore in the second step of the bromine-addition mechanism, shown in Equation 10-2, the bromoethyl cation is expected to combine rapidly with bromide ion to give the dibromo compound. However, if other nucleophiles, such as \(\ce{Cl}^\ominus\) or \(\ce{CH_3OH}\), are present in solution, they should be able to compete with bromide ion for the cation, as in Equations 10-3 and 10-4, and mixtures of products will result: To account for the observation that all of these reactions result in antarafacial addition, we must conclude that the first and second steps take place from . The simple carbocation intermediate of Equation 10-1 does not account for formation of the antarafacial-addition product. The results with \(S_\text{N}1\) reactions (Section 8-6) and the atomic-orbital representation (see ) predict that the bonds to the positively charged carbon atom of a carbocation should lie in a plane. Therefore, in the second step of addition of bromine to cycloalkenes, bromide ion could attack either side of the planar positive carbon to give a mixture of - and -1,2-dibromocyclohexanes. Nonetheless, antarafacial addition occurs exclusively: To account for the stereospecificity of bromine addition to alkenes, it has been suggested that in the initial electrophilic attack of bromine a cyclic intermediate is formed that has bromine bonded to carbons of the double bond. Such a "bridged" ion is called a because the bromine formally carries the positive charge: An \(S_\text{N}2\)-type of attack of bromide ion, or other nucleophile, at carbon on the side to the bridging group then results in formation of the antarafacial-addition product: We may seem to have contradicted ourselves because Equation 10-1 shows a carbocation to be formed in bromine addition, but Equation 10-5 suggests a bromonium ion. Actually, the formulation of intermediates in alkene addition reactions as "open" ions or as cyclic ions is a controversial matter, even after many years of study. Unfortunately, it is not possible to determine the structure of the intermediate ions by any direct physical method because, under the conditions of the reaction, the ions are so reactive that they form products more rapidly than they can be observed. However, it is possible to generate stable bromonium ions, as well as the corresponding chloronium and iodonium ions. The technique is to use low temperatures in the absence of any strong nucleophiles and to start with a 1,2-dihaloalkane and antimony pentafluoride in liquid sulfur dioxide: The \(\ce{C_2H_4Br}^\oplus\) ions produced in this way are relatively stable and have been shown by nmr to have the cyclic halonium ion structure. There is a further aspect of polar additions to alkenes that we should consider, namely, that electrophilic reagents form loose complexes with the \(\pi\) electrons of the double bonds of alkenes to reaction by addition. Complexes of this type are called (or ). Formation of a complex between iodine and cyclohexene is demonstrated by the fact that iodine dissolves in cyclohexene to give a solution, whereas its solutions in cyclohexane are . The brown solution of iodine in cyclohexene slowly fades as addition occurs to give colorless -1,2-diiodocyclohexane. Precise Lewis structures cannot be written for charge-transfer complexes, but they commonly are represented as with the arrow denoting that electrons of the double bond are associated with the electrophile. These complexes probably represent the first stage in the formation of addition products by a sequence such as the following for bromine addition: We have seen that electrophiles can react with alkenes to form carbon-halogen bonds by donating positive halogen, \(\ce{Br}^\oplus\), \(\ce{Cl}^\oplus\), or \(\ce{I}^\oplus\). Likewise, carbon-hydrogen bonds can be formed by appropriately strong proton donors, which, of course, are typically strong proton acids. These acids are more effective in the absence of large amounts of water because water can compete with the alkene as a proton acceptor (also see ). Hydrogen chloride addition to ethene occurs by way of a proton-transfer step to give the ethyl cation and a chloride ion (Equation 10-6) followed by a step in which the nucleophilic chloride ion combines with the ethyl cation (Equation 10-7): All of the hydrogen halides \(\ce{HF}\), \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\)) will add to alkenes. Addition of hydrogen fluoride, while facile, is easily reversible. However, a solution of \(70\%\) anhydrous hydrogen fluoride and \(30\%\) of the weak organic base, pyridine, which is about 1/10,000 times as strong as ammonia, works better, and with cyclohexene gives fluorocyclohexane. With hydrogen iodide, care must be taken to prevent \(\ce{I_2}\) addition products resulting from iodine formed by oxidation reactions such as \[4 \ce{HI} + \ce{O_2} \rightarrow 2 \ce{I_2} + 2 \ce{H_2O}\] With hydrogen bromide, radical-chain addition may intervene unless the reaction conditions are controlled carefully (this will be discussed in ). The stereochemistry of addition depends largely on the structure of the alkene, but for simple alkenes and cycloalkenes, addition occurs predominantly in an antarafacial manner. For example, hydrogen bromide reacts with 1,2-dimethylcyclohexene to give the antarafacial addition product: We mentioned previously that the hydration of alkenes required a strong acid as a catalyst, because water itself is too weak an acid to initiate the proton-transfer step. However, if a small amount of a strong acid such as sulfuric acid is present, hydronium ions, \(\ce{H_3O}^\oplus\), are formed in sufficient amount to protonate reasonably reactive alkenes, although by no means as effectively as does concentrated sulfuric acid. The carbocation formed then is attacked rapidly by a nucleophilic water molecule to give the alcohol as its conjugate acid,\(^2\) which regenerates hydronium ion by transferring a proton to water. The reaction sequence follows for 2-methylpropene: In this sequence, the acid acts as a because the hydronium ion used in the proton addition step is regenerated in the final step. Sulfuric acid (or phosphoric acid) is preferred as an acid catalyst for addition of water to alkenes because the conjugate base, \(\ce{HSO_4-}\) (or \(\ce{H_2PO_4-}\)), is a poor nucleophile and does not interfere in the reaction. However, if the water concentration is kept low by using concentrated acid, addition occurs to give sulfate (or phosphate) esters. The esters formed with sulfuric acid are either alkyl acid sulfates \(\ce{R-OSO_3H}\) or dialkyl sulfates \(\ce{(RO)_2SO_2}\). In fact, this is one of the major routes used in the commercial production of ethanol and 2-propanol. Ethen and sulfuric acid give ethyl hydrogen sulfate, which reacts readily with water in a second step to give ethanol: One of the more confusing features of organic chemistry is the multitude of conditions that are used to carry out a given kind of reaction, such as the electrophilic addition of proton acids to different alkenes. Strong acids, weak acids, water, no water - Why can't there be a standard procedure? The problem is that alkenes have very different tendencies to accept protons. In the vapor phase, \(\Delta H^0\) for addition of a proton to ethene is about \(35 \: \text{kcal}\) more positive than for 2-methylpropene, and although the difference should be smaller in solution, it still would be large. Therefore we can anticipate (and we find) that a much more powerful proton donor is needed to initiate addition of an acid to ethene than to 2-methylpropene. But why not use in all cases a strong enough acid to protonate alkene one might want to have a proton acid add to? Two reasons: First, strong acids can induce undesirable side reactions, so that one usually will try not to use a stronger acid than necessary; second, very strong acid may even prevent the desired reaction from occurring! In elementary chemistry, we usually deal with acids in more or less dilute aqueous solution and we think of sulfuric, hydrochloric, and nitric acids as being similarly strong because each is essentially completely disassociated in dilute water solution: \[\ce{HCl} + \ce{H_2O} \overset{\longrightarrow}{\leftarrow} \ce{H_3O}^\oplus + \ce{Cl}^\ominus\] This does not mean they actually are equally strong acids. It means only that each of the acids is sufficiently strong to donate all of its protons to water. We can say that water has a "leveling effect" on acid strengths because as long as an acid can donate its protons to water, the solution has but one acid "strength" that is determined by the \(\ce{H_3O}^\oplus\) concentration, because \(\ce{H_3O}^\oplus\) is where the protons are. Now, if we use poorer proton acceptors as solvent we find the proton-donating powers of various "strong" acids begin to spread out immensely. Furthermore, new things begin to happen. For example, ethene is not hydrated appreciably by dilute aqueous acid; it just is too hard to transfer a proton from hydronium ion to ethene. So we use concentrated sulfuric acid, which is strong enough to add a proton to ethene. But now we don't get hydration, because any water that is present in concentrated sulfuric acid is virtually all converted to \(\ce{H_3O}^\oplus\), which is non-nucleophilic! \[\ce{H_2SO_4} + \ce{H_2O} \rightarrow \ce{H_3O}^\oplus + \ce{HSO_4-}\] However, formation of \(\ce{H_3O}^\oplus\) leads to formation of \(\ce{HSO_4-}\), which has enough nucleophilic character to react with the \(\ce{CH_3CH_2+}\) to give ethyl hydrogen sulfate and this is formed instead of the conjugate acid of ethanol ( ). The epitome of the use of stronger acid and weaker nucleophile is with liquid \(\ce{SO_2}\) (bp \(\sim 10^\text{o}\)) as the solvent and \(\ce{HBF_6}\) as the acid. This solvent is a very poor proton acceptor (which means that its conjugate acid is a very good proton donor) and \(\ce{SbF_6-}\) is an extremely poor nucleophile. If we add ethene to such a solution, a stable solution of \(\ce{CH_3CH_2+} \ce{SbF_6-}\) is formed. The reason is that there is no better proton acceptor present than \(\ce{CH_2=CH_2}\) and no nucleophile good enough to combine with the cation. The conversion of fumaric acid to malic acid is an important biological hydration reaction. It is one of a cycle of reactions (Krebs citric acid cycle) involved in the metabolic combustion of fuels (amino acids and carbohydrates) to \(\ce{CO_2}\) and \(\ce{H_2O}\) in a living cell. \(^1\)An alternative to Equation 10-1 would be to have \(\ce{Br_2}\) ionize to \(\ce{Br}^\oplus\) and \(\ce{Br}^\ominus\), with a subsequent attack of \(\ce{Br}^\oplus\) on the double bond to produce the carbocation. The fact is that energy required for such an ionization of \(\ce{Br_2}\) is prohibitively large even in water solution \(\left( \Delta H^0 \geq 80 \: \text{kcal} \right)\). One might well wonder why Equation 10-1 could possibly be more favorable. The calculated \(\Delta H^0\) for \(\ce{CH_2=CH_2} + \ce{Br_2} \rightarrow \cdot \ce{CH_2-CH_2Br} + \ce{Br} \cdot\) is \(+41 \: \text{kcal}\), which is only slightly more favorable than the \(\Delta H^0\) for \(\ce{Br_2} \rightarrow 2 \ce{Br} \cdot\) of \(46.4 \: \text{kcal}\). However, available thermochemical data suggest that the ease of transferring an electron from \(\cdot \ce{CH_2CH_2Br}\) to \(\ce{Br} \cdot\) to give \(^\oplus \ce{CH_2CH_2Br} + \ce{Br}^\ominus\) is about \(80 \: \text{kcal}\) more favorable than \(2 \ce{Br} \cdot \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). Thus the overall \(\Delta H^0\) of Equation 10-1 is likely to be about \(85 \: \text{kcal}\) more favorable than \(\ce{Br_2} \rightarrow \ce{Br}^\oplus + \ce{Br}^\ominus\). \(^2\)The terms and are very convenient to designate substances that are difficult to name simply as acids, bases, or salts. The conjugate acid of a compound \(\ce{X}\) is \(\ce{XH}^\oplus\) and the conjugate base of \(\ce{HY}\) is \(\ce{Y}^\ominus\). Thus \(\ce{H_3O}^\oplus\) is the conjugate acid of water, while \(\ce{OH}^\ominus\) is its conjugate base. Water itself is then both the conjugate base of \(\ce{H_3O}^\oplus\) and the conjugate acid of \(\ce{OH}^\ominus\) and (1977)

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There are many factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulsion from ions of opposite charge and ions of the same charge. The Madelung constant is a property of the crystal structure and depends on the lattice parameters, anion-cation distances, and molecular volume of the crystal. Before considering a three-dimensional crystal lattice, we shall discuss the calculation of the energetics of a linear chain of ions of alternate signs (Figure \(\Page {1}\)). Let us select the positive sodium ion in the middle (at \(x=0\)) as a reference and let \(r_0\) be the shortest distance between adjacent ions (the sum of ionic radii). The Coulomb energy of the other ions in this 1D lattice on this sodium atom can be decomposed by proximity (or "shells"). and so on. Thus the total energy due to all the ions in the linear array is \[ E = - \dfrac{2e^2}{4 \pi \epsilon_o r_o} + \dfrac{2e^2}{4 \pi \epsilon_o (2r_o)} - \dfrac{2e^2}{4 \pi \epsilon_o (3r_o)} - \ldots\] or \[ E= \dfrac{e^2}{4 \pi \epsilon_o r_o} \left[ 2 \left (1 -\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \ldots \right) \right] \label{eq6}\] We can use the following Maclaurin expansion \[ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}- \frac{x^3}{4} + \cdots\] to simplify the sum in the parenthesis of Equation \ref{eq6} as \(\ln (1+ 1)\) to obtain \[ \begin{align} E &= \dfrac{e^2}{4 \pi \epsilon_o r_o} \left[ 2 \ln 2 \right] \label{eq7} \\[4pt] &= \dfrac{e^2}{4 \pi \epsilon_o } M \end{align} \] The first factor of Equation \ref{eq7} is the Coulomb energy for a single pair of sodium and chloride ions, while the \(2 \ln 2\) factor is the (\(M \approx 1.38 \)) per molecule. The Madelung constant is named after Erwin Medelung and is a geometrical factor that depends on the arrangement of ions in the solid. If the lattice were different (when considering 2D or 3D crystals), then this constant would naturally differ. In three dimensions the series does present greater difficulty and it is not possible to sum the series conveniently as in the case of one-dimensional lattice. As an example, let us consider the the \(\ce{NaCl}\) crystal. In the following discussion, assume \(r\) be the distance between \(\ce{Na^{+}}\) and \(\ce{Cl^-}\) ions. The nearest neighbors of \(\ce{Na^{+}}\) are six \(\ce{Cl^-}\) ions at a distance 1 , 12 \(\ce{Na^{+}}\) ions at a distance 2 , eight \(\ce{Cl^-}\) ions at 3 , six \(\ce{Na^{+}}\) ions at 4 , 24 \(\ce{Na^{+}}\) ions at 5 , and so on. Thus, the electrostatic potential of a single ion in a crystal by approximating the ions by point charges of the surrounding ions: \[ E_{ion-lattice} = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{12.5.4}\] For NaCl is a poorly converging series of interaction energies: \[ M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{21.5.5}\] with The Madelung constant depends on the structure type and Equation \(\ref{21.5.5}\) is applicable only for the sodium chloride (ei.g, rock salt) lattice geometry. Other values for other structural types are given in Table \(\Page {2}\). \(A\) is the number of anions coordinated to cation and \(C\) is the numbers of cations coordinated to anion. is the number of anions coordinated to cation and is the numbers of cations coordinated to anion. There are other factors to consider for the evaluation of lattice energy and the treatment by Max Born and Alfred Lande led to the formula for the evaluation of lattice energy for a mole of . The Born–Landé equation (Equation \(\ref{21.5.6}\)) is a means of calculating the lattice energy of a crystalline ionic compound and derived from the electrostatic potential of the ionic lattice and a repulsive potential energy term \[ U= \dfrac{N_A M Z^2e^2}{4\pi \epsilon_o r} \left( 1 - \dfrac{1}{n} \right) \label{21.5.6}\] where Estimate the lattice energy for \(\ce{NaCl}\). Using the values giving in the discussion above, the estimation is given by \[\begin{align*} U_{NaCl} &= \dfrac{(6.022 \times 10^{23} /mol) (1.74756 ) (1.6022 \times 10 ^{-19})^2 (1.747558)}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m)} \left( 1 - \dfrac{1}{9.1} \right) \nonumber \\[4pt] &= - 756 \,kJ/mol\nonumber \end{align*} \nonumber\] Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Hable cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment.

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As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member. As shown in Table \(\Page {1}\), the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns (n − 1)d valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas: \[2M_{(s)} + 6H_2O_{(l)} \rightarrow 2M(OH)_{3(s)} + 3H_{2(g)}\label{Eq1}\] The chemistry of the group 3 metals is almost exclusively that of the M ion; the elements are powerful reductants. Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M (aq). The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX . The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M O ), which react with H O or CO to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters. Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns (n − 1)d valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti: \[2FeTiO_{3(s)} + 6C_{(s)} + 7Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + 2FeCl_{3(g)} + 6CO_{(g)}\label{Eq2}\] followed by reduction of the tetrachlorides with an active metal such as Mg. The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states. In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO -containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines. Consistent with the periodic trends shown in Figure 23.2, the group 4 metals become denser, higher melting, and more electropositive down the column (Table \(\Page {1}\)). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns (n − 1)d valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti are usually powerful reductants. In fact, the Ti (aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas. Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX ), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl is an ionic salt, whereas TiCl is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO ) and dichalcogenides (MY ). Industrially, TiO , which is used as a white pigment in paints, is prepared by reacting TiCl with oxygen at high temperatures: \[TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} + 2Cl_{2(g)}\label{Eq3}\] The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure \(\Page {1}\)). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH ), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB ), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity. Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores. Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V C , which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V O , an important catalyst for the industrial conversion of SO to SO in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb Zr and Nb Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones. The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4. As indicated in Table \(\Page {1}\), the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M O ), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H O) ] ; the blue-green [V(H O) ] ion is acidic, dissociating to form small amounts of the [V(H O) (OH)] ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H O) VO] , which contains a formal V=O bond (Figure \(\Page {2}\)). Consistent with its covalent character, V O is acidic, dissolving in base to give the vanadate ion ([VO ] ), whereas both Nb O and Ta O are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric. Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO ] ion; V(IV) is the blue vanadyl ion [VO] ; and V(III) and V(II) exist as the hydrated V (blue-green) and V (violet) ions, respectively. Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY ), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools. As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS ) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr O ), which is oxidized to the soluble [CrO ] ion under basic conditions and reduced successively to Cr O and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr O in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO , which is then reduced with hydrogen to give the metal. The metals become increasing polarizable across the d block. Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (Table \(\Page {2}\)). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr , which is a powerful reductant, to CrO , a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known. The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states. As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF is unstable at temperatures above −100°C, whereas MoF and WF are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX (X is Cl, Br, or I), while molybdenum forms MoCl , MoBr , and MoI , and tungsten gives WCl , WBr , and WI . Both Mo and W react with oxygen to form the covalent trioxides (MoO and WO ), but Cr reacts to form only the so-called sesquioxide (Cr O ). Chromium will form CrO , which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO ] ). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr O ) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo O ] ion, whose structure is as follows: An isopolymolybdate cluster. The [Mo O ] ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution. Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS ; it has a layered structure similar to that of TiS (Figure \(\Page {1}\)), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS and WS are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS particles. As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits. Continuing across the periodic table, we encounter the group 7 elements (Table \(\Page {2}\)). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, Tc (m for metastable), has become an important biomedical tool for imaging internal organs. Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline. All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (Figure \(\Page {3}\)). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d electron configuration, the aqueous Mn ion, with a 3d valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound. Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn (aq) ion is pale pink; Mn(OH) , which contains Mn(III), is a dark brown solid; MnO , which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO ] and the purple permanganate ion [MnO ] , respectively. Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F gives only MnF , a high-melting, red-purple solid, whereas Re reacts with F to give ReF , a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl , a weaker oxidant than F , gives MnCl and ReCl . Reaction of Mn with oxygen forms only Mn O , a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe O ). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M O ), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn O , an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO ] is a potent oxidant, whereas [TcO ] and [ReO ] are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS , as well as more complex heptasulfides (M S ). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties. The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced. In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert. The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3. The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age. Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores. Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold. Some properties of the elements in groups 8–10 are summarized in Table \(\Page {3}\). As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state. We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF . In contrast to Fe, Ru and Os form a series of fluorides up to RuF and OsF . The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe O , and the mixed-valent Fe O (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO ) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block. Higher oxidation states become less stable across the d-block, but more stable down a group. Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S ), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S ). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character. The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe C), which is used to strengthen steel. At high temperatures, Fe C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH . Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H but essentially impermeable to all other gases, including He. Consequently, diffusion of H through Pd is an effective method for separating hydrogen from other gases. The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times. Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe ). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer. Some properties of the coinage metals are listed in Table \(\Page {4}\). The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns (n − 1)d valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s 5d valence electron configuration. All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O at high temperatures to produce Cu O and with sulfur to form Cu S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO and in hot concentrated H SO , while Au dissolves in the 3:1 HCl:HNO mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN) ] ions, a reaction that is used to separate gold from its ores. Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au. All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au and Cu ). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF . In contrast, all the gold trihalides (AuX ) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements. We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries. As shown in Table \(\Page {4}\), the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns (n − 1)d valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg . The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble. All the possible group 12 dihalides (MX ) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl ). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO and H SO . All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur. For each reaction, explain why the indicated products form. : balanced chemical equation Asked for: why the indicated products form Refer to the periodic trends in this section. : Predict the products of each reactions and then balance each chemical equation. The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M O ). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr O is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands.

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In our survey of the dioxygen chemistry of iron and copper species in earlier subsections, three general functions for the protein matrix became apparent: provision of ligand(s) in an appropriate stereochemistry; protection of the metal-dioxygen moiety from oxidation and competitive ligands; and modulation of dioxygen affinity through nonbonded interactions with distal groups. In the hemoglobin family the heme group is anchored in a cleft in the globin chain by an imidazole ligand from a histidine residue (the proximal histidine). The other (distal) side of the heme plane is more or less open to accommodate a small sixth ligand (see Figure 4.2). For hemerythrin and hemocyanin the requirements of the protein chain are more severe. In contrast to the hemoglobin family, all but two of the ligands are provided by the protein chain, and in addition the metal ions are encapsulated as a pair. The exogenous ligands for hemerythrin are a \(\mu\)-(hydr)oxo moiety and dioxygen or anions (depending on oxidation state); for hemocyanin, the identity of the second exogenous ligand, if there is one at all, is still unclear. Although hemerythrin has a distinctive cofacial bioctahedral structure (Figure 4.10) that would appear to be very difficult to assemble in the absence of the protein, it turns out that with a variety of tridentate ligands the (\(\mu\)-oxo)bis(\(\mu\)-carboxylato)diiron(III) core may be assembled rather easily. Thus, this core appears to be a thermodynamically very stable structural motif. Such a synthesis has been termed "self-assembly" and appears to be a common phenomenon in biological systems. The low-temperature assembly of bis-copper(II)-\(\mu\)-peroxo complexes (models for oxyhemocyanin) from mononuclear copper(l) compounds provides other examples of this phenomenon. The immobilization of the heme group inside the protein prevents (i) the bimolecular contact of an FeO species with an Fe species (Reaction 4.29b), the key step in the irreversible oxidation of Fe porphyrins; (ii) the facile access of nucleophiles that would cause autoxidation (Reactions 4.30 and 4.31); (iii) the oxygenase activity (Reaction 4.32) that is the normal function of other hemoproteins, such as cytochrome P-450, horseradish peroxidase, catalase, etc.; and (iv) the self-oxygenase activity that has been observed in some iron(II) systems that bind dioxygen, activating it for destruction of the ligand itself. Avoiding these last two fates also appears to be very important in the active site of hemocyanin. Finally (v), the globin chain serves to restrain the binding of the distal histidine to give a six-coordinate hemochrome (Reaction 4.33), at least at room temperature. Thus, unoxygenated hemoglobin is held in a five-coordinate state, allowing a rapid rate of oxygen binding and greater oxygen affinity—hemochromes such as Fe(TPP)(Py) are impervious to oxygenation and subsequent oxidation. The protein chain in hemoglobin may place restraints on the iron-to-proximal histidine bond. On the other side of the heme, the distal histidine and occluded water molecules may hydrogen-bond to the coordinated dioxygen and force ligands to adopt geometries that are different from those observed in the absence of steric hindrances. The conformation of the porphyrin skeleton may also be perturbed by the protein chain. Clearly, it is the protein chain that bestows the property of cooperativity on oligomeric oxygen carriers.

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In our survey of the dioxygen chemistry of iron and copper species in earlier subsections, three general functions for the protein matrix became apparent: provision of ligand(s) in an appropriate stereochemistry; protection of the metal-dioxygen moiety from oxidation and competitive ligands; and modulation of dioxygen affinity through nonbonded interactions with distal groups. In the hemoglobin family the heme group is anchored in a cleft in the globin chain by an imidazole ligand from a histidine residue (the proximal histidine). The other (distal) side of the heme plane is more or less open to accommodate a small sixth ligand (see Figure 4.2). For hemerythrin and hemocyanin the requirements of the protein chain are more severe. In contrast to the hemoglobin family, all but two of the ligands are provided by the protein chain, and in addition the metal ions are encapsulated as a pair. The exogenous ligands for hemerythrin are a \(\mu\)-(hydr)oxo moiety and dioxygen or anions (depending on oxidation state); for hemocyanin, the identity of the second exogenous ligand, if there is one at all, is still unclear. Although hemerythrin has a distinctive cofacial bioctahedral structure (Figure 4.10) that would appear to be very difficult to assemble in the absence of the protein, it turns out that with a variety of tridentate ligands the (\(\mu\)-oxo)bis(\(\mu\)-carboxylato)diiron(III) core may be assembled rather easily. Thus, this core appears to be a thermodynamically very stable structural motif. Such a synthesis has been termed "self-assembly" and appears to be a common phenomenon in biological systems. The low-temperature assembly of bis-copper(II)-\(\mu\)-peroxo complexes (models for oxyhemocyanin) from mononuclear copper(l) compounds provides other examples of this phenomenon. The immobilization of the heme group inside the protein prevents (i) the bimolecular contact of an FeO species with an Fe species (Reaction 4.29b), the key step in the irreversible oxidation of Fe porphyrins; (ii) the facile access of nucleophiles that would cause autoxidation (Reactions 4.30 and 4.31); (iii) the oxygenase activity (Reaction 4.32) that is the normal function of other hemoproteins, such as cytochrome P-450, horseradish peroxidase, catalase, etc.; and (iv) the self-oxygenase activity that has been observed in some iron(II) systems that bind dioxygen, activating it for destruction of the ligand itself. Avoiding these last two fates also appears to be very important in the active site of hemocyanin. Finally (v), the globin chain serves to restrain the binding of the distal histidine to give a six-coordinate hemochrome (Reaction 4.33), at least at room temperature. Thus, unoxygenated hemoglobin is held in a five-coordinate state, allowing a rapid rate of oxygen binding and greater oxygen affinity—hemochromes such as Fe(TPP)(Py) are impervious to oxygenation and subsequent oxidation. The protein chain in hemoglobin may place restraints on the iron-to-proximal histidine bond. On the other side of the heme, the distal histidine and occluded water molecules may hydrogen-bond to the coordinated dioxygen and force ligands to adopt geometries that are different from those observed in the absence of steric hindrances. The conformation of the porphyrin skeleton may also be perturbed by the protein chain. Clearly, it is the protein chain that bestows the property of cooperativity on oligomeric oxygen carriers.

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Some forces that interact within pure liquids are also present during mixtures and solutions. Forces such as Cohesive as well as Adhesive forces still apply to mixtures; however, more importantly we focus on the interaction between different molecules. Why is oil only soluble in benzene and not water? Why do only "like" molecules dissolve in "like" molecules? Before we go on to the more specific mechanisms of mixing, let's discuss its process. Mixing is a spontaneous process that increases the entropy of the solution. In order to form a mixture of homogenous solutions by distributing the solute molecules evenly within the solvent molecules, heat transfers are inevitable. This heat transfer is denoted ΔH for our general comprehension. ΔH is the change in heat energy found by subtracting the enthalpy of the reactant from that of the product: \[H_{products} - H_{reactants}= ΔH_{soln}.\] What then is the significance of \(ΔH_{soln}\)? It presents a clear indication of the magnitude as well as direction of the heat transfer so that when: What we have to supply for our understanding for this equation is that the extra energy is seized either from or give to the surrounding. And to ascertain the enthalpy of solution, we take the three step approach in enthalpy when a solute is mixed with the solvent. note that usually \(ΔH_1\) and \(ΔH_2\) are opposite in sign as \(ΔH_3\). Separating the solute and the solvent solutions alone are usually endothermic reactions in that their cohesive forces are broken while letting the molecules to react freely is an exothermic reaction. The figure below explains pictorially how positive and negative \(ΔH\) can be obtained through the three step process of mixing solution. Ideal solution is the mixture that has little to no net intermolecular interactions that differentiates it from its ideal behavior. Thus if the intermolecular forces of attraction are the same and have the same strength, both the solvent and solute will mix at random. This solution is called an , which means that \(ΔH_{soln} = 0\). If the intermolecular forces of attraction of different molecules are greater than the forces of attraction of like molecules, then it is called a nonideal solution. This will result in an exothermic process (\(ΔH_{soln}<0\)). If the intermolecular forces of attraction of different molecules are a bit weaker than the forces of nonideal solution, has bigger enthalpy value than pure components, and it goes through an endothermic process. Lastly, if the intermolecular forces of attraction of different molecules is a lot weaker than the forces of attraction of like molecules, the solution becomes a heterogeneous mixture (e.g., water and olive oil). The epitome of intermolecular forces in solution is the miracle of solubility, because when a matter precipitates it no longer interacts with the solvent. So what is the attraction between "like" molecules that makes them attract to each other? Let's take a phospholipid, the building block of a cell's membrane, as an example. This molecule is amphipathic, meaning that it is both hydrophilic and hydrophobic. Beginning with the structure of a phospholipid, it has a polar head which is hydrophilic and a nonpolar tail which is hydrophobic as the picture below. How can a single molecule be both polar molecule loving and polar molecule disliking at the same time? This is because at the polar head, the phosphate has a net negative charge thus attracting the partial positive charge of the hydrogen molecules of water. Its nonpolar tails on the other hand, is a very organized form of hydrocarbon, consisting of no net charges. The tail is then repelled by water as it struggles to fit between the partial positive and partial negative of the water molecule. Another side effect of the interactions of molecules is reflected by the use of the activity coefficient during thermodynamic equilibrium constant calculations. This constant differentiates ideal and nonideal solutions so that interactions for solution equilibrium can be more accurately estimated. Most versions of the equilibrium constant K utilizes activity instead of concentration so that the units would disappear more fluently. For an ideal solution, the activity coefficient is 1 [x]/ Celcius, thus when the concentration is dived by it to yield activity, it is unaltered. Based on the concept of intermolecular interactions, ascertain the reason behind freezing-point depression and boiling-point elevation. When an ion is added into solution, it exerts an intermolecular force which binds loosely to the water molecules in solution. This weak force then increases the energy necessary to break each molecule loose, thus increasing temperature in relationship to vapor pressure. It now takes more energy input to obtain the same vapor pressure, thus elevating the boiling point. For freezing-point depression, the same force that is holding the water molecules from evaporating is holding them against being placed into an organized solid form. It now takes more energy to form the weak bonds between each water molecule because the intermolecular forces between water and the ions first have to be overcome, hence reducing the freezing point. Give examples that present the involvement of intermolecular forces thus differentiating ideal from nonideal solutions. Gibbs free energy (relating ΔG with ΔG ), calculating thermodynamic equilibrium constants, boiling-point elevation, and freezing-point depression

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Most alkenes react readily with ozone \(\left( \ce{O_3} \right)\), even at low temperatures, to yield cyclic peroxidic derivatives known as . For example, These substances, like most compounds with peroxide \(\left( \ce{O-O} \right)\) bonds, may explode violently and unpredictably. Therefore ozonizations must be carried out with appropriate caution. The general importance of these reactions derives not from the ozonides, which usually are not isolated, but from their subsequent products. The ozonides can be converted by hydrolysis with water and reduction, with hydrogen (palladium catalyst) or with zinc and acid, to carbonyl compounds that can be isolated and identified. For example, 2-butene gives ethanal on ozonization, provided the ozonide is destroyed with water and a reducing agent which is effective for hydrogen peroxide: An alternative procedure for decomposing ozonides from di- or trisubstituted alkenes is to treat them with methanol \(\left( \ce{CH_3OH} \right)\). The use of this reagent results in the formation of an aldehyde or ketone and a carboxylic acid: The overall ozonization reaction sequence provides an excellent means for locating the positions of double bonds in alkenes. The potentialities of the method may be illustrated by the difference in reaction products from the 1- and 2-butenes: Ozonization of alkenes has been studied extensively for many years, but there is still disagreement about the mechanism (or mechanisms) involved because some alkenes react with ozone to give oxidation products other than ozonides. It is clear that the ozonide is not formed directly, but by way of an unstable intermediate called a . the molozonide then either isomerizes to the "normal" ozonide or participates in other oxidation reactions. Although the structure of normal ozonides has been established beyond question, that of the molozonide, which is very unstable even at \(-100^\text{o}\), is much less certain. The simplest and most widely accepted mechanism involves formation of a molozonide by a direct of ozone to the double bond.\(^1\) Isomerization of the molozonide appears to occur by a fragmentation-recombination reaction, as shown in Equations 11-7 and 11-8: Several oxidizing reagents react with alkenes under mild conditions to give, as the overall result, addition of hydrogen peroxide as \(\ce{HO-OH}\). Of particular importance are alkaline permanganate \(\left( \ce{MnO_4^-} \right)\) and osmium tetroxide \(\left( \ce{OsO_4} \right)\), both of which react in an initial step by a suprafacial cycloaddition mechanism like that postulated for ozone. Each of these reagents produces -1,2-dihydroxy compounds (diols) with cycloalkenes: Osmium tetroxide is superior to permanganate in giving good yields of diol, but its use is restricted because it is a very costly and very toxic reagent. Alkenes can be oxidized with peroxycarboxylic acids, \(\ce{RCO_3H}\), to give oxacyclopropanes (oxiranes, epoxides), which are three-membered cyclic ethers: The reaction, known as , is valuable because the oxacyclopropane ring is cleaved easily, thereby providing a route to the introduction of many kinds of functional groups. In fact, oxidation of alkenes with peroxymethanoic acid (peroxyformic acid), prepared by mixing methanoic acid and hydrogen peroxide, usually does not stop at the oxacyclopropane stage, but leads to ring-opening and the subsequent formation of a diol: This is an alternative scheme for the hydroxylation of alkenes (see ). However, the overall stereochemistry is opposite to that in permanganate hydroxylation. For instance, cyclopentene gives -1,2-cylcopentanediol. First the oxirane forms by suprafacial addition and then undergoes ring opening to give the trans product: The ring opening is a type of \(S_\text{N}2\) reaction. Methanoic acid is sufficiently acidic to protonate the ring oxygen, which makes it a better leaving group, thus facilitating nucleophilic attack by water. The nucleophile always attacks from the side remote from the leaving group: The peroxyacids that are used in the formation of oxacyclopropanes include peroxyethanoic \(\left( \ce{CH_3CO_3H} \right)\), peroxybenzoic \( \left( \ce{C_6H_5CO_3H} \right)\), and trifluoroperoxyethanoic \(\left( \ce{CF_3CO_3H} \right)\) acids. A particularly useful peroxyacid is 3-chloroperoxybenzoic acid, because it is relatively stable and is handled easily as the crystalline solid. The most reactive reagent is trifluoroperoxyethanoic acid, which suggests that the peroxyacid behaves as an electrophile (the electronegativity of fluorine makes the \(\ce{CF_3}\) group strongly electron-attracting). The overall reaction can be viewed as a , in which the proton on oxygen is transferred to the neighboring carbonyl oxygen more or less simultaneously with formation of the three-membered ring: A reaction of immense industrial importance is the formation of oxacyclopropane itself (most often called ethylene oxide) by oxidation fo ethene with oxygen over a silver oxide catalyst at \(300^\text{o}\): Oxacyclopropane is used for many purposes, but probably the most important reaction is ring opening with water to give 1,2-ethanediol (ethylene glycol, bp \(197^\text{o}\)). This diol, mixed with water, is employed widely in automotive cooling systems to provide both a higher boiling and lower freezing coolant than water alone: Propene and higher alkenes are not efficiently epoxidized by oxygen and \(\ce{Ag_2O}\) in the same way as ethene because of competing attack at other than the double-bond carbons. \(^1\)The ozone structure shown here with single electrons having paired spins on the terminal oxygens accords both with the best available quantum mechanical calculations and the low dipole moment of ozone, which is not consonant with the conventional \(\ce{O=} \overset{\oplus}{\ce{O}} - \overset{\ominus}{\ce{O}}\) structure. See W. A. Goddard III, T. H. Dunning, Jr., W. J. Hunt, and P. J. Hay, , 368 (1973). and (1977)

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states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the gases in the mixture. the different molecules in a mixture of gases are so far apart that they act independently; they do not react with each other. The pressure of an ideal gas is determined by its collisions with the container, not collisions with molecules of other substances since there are no other collisions. A gas will expand to fill the container it is in without affecting the pressure of another gas. So it can be concluded that the pressure of a certain gas is based on the number of moles of that gas and the volume and temperature of the system. Since the gases in a mixture of gases are in one container, the Volume (V) and Temperature (T) for the different gases are the same as well. Each gas exerts its own pressure on the system, which can be added up to find the total pressure of the mixture of gases in a container. This is shown by the equation \[P_{total} = P_A + P_B + ... \tag{1}\] We have, from the \[PV = nRT \tag{2}\] If we know the molar composition of the gas, we can write \[n_{total} = n_a + n_b + ... \tag{3}\] Again, based on the kinetic theory of gases and the ideal gas law, Dalton’s law can also be applied to the number of moles so that the total number of moles equals the sum of the number of moles of the individual gases. Here, the pressure, temperature and volume are held constant in the system. The total volume of a gas can be found the same way, although this is not used as much. This yields the equation, \[P_{total} V=n_{total} RT \tag{4}\] We can rearrange this equation to find the total number of moles. Sometimes masses of each sample of gas are given and students are asked to find the total pressure. This can be done by converting grams to moles and using Dalton's law to find the pressure. From the partial pressure of a certain gas and the total pressure of a certain mixture, the mole ratio, called Xi, of a gas can be found. The mole ratio describes what fraction of the mixture is a specific gas. For example, if oxygen exerts 4 atm of pressure in a mixture and the total pressure of the system is 10 atm, the mole ratio would be 4/10 or 0.4. The mole ratio applies to pressure, volume, and moles as seen by the equation below. This also means that 0.4 moles of the mixture is made up of gas i. \[X_i=\dfrac{P_i}{P_{tot}}=\dfrac{n_i}{n_{tot}}=\dfrac{V_i}{V_{tot}} \tag{5}\] The mole ratio, (\(X_i\)) is often used to determine the composition of gases in a mixture. The sum of the mole ratios of each gas in a mixture should always equal one since they represent the proportion of each gas in the mixture. The Law of Partial Pressures is commonly applied in looking at the pressure of a closed container of gas and water. The total pressure of this system is the pressure that the gas exerts on the liquid. The gas is made up of whatever sample of gas there is plus the evaporated water. The pressure of the gas on the liquid consists of the pressure of the evaporated water and the pressure of the gas collected. Based on Dalton’s law, the pressure of the gas collected can be calculated by subtracting the pressure of the water vapor from the total pressure. Real gases are gases that do not behave ideally. That is, they violate one or more of the rules of the kinetic theory of gases. Real gases behave ideally when the gases are at low pressure and high temperature. Therefore at high pressures and low temperatures, Dalton’s law is not applicable since the gases are more likely to react and change the pressure of the system. For example, if there are forces of attraction between the molecules, the molecules would get closer together and the pressure would be adjusted because the molecules are interacting with each other. 1: The law of partial pressures also applies to the total number of moles if the other values are constant, so 4 mol Hydrogen+8 mol Oxygen+12 mol Helium+6 mol Nitrogen= 4. 5.

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Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth is introduced. Monodentate ligands bind through only one donor atom. Monodentate means "one-toothed." The halides, phosphines, ammonia and amines seen previously are monodentate ligands. Bidentate ligands bind through two donor sites. Bidentate means "two-toothed." An example of a bidentate ligand is ethylenediamine. It can bind to a metal via two donor atoms at once. Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the " ." This term comes from the Greek chelos, meaning "crab." A crab does not have any teeth at all, but it does have two claws for tightly holding onto something for a couple of reasons. A very simple analogy is that, if you are holding something with two hands rather than one, you are not as likely to drop it. In the examples previously disccussed, each ligand only forms one bond with the central metal ion to give the complex ion. Such a ligand is said to be unidentate. That means literally that it only has one tooth! It only has one pair of electrons that it can use to bond to the metal - any other lone pairs are pointing in the wrong direction. Some ligands, however, have rather more teeth! These are known generally as multidentate or polydentate ligands, but can be broken down into a number of different types. Bidentate ligands have two lone pairs, both of which can bond to the central metal ion. The two commonly used examples are 1,2-diaminoethane (old name: ethylenediamine - often given the abbreviation "en"), and the ethanedioate ion (old name: oxalate). In the ethanedioate ion, there are lots more lone pairs than the two shown, but these are the only ones we are interested in. You can think of these bidentate ligands rather as if they were a pair of headphones, carrying lone pairs on each of the "ear pieces". These will then fit snuggly around a metal ion. You might find this abbreviated to \([Ni(en)_3]^{2+}\). The structure of the ion looks like this In this case, the "ear pieces" are the nitrogen atoms of the NH2 groups - and the "bit that goes over your head " is the \(-CH_2CH_2-\) group. If you were going to draw this in an exam, you would obviously want to draw it properly - but for learning purposes, drawing all the atoms makes the diagram look unduly complicated! Notice that the arrangement of the bonds around the central metal ion is exactly the same as it was with the ions with 6 water molecules attached. The only difference is that this time each ligand uses up two of the positions - at right angles to each other. Because the nickel is forming 6 co-ordinate bonds, the coordination number of this ion is 6, despite the fact that it is only joined to 3 ligands. Coordination number counts the number of bonds, not the number of ligands. This is the complex ion formed by attaching 3 ethanedioate (oxalate) ions to a chromium(III) ion. The shape is exactly the same as the previous nickel complex. The only real difference is the number of charges. The original chromium ion carried 3+ charges, and each ethanedioate ion carried -2, i.e., \[ (+3) + (3 \times -2) = -3. \nonumber \] The structure of the ion looks like this: Again, if you drew this in an exam, you would want to show all the atoms properly. If you need to be able to do this, practice drawing it so that it looks clear and tidy! Refer back to the diagram of the ethanedioate ion further up the page to help you. A quadridentate ligand has four lone pairs, all of which can bond to the central metal ion. An example of this occurs in haemoglobin (American: hemoglobin). The functional part of this is an iron(II) ion surrounded by a complicated molecule called heme. This is a sort of hollow ring of carbon and hydrogen atoms, at the center of which are 4 nitrogen atoms with lone pairs on them. Heme is one of a group of similar compounds called porphyrins. They all have the same sort of ring system, but with different groups attached to the outside of the ring. You aren't going to need to know the exact structure of the haem at this level. We could simplify the heme with the trapped iron ion as: Each of the lone pairs on the nitrogen can form a co-ordinate bond with the iron(II) ion - holding it at the center of the complicated ring of atoms. The iron forms 4 co-ordinate bonds with the heme, but still has space to form two more - one above and one below the plane of the ring. The protein globin attaches to one of these positions using a lone pair on one of the nitrogens in one of its amino acids. The interesting bit is the other position. The water molecule which is bonded to the bottom position in the diagram is easily replaced by an oxygen molecule (again via a lone pair on one of the oxygens in \(O_2\)) - and this is how oxygen gets carried around the blood by the haemoglobin. When the oxygen gets to where it is needed, it breaks away from the haemoglobin which returns to the lungs to get some more. You probably know that carbon monoxide is poisonous because it reacts with hemeoglobin. It bonds to the same site that would otherwise be used by the oxygen - but it forms a very stable complex. The carbon monoxide doesn't break away again, and that makes that hemeoglobin molecule useless for any further oxygen transfer. A hexadentate ligand has 6 lone pairs of electrons - all of which can form co-ordinate bonds with the same metal ion. The best example is . The diagram shows the structure of the ion with the important atoms and lone pairs picked out. The EDTA ion entirely wraps up a metal ion using all 6 of the positions that we have seen before. The co-ordination number is again 6 because of the 6 co-ordinate bonds being formed by the central metal ion. The diagram below shows this happening with a copper(II) ion. Here is a simplified version. Make sure that you can see how this relates to the full structure above. The overall charge, of course, comes from the 2+ on the original copper(II) ion and the 4- on the \(EDTA^{4-}\) ion. Jim Clark ( )

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In this chapter we used a very simple model called the particle-in-a-box model or the infinite-potential-well model to obtain very crude approximate wavefunctions for pi electrons in cyanine dye molecules. With the particle in the box model, we can estimate the wavelengths at which the peaks occur in the absorption spectra from estimated bond lengths, or we can use the wavelength information to determine average bond lengths in a series of dye molecules. By evaluating the transition moment integral, we can explain the relative intensities of these peaks and obtain selection rules for the spectroscopic transitions. The selection rules also can be deduced from qualitative symmetry considerations. This model assumes the electrons are independent of each other and uses a particularly simple form for the potential energy of the electrons. The model also assumes that the atomic nuclei are fixed in space, i.e. the molecule is not vibrating or rotating. This latter assumption, which is known as the Crude Born-Oppenheimer Approximation, will be discussed in a later chapter. The physical basis for this approximation is the fact that the mass of the electron is much smaller than the mass of an atomic nucleus. The electrons therefore respond to forces or are accelerated by forces much faster than the nuclei (remember a = f/m) so the electron motion in a molecule can be examined by assuming that the nuclei are stationary. We did not discuss the widths and shapes of the peaks. Contributions to the line widths and shapes come from motion of the nuclei; which we will consider later. Nuclei in a molecule vibrate, i.e. move relative to each other, and rotate around the center of mass of the molecule. The rotational and vibrational motion, as well as interaction with the solvent, which also is neglected, produce the spectral band widths and shapes and even affect the position of the absorption maximum. When light is absorbed the vibrational and rotational energy of the molecule can change along with the change in the electronic energy. Line widths and shapes therefore depend upon the absorption of different amounts of vibrational and rotational energy. Actually, in a condensed phase, molecular rotation is hindered. This hindered rotation is called libration. An outcome of our examination of the cyanine dye wavefunctions was a glimpse at three fundamental properties of quantum mechanical systems: orthogonality of wavefunctions, the Heisenberg Uncertainty Principle, and the zero-point energy of bound systems. We also observed that quantum numbers result from the boundary conditions used to describe the physical system. Another observation was that the energy levels for the particle in the box get further apart as the quantum number n increases, but closer together as the size of the box increases. Lastly, the spectra we observe occur because of the interaction of molecules with electromagnetic radiation and the resulting transition of the molecule from one energy level to a higher energy level.

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We learned earlier in 20-5 that entropy, \(S\), is related to the number of microstates, \(W\), in an ensemble with \(A\) systems: \[S_{ensemble}=k_B \ln{W} \label{eq1} \] and \[W=\frac{A!}{\prod_j{a_j}!} \label{eq2} \] Combining Equations \ref{eq1} and \ref{eq2} to get: \[\begin{split} S_{ensemble} &= k_B \ln{\frac{A!}{\prod_j{a_j}!}} \\ &= k_B \ln{A!}-k_B\sum_j{\ln{a_j}!} \end{split} \nonumber \] Using \[\ln{A!} \approx A\ln{A}-A \nonumber \] We obtain: \[S_{ensemble} = k_B A \ln{A}-k_BA-k_B\sum_j{a_j\ln{a_j}}+k_B\sum{a_j} \nonumber \] Since: \[A=\sum{a_j} \nonumber \] The expression simplifies to: \[S_{ensemble} = k_B A \ln{A}-k_B\sum_j{a_j\ln{a_j}} \nonumber \] We can make use of the fact that the number of microstates in state \(j\) is equal to the total number of microstates multiplied by the probability of finding the system in state \(j\), \(p_j\): \[a_j=p_jA \nonumber \] Plugging in, we obtain \[\begin{split}S_{ensemble} &= k_B A \ln{A}-k_B\sum_j{p_jA\ln{p_jA}} \\ &= k_B A \ln{A}-k_B\sum_j{p_jA\ln{p_j}}-k_B\sum_j{p_jA\ln{A}} \end{split} \nonumber \] Since \(A\) is a constant and the sum of the probabilities of finding the system in state \(j\) is always 1: \[\sum{p_j}=1 \nonumber \] The first and last term cancel out: \[S_{ensemble} = -k_BA\sum_j{p_j\ln{p_j}} \nonumber \] We can use that the entropy of the system is the entropy of the ensemble divided by the number of systems: \[S_{system}=S_{ensemble}/A \nonumber \] Dividing by \(A\), we obtain: \[S_{system} = -k_B\sum_j{p_j\ln{p_j}} \nonumber \] We can differentiate this equation and dropping the subscript: \[dS = -k_B\sum_j{\left(dp_j+\ln{p_j}dp_j\right)} \nonumber \] Since \(\sum_j{p_j}=1\), the derivative \(\sum_j{dp_j}=0\): \[dS = -k_B\sum_j{\ln{p_j}dp_j} \nonumber \] In short: \[\sum_j{\ln{p_j}dp_j}=-\frac{\delta q_{rev}}{k_BT} \nonumber \] Plugging in: \[dS = \frac{\delta q_{rev}}{T} \nonumber \]

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The transition metals, groups 3–12 in the periodic table, are generally characterized by partially filled d subshells in the free elements or their cations. (Although the metals of group 12 do not have partially filled d shells, their chemistry is similar in many ways to that of the preceding groups, and we therefore include them in our discussion.) Unlike the and elements, the transition metals exhibit significant horizontal similarities in chemistry in addition to their vertical similarities. The valence electron configurations of the first-row transition metals are given in Table \(\Page {1}\). As we go across the row from left to right, electrons are added to the 3d subshell to neutralize the increase in the positive charge of the nucleus as the atomic number increases. With two important exceptions, the 3d subshell is filled as expected based on the aufbau principle and Hund’s rule. Unexpectedly, however, chromium has a 4s 3d electron configuration rather than the 4s 3d configuration predicted by the aufbau principle, and copper is 4s 3d rather than 4s 3d . In Chapter 7, we attributed these anomalies to the extra stability associated with half-filled subshells. Because the ns and (n − 1)d subshells in these elements are similar in energy, even relatively small effects are enough to produce apparently anomalous electron configurations. In the second-row transition metals, electron–electron repulsions within the 4d subshell cause additional irregularities in electron configurations that are not easily predicted. For example, Nb and Tc, with atomic numbers 41 and 43, both have a half-filled 5s subshell, with 5s 4d and 5s 4d valence electron configurations, respectively. Further complications occur among the third-row transition metals, in which the 4f, 5d, and 6s orbitals are extremely close in energy. Although La has a 6s 5d valence electron configuration, the valence electron configuration of the next element—Ce—is 6s 5d 4f . From this point through element 71, added electrons enter the 4f subshell, giving rise to the 14 elements known as the lanthanides. After the 4f subshell is filled, the 5d subshell is populated, producing the third row of the transition metals. Next comes the seventh period, where the actinides have three subshells (7s, 6d, and 5f) that are so similar in energy that their electron configurations are even more unpredictable. As we saw in the s-block and p-block elements, the size of neutral atoms of the d-block elements gradually decreases from left to right across a row, due to an increase in the effective nuclear charge (Z ) with increasing atomic number. In addition, the atomic radius increases down a group, just as it does in the s and p blocks. Because of the lanthanide contraction, however, the increase in size between the 3d and 4d metals is much greater than between the 4d and 5d metals (Figure 23.1).The effects of the lanthanide contraction are also observed in ionic radii, which explains why, for example, there is only a slight increase in radius from Mo to W . As you learned previously, electrons in (n − 1)d and (n − 2)f subshells are only moderately effective at shielding the nuclear charge; as a result, the effective nuclear charge experienced by valence electrons in the d-block and f-block elements does not change greatly as the nuclear charge increases across a row. Consequently, the ionization energies of these elements increase very slowly across a given row (Figure \(\Page {2}\)). In addition, as we go from the top left to the bottom right corner of the d block, electronegativities generally increase, densities and electrical and thermal conductivities increase, and enthalpies of hydration of the metal cations decrease in magnitude, as summarized in Figure \(\Page {2}\). Consistent with this trend, the transition metals become steadily less reactive and more “noble” in character from left to right across a row. The relatively high ionization energies and electronegativities and relatively low enthalpies of hydration are all major factors in the noble character of metals such as Pt and Au. The similarity in ionization energies and the relatively small increase in successive ionization energies lead to the formation of metal ions with the same charge for many of the transition metals. This in turn results in extensive horizontal similarities in chemistry, which are most noticeable for the first-row transition metals and for the lanthanides and actinides. Thus all the first-row transition metals except Sc form stable compounds that contain the 2+ ion, and, due to the small difference between the second and third ionization energies for these elements, all except Zn also form stable compounds that contain the 3+ ion. The relatively small increase in successive ionization energies causes most of the transition metals to exhibit multiple oxidation states separated by a single electron. Manganese, for example, forms compounds in every oxidation state between −3 and +7. Because of the slow but steady increase in ionization potentials across a row, high oxidation states become progressively less stable for the elements on the right side of the d block. The occurrence of multiple oxidation states separated by a single electron causes many, if not most, compounds of the transition metals to be paramagnetic, with one to five unpaired electrons. This behavior is in sharp contrast to that of the p-block elements, where the occurrence of two oxidation states separated by two electrons is common, which makes virtually all compounds of the p-block elements diamagnetic. Due to a small increase in successive ionization energies, most of the transition metals have multiple oxidation states separated by a single electron. Most compounds of transition metals are paramagnetic, whereas virtually all compounds of the p-block elements are diamagnetic. The electronegativities of the first-row transition metals increase smoothly from Sc (χ = 1.4) to Cu (χ = 1.9). Thus Sc is a rather active metal, whereas Cu is much less reactive. The steady increase in electronegativity is also reflected in the standard reduction potentials: thus E° for the reaction M (aq) + 2e → M (s) becomes progressively less negative from Ti (E° = −1.63 V) to Cu (E° = +0.34 V). Exceptions to the overall trends are rather common, however, and in many cases, they are attributable to the stability associated with filled and half-filled subshells. For example, the 4s 3d electron configuration of zinc results in its strong tendency to form the stable Zn ion, with a 3d electron configuration, whereas Cu , which also has a 3d electron configuration, is the only stable monocation formed by a first-row transition metal. Similarly, with a half-filled subshell, Mn (3d ) is much more difficult to oxidize than Fe (3d ). The chemistry of manganese is therefore primarily that of the Mn ion, whereas both the Fe and Fe ions are important in the chemistry of iron. The transition metals form cations by the initial loss of the ns electrons of the metal, even though the ns orbital is lower in energy than the (n − 1)d subshell in the neutral atoms. This apparent contradiction is due to the small difference in energy between the ns and (n − 1)d orbitals, together with screening effects. The loss of one or more electrons reverses the relative energies of the ns and (n − 1)d subshells, making the latter lower in energy. Consequently, all transition-metal cations possess d valence electron configurations, as shown in Table 23.2 for the 2+ ions of the first-row transition metals. All transition-metal cations have d electron configurations; the ns electrons are always lost before the (n − 1)d electrons. The most common oxidation states of the first-row transition metals are shown in Table \(\Page {3}\). The second- and third-row transition metals behave similarly but with three important differences: The highest possible oxidation state, corresponding to the formal loss of all valence electrons, becomes increasingly less stable as we go from group 3 to group 8, and it is never observed in later groups. In the transition metals, the stability of higher oxidation states increases down a column. Binary transition-metal compounds, such as the oxides and sulfides, are usually written with idealized stoichiometries, such as FeO or FeS, but these compounds are usually cation deficient and almost never contain a 1:1 cation:anion ratio. Thus a substance such as ferrous oxide is actually a nonstoichiometric compound with a range of compositions. The acid–base character of transition-metal oxides depends strongly on the oxidation state of the metal and its ionic radius. Oxides of metals in lower oxidation states (less than or equal to +3) have significant ionic character and tend to be basic. Conversely, oxides of metals in higher oxidation states are more covalent and tend to be acidic, often dissolving in strong base to form oxoanions. Two of the group 8 metals (Fe, Ru, and Os) form stable oxides in the +8 oxidation state. Identify these metals; predict the stoichiometry of the oxides; describe the general physical and chemical properties, type of bonding, and physical state of the oxides; and decide whether they are acidic or basic oxides. : group 8 metals : identity of metals and expected properties of oxides in +8 oxidation state : Refer to the trends outlined in Figure 23.1, Figure 23.2, Table 23.1, Table 23.2, and Table 23.3 to identify the metals. Decide whether their oxides are covalent or ionic in character, and, based on this, predict the general physical and chemical properties of the oxides. : The +8 oxidation state corresponds to a stoichiometry of MO . Because the heavier transition metals tend to be stable in higher oxidation states, we expect Ru and Os to form the most stable tetroxides. Because oxides of metals in high oxidation states are generally covalent compounds, RuO and OsO should be volatile solids or liquids that consist of discrete MO molecules, which the valence-shell electron-pair repulsion (VSEPR) model predicts to be tetrahedral. Finally, because oxides of transition metals in high oxidation states are usually acidic, RuO and OsO should dissolve in strong aqueous base to form oxoanions Predict the identity and stoichiometry of the stable group 9 bromide in which the metal has the lowest oxidation state and describe its chemical and physical properties. Because the lightest element in the group is most likely to form stable compounds in lower oxidation states, the bromide will be CoBr . We predict that CoBr will be an ionic solid with a relatively high melting point and that it will dissolve in water to give the Co (aq) ion. The transition metals are characterized by partially filled d subshells in the free elements and cations. The ns and (n − 1)d subshells have similar energies, so small influences can produce electron configurations that do not conform to the general order in which the subshells are filled. In the second- and third-row transition metals, such irregularities can be difficult to predict, particularly for the third row, which has 4f, 5d, and 6s orbitals that are very close in energy. The increase in atomic radius is greater between the 3d and 4d metals than between the 4d and 5d metals because of the lanthanide contraction. Ionization energies and electronegativities increase slowly across a row, as do densities and electrical and thermal conductivities, whereas enthalpies of hydration decrease. Anomalies can be explained by the increased stabilization of half-filled and filled subshells. Transition-metal cations are formed by the initial loss of ns electrons, and many metals can form cations in several oxidation states. Higher oxidation states become progressively less stable across a row and more stable down a column. Oxides of small, highly charged metal ions tend to be acidic, whereas oxides of metals with a low charge-to-radius ratio are basic.

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This page takes an introductory look at two areas of bonding theory needed for a proper understanding of how organic compounds absorb some of the UV or visible light that passes through them. It looks simply at anti-bonding orbitals, and what is meant by conjugation in compounds and how it contributes to the delocalization of electrons. I am assuming that you know how a simple covalent bond between two atoms forms. Half-filled atomic orbitals on each atom overlap in space to form a new orbital (a molecular orbital) containing both electrons. In the case of two hydrogen atoms, each has one electron in the 1s orbital. These come together to make a new orbital surrounding both of the hydrogen nuclei. It is important to understand exactly what this molecular orbital means. The two electrons are most likely to be found in this region of space - and the most likely place to find them within this space is on the line between the two nuclei. The molecule holds together because both nuclei are strongly attracted to this same pair of electrons. This most simple of bonds is called a sigma bond - a sigma bond is one where the electron pair is most likely to be found on the line between the two nuclei. However . . . This is all a bit of a simplification! Molecular orbital theory demands that if you start with two atomic orbitals, you must end up with two molecular orbitals - and we seem to be only producing one. A second molecular orbital is formed, but in most cases (including the hydrogen molecule) it is left empty of electrons. It is described as an anti-bonding orbital. The anti-bonding orbital has a quite different shape and energy from the bonding orbital. The next diagram shows the relative shapes and energies of the various atomic and molecular orbitals when two hydrogen atoms combine. An anti-bonding orbital is always shown by the use of a star after its symbol. Notice that when a bonding orbital forms, it is at a lower energy than the original atoms. Energy is released when the bonding orbital is formed, and the hydrogen molecule is more energetically stable than the original atoms. However, an anti-bonding orbital is less energetically stable than the original atoms. A bonding orbital is stable because of the attractions between the nuclei and the electrons. In an anti-bonding orbital there are no equivalent attractions - instead you get repulsions. There is very little chance of finding the electrons between the two nuclei - and in fact half-way between the nuclei there is zero chance of finding them. There is nothing to stop the two nuclei from repelling each other apart. So in the hydrogen case, both of the electrons go into the bonding orbital, because that produces the greatest stability - more stable than having separate atoms, and a lot more stable than having the electrons in the anti-bonding orbital. You might reasonably say that helium can't form an He molecule because it doesn't have any unpaired electrons to share. Fine! But let's also look at it from the point of molecular orbital theory. The diagram for helium is just a small modification of the last one. This time we have a total of 4 electrons in the original atomic orbitals. Two atomic orbitals have to form two molecular orbitals. That means that this time, we would have to use both the bonding and anti-bonding molecular orbitals to accommodate them. But any gain in energetic stability due to the formation of the bonding orbitals would be countered by the loss of energetic stability because of the anti-bonding ones. There is no energetic advantage in He forming - and so it doesn't. You are probably familiar with a picture of the double bond in ethene shown as: The pi bond shown in red is, of course, a normal bonding orbital. It was formed by sideways overlap between a half-filled p-orbital on each of the two carbon atoms. Remember that the two red shapes shown in the diagram are part of the same pi bonding orbital. But if you overlap two atomic orbitals, you must get two molecular orbitals according to molecular orbital theory. The second one is an anti-bonding pi orbital - and we never draw it under normal circumstances. The anti-bonding pi orbital is (just like the anti-bonding sigma one) at a higher energy than the bonding orbital - and so isn't used to hold electrons. Both of the electrons in the pi bond are found in the pi bonding orbital. The next diagram gives a general impression of how the energies of various types of orbital relate to each other in the sort of compounds we will be looking at when we try to explain the absorption of light. It is not to scale. You will find that a new sort of orbital has crept into the diagram - labelled "n" (for non-bonding). The sort of non-bonding orbitals that we will be interested in contain lone pairs of electrons on, for example, oxygen, nitrogen and halogen atoms. So . . . think of non-bonding orbitals as those containing lone pairs of electrons at the bonding level. When light passes through a compound, some of the energy in the light kicks an electron from one of the bonding or non-bonding orbitals into one of the anti-bonding ones. The energy gaps between these levels determine the frequency (or wavelength) of the light absorbed, and those gaps will be different in different compounds. This is covered in detail on another page. We are going to leave explaining what conjugation is for a while - it is necessary to look at some more bonding first. To understand about conjugated double bonds, you first need to be sure that you understand simple double bonds. Ethene contains a simple double bond between two carbon atoms, but the two parts of this bond are different. Part of it is a simple sigma bond formed from end-to-end overlap between orbitals on each carbon atom, and part is caused by sideways overlap between a p-orbital on each carbon. The important diagram is the one leading up to the formation of the pi bond - where the two p-orbitals are overlapping sideways: . . . giving the familiar pi bond. Buta-1,3-diene has the structure: Now picture the formation of the various molecular orbitals as if you were thinking about two ethene molecules joined together. You would have sigma bonds formed by the end-to-end overlap of various orbitals on the carbons and hydrogens. That would leave you with a p-orbital on each carbon atom. Those p-orbitals will overlap sideways - all of them! A system of delocalised pi bonds is formed, similar to the benzene case that you are probably familiar with. The diagram shows one of those molecular orbitals. To stress again - the diagram shows only one of the delocalised molecular orbitals. Remember that both of the red bits in the diagram are part of the same orbital. The interaction of the two double bonds with each other to produce a delocalised system of pi electrons over all four atoms is known as conjugation. Conjugation in this context literally means "joining together". In reality, if you start by overlapping four atomic orbitals, you will end up with four molecular orbitals. The four electrons will go into the two lowest energy of these - two in each. That means that you get two pi bonding orbitals. We just draw one of these for simplicity - the other one has a different shape. There are also two pi anti-bonding orbitals, but these are normally empty. For most purposes, we ignore these entirely - although not for this topic because energy from light can promote electrons from a pi bonding orbital into one of the anti-bonding orbitals (as you will see on the next page). You can recognize the presence of conjugated double bonds in a molecule containing more than one double bond because of the presence of alternating double and single bonds. The double bonds don't have to be always between carbon atoms. All of the following molecules contain conjugated double bonds, although in the last case, the conjugation doesn't extend over the whole molecule: However, although the next molecule contains two double bonds, they aren't conjugated. They are separated by two single bonds. The reason why it is important to have the double and single bonds alternating is that this is the only way you can get all the p-orbitals overlapping sideways. In the last case, you will get sideways overlap at each end of the molecule to get two individual pi bonds. But the extra single bond in the middle stops them from interacting with each other. You are almost certainly familiar with the delocalization which occurs in a benzene ring. If you think about benzene using the Kelulé structure, you have a perfect system of alternating single and double bonds around the molecule. These conjugate to give the familiar delocalised pi system. Once again, remember that this only shows one of the molecular orbitals formed. There will actually be three bonding pi orbitals and three anti-bonding ones - because they arise from combining a total of six atomic orbitals. The extra bonding orbitals aren't usually drawn. delocalization can also extend beyond pi bonds to include lone pairs on atoms like nitrogen or oxygen. Two simple examples of this are phenylamine (aniline) and phenol. Writing these using the Kekulé structure for benzene: You can see the alternating double and single bonds around the benzene ring. This conjugation leads to the familiar delocalised electron system in benzene which we usually show as: But the delocalization doesn't stop at the ring. It extends out to the nitrogen or oxygen atoms. In the phenylamine case, there is a lone pair on the nitrogen atom which can overlap with the ring electrons . . . . . . leading to delocalization which takes in both ring and nitrogen. Exactly the same thing happens with phenol. One of the oxygen lone pairs overlaps with the ring electrons. The other one is pointing in the wrong direction to get involved. So . . . if you are trying to work out how far delocalization extends in a molecule, don't forget to look for atoms with lone pairs that might get involved in the delocalization. Look especially for benzene rings with groups attached which contain double bonds. Let's start with a couple of simple ones - phenylethene (styrene) and benzaldehyde. In each case, you've got delocalization over the ring. Does it extend to the attached group? Do you have anything like alternating single and double bonds? Yes, you do. You have the double bond in the side group, then a single bond, then the ring delocalization. Looking at this in the phenylethene case, and imagining the arrangement of orbitals just before delocalization over the side group: You can see that the double bond and ring electrons will overlap to form a delocalised system looking something like this: The benzaldehyde case is very similar, except that this time instead of the CH group at the end there is an oxygen with two lone pairs. The delocalization is just the same. Be careful, though! Remember that to get this extended delocalization, any double bond in the side chain must be able to conjugate with the ring electrons - the two bits must be close enough to join together. Molecules like those in the next diagram don't have the delocalization extending out into the side chain. The extra CH group prevents the necessary sideways overlap between the p orbitals of the double bond and the ring electrons. The other side group which it would be useful to know about is the nitro group, NO - for example in nitrobenzene. The bonding in the nitro group is surprisingly awkward to work out. It is often shown with a double bond between the nitrogen and one of the oxygens, and a co-ordinate (dative covalent) bond to the other. This structure is actually misleading. Both nitrogen-oxygen bonds are identical and the group already has delocalization. This is often shown as: The dotted half-circle suggests the delocalization. Think of it as being like the circle you draw in the middle of the benzene hexagon. This delocalization is just one single bond away from the ring delocalization. You get conjugation between the two, and the delocalization takes in the whole molecule. Another case we need to look at (because it occurs in a molecule we'll explore on the next page) is an SO group attached to a benzene ring. It looks as if the double bonds are in just the right position relative to the ring for the delocalization to extend out over this group. However, I have been told on good authority that the delocalization doesn't extend from the side group into the ring. I can't, though, find any reference to this anywhere on the web or in the textbooks that I have available. The bonding in the sulfonate group isn't at all easy to describe in orbital terms. In fact, it is most easily explained in terms of , but introducing that now is just complicating things pointlessly. When you are trying to work out how far delocalization extends in a molecule, look for: And finally . . . why does all this matter? The wavelength of UV or visible light absorbed by organic compounds depends largely on the extent of delocalization in the molecules. That means that you may well have to look at an unfamiliar molecule and make a reasonable estimate of whether it is highly delocalised or not very delocalised. It will always be pretty clear-cut at this level, so don't worry too much about it!

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Electron configuration describes the distribution of electrons among different orbitals (including shells and subshells) within atoms and molecules. here are four principle (s, p, d, and f) which are filled according to the energy level and valence electrons of the element. All four orbitals can hold different number of electrons. The s-orbital can hold 2 electrons, and the other three orbitals can hold up to 6, 10, and 14 electrons, respectively. The s-orbital primarily denotes group 1 or group 2 elements, the p-orbital denotes group 13, 14, 15, 16, 17, or 18 elements, and the f-orbital denotes the and group. The main focus of this module however will be on the electron configuration of transition metals, which are found in the d-orbitals (d-block). The electron configuration of transition metals is special in the sense that they can be found in numerous oxidation states. Although the elements can display many different oxidation states, they usually exhibit a common oxidation state depending on what makes that element most stable. For this module, we will work only with the first row of transition metals; however the other rows of transition metals generally follow the same patterns as the first row.   In the first row of the transition metals, the ten elements that can be found are:   Below is a table of the oxidation states that the transition metals can or cannot form. As stated in the boxes, the “No” indicates that the elements are not found with that oxidation state. The “Rare” signifies the oxidation states that the elements are rarely found in. Lastly, the “ ” identifies the oxidation states that the elements readily found in. The electron configuration for the consists of 4s and 3d subshells with an argon (noble gas) core. This applies to the first row transition metals, adjustments will be necessary when writing the electron configuration for the other rows of transition metals. The noble gas before the first row of transition metals would be the core written with brackets around the element symbol (i.e. [Ar] would be used for the transition metals), and the electron configuration would follow a [Ar] ns nd format. In the case of first row transition metals, the electron configuration would simply be [Ar] 4s 3d . The energy level, "n", can be determined based on the periodic table, simply by looking at the row number in which the element is in. However, there is an exception for the d-block and f-block, in which the energy level, "n" for the d block is "n-1" ("n" minus 1) and for the f block is "n-2" (See following periodic table for clarification). In this case, the "x" in ns and nd is the number of electrons in a specific orbital (i.e. s-orbitals can hold of 2 electrons, p-orbitals can hold 6 electrons, d-orbitals can hold 10 electrons, and f-orbitals can hold 14 electrons). To determine what "x" is, simply count the number of boxes that you come across before reaching the element you are attempting to determine the electron configuration for. For example, if we want to determine the electron configuration for Cobalt (Co) at ground state, we would first look at the row number, which is 4 according to the periodic table below; meaning . In addition, since we know that the energy level for the d orbital is "n-1", therefore in this case. Thus, the electron configuration for Cobalt at ground state would simply be Co: [Ar] 4s 3d . The reason why it is 3d can be explained using the periodic table. As stated, you could simply count the boxes on the periodic table, and since Cobalt is the 7th element of the first row transition metals, we get Co: [Ar] 4s 3d . The following image shows the order for filling the subshells: In the ground state, the electron configuration of the transition metals follows the format, ns nd . As for the electron configuration for transition metals that are charged (i.e. Cu ), the electrons from the s orbital will be moved to the d-orbital to form either ns nd or ns nd . Vanadium at Ground State (Neutral): V: 5 electrons = [Ar] 4s 3d Vanadium with an Oxidation State of +4: V : [Ar] 4s 3d Or you can also write it as V : [Ar] 3d Above is a video showing how to write the electron configuration for Nickel (Ni) and Zirconium (Zr) from the d-block. Nickel at Ground State: Ni: 8 electrons = [Ar] 4s 3d Nickel with an Oxidation State of +2: Ni : [Ar] 4s 3d Or simply Ni : [Ar] 3d Rhodium at Ground State: Rh: 7 electrons = [Kr] 5s 4d Rhodium with an Oxidation State of +3: Rh : [Kr] 5s 4d Or simply Rh : [Kr] 4d Note: Osmium is stable with oxidation states of +2, +3, +4, as well as +8. Osmium at Ground State: Os: 6 electrons = [Xe] 6s 4f 5d Osmium with an Oxidation State of +2: Os: [Xe] 4f 5d Osmium with an Oxidation State of +3: Os: [Xe] 4f 5d For fourth row transition metals, the electron configuration is very similar to the electron configuration of the third row transition metals. However, for a fourth row transition metal, you would follow the format of [Rn] 7sx5fx6dx rather than the third row transition metal formating of [Xe] 6sx4fx5dx. To see an example of an element from the second row or third row transition metals, see "Electron Configuration of a Second Row Transition Metal (Rhodium)" and "Electron Configuration of a Third Row Transition Metal (Osmium)".) A) V B) V C) V D) Cr E) Cr F) Cr G) Mn H) Mn I) Mn J) Mn K) Mn L) Fe M) Fe N) Co O) Co P) Cu Q) Zn See File Attachment for Solutions. (You will probably need Adobe Reader to open the PDF file.)

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Sulfur is a chemical element that is represented with the chemical symbol "S" and the atomic number 16 on the periodic table. Because it is 0.0384% of the Earth's crust, sulfur is the seventeenth most abundant element following strontium. Sulfur also takes on many forms, which include elemental sulfur, organo-sulfur compounds in oil and coal, H S(g) in natural gas, and mineral sulfides and sulfates. This element is extracted by using the Frasch process (discussed below), a method where superheated water and compressed air is used to draw liquid sulfur to the surface. Offshore sites, Texas, and Louisiana are the primary sites that yield extensive amounts of elemental sulfur. However, elemental sulfur can also be produced by reducing H S, commonly found in oil and natural gas. For the most part though, sulfur is used to produce SO (g) and H SO . Known from ancient times (mentioned in the Hebrew scriptures as brimstone) sulfur was classified as an element in 1777 by Lavoisier. Pure sulfur is tasteless and odorless with a light yellow color. Samples of sulfur often encountered in the lab have a noticeable odor. Sulfur is the tenth most abundant element in the known universe. Sulfur has an atomic weight of 32.066 grams per mole and is part of group 16, the oxygen family. It is a nonmetal and has a specific heat of 0.706 J g C . The electron affinity if 200 kJ mol and the electronegativity is 2.58 (unitless). Sulfur is typically found as a light-yellow, opaque, and brittle solid in large amounts of small orthorhombic crystals. Not only does sulfur have twice the density of water, it is also insoluble in water. On the other hand, sulfur is highly soluble in carbon disulfide and slightly soluble in many common solvents. Sulfur can also vary in color and blackens upon boiling due to carbonaceous impurities. Even as little as 0.05% of carbonaceous matter darkens sulfur significantly. Most sulfur is recovered directly as the element from underground deposits by injecting super-heated water and piping out molten sulfur (sulfur melts at 112 C). Compared to other elements, sulfur has the most allotropes. While the S ring is the most common allotrope, there are 6 other structures with up to 20 sulfur atoms per ring. While has fewer allotropes than sulfur, including \(\ce{O}\), \(\ce{O_2}\), \(\ce{O_3}\), \(\ce{O_4}\), \(\ce{O_8}\), metallic \(\ce{O}\) (and four other solid phases), many of these actually have a corresponding sulfur variant. However, sulfur has more tendency to (t ). Here are the values of the single and double bond enthalpies: \[\begin{array}{c|r} \ce {O-O} & \ce{142\ kJ/mol} \\ \ce {S–S} & \ce{268\ kJ/mol} \\ \ce {O=O} & \ce{499\ kJ/mol} \\ \ce {S=S} & \ce{352\ kJ/mol} \\ \end{array}\] This means that \(\ce{O=O}\) is stronger than \(\ce{S=S}\), while \(\ce{O–O}\) is weaker than \(\ce{S–S}\). So, in sulfur, single bonds are favored and catenation is easier than in oxygen compounds. It seems that the reason for the weaker \(\ce{S=S}\) double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the \(p\) orbitals overlap is small and the \(\pi\) bond is weak. This is attested by looking down the periodic table: \(\ce{Se=Se}\) has an even weaker bond enthalpy of \(\ce{272 kJ/mol}\). What happens when the solid sulfur melts? The \(\ce{S8}\) molecules break up. When suddenly cooled, long chain molecules are formed in the plastic sulfur which behave as rubber. Plastic sulfur transform into rhombic sulfur over time. Reading the following reactions, figure out and notice the change of the oxidation state of \(\ce{S}\) in the reactants and products. Common oxidation states of sulfur are -2, 0, +4, and +6. Sulfur (brimstone, stone that burns) reacts with \(\ce{O2}\) giving a blue flame (Figure \(\Page {1}\)): \[\ce{S + O_2 \rightarrow SO_2}\] \(\ce{SO2}\) is produced whenever metal sulfide is oxidized. It is recovered and oxidized further to give \(\mathrm{SO_3}\), for production of \(\mathrm{H_2SO_4}\). \(\mathrm{SO_2}\) reacts with \(\mathrm{H_2S}\) to form \(\mathrm{H_2O}\) and \(\ce{S}\). \[\mathrm{2 SO_2 + O_2 \rightleftharpoons 2 SO_3}\] \[\mathrm{SO_3 + H_2O \rightleftharpoons H_2SO_4} \;\;(\text{a valuable commodity})\] \[\mathrm{SO_3 + H_2SO_4 \rightleftharpoons H_2S_2O_7} \;\;\; (\text{pyrosulfuric acid})\] Sulfur reacts with sulfite ions in solution to form thiosulfate, \[\ce{S + SO_3^{2-} -> S_2O_3^{2-}}\] but the reaction is reversed in an acidic solution. There are many different stable sulfur oxides, but the two that are commonly found are sulfur dioxide and sulfur trioxide. Sulfur dioxide is a commonly found oxide of sulfur. It is a colorless, pungent, and nonflammable gas. It has a density of 2.8 kg/m and a melting point of -72.5 C. Because organic materials are more soluble in \(SO_2\) than in water, the liquid form is a good solvent. \(SO_2\) is primarily used to produce \(SO_3\). The direct combustion of sulfur and the roasting of metal sulfides yield \(SO_2\) via : \[\underbrace{S(s) + O_2(g) \rightarrow SO_2(g)}_{\text{Direct combustion}}\] \[\underbrace{2 ZnS(s) + 3 O_2(g) \rightarrow 2 ZnO(s) + 2 SO_2(g)}_{\text{Roasting of metal sulfides}}\] Sulfur trioxide is another one of the commonly found oxides of sulfur. It is a colorless liquid with a melting point of 16.9 C and a density of kg/m . \(SO_3\) is used to produce sulfuric acid. \(SO_2\) is used in the synthesis of \(SO_3\): \[\underbrace{2 SO_2 (g) + O_2(g) \rightleftharpoons 2 SO_3(g)}_{\text{Exothermic, reversible reaction}}\] This reaction needs a catalyst to be completed in a reasonable amount of time with \(V_2O_5\) is the catalyst most commonly used. Perhaps the most significant compound of sulfur used in modern industrialized societies is sulfuric acid (\(H_2SO_4\)). Sulfur dioxide (\(SO_2\)) finds practical applications in bleaching and refrigeration but it is also a nuisance gas resulting from the burning of sulfurous coals. Sulfur dioxide gas then reacts with the water vapor in the air to produce a weak acid, sulfurous acid (\(H_2SO_3\)) which contributes to the acid rain problem. \[C_{12}H_{22}O_{11}(s) \rightarrow 12 C(s) + 11 H_2O(l)\] (Concentrated sulfuric acid used in forward reaction to remove H and O atoms.) O (g) + 2 SO (aq) \rightarrow 2 SO (aq) (Reducing agent) 2 H S(g) + 2 H (aq) + SO (aq) \rightarrow 3 H O(l) + 3 S(s) (Oxidizing agent) H SO is a diprotic acid that acts as a weak acid in both steps and H SO is also a diprotic acid but acts as a strong acid in the first step and a weak acid in the second step. Acids like NaHSO and NaHSO are called acid salts because they are the product of the first step of these diprotic acids. Boiling elemental sulfur in a solution of sodium sulfite yields . Not only are thiosulfates important in photographic processing, but they are also common analytical reagents used with iodine (like in the following two reactions). \[2 Cu^{2+}_{(aq)} + 5 I^-_{(aq)} \rightarrow 2 CuI_{(s)} + I^-_{3(aq)} \] \[I^-_{3(aq)} + 2 S_2O^{2-}_{3(aq)} \rightarrow 3 I^-_{(aq)} + S_4O^{2-}_{6(aq)}\] with excess triiodide ion titrated with Na S O (aq). Other than sulfuric acid, perhaps the most familiar compound of sulfur in the chemistry lab is the foul-smelling hydrogen sulfide gas, \(H_2S\), which smells like rotten eggs. \[SCl_2 + 2CH_2CH_2 \rightarrow S(CH_2CH_2Cl)_2 \] Sulfur can be mined by the Frasch process. This process has made sulfur a high purity (up to 99.9 percent pure) chemical commodity in large quantities. Most sulfur-containing minerals are metal sulfides, and the best known is perhaps pyrite (\(\mathrm{FeS_2}\), known as fools gold because of its golden color). The most common sulfate containing mineral is gypsum, \(\mathrm{CaSO_4 \cdot 2H_2O}\), also known as plaster of Paris. The Frasch process forces (99.5% pure) sulfur out by using hot water and air. In this process, superheated water is forced down the outermost of three concentric pipes. Compressed air is pumped down the center tube, and a mixture of elemental sulfur, hot water, and air comes up the middle pipe. Sulfur is melted with superheated water (at 170 °C under high pressure) and forced to the surface of the earth as a slurry. Sulfur is mostly used for the production of sulfuric acid, \(\ce{H2SO4}\). Most sulfur mined by Frasch process is used in industry for the manufacture of sulfuric acid. Sulfuric acid, the most abundantly produced chemical in the United States, is manufactured by the . Most (about 70%) of the sulfuric acid produced in the world is used in the fertilizer industry. Sulfuric acid can act as a strong acid, a dehydrating agent, and an oxidizing agent. Its applications use these properties. Sulfur is an essential element of life in sulfur-containing proteins. Sulfur has many practical applications. As a fungicide, sulfur is used to counteract apple scab in organically farmed apple production. Other crops that utilize sulfur fungicides include grapes, strawberries, and many vegetables. In general, sulfur is effective against mildew diseases and black spot. Sulfur can also be used as an organic insecticide. Sulfites are frequently used to bleach paper and preserve dried fruit. The vulcanization of rubber includes the use of sulfur as well. Cellophane and rayon are produced with carbon disulfide, a product of sulfur and methane. Sulfur compounds can also be found in detergents, acne treatments, and agrichemicals. Magnesium sulfate (epsom salt) has many uses, ranging from bath additives to exfoliants. Sulfur is being increasingly used as a fertilizer as well. Because standard sulfur is hydrophobic, it is covered with a surfactant by bacteria before oxidation can occur. Sulfur is therefore a slow-release fertilizer. Lastly, sulfur functions as a light-generating medium in sulfur lamps. Concentrated sulfuric acid was once one of the most produced chemicals in the United States, the majority of the H SO that is now produced is used in fertilizer. It is also used in oil refining, production of titanium dioxide, in emergency power supplies and car batteries. The mineral gypsum is calcium sulfate dihydrate is used in making plaster of Paris. Over one million tons of aluminum sulfate is produced each year in the United States by reacting H SO and Al O . This compound is important in water purification. Copper sulfate is used in electroplating. Sulfites are used in the paper making industry because they produce a substance that coats the cellulose in the word and frees the fibers of the wood for treatment. Particles, SO (g), and H SO mist are the components of . Because power plants burn coal or high-sulfur fuel oils, SO (g) is released into the air. When catalyzed on the surfaces of airborne particles, SO can be oxidized to SO . A reaction with NO works as well as shown in the following reaction: \[ SO_{2(g)} + NO_{2(g)} \rightarrow SO_{3(g)} + NO_{(g)}\] H SO mist is then produced after SO reacts with water vapor in the air. If H SO reacts with airborne NH , (NH ) SO is produced. When SO (g) and H SO reach levels that exceed 0.10 ppm, they are potentially harmful. By removing sulfur from fuels and controlling emissions, and industrial smog can be kept under control. Processes like the have been presented to remove SO from smokestack gases.

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In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution. By convention, activity is treated as a dimensionless quantity, although its value depends on customary choices of standard state for the species. The activity of pure substances in condensed phases (solid or liquids) is normally taken as unity (the number 1). Activity depends on temperature, pressure and composition of the mixture, among other things. For gases, the activity is the effective partial pressure, and is usually referred to as . The difference between activity and other measures of composition arises because molecules in non-ideal gases or solutions interact with each other, either to attract or to repel each other. The activity of an ion is particularly influenced by its surroundings. Compare the pressures predicted for one mole of ethane at 298.15 K under the following equations of states: What is the deviation of the two? a) Ideal gas law equation of state: Calculate the pressure of 1.000 mole of ethane at 298.15 K in a 1.000 L flask using the ideal gas law. \[\begin{align} P_{ideal}V &=nRT \label{ideal gas} \\[4pt] P_{ideal} &=\dfrac{nRT}{V} \nonumber \\[4pt] &= \dfrac{ (1\;mole)(0.0821 L \; atm \; K^{-1} \; mol^{-1}) ( 298.15 \;K)}{1 \;L} \nonumber \\[4pt] &= 24.47\; atm \nonumber \end{align} \] b) Van der Waal's equation of state: Calculate the pressure of 1.00 mole of ethane at 298 K in a 1.00 L flask using the van der Waals equation. The van der Waals constants for ethane can found in . \[\begin{align} \left(P_{vdW} + \dfrac{an^2}{V^2}\right) (V − nb) &=nRT \label{vdW equation} \\[4pt] P_{vdW} &= \dfrac{nRT}{V − nb} - \dfrac{an^2}{V^2} \nonumber \\[4pt] &= 20.67\; atm \nonumber \end{align}\] Calculate the percent error between the \(P_{ideal}\) and the \(P_{vdW}\) \[ Error = \dfrac{P_{vdW} - P_{ideal}}{P_{vdW}} = 18.36\% \nonumber\] An error this large is often when carrying out a gas-phase reaction or designing a vessel in which to carry out such a reaction. There are two ways to deal with real systems that deviate appreciably from ideal conditions: In a similar fashion, the difference between the calculated solute concentrations in an ideal solution and in a real solution can lead to wide variations in experimental results. The following three examples compare the results obtained when formal concentrations are used (assuming ideality) and when activities are used (assuming non-ideality). Just like gases, "ideal solutions" have certain predictable physical properties (e.g. ) that real solutions often deviate from. As with the van der Waal equation in Example 1, this deviation originates from solute-solvent, solvent-solvent and solute-solute . The magnitude of this non-ideality naturally greater with higher with solute concentrations and with greater (e.g., ions vs. non-charged species). Lewis introduced idea of 'effective concentration' or 'activity' to deal with this problem by allowing an "ideal solution" description for non-ideal solutions. Since the Van der Waals equation describes real gases instead of the ideal gases law, can be used in place of concentration to describe the behavior of real solutions vs. ideal solutions. The activity of a substance (abbreviated as ) describes the effective concentration of that substance in the reaction mixture. Activity takes into account the non-ideality of the reaction mixture, including solvent-solvent, solvent-solute, and solute-solute interactions. Thus, activity provides a more accurate description of how all of the particles act in solution. For very dilute solutions, the activities of the substances in the solution closely approach the formal concentration (what the calculated concentration should be based on how much substance was measured out.) As solutions get more concentrated, the activities of all of the species tend to be smaller than the formal concentration. The decrease in activity as concentration increases is much more pronounced for ions than it is for neutral solutes. Activities are actually ratios that compare an effective pressure or an effective concentration to a standard state pressure or concentration (the correct term for the effective pressure is ). There are several ways to define standard states for the different components of a solution, but a common system is Thus, when we discuss the activity of a gas, we actually are discussing the ratio of the effective pressure to the standard state pressure: \[a_{gas} = \dfrac{P}{P^º} \label{1}\] Likewise, the activity of a solute in solution would be: \[a_{solute} = \dfrac{C}{C^º} \label{2}\] For all solids, the activity is a ratio of the concentration of a pure solid to the concentration of that same pure solid \[a_{solid} = \dfrac{C_{\text{effective solid}}}{C^º_{\text{effective solid}}} = 1 \label{3}\] For all liquids, the activity is a ratio of the concentration of a pure liquid to the concentration of that same pure liquid: \[a_{liquid} =\dfrac{C_{\text{effective liquid}}}{C^º_{\text{effective liquid}}} = 1 \label{4}\] For most experimental situations, solutions are assumed to be dilute with respect to the solvent. This assumption implies the solvent can be approximated with pure liquid. According to , the vapor pressure of the solvent in a solution is equal to the mole fraction of the solvent in the solution times the vapor pressure of the pure solvent: \[\chi= \dfrac{P}{P^º} \label{5} \] The mole fraction of solvent in a dilute solution is approximately 1, so the vapor pressure of the solution is essentially identical to the vapor pressure of the pure solvent. This means that the activity of a solvent in dilute solution will always has a value of 1, with no units. Activity indicates how many particles "appear" to be present in the solution, which is different from how many actually are present. Hence, activity is a "fudge factor" to ideal solutions that correct the true concentration. The activity of a substance can be estimated from the nominal concentration of that substance (C) by using an activity coefficient, \(\gamma\): \[a = \gamma \cdot [C] \label{6}\] The value of \(\gamma\) depends upon the substance, the temperature, and the concentration of all solute particles in the solution. The lower the concentration of all solute particles in the solution, the closer the value of \(\gamma\) for each solute approaches 1: \[\lim_{[C] \rightarrow 0} \gamma \rightarrow 1 \label{6a}\] Therefore, as \(\gamma\) approaches 1, the value of \(a\) for the solute approaches C. \[\lim_{\gamma \rightarrow 1} a \rightarrow [C] \label{6b}\] The activity coefficient for a nonvolatile, neutral solute is often estimated by non-linear curve fitting, taking into account the molality of the solute and the activity of the solvent (usually its vapor pressure). In most situations, it is more practical to look up the values of the activity coefficient for a given solute than it is to carry out the curve fitting. Estimating the activity coefficient of electrolytes (solutes that dissolve or react with the solvent to form ions) depends upon the number of ions formed by the dissociation of the solute in solution or the reaction of the solute with the solution, because each ion formed is dealt with individually. In a theoretical, infinitely dilute ideal solution, an electrolyte would dissociate or react completely to form an integer number of independent ions. For example, 1 mole of NaCl would dissociate to form 2 moles of ions (1 mole of Na ions and 1 mole of Cl ions). In reality, it is found that electrolytes almost always act as if they contain moles of ions than expected based on the formal concentration. This non-ideality is attributed to the degree of dissociation/reaction of the solute, to the solute-solvent interactions such as complex ion formation, and to the solute-solute interactions such as ion pairing. An activity coefficient incorporates the particle interactions into a single term that modifies the formal concentration to give an estimate of the effective concentration, or activity, of each ion. At infinite dilution, \(gamma\) is solely determined by the Debye-Hückel limiting law (Equation \(\ref{Debye Equation}\)) and depends only on the number and charges of the cations and anions. This means that the same limiting mean ionic activity coefficient is found for sodium chloride and potassium chloride and that also the values for the 2-1 and 1-2 salts sodium sulfate and calcium chloride are identical. At higher electrolyte concentrations though, these values change very strongly and are usually modeled using empirical parameters regressed to the experimental data. Electrolytes almost always act as if they contain moles of ions than expected based on the formal concentration. Single ion activity coefficients are calculated using various forms of the : \[\log \gamma = \dfrac{-0.51 z^2 \sqrt{\mu}}{1+ \dfrac{\alpha\sqrt{\mu}}{305}} \label{Debye Equation}\] This equation takes into account the solution environment as well as the individual characteristics of the specific ion of interest. It is not difficult to calculate single ion activity coefficients, but tables of these activity coefficients for many common ions in solutions of various concentrations are available (e.g., Table \(\Page {1}\)). The states that a reaction at a constant temperature will proceed spontaneously and predominantly in one direction until a constant ratio of concentrations of products and reactants is obtained. For the generic reaction \[aA + bB \rightleftharpoons cC + dD \label{8}\] the ratio of concentrations (called the mass action expression or equilibrium constant expression) is \[\dfrac{[C]^c[D]^d}{[A]^a[B]^b} = Q \label{9}\] where [ ] represents concentration in \[\dfrac{\text{moles of solute}}{\text{Liter of solution}} \label{10}\] This ratio can take on any value greater than zero, depending on the reaction conditions. Thus, it is often called the , Q. The term “instantaneous” signifies that the reaction will (and is) proceeding spontaneously to reach a constant ratio of products and reactants. When the reaction attains that constant ratio of products and reactants, it has reached a state of dynamic equilibrium, and the ratio of concentrations can be represented by the symbol K, the equilibrium constant: \[\dfrac{[C]_{eq}^c[D]_{eq}^d}{[A]_{eq}^a[B]_{eq}^b} = K \label{11}\] Laws of mass action and equilibrium constants are discussed in most general chemistry textbooks, but they are often discussed as if they were describing ideal systems. For instance, if all of the substances are gases, partial pressures are used in the mass-action expression. If the substances are in solution, molarities are used in the mass-action expression. To be thermodynamically correct, however, the of the substances must be compared in the mass-action expression. Activities are needed for precise work because, unlike concentrations, activities contain information about the effects of the solvent and other surrounding particles on the behavior of the particles of interest. Using any unit of comparison other than activities will give an incorrect value for K, but it is assumed that the approximate value is close enough to the true value for most situations. Many list K values to 2-3 significant digits, but this degree of precision is valid only under the exact experimental conditions used to obtain those values. Given all of the above information on activities, it is now possible to show how a true mass-action equation involving activities can be approximated by a mass-action equation involving molarites. It should be noted that just as the activities were unitless ratios, the molalities and molarities that appear in the following approximations should also be thought of as unitless ratios of concentrations divided by the standard state concentration. Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction: \[NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)}\] \[Q = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}a_{H_2O}} = \dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}(1)}\] \[Q=\dfrac{a_{NH_4^+}a_{OH^-}}{a_{NH_3}}=\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}}\] \[Q =\dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{m_{NH_4^+}m_{OH^-}}{m_{NH_3}} \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+,OH^-]}{[NH_3]}\] \[Q \approx \dfrac{\gamma_{NH_4^+}\gamma_{OH^-}}{\gamma_{NH_3}} \dfrac{[NH_4^+,OH^-]}{[NH_3]}\approx \dfrac{[NH_4^+,OH^-]}{[NH_3]}\] Therefore the final mass action equation typically used for this reaction is \[Q \approx \dfrac{[NH_4^+,OH^-]}{[NH_3]}\] Starting with the mass action equation in terms of activities, show all the approximations needed to obtain the mass action equation in terms of molar concentrations for the reaction: \[Ba(OH)_2 \cdot 8H_2O(s) + 2NH_4NO_3(s) \rightleftharpoons Ba^{2+}(aq) + 2NO_3^-(aq) + 2NH_3(aq) + 10H_2O(l)\] \[Q = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2a_{H_2O}^{10}}{a_{Ba(OH)_2·8H_2O}a_{NH_4NO_3}} = \dfrac{a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2·(1)}{(1)·(1)} = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2 \] \[Q = a_{Ba^{2+}}a_{NO_3^-}^2a_{NH_3}^2 = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2\gamma_{NH_3}^2m_{Ba^{2+}}m_{NO_3^-}^2m_{NH_3}^2\] \[Q = \gamma_{Ba^{2+}} \gamma_{NO_3^-}^2 \gamma_{NH_3}^2m_{Ba^{2+}} {m_{NO_3^-}}^2 m_{NH_3}^2 \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2 [Ba^{2+}] [NO_3^-]^2 [NH_3]^2\] \[Q \approx \gamma_{Ba^{2+}}\gamma_{NO_3^-}^2 \gamma_{NH_3}^2[Ba^{2+},NO_3^-]^2[NH_3]^2 \approx [Ba^{2+},NO_3^-]^2[NH_3]^2\] \(Q \approx [Ba^{2+},NO_3^-]^2[NH_3]^2\) Determine the molar solubility \(\dfrac {moles}{Liter}\) of the slightly soluble solid, BaSO , inPure Water and An Aqueous 0.1 M NaCl Solution? Pure Water Barium sulfate is a solid that is slightly soluble in water, with a K value of 1.1 x 10 : \(BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) \) This solution is dilute enough that the Ba , SO , OH , and H ions will not affect each other greatly, thus the activity of the ions closely approaches their formal concentration. In this nearly ideal aqueous solution, the mass action expression would be \[1.1 \times 10^{-10} = a_{Ba} · a_{SO_4} \approx [Ba^{2+},SO_4^{2-}]\] Remember that BaSO is a solid, with an activity equal to 1. At equilibrium, the [Ba ] = [SO ] = 1.05 x 10 M. An Aqueous 0.1 M NaCl Solution A saturated aqueous solution of BaSO that also is 0.1 M in NaCl is no longer near to ideality. The Na and Cl ions surround the Ba and SO ions and prevent these ions from being able to reform solid BaSO readily as they did in pure water. The activities of the Ba and the SO ions will be lower than their formal concentrations. However, the product of the activities must still be equal to the true (thermodynamic) equilibrium constant. \[1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = \gamma_{Ba}[Ba^{2+}]·\gamma_{SO_4}[SO_4^{2-}]\] With the total amount of Na , Cl , Ba , SO , OH , and H ions in the solution, \(\gamma_{Ba}\) = 0.38 and \(\gamma_{SO_4}\) = 0.355. \[1.1 x 10^{-10} = a_{Ba}·a_{SO_4} = (0.38)[Ba^{2+}](0.355)[SO_4^{2-}]\] \[8.2 x 10^{-10} =[Ba^{2+},SO_4^{2-}]\] \[[Ba^{2+}] = [SO_4^{2-}] = 2.9 x 10^{-5}\] The net result is that more solid BaSO will dissolve in the 0.1M NaCl solution than in water, and the experimental equilibrium constant will seem to be larger than the thermodynamic equilibrium constant. The van 't Hoff factor, , is a term that often appears in calculations to account for the fact that electrolytes will form two or more moles of ions per every mole of electrolyte. In most cases, the solutions are treated as if they are ideal, in which case i will equal an integer representing the total number of independent ions per one formula unit of the solute (Table 2). The van 't Hoff factor is actually rarely an integer, and was, in fact, developed to take into account the non-ideality of solutes. Tables listing the i values for specific compounds in specific solutions are available, but it is also possible to use activities to estimate to effective concentrations of ions in solution for use in colligative property calculations. What is the freezing point of a 0.1 m BaCl aqueous solution? The calculation for an ideal solution would be \(\Delta T = mki\), where \(m = 0.1 molal\), \(k = 1.86\dfrac{ºC}{molal}\), and \(i = 3\). The resulting \(\Delta T\) is \[\Delta T = (0.1 molal)(1.86\dfrac{ºC}{molal})(3) = 0.558ºC\] if non-ideality is assumed, the calculation becomes \[\Delta T = \gamma_{Ba} \cdot (\gamma_{Cl})^2•(m_{Ba})(m_{Cl})^2\] Substituting in the estimated \(\gamma\) values of \(\gamma_{Ba} = 0.38\) and\(\gamma_{Cl} = 0.755\), the ion activities are and the \(\Delta T\) is \[\Delta T = (0.038 + 0.151)(1.86 ºC) = 0.351 ºC\] The \(\Delta T\) obtained using activities is lower than the \(\Delta T\) obtained when using an integer value for because the activity values take into account the fact that the ions in the solution are not able to act as free and independent particles because of their interactions with each other and with the solvent,

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Chemists are committed to the idea that a balanced chemical equation such as \[2 Fe + O_2 \rightarrow 2 FeO \label{1} \] not only tells how many atoms or molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved. Equation (1) says that 1 Fe atom can react with 2 O to give 2 of FeO. Here we're using the term "formula unit" to indicate that the substance may not be a molecule, but rather an ionic compound or ["network crystal"]. A "formula unit" gives the composition of the substance without specifying the type of bonding. The equation also says that 1 mol of Fe atoms would react with 2 O yielding 2 FeO. We have become so accustomed to the idea of atoms, that it seems logical that equations should represent whole numbers of atoms. We now talk about the stoichiometric ratios of atoms when we want to indicate that they must combine in small whole number ratios, like 1:1 in FeO. The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. But the idea of whole number stoichiometric ratios was strongly opposed in the early nineteenth century. This is understandable, because the Law of Definite Proportions is quite anti-intuitive. After all, it seems that you can generally mix things in virtually any ratio to get desired results. Since Dalton's atomic theory implied that atoms should combine in definite ratios, many of his contemporaries opposed the atomic theory. Dalton's theory explained Proust's earlier Law of constant composition (Law of Definite Proportions or "Proust's Law"(1797)) , There are in fact many important compounds that are , including the product of Equation (1)! They are often called "Berthollides" after French inorganic chemist Claude Louis Berthollet (1748–1822), who attacked Dalton's atomic theory (1803-5) and Proust's Law . N. S. Kurnakov introduced the terms "Daltonides" and "Berthollides" in 1912–14 to designate chemical compounds of constant composition (Daltonides) and variable composition (Berthollides) . Claude Louis Berhollet. Opponent of the Law of Constant Composition Joseph Proust (1754-1826). Discoverer of Law of Constant Composition This episode in the history of chemistry is a good example of the claim by philosophers of science that counterexamples to an accepted theory may not be recognized. The concept of stoichiometry was used to cover up the most glaring exceptions, and, for more than a century, substances that could not be made to fit these new rules were ignored until Kurnakov called attention to Berthollides . Berthollet opposed Dalton's atomic theory, pointing to many reactions, like the reaction of many transition metals with oxygen, that are not stoichiometric. Some of his writings are available . The reaction in Equation (1) often does not give FeO. In nature, FeO is the mineral wüstite, which has the actual stoichiometry closer to Fe O. For each "missing" Fe ion, the crystal contains two Fe ions to supply the missing 2+ charge for charge balance . In the diagram below, one vacancy (Fe ) compensated by two substitutions (Fe ) at lateice points. The "Frenkel Pair" occurs when a lattice ion is replaced by an interstitial ion in some nonstoichiometric compounds. The composition of a non-stoichiometric compound may vary only slightly, as in wüstite where the formula may be written as Fe O, where x is a small number (~0.05). In some cases, like copper sulphides, the variation can be much larger . For practical purposes, the term describes materials where the non-stoichiometry is at least 1% of the ideal composition. To contrast stoichiometric with nonstoichiometric compounds, refer to Equation (2), \[3 Fe + 2 O_2 \rightarrow Fe_3O_4 \label{2} \] In addition to atom ratios, it also tells us that 2 × 3 mol = 6 mol Fe will react with 2 × 2 mol = 4 mol O , and that ½ × 3 mol = 1.5 mol Fe requires only ½ × 2 mol = 1 mol O . In other words, the equation indicates that exactly 2 mol O must react 3 mol Fe consumed. For the purpose of calculating how much O is required to react with a certain amount of Fe therefore, the significant information contained in Eq. (2) is the \[\dfrac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}} \nonumber \] We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Eq. (2), \[\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}} \tag{3}\] Under normal circumstances, the stoichiometric ratio holds closely (Fe O is a combination of Fe O and FeO, so it may incorporate the nonstoichiometry of FeO described above). But unusual geological or synthetic conditions lead to other stoichiometries. For example, a technique called or "Sputtering" involves heating one reactant in a vacuum with a small amount of the second reactant present as a vapor, as shown in the Figure at right. The product is collected on a cold substrate. This leads to berthollides when an iron sample is sputtered in the presence of water vapor. Oxide compositions ranging from Fe O to Fe O , were obtained, depending on the temperature and pressure . They would have stoichiometric ratios ranging from \(\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}\) for Equation (2). to \(\frac{\text{1.5 mol O}_{\text{2}}}{\text{2 mol Fe}}\) for the reaction 2 Fe + 1.5 O → Fe O , and most of the ratios would not be easily reduced to whole number ratios. It is remarkable that , so this is truly an important area of chemistry. Derive all possible stoichiometric ratios from Eq. (2) Any ratio of amounts of substance given by coefficients in the equation may be used: \[\text{S}\left( \frac{\text{Fe}}{\text{O}_{2}} \right)=\frac{\text{3 mol Fe}}{\text{2 mol O}_{\text{2}}}~~~~~~\text{S}\left( \frac{\text{Fe}_{3}\text{O}_{4}}{\text{Fe}} \right)=\frac{\text{1 mol Fe}_{3}\text{O}_{4}}{\text{3 mol Fe}}\] \[\text{S}\left( \frac{\text{O}_{2}}{\text{Fe}_{3}\text{O}_{4}} \right)=\frac{\text{2 mol O}_{2}}{\text{1 mol Fe}_{2}\text{O}_{4}}~~~~~\text{S}\left( \frac{\text{Fe}}{\text{Fe}_{\text{3}}\text{O}_{4}} \right)=\frac{\text{3 mol Fe}}{\text{1 mol Fe}_{\text{3}}\text{O}_{4}}\] . Using Eq. (2) as an example, this means that the ratio of the amount of O consumed to the amount of Fe consumed must be the stoichiometric ratio S(O /Fe): \(\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{Fe}_\text{ consumed}}}=\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}\) Similarly, the ratio of the amount of Fe O produced to the amount of Fe consumed must be S(Fe O /Fe): \[\frac{n_{\text{Fe}_{3}\text{O}_{4}}\text{ produced}}{n_{\text{Fe}\text{ consumed}}}\] Note that in the word Eq. (4a) and the symbolic Eq. (4b), X and Y may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. Find the amount of Fe O produced when 3.68 mol Fe is consumed according to Eq. (2). The amount of Fe O produced must be in the stoichiometric ratio S(Fe O /Fe) to the amount of Fe consumed: \[\text{S}\left( \frac{\text{Fe}_{3}\text{O}_{4}}{\text{Fe}} \right)\] Multiplying both sides , by we have \[n_{\text{Fe}_{3}\text{O}_{4}\text{ produced}}\] This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to , where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (4) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form \(\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}\) or symbolically. When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol Fe cancels 1 mol Fe but does not cancel 1 mol Fe O . The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. Add example text here. Add example text here. Calculate the mass of Oxygen (O ) consumed when 3.68 mol Fe reacts according to Equation (2). The problem asks that we calculate the mass of O consumed. As we learned in , the molar mass can be used to convert from the amount of O to the mass of O . Therefore this problem in effect is asking that we calculate the amount of O consumed from the amount of Fe consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio \[\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}\] \[n_{\text{O}_{\text{2}}}\text{ consumed}\] The of O is \(\text{m}_{\text{O}_{\text{2}}}=\text{2}\text{.45 mol O}_{\text{2}}\times \frac{\text{32}\text{.0 g O}_{\text{2}}}{\text{1 mol O}_{\text{2}}}=\text{78.5 g O}_{\text{2}}\) With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of Fe to moles of O and the molar mass will convert moles of O to grams of O . A schematic road map for the one-step calculation can be written as \(n_{\text{Fe}}~~\xrightarrow{S\text{(O}_{\text{2}}\text{/Fe}\text{)}}~~n_{\text{O}_{\text{2}}}~~\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}\) Thus \(\text{m}_{\text{O}_{\text{2}}}=\text{3.68 mol Fe}~~\times ~~\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}~~\times~~\frac{\text{32.0 g}}{\text{1 mol O}_{\text{2}}}=\text{78.5 g O}_{2}\) These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations of the mass of product. We noted above that when iron is sputtered in the presence of water vapor, a range of products from Fe O to Fe O (common rust) may be produced. Prepare a table similar to the one above for the reaction in which 0.200 g of Fe is converted to Fe O . First, write a balanced equation 2 Fe + 3 H O → Fe O + 3 H The problem gives the mass of Fe. Thinking the problem through before trying to solve it, we realize that the molar mass of Fe could be used to calculate the amount of Fe consumed. Then we need a stoichiometric ratio to get the amount of O consumed and the amount of Fe O and H produced. Finally, the molar masses of O , Fe O , and H permit calculation of the mass of O , Fe O , and H . We might start the table by entering the given mass, and the molar masses which we calculate from atomic weight tables: Now we can calculate the amount of Fe present: \[\text{n (mol)}~~ = ~~\frac{\text{m (g)}}{\text{M (g/mol)}}\] \[\text{n (mol)} ~~=~~\frac{\text{0.200 g Fe}}{\text{55.847 g/mol}}~~=~~\text{0.00358 mol Fe}\] Then the stoichiometric ratios are used to calculate the amounts of water and products: \[\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{Fe}} \right)=\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{2 mol Fe}}\] So the amount of water required is 0.00358 mol Fe x (3 mol H O / 2 mol Fe) = 0.00537 mol H O Similarly, we use the stoichiometric ratio to calculate the amount of product: \[\text{S}\left( \frac{\text{Fe}_{\text{2}}\text{O}_{3}}{\text{Fe}} \right)=\frac{\text{1 mol Fe}_{\text{2}}\text{O}_{3}}{\text{2 mol Fe}}\] So the amount of Fe O produced is 0.00358 mol Fe x (1 mol Fe O / 2 mol Fe) = 0.00179 mol Fe O . The amount of H produced must be the same as the amount of water consumed, since they are in the ratio 3:3 from the equation. We can add these to the table: Finally, we can use the molar masses to convert from amounts (in mol) to masses (in g): Note that the sum of the masses of reactants equals the sum of the masses of the products.

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Few compounds in which a carbon–hydrogen bond must serve as the hydrogen-atom source are reactive enough to function as hydrogen-atom transfers in radical reactions of carbohydrates. The reason for this is that when less reactive donors are used, other reactions become competitive. Even com­pounds with quite reactive C–H bonds are poor hydrogen-atom transfers when compared to tri- -butyltin hydride or tris(trimethylsilyl)silane. One com­pound that does have the necessary reac­tivity, but just barely, is 2‑propanol. When reaction of the xanthate is conducted with 2‑propanol as the solvent, hydrogen-atom abstraction is in spirited com­petition with xanthate-di­thio­car­bonate rearrangement (eq 15). This competition exists because hydrogen-atom abstraction by the carbo­hydrate radical R· is slow enough that addition of R· to another molecule of the xanthate has a comparable rate (Scheme 5). The adduct radical formed by this addition fragments to give the dithio­car­bon­ate and a carbo­hydrate radical (R·). The xanthate reacts to form the corresponding deoxy sugar in 85% yield (eq 16). In this reaction cyclohexane functions as the hydrogen-atom transfer. Since cyclohexane is not a noticeably better hydrogen-atom transfer than 2-propanol, it is initially surprising that no dithiocarbonate is formed from even though (as described in the previous section) dithiocarbonate formation is significant in reaction of the xanthate ( ). The structural difference between the starting materials ( and ) in these two reactions accounts for their difference in reactivity. Un­like , the xanthate has a sulfur atom directly attached to the carbohydrate portion of the molecule. This means that when the carbo­hydrate radical R· adds to , the options available to the adduct radical are either regen­er­at­ing the starting materials or expelling an unsta­bil­ized, primary radical (Scheme 6). Not surprisingly, no dithiocarbonate from primary radical expulsion is observed; therefore, the only operative pathway for the radical is reforming of R· and the xan­thate . Each regeneration of R· creates a new oppor­tun­ity for it to abstract a hydrogen atom. With these multiple opportun­ities even a marginally effective hydrogen-atom transfer event­ually is able to react with R· to produce the hydrogen-abstraction product RH. Even though the yield is good, the reaction shown in ­is not an attractive option for deoxy sugar synthesis because it requires reaction of the carbohydrate to replace a C–O bond with a C–S bond before conducting the Barton-McCombie reaction. The additional steps neces­sary for this conver­sion add to the effort required for deoxygen­ation. Silylated cyclohexadienes, such as , are effective hydrogen-atom transfers in Barton-McCombie reactions (eq 17). Compound has the advantage of being a solid material that can be easily stored and handled. Although this compound ( ) is an order of magnitude less reactive than (Me Si) SiH ( ), it is able to support chain reactions. The propagation steps in a proposed mech­an­ism for replacement of an -phenoxythiocarbonyl group with a hydrogen atom sup­plied by are given in Scheme 7.

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There are two important statistics associated with any probability distribution, the of a distribution and the of a distribution. The mean is defined as the expected value of the random variable itself. The Greek letter \(\mu\) is usually used to represent the mean. If \(f\left(u\right)\) is the cumulative probability distribution, the mean is the expected value for \(g\left(u\right)=u\). From our definition of expected value, the mean is \[\mu =\int^{\infty }_{-\infty }{u\left(\frac{df}{du}\right)du}\] The variance is defined as the expected value of \({\left(u-\mu \right)}^2\). The variance measures how dispersed the data are. If the variance is large, the data are—on average—farther from the mean than they are if the variance is small. The is the square root of the variance. The Greek letter \(\sigma\) is usually used to denote the standard deviation. Then, \(\sigma^2\) denotes the variance, and \[\sigma^2=\int^{\infty }_{-\infty }{{\left(u-\mu \right)}^2\left(\frac{df}{du}\right)du}\] If we have a small number of points from a distribution, we can estimate \(\mu\) and \(\sigma\) by approximating these integrals as sums over the domain of the random variable. To do this, we need to estimate the probability associated with each interval for which we have a sample point. By the argument we make in , the best estimate of this probability is simply \({1}/{N}\), where \(N\) is the number of sample points. We have therefore \[\mu =\int^{\infty }_{-\infty }{u\left(\frac{df}{du}\right)du\approx \sum^N_1{u_i\left(\frac{1}{N}\right)=\overline{u}}}\] That is, the best estimate we can make of the mean from \(N\) data points is \(\overline{u}\), where \(\overline{u}\) is the ordinary arithmetic average. Similarly, the best estimate we can make of the variance is \[ \sigma^2 = \int_{- \infty}^{ \infty} (u - \mu )^2 \left( \frac{df}{du} \right) du \approx \sum_{i=1}^N (u_i - \mu )^2 \left( \frac{1}{N} \right)\] Now a complication arises in that we usually do not know the value of \(\mu\). The best we can do is to estimate its value as \(\mu \approx \overline{u}\). It turns out that using this approximation in the equation we deduce for the variance gives an estimate of the variance that is too small. A more detailed argument (see ) shows that, if we use \(\overline{u}\) to approximate the mean, the best estimate of \(\sigma^2\), usually denoted \(s^2\), is \[estimated\ \sigma^2=s^2=\sum^N_{i=1}{{\left(u_i-\overline{u}\right)}^2\left(\frac{1}{N-1}\right)}\] Dividing by \(N-1\), rather than \(N\), compensates exactly for the error introduced by using \(\overline{u}\) rather than \(\mu\). The mean is analogous to a center of mass. The variance is analogous to a moment of inertia. For this reason, the variance is also called . To show these analogies, let us imagine that we draw the probability density function on a uniformly thick steel plate and then cut along the curve and the \(u\)-axis (Figure \(\Page {1}\)). Let \(M\) be the mass of the cutout piece of plate; \(M\) is the mass below the probability density curve. Let \(dA\) and \(dm\) be the increments of area and mass in the thin slice of the cutout that lies above a small increment, \(du\), of \(u\). Let \(\rho\) be the density of the plate, expressed as mass per unit area. Since the plate is uniform, \(\rho\) is constant. We have \(dA=\left({df}/{du}\right)du\) and \(dm=\rho dA\) so that \[dm=\rho \left(\frac{df}{du}\right)du\] The mean of the distribution corresponds to a vertical line on this cutout at \(u=\mu\). If the cutout is supported on a knife-edge along the line \(u=\mu\), gravity induces no torque; the cutout is balanced. Since the torque is zero, we have \[0=\int^M_{m=0}{\left(u-\mu \right)dm=\int^{\infty }_{-\infty }{\left(u-\mu \right)\rho \left(\frac{df}{du}\right)du}}\] Since \(\mu\) is a constant property of the cut-out, it follows that \[\mu =\int^{\infty }_{-\infty }{u\left(\frac{df}{du}\right)}du\] The cutout’s moment of inertia about the line \(u=\mu\) is \[\begin{aligned} I & =\int^M_{m=0}{{\left(u-\mu \right)}^2dm} \\ ~ & =\int^{\infty }_{-\infty }{{\left(u-\mu \right)}^2\rho \left(\frac{df}{du}\right)du} \\ ~ & =\rho \sigma^2 \end{aligned}\] The moment of inertia about the line \(u-\mu\) is simply the mass per unit area, \(\rho\), times the variance of the distribution. If we let \(\rho =1\), we have \(I=\sigma^2\). We define the mean of \(f\left(u\right)\) as the expected value of \(u\). It is the value of \(u\) we should “expect” to get the next time we sample the distribution. Alternatively, we can say that the mean is the best prediction we can make about the value of a future sample from the distribution. If we know \(\mu\), the best prediction we can make is \(u_{predicted}=\mu\). If we have only the estimated mean, \(\overline{u}\), then \(\overline{u}\) is the best prediction we can make. Choosing \(u_{predicted}=\overline{u}\) makes the difference,\(\ \left|u-u_{predicted}\right|\), as small as possible. These ideas relate to another interpretation of the mean. We saw that the variance is the second moment about the mean. The first moment about the mean is \[ \begin{aligned} 1^{st}\ moment & =\int^{\infty }_{-\infty }{\left(u-\mu \right)}\left(\frac{df}{du}\right)du \\ ~ & =\int^{\infty }_{-\infty }{u\left(\frac{df}{du}\right)du}-\mu \int^{\infty }_{-\infty }{\left(\frac{df}{du}\right)du} \\ ~ & =\mu -\mu \\ ~ & =0 \end{aligned}\] Since the last two integrals are \(\mu\) and 1, respectively, . We could have defined the mean as the value, \(\mu\), for which the first moment of \(u\) about \(\mu\) is zero. The first moment about the mean is zero. The second moment about the mean is the variance. We can define third, fourth, and higher moments about the mean. Some of these higher moments have useful applications.

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Aluminum oxide, with the chemical formula \(Al_2O_3\), is an amphoteric oxide and is commonly referred to as alumina. Corundum (α-aluminum oxide), emery, sapphire, amethyst, topaz, as well as many other names are reflecting its widespread occurrence in nature and industry. Corundum is the most common naturally occurring crystalline form of aluminum oxide. Rubies and sapphires are gem-quality forms of corundum, which owe their characteristic colors to trace impurities. Rubies are given their characteristic deep red color and their laser qualities by traces of chromium. Sapphires come in different colors given by various other impurities, such as iron and titanium. Its most significant use is in the production of aluminum metal, although it is also used as an abrasive due to its hardness and as a refractory material due to its high melting point. Aluminum oxide is an electrical insulator but has a relatively high thermal conductivity (\(30\, W m^{-1} K^{-1}\)) for a ceramic material. It is thus used as insulating material in power electronics. Aluminum oxide is responsible for resistance of metallic aluminum to weathering. Since metallic aluminum is very reactive with atmospheric oxygen, a thin passivation layer of alumina (4 nm thickness) forms on any exposed aluminum surface, protecting the metal from further . The thickness and properties of this oxide layer can be enhanced using a process called anodizing. The production of aluminum oxide is mainly from bauxite (the main aluminum ore), which is a mixture of various minerals including gibbsite (\(Al(OH)_3\), boehmite (\(\gamma-AlO(OH)\)), and diaspore (\(\alpha-AlO(OH)\)) along with impurities of iron oxides, quartz, and silicates. Bauxite is purified by the which is the principal industrial refining process. As bauxite contains only about 40 to 50% of alumina, the rest has been removed. This is achieved by washing bauxite with hot sodium hydroxide, which dissolves the alumina by converting it to aluminum hydroxide which forms a solution in a strong base: \[Al_2O_3 + 2 OH^- + 3 H_2O \rightarrow 2 [Al(OH)_4]^-\] The other components of the bauxite do not dissolve and are filtered off (the residues usually form a red sludge which presents a disposal problem, since it contains, for example, arsen and cadmium(1)). Next the solution is cooled which causes precipitation of a fluffy solid (aluminum hydroxide). The aluminum hydroxide is then heated to 1050°C which causes it to decompose into aluminum oxide and water: \[2 Al(OH)_3 \rightarrow Al_2O_3 + 3 H_2O\] An accident at an Hungarian alumina plant at Devecser caused several fatalities and many injuries. The wall of a waste-retaining pond broke, releasing a torrent of the toxic red sludge down a local stream. The sludge stream can be seen on a NASA satellite photo (see ).

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We can use these ideas to create a plot that approximates the cumulative probability distribution function given any set of \(N\) measurements of a random variable \(u\). To do so, we put the \(u_i\) values found in our \(N\) measurements in order from smallest to largest. We label the ordered values \(u_1\), \(u_2\), ,\(u_i\), ,\(u_N\), where \(u_1\) is the smallest. By the argument that we develop in the previous section, the probability of observing a value less than \(u_1\) is about \({1}/{\left(N+1\right)}\). If we were to make a large number of additional measurements, a fraction of about \({1}/{\left(N+1\right)}\) of this large number of additional measurements would be less than \(u_1\). This fraction is just \(f\left(u_1\right)\), so we reason that \(f\left(u_1\right)\approx {1}/{\left(N+1\right)}\). The probability of observing a value between \(u_1\) and \(u_2\) is also about \({1}/{\left(N+1\right)}\); so the probability of observing a value less than \(u_2\) is about \({2}/{\left(N+1\right)}\), and we expect \(f\left(u_2\right)\approx {2}/{\left(N+1\right)}\). In general, the probability of observing a value between \(u_{i-1}\) and \(u_i\) is also about \({1}/{\left(N+1\right)}\), and the probability of observing a value less than \(u_i\) is about \({i}/{\left(N+1\right)}\). In other words, we expect the cumulative probability distribution function for \(u_i\) to be such that the \(i^{th}\) smallest observation corresponds to \(f\left(u_i\right)\approx {i}/{\left(N+1\right)}\). The quantity \({i}/{\left(N+1\right)}\) is often called the of the \(i^{th}\) data point. Figure 11 is a sketch of the sigmoid shape that we usually expect to find when we plot \({i}/{\left(N+1\right)}\) versus the \(i^{th}\) value of \(u\). This plot approximates the cumulative probability distribution function, \(f\left(u\right)\). We expect the sigmoid shape because we expect the observed values of \(u\) to bunch up around their average value. (If, within some domain of \(u\) values, all possible values of \(u\) were equally likely, we would expect the difference between successive observed values of \(u\) to be roughly constant, which would make the plot look approximately linear.) At any value of \(u\), the slope of the curve is just the probability-density function, \({df\left(u\right)}/{du}\). These ideas mean that we can test whether the experimental data are described by any particular mathematical model, say \(F\left(u\right)\). To do so, we use the mathematical model to predict each of the rank probability values: \({1}/{\left(N+1\right)}\), \({2}/{\left(N+1\right)}\), , \({i}/{\left(N+1\right)}\), , \({N}/{\left(N+1\right)}\). That is to say, we calculate \(F\left(u_1\right)\), \(F\left(u_2\right)\), , \(F\left(u_i\right)\), , \(F\left(u_N\right)\); if \(F\left(u\right)\) describes the data well, we will find, for all \(i\), \(F\left(u_i\right)\approx {i}/{\left(N+1\right)}\). Graphically, we can test the validity of the relationship by plotting \({i}/{\left(N+1\right)}\) \(F\left(u_i\right)\). If \(F\left(u\right)\) describes the data well, this plot will be approximately linear, with a slope of one. In , we introduce the , which is a mathematical model that describes a great many sources of experimental observations. The normal distribution is a distribution function that involves two parameters, the mean, \(\mu\), and the standard deviation, \(\sigma\). The ideas we have discussed can be used to develop a particular graph paper—usually called normal probability paper. If the data are normally distributed, plotting them on this paper produces an approximately straight line. We can do essentially the same test without benefit of special graph paper, by calculating the average, \(\overline{u}\approx \mu\), and the estimated standard deviation, \(s\approx \sigma\), from the experimental data. (Calculating \(\overline{u}\) and \(s\) is discussed below.) Using \(\overline{u}\) and \(s\) as estimates of \(\mu\) and \(\sigma\), we can find the model-predicted probability of observing a value of the random variable that is less than \(u_i\). This value is \(f\left(u_i\right)\) for a normal distribution whose mean is \(\overline{u}\) and whose standard deviation is \(s\). We can find \(f\left(u_i\right)\) by using standard tables (usually called the in mathematical compilations), by numerically integrating the normal distribution’s probability density function, or by using a function embedded in a spreadsheet program, like Excel\({}^{\circledR }\). If the data are described by the normal distribution function, this value must be approximately equal to the rank probability; that is, we expect \(f\left(u_i\right)\approx {i}/{\left(N+1\right)}\). A plot of \({i}/{\left(N+1\right)}\) \(f\left(u_i\right)\) will be approximately linear with a slope of about one.

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Statistical mechanics makes the connection between macroscopic dynamics and equilibriums states based on microscopic dynamics. For example, while thermodynamics can manipulate equations of state and fundamental relations, it cannot be used to derive them. Statistical mechanics can derive such equations and relations from first principles. Before we study statistical mechanics, we need to introduce the concept of the density, referred to in classical mechanics as density function, and in quantum mechanics as density operator or density matrix. The key idea in statistical mechanics is that the system can have “microstates,” and these microstates have a probability. For example, there may be a certain probability that all gas atoms are in a corner of the room, and this is probably much lower than the probability that they are evenly distributed throughout the room. Statistical mechanics deals with these probabilities, rather than with individual particles. Three general contexts of probability are in common use: a. Discrete systems: and example would be rolling dice. A die has 6 faces, each of which is a “microstate.” Each outcome is equally likely if the die is not loaded, so \[p_{i}=\dfrac{1}{W}=\dfrac{1}{6}\] is the probability of being in microstate “i” where =6 is the total number of microstates. b. Classical systems: here the microstate has to be specified by the positions and the momenta p of all particles in the system, so \(\rho\left ( x_{i},p_{i},t \right )\) is the time dependent probability of finding the particles at and . Do not confuse the momentum here with the probability in a.! It should be clear from the context. is the classical density function. c. Quantum systems: here the microstate is specified by the density operator or density matrix \(\hat{\rho}\left ( t \right ) \). The probability that the system is in quantum state “i” with state |i> is given by \[p_{i}(t)=\int d x \Psi_{i}^{*}(x) \hat{\rho}(x, t) \Psi_{i}(x)=\langle i|\hat{\rho}(t)| i\rangle\]. Of course the probability does not have to depend on time if we are in an equilibrium state. In all three cases, statistical mechanics attempts to evaluate the probability from first principles, using the Hamiltonian of the closed system. In an equilibrium state, the probability does not depend on time, but still depends on x and p (classical) or just x (quantum). We need to briefly review basic concepts in classical and quantum dynamics to see how the probability evolves in time, and when it does not evolve (reach equilibrium). Definition of phase space: Phase space is the 6n dimensional space of the 3n coordinates and 3n momenta of a set of n particles, which, taken together constitute the system. Definition of a trajectory: The dynamics of a system of degrees of freedom ( = 6n for n particles in 3-D) are specified by a trajectory {x (t), p (t)}, = 1 3n in phase space. Note: a system of n particles is phase space is defined by a single point {x (t), x (t), x (t), p (t), p (t), p (t)} that evolves in time. For that specific system, the density function is a delta function in time centered at the single point, and moving along in time as the phase spacve trajectory trajectory moves along. The phase space trajectory is not to be confused with the 3-D trajectories of individual particles. In statistical mechanics, the system must satisfy certain constraints: (e.g. all (t) must lie within a box of volume ; all speeds must be less than the speed of light; etc). The density function \(\rho\left ( x_{i},p_{i},t_{i};constraints \right )\) that we usually care about in statistical mechanics is the average probability density of systems satisfying the constraints. We call the group of systems satisfying the same constraints an “ensemble”, and the average density matrix the “ensemble denisty matrix”: \[\rho=\dfrac{1}{W} \sum_{i=1}^{W} \rho_{i}\], where the sum is over all W possible systems in the microstates “ ” satisfying the constraint. Unlike an individual , which is a moving spike in phase space, sums over all microstates and looks continuous (at least after a small amount of smoothing). For example, consider a single atom with position and momentum {x, p} in a small box. Each different for each microstate “m” is a spike moving about in the phase space {x, p}. Averaging over all such microstates yields a that is uniformly spread over the positions in the box (independent of position x), with a Maxwell-Boltzmann distribution of velocities: \( \rho \sim e^{-p^{2}/2mk_{B}T} \) (assuming the walls of the box can equilibrate the particle to a temperature T). How does evolve in time? Each coordinate q = x evolves according to Newton’s law, which can be recast as \[ \begin{gathered} F_{i}=m \ddot{x}_{i} \\ -\dfrac{\partial V}{\partial x_{i}}=\dfrac{d}{d t}\left(m \dot{x}_{i}\right)=\dfrac{d}{d t}\left(\dfrac{\partial K}{\partial \dot{x}_{i}}\right) \end{gathered} \] if the force is derived from a potential and where \[K=\dfrac{1}{2} \sum_{i} m_{i} \dot{x}_{i}^{2}=\dfrac{1}{2} \sum_{i} \dfrac{p_{i}^{2}}{2 m_{i}}\] is the kinetics energy. We can define the Lagrangian = – and rewrite the above equation as \[\dfrac{d}{d t}\left(\dfrac{\partial L\left(x_{i}, \dot{x}_{i}, t\right)}{\partial \dot{x}_{i}}\right)-\dfrac{\partial L}{\partial x_{i}}=0\]. One can prove using variational calculus (see appendix A) that this differential equation (Lagrange’s equation) is valid in any coordinate system, and is equivalent to the statement \[\min \left\{S=\int_{t_{0}}^{t_{f}} L\left(x_{i}, \dot{x}_{i}\right) d t\right\}\] where is the action ( to be confused with the entropy!). Let’s say the particle moves from from positions x (t ) at t=t to x (t ) at t . Guess a trajectory x (t). The trajectory x (t) can be used to compute velocity \( = )\partial x_{i}/\partial t =\dot{x} \;and\;L\left ( x_{i},\dot{x}_{i} \right )\). The actual trajectory followed by the classical particles is the one that minimizes the above integral, called “the action.” \[\dfrac{\partial L}{\partial \dot{x}_{i}}=p_{i} \quad\left(p_{i}=m \dot{x}_{i}\right)\] Thinking of \(L\) as \(L\left(\dot{x}_{i}\right)\) and of \(p_{i}\) as the derivatives, we can Legendre transform to a new representation \(H\) \[-H \equiv L-\sum_{i=1}^{N} \dot{x}_{i} p_{i} .\] It will become obvious shortly why we define \(H\) with a minus sign. According to the rules for Legendre transforms, \[\dfrac{\partial H}{\partial p_{i}}=\dot{x}_{i} \text { and } \dfrac{\partial H}{\partial \dot{x}_{i}}=p_{i}\] These are Hamilton's equation of motion for a trajectory in phase space. They are equivalent to solving Newton's equation. Evaluating \(H\), \[\begin{aligned} &H\left(\dot{x}_{i}, p_{i}\right)=-K+V+\sum_{i} \dfrac{p_{i}}{m_{i}} p_{i} \\ &=K+V \end{aligned}\] The Hamiltonian is sum of kinetic and potential energy, i.e. the total energy, and is conserved if \(H\) is not explicitly time-dependent: \[\dfrac{d H(x, p)}{d t}=\dfrac{\partial H}{\partial x} \dfrac{d x}{d t}+\dfrac{\partial H}{\partial p} \dfrac{d p}{d t}=\dfrac{\partial H}{\partial x} \dfrac{d H}{d p}-\dfrac{\partial H}{\partial p} \dfrac{d H}{d x}=0 .\] Thus Newton's equations conserve energy. This is because all particles are accounted for in the Hamiltonian \(H\) (closed system). Note that Lagrange's and Hamilton's equations hold in any coordinate system, so from now on we will write \(H\left(q_{i}, p_{i}\right)\) instead of using \(x_{i}\) (cartesian coordinates). Let \(\hat{A}\left(q_{i}, p_{i}, t\right)\) be any dynamical variable (many \(\hat{A} \mathrm{~s}\) of interest do not depend explicitly on t, but we include it here for generality). Then \[\dfrac{d \hat{A}}{d t}=\sum_{i} \dfrac{\partial \hat{A}}{\partial q_{i}} \dfrac{d q_{i}}{d t}+\dfrac{\partial \hat{A}}{\partial p_{i}} \dfrac{d p_{i}}{d t}+\dfrac{\partial \hat{A}}{\partial t}=\sum_{i}\left(\dfrac{\partial \hat{A}}{\partial q_{i}} \dfrac{d H}{d p_{i}}-\dfrac{\partial \hat{A}}{\partial p_{i}} \dfrac{d H}{d q_{i}}\right)+\dfrac{\partial \hat{A}}{\partial t}=[\hat{A}, H]_{p}+\dfrac{\partial \hat{A}}{\partial t}\] gives the time dependence of \(\hat{A}\). [] \(]_{\mathrm{P}}\) is the Poisson bracket. Now consider the ensemble density \(\rho\left ( x_{i},p_{i},t \right )\) as a specific example of a dynamical variable. Because trajectories cannot be destroyed, we can normalize \[\iint d q_{i} d p_{i} \rho\left(q_{i}, p_{i}, t\right)=1\] Integrating the probability over all state space, we are guaranteed to find the system somewhere subject to the constraints. Since the above integral is a constant, we have \[\begin{aligned} \dfrac{d}{d t} \iint d q_{i} d p_{i} \rho=0=& \iint d q_{i} d p_{i} \dfrac{d \rho}{d t}=\iint d q_{i} d p_{i}\left\{[\rho, H]+\dfrac{\partial \rho}{\partial t}\right\}=0 \\ & \Rightarrow \dfrac{\partial \rho}{\partial t}=-[\rho, H]_{p} \end{aligned}\] This is the Liouville equation, it describes how the density propagates in time. To calculate the average value of an observable \(\hat{A}\left(q_{i}, p_{i}\right)\) in the ensemble of systems described by \(\rho\), we calculate \[A(t)=\iint d q_{i} d p_{i} \rho\left(q_{i}, p_{i}, t\right) \hat{A}\left(q_{i}, p_{i}\right)=\langle A\rangle_{\text {ens }}\] Thus if we know \(\rho\), we can calculate any average observable. For certain systems which are left unperturbed by outside influences (closed systems) \[\lim _{t \rightarrow \infty} \dfrac{\partial \rho}{\partial t}=0 \text {. }\] \(\rho\) reaches an equilibrium distribution \(\rho_{e q}\left(q_{i}, p_{i}\right)\) and \[\left[\rho_{e q}, H\right]_{p}=0 \text { (definition of equilibrium) }\] In such a case, \(A(\mathrm{t}) \rightarrow \mathrm{A}\), the equilibrium value of the observable. The goal of equilibrium statistical mechanics is to find the values A for a \(\rho\) subject to certain imposed constraints; e.g. \(\rho_{e q}\left(q_{i}, p_{i}, U, V\right)\) is the set of all possible system trajectories such that \(U\) and \(V\) are constant. The more general goal of non-equilibrium statistical mechanics is to find \(A(\mathrm{t})\) given an initial condition \(\rho_{0}\left(q_{i}, p_{i}\right.\); constraints). So much for the classical picture. Now let us rehearse the whole situation again for quantum mechanics. The quantum formulation is the one best suited to systems where the energy available to a degree of freedom becomes comparable or smaller than the characteristic energy gap of the degree of freedom. Classical and quantum formulations are highly analogous. A fundamental quantity in quantum mechanics is the density operator \[\hat{\rho}_{i}(t)=\left|\psi_{i}(t)\right\rangle\left\langle\psi_{i}(t)|=| t\right\rangle\langle t| \text {. }\] This density operator projects onto the microstate "i" of the system, \(\left|\psi_{i}(t)\right\rangle\). If we have an ensemble of W systems, we can define the ensemble density operator \(\hat{\rho}\) subject to some constraints as \[\hat{\rho}(t ; \text { constraints })=\dfrac{1}{W} \sum_{I=1}^{\mathrm{W}} \hat{\rho}_{I}(t ; \text { constraints })\] For example, let the constraint be \(U=\) const. Then we would sum over all microstates that are degenerate at the same energy \(U\). This average is analogous to averaging the classical probability density over microstates subject to constraints. To obtain the equation of motion for \(\hat{\rho}\), we first look at the wavefunction. Its equation of motion is \[H \psi=i \hbar \dfrac{\partial}{\partial T} \psi\] which by splitting it into \(\psi_{r}\) and \(i \psi_{i}\), can be written \[\dfrac{1}{\hbar} \hat{H} \psi_{r}=\psi_{i} \text { and } \dfrac{1}{\hbar} \hat{H} \psi_{i}=-\psi_{r} .\] Using any complete basis \(H\left|\varphi_{j}\right\rangle=E_{j}\left|\varphi_{j}\right\rangle\), the trace of \(\hat{\rho}\) is conserved \[\operatorname{Tr}\left\{\rho_{i}(t)\right\}=\sum_{j}<\varphi_{j}\left|\psi_{i}(t)\right\rangle\left\langle\psi_{i}(t) \mid \varphi_{j}\right\rangle=\sum_{j}\left\langle\psi_{i} \mid \varphi_{j}\right\rangle\left\langle\varphi_{j} \mid \psi_{i}\right\rangle=\left\langle\psi_{i}(t) \mid \psi_{i}(t)\right\rangle=1\] if \(\psi_{j}(t)\) is normalized, or \[\operatorname{Tr}\left\{\hat{\rho}_{i}\right\}=1 \text {, and } \operatorname{Tr}\{\hat{\rho}\}=1 \text {. }\] This basically means that probability density cannot be destroyed, in analogy to trajectory conservation. Note that if \[\begin{gathered} \hat{\rho}_{i}=|\psi\rangle\langle\psi| \\ \Rightarrow \hat{\rho}_{i}^{2}=\hat{\rho}_{i} \text { so } \operatorname{Tr}\left(\hat{\rho}_{i}^{2}\right)=1 \end{gathered}\] A state described by a wavefunction \(\mid \Psi>\) that satisfies the latter equation is a pure state. Most states of interest in statistical mechanics are NOT pure states. If \[\hat{\rho}=\dfrac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{i},\] the complex off-diagonal elements tend to cancel because of random phases and \(\operatorname{Tr}\left(\hat{\rho}^{2}\right)<1\), this an impure state. Let \(\psi=c_{0}^{i}|0\rangle+c_{1}^{i}|1\rangle\) be an arbitrary wavefunction for a two-level system. \[\Rightarrow \rho=|\Psi\rangle\left\langle\Psi\left|=c_{0}^{i} c_{0}^{i^{*}}\right| 0\right\rangle\left\langle 0\left|+c_{0}^{i} c_{1}^{i^{*}}\right| 0\right\rangle\left\langle 1\left|+c_{1}^{i} c_{0}^{i^{*}}\right| 1\right\rangle\left\langle 0\left|+c_{1}^{i} c_{1}^{i^{*}}\right| 1\right\rangle\langle 1|\] or, in matrix form, \[\hat{\rho}_{i}=\left(\begin{array}{cc}\left|c_{0}^{i}\right|^{2} & c_{0}^{i} c_{1}^{i^{*}} \\c_{0}^{i} c_{1}^{i^{*}} & \left|c_{1}^{i}\right|^{2} \end{array}\right) \quad T_{r}\left(\hat{\rho}_{i}\right)=\left|c_{0}^{i}\right|^{2}+\left|c_{1}^{i}\right|^{2}=1\] Generally, macroscopic constraints (volume, spin population, etc.) do not constrain the phases of \(c_{0}=\left|c_{0}\right| e^{i \varphi_{0}}\) and \(c_{1}=\left|c_{1}\right| e^{i \varphi_{1}}\). Thus, ensemble averaging \[\begin{aligned} &\hat{\rho}=\lim _{W \rightarrow \infty} \dfrac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{I}=\lim _{W \rightarrow \infty}\left(\begin{array}{cc} \left|c_{0}\right|^{2} & \dfrac{e^{i \bar{\varphi}}}{\sqrt{W}} \\ \dfrac{e^{-i \bar{\varphi}}}{\sqrt{W}} & \left|c_{1}\right|^{2} \end{array}\right)=\left(\begin{array}{cc} \left|c_{0}\right|^{2} & 0 \\ 0 & \left|c_{1}\right|^{2} \end{array}\right) \\ &\Rightarrow T_{r}\left(\hat{\rho}^{2}\right)=\left|c_{0}\right|^{4}+\left|c_{1}\right|^{4}<1 \end{aligned}\] Such a state is known as an impure state. To obtain the equation of motion of \(\hat{\rho}\), use the time-dependent Schrödinger equation in propagator form, \[\left|\psi_{i}(t)\right\rangle=e^{-\dfrac{i}{\hbar} \hat{H} t}\left|\psi_{i}(0)\right\rangle\] and the definition of \(\rho\), \[\hat{\rho}_{i}(t)=\left|\psi_{i}(t)\right\rangle\left\langle\psi_{i}(t)\right|\] to obtain the equation of motion \[\begin{aligned} \dfrac{\partial}{\partial t} \hat{\rho}_{i}(t) &=\dfrac{\partial}{\partial t}\left\{e^{-\dfrac{i}{\hbar} \hat{H} t}\left|\psi_{i}(0)\right\rangle\left\langle\psi_{i}(0)\right| e^{+\dfrac{i}{\hbar} \hat{H} t}\right\} \\ &=-\dfrac{i}{\hbar} \hat{H} e^{-\dfrac{i}{\hbar} \hat{H} t}\left|\psi_{i}(0)\right\rangle\left\langle\psi_{i}(0)\left|e^{+\dfrac{i}{\hbar} \hat{H} t}+e^{-\dfrac{i}{\hbar} \hat{H} t}\right| \psi_{i}(0)\right\rangle\left\langle\psi_{i}(0)\right| e^{+\dfrac{i}{\hbar} \hat{H} t}\left(+\dfrac{i}{\hbar} \hat{H}\right) \\ &=-\dfrac{i}{\hbar} H \rho_{i}+\dfrac{i}{\hbar} \rho_{i} H \\ &=\dfrac{1}{i \hbar}\left[\rho_{i}, H\right] \end{aligned}\] This is known as the Liouville-von Neumann equation. The commutator defined in the last line is equivalent to the Poisson bracket in classical dynamics. Summing over all microstates to obtain the average density operator \[\dfrac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{i} \Rightarrow \dfrac{\partial \hat{\rho}}{\partial t}=-\dfrac{1}{i \hbar}[\hat{\rho}, \hat{H}]\] This von-Neumann equation is the quantum equation of motion for \(\hat{\rho}\). If \(\rho\) represent an impure state, this propagation cannot be represented by the time-dependent Schrödinger equation. We are interested in average values of observables in an ensemble of systems. Starting with a pure state, \[A(t)=\left\langle\psi_{i}(t)|\hat{A}| \psi_{i}(t)\right\rangle=\sum_{j}\left\langle\psi_{i}(t) \mid \varphi_{j}\right\rangle\left\langle\varphi_{j}|\hat{A}| \psi_{i}(t)\right\rangle=\sum_{j}\left\langle\varphi_{j}|\hat{A}| \psi_{i}(t)\right\rangle\left\langle\psi_{i}(t) \mid \varphi_{j}\right\rangle=T_{r}\left\{\hat{A} \hat{\rho}_{i}(t)\right\}\] Summing over ensembles, \[\dfrac{1}{W} \sum_{i=1}^{W} T_{r}\left\{\hat{A} \rho_{i}\right\}=T_{r}\left\{\hat{A} \dfrac{1}{W} \sum_{i=1}^{W} \hat{\rho}_{i}\right\} \Rightarrow A(t)=T_{r}\{\hat{A} \hat{\rho}(t)\}\] In particular, \(P_{j}=T_{r}\left\{\hat{P}_{j} \hat{\rho}(t)\right\}=\operatorname{Tr}\left\{\left|\varphi_{j}\right\rangle\left\langle\varphi_{j}\right| \hat{\rho}(t)\right\}=\left\langle\varphi_{j}|\hat{\rho}(t)| \varphi_{j}\right\rangle=\rho_{j j}(t) \quad\) is \(\quad\) the probability of being in state \(\mathrm{j}\) at time \(\mathrm{t}\). Finally, \(\hat{\rho}\) may evolve to long-time solutions \(\hat{\rho}_{e q}\) such that \(\dfrac{\partial \rho}{\partial t}=0 \Rightarrow\) \[\left[\hat{\rho}_{e q}, H\right]=0 \text{(condition for equilibrium)} \]. In that case, the density matrix has relaxed to the equilibrium density matrix, which no longer evolves in time. Consider a two-level system again. Let \[H|j\rangle=E|j\rangle \text { for } j=0,1 \text { or } H=\left(\begin{array}{cc} E_{0} & 0 \\ 0 & E_{1} \end{array}\right) ; \text { if } \hat{p}=\left(\begin{array}{cc} \rho_{00} & 0 \\ 0 & \rho_{11} \end{array}\right) \Rightarrow[\hat{\rho}, H]=0\] Thus a diagonal density matrix of a closed system does not evolve. The equilibrium density matrix must be diagonal so \(\left[\rho_{e q}, H\right]=0\). This corresponds to a completely impure state. Note that unitary evolution cannot change the purity of any closed system. Thus, the density matrix of a single closed system cannot evolve to diagonality unless \[ \hat{\rho}=\dfrac{1}{W} \sum_{i} \rho_{i} \] for the ensemble is already diagonal. In reality, single systems still decohere because they are open to the environment: let i denote the degrees of freedom of the system, and \(\mathrm{j}\) of a bath (e.g. a heat reservoir). \(\hat{\rho}\) depends on both \(i\) and \(j\) and can be written as a matrix \(\hat{\rho}_{i j, i^{1} j^{1}}(t)\). For example, for a two-level system coupled to a large bath, indices i,j only go from 1 to 2 , but indices i' and j' could go to \(10^{20}\). We can average over the bath by letting \[\hat{\rho}^{(r e d)}=\operatorname{Tr}_{j}\{\hat{\rho}\}\] \(\hat{\rho}^{(r d)}\) only has matrix elements for the system degrees of freedom, e.g. for a two level system in contact with a bath of \(10^{20}\) states, \(\hat{\rho}^{(c e d)}\) is still a \(2 \times 2\) matrix. We will show later that for a bath is at constant \(T\). \[\hat{\rho}^{(r e d)} \rightarrow \rho_{e q}=\left(\begin{array}{cc} \dfrac{e^{-E_{0} / k_{B} T}}{Q} & 0 \\ 0 & \dfrac{e^{-E_{1} / k_{B} T}}{Q} \end{array}\right)\] Note that "reducing" was not necessary in the classical discussion because quantum coherence and phases do not exist there. To summarize analogous entities: Eq. of motion for traj. ψ To illustrate basic ideas, we will often go back to simple discrete models with probability for each microstate \(p_{i}\), but to get accurate answers, one may have to work with the full classical or quantum probability \(\rho\).

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This page describes how to perform a flame test for a range of metal ions, and briefly discusses how the flame color arises. Flame tests are used to identify the presence of a relatively small number of metal ions in a compound. Not all metal ions give flame colors. For compounds, flame tests are usually by far the easiest way of identifying which metal you have got. For other metals, there are usually other easy methods that are more reliable - but the flame test can give a useful hint as to where to look. There will, in fact, always be a trace of orange in the flame if you use nichrome. Platinum is much better to use but is much, much more expensive. If you have a particularly dirty bit of nichrome wire, you can just chop the end off. You do not do that with platinum! Dilute hydrochloric acid can be used instead of concentrated acid for safety reasons, but does not always give such intense flame colors. The colors in Table \(\Page {1}\) are just a guide. Almost everybody sees and describes colors differently. I have, for example, used the word "red" several times to describe colors that can be quite different from each other. Other people use words like "carmine" or "crimson" or "scarlet", but not everyone knows the differences between these words - particularly if their first language is not English. What do you do if you have a red flame color for an unknown compound and do not know which of the various reds it is? Get samples of known lithium, strontium (etc) compounds and repeat the flame test, comparing the colors produced by one of the known compounds and the unknown compound side by side until you have a good match. If you excite an atom or an ion by very strong heating, electrons can be promoted from their normal unexcited state into higher orbitals. As they fall back down to lower levels (either in one go or in several steps), energy is released as light. Each of these jumps involves a specific amount of energy being released as light energy, and each corresponds to a particular wavelength (or frequency). As a result of all these jumps, a spectrum of lines will be produced, some of which will be in the visible part of the spectrum. The color you see will be a combination of all these individual colors. In the case of sodium (or other metal) , the jumps involve very high energies and these result in lines in the UV part of the spectrum which your eyes can't see. The jumps that you can see in flame tests come from electrons falling from a higher to a lower level in the metal . So if, for example, you put sodium chloride which contains sodium ions, into a flame, where do the atoms come from? In the hot flame, some of the sodium ions regain their electrons to form neutral sodium atoms again. A sodium atom in an unexcited state has the structure 1s 2s 2p 3s , but within the flame there will be all sorts of excited states of the electrons. Sodium's familiar bright orange-yellow flame color results from promoted electrons falling back from the 3p level to their normal 3s level. The exact sizes of the possible jumps in energy terms vary from one metal to another. That means that each different metal will have a different pattern of spectral lines, and so a different flame color. Flame colors are produced from the movement of the electrons in the metal ions present in the compounds. For example, a sodium ion in an unexcited state has the electron configuration 1s 2s 2p . When heated, the electrons gain energy and can be excited into any of the empty higher-energy orbitals—7s, 6p, 4d, or any other, depending on the amount of energy a particular electron happens to absorb from the flame. Because the electron is now at a higher and more energetically unstable level, it falls back down to the original level, but not necessarily in one transition. The electron transitions which produced lines in the visible spectrum involved atoms rather than ions. Jim Clark ( )

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Entropy is a state function, so we can calculate values for a process using any path. This allows us to calculate the entropy change of a chemical reaction using standard entropies. Specifically, we sum the entropies of the products and subtract the entropies of the reactions: \[\Delta_{rxn}S^\circ = \sum_{\text{Products}}{v_i S^\circ_i} - \sum_{\text{Reactants}}{v_i S^\circ_i} \nonumber \] Where \(v_i\) is the stoichiometric coefficient. Let's look at the combustion of methane: \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} + \ce{CO_2}\left(g\right) \nonumber \] The standard entropies are: The entropy for the combustion of methane is: \[\Delta S^\circ = \left[ 2 \left( 188.84 \right) + 1 \left( 213.70 \right) \right] - \left[ 1 \left( 186.25 \right) + 2 \left( 205.15 \right) \right] = -5.17 \: \frac{\text{J}}{\text{mol} \cdot \text{K}} \nonumber \]

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A change is said to occur reversibly when it can be carried out in a series of infinitesimal steps, each one of which can be undone by making a similarly minute change to the conditions that bring the change about. For example, the reversible expansion of a gas can be achieved by reducing the external pressure in a series of infinitesimal steps; reversing any step will restore the system and the surroundings to their previous state. Similarly, heat can be transferred reversibly between two bodies by changing the temperature difference between them in infinitesimal steps each of which can be undone by reversing the temperature difference. The most widely cited example of an change is the free expansion of a gas into a vacuum. Although the system can always be restored to its original state by recompressing the gas, this would require that the surroundings perform work on the gas. Since the gas does no work on the surrounding in a free expansion (the external pressure is zero, so \(PΔV = 0\),) there will be a permanent change in the surroundings. Another example of irreversible change is the conversion of mechanical work into frictional heat; there is no way, by reversing the motion of a weight along a surface, that the heat released due to friction can be restored to the system. These diagrams show the same expansion and compression ±ΔV carried out in different numbers of steps ranging from a single step at the top to an "infinite" number of steps at the bottom. As the number of steps increases, the processes become less irreversible; that is, the difference between the work done in expansion and that required to re-compress the gas diminishes. In the limit of an ”infinite” number of steps (bottom), these work terms are identical, and both the system and surroundings (the “world”) are unchanged by the expansion-compression cycle. In all other cases the system (the gas) is restored to its initial state, A reversible change is one carried out in such as way that, when undone, both the system and surroundings (that is, the world) remain unchanged. It should go without saying, of course, that any process that proceeds in infinitesimal steps would take infinitely long to occur, so thermodynamic reversibility is an idealization that is never achieved in real processes, except when the system is already at equilibrium, in which case no change will occur anyway! So why is the concept of a reversible process so important? The answer can be seen by recalling that the change in the internal energy that characterizes any process can be distributed in an infinity of ways between heat flow across the boundaries of the system and work done on or by the system, as expressed by the \[ΔU = q + w. \label{first} \] Each combination of and represents a different pathway between the initial and final states. It can be shown that as a process such as the expansion of a gas is carried out in successively longer series of smaller steps, the absolute value of approaches a minimum, and that of approaches a maximum that is characteristic of the particular process. Thus when a process is carried out reversibly, the -term in Equation \ref{first} has its greatest possible value, and the -term is at its smallest. These special quantities \(w_{max}\) and \(q_{min}\) (which we denote as \(q_{rev}\) and pronounce “q-reversible”) have unique values for any given process and are therefore state functions. Changes in entropy (\(ΔS\)), together with changes in enthalpy (\(ΔH\)), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (\(P_{ext} = 0\)) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change. Because work done during the expansion of a gas depends on the opposing external pressure (\(w = P_{ext}ΔV\)), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: \(w_{rev} ≥ w_{irrev}\). Whether a process is reversible or irreversible, the first law of thermodynamics holds (Equation \ref{first}). Because \(U\) is a state function, the magnitude of \(ΔU\) does not depend on reversibility and is independent of the path taken. So \[\begin{align} ΔU &= q_{rev} + w_{rev} \\[4pt] &= q_{irrev} + w_{irrev} \label{Eq1} \end{align}\] Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: \[w_{rev} \ge w_{irrev}\] In other words, \(ΔU\) for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (\(q_{rev}\)) to define entropy quantitatively. For a process that reversibly exchanges a quantity of heat \(q_{rev}\) with the surroundings, the entropy change is defined as \[ \color{red} \Delta S = \dfrac{q_{rev}}{T} \label{23.2.1}\] This is the basic way of evaluating \(ΔS\) for such as phase changes, or the isothermal expansion of a gas. For processes in which the temperature is not constant such as heating or cooling of a substance, the equation must be integrated over the required temperature range, as discussed below. This is a rather fine point that you should understand: although transfer of heat between the system and surroundings is impossible to achieve in a truly reversible manner, this idealized pathway is only crucial for the definition of \(ΔS\); by virtue of its being a state function, the same value of \(ΔS\) will apply when the system undergoes the same net change via any pathway. For example, the entropy change a gas undergoes when its volume is doubled at constant temperature will be the same regardless of whether the expansion is carried out in 1000 tiny steps (as reversible as patience is likely to allow) or by a single-step (as irreversible a pathway as you can get!) expansion into a vacuum (Figure \(\Page {1}\)). Entropy is an quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy. Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical sense this is correct, since the spreading and sharing of thermal energy does have the effect of randomizing the disposition of thermal energy within a system. But to simply equate entropy with “disorder” without further qualification is extremely misleading because it is far too easy to forget that Carrying these concepts over to macro systems may yield compelling analogies, but it is no longer science. It is far better to avoid the term “disorder” altogether in discussing entropy. As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy of solute (the entropy) will of course always increase as the solution becomes more dilute. For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by Note: If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course. Because the pressure of an ideal gas is to its volume, i.e., \[P = \dfrac{nRT}{V}\] we can easily alter Equation \ref{23.2.4} to express the entropy change associated with a change in the pressure of an ideal gas: we can expressing the entropy change directly in concentrations, we have the similar relation Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture. Because the quantity of heat transferred (\(q_{rev}\)) is directly proportional to the absolute temperature of an object (\(T\)) (\(q_{rev} \propto T\)), the hotter the object, the greater the amount of heat transferred. Moreover, adding heat to a system increases the kinetic energy of the component atoms and molecules and hence their disorder (\(ΔS ∝ q_{rev}\)). Combining these relationships for any reversible process, \[q_{\textrm{rev}}=T\Delta S\] and Because the numerator (q ) is expressed in units of energy (joules), the units of ΔS are joules/kelvin (J/K). Recognizing that the work done in a reversible process at constant pressure is \[w_{rev} = −PΔV,\] we can express Equation \(\ref{Eq1}\) as follows: \[\begin{align} ΔU &= q_{rev} + w_{rev} \\[4pt] &= TΔS − PΔV \label{Eq3} \end{align} \] Thus the change in the internal energy of the system is related to the change in entropy, the absolute temperature, and the \(PV\) work done. To illustrate the use of Equation \(\ref{Eq2}\) and Equation \(\ref{Eq3}\), we consider two reversible processes before turning to an irreversible process. When a sample of an ideal gas is allowed to expand reversibly at constant temperature, heat must be added to the gas during expansion to keep its T constant (Figure \(\Page {5}\)). The internal energy of the gas does not change because the temperature of the gas does not change; that is, ΔU = 0 and q = −w . During expansion, ΔV > 0, so the gas performs work on its surroundings: \[w_{rev} = −PΔV < 0.\] According to Equation \(\ref{Eq3}\), this means that \(q_{rev}\) must increase during expansion; that is, the gas must absorb heat from the surroundings during expansion, and the surroundings must give up that same amount of heat. The entropy change of the is therefore \[ΔS_{sys} = +\dfrac{q_{rev}}{T}\] and the entropy change of the is \[ΔS_{surr} = −\dfrac{q_{rev}}{T}.\] The corresponding change in entropy of the universe is then as follows: \[\begin{align} \Delta S_{\textrm{univ}} &=\Delta S_{\textrm{sys}}+\Delta S_{\textrm{surr}} \\[4pt] &=\dfrac{q_{\textrm{rev}}}{T}+\left(-\dfrac{q_\textrm{rev}}{T}\right) \\[4pt] &= 0 \label{Eq4} \end{align} \] Thus no change in \(ΔS_{univ} \) has occurred. In the initial state (top), the temperatures of a gas and the surroundings are the same. During the reversible expansion of the gas, heat must be added to the gas to maintain a constant temperature. Thus the internal energy of the gas does not change, but work is performed on the surroundings. In the final state (bottom), the temperature of the surroundings is lower because the gas has absorbed heat from the surroundings during expansion. Now consider the reversible melting of a sample of ice at 0°C and 1 atm. The enthalpy of fusion of ice is 6.01 kJ/mol, which means that 6.01 kJ of heat are absorbed reversibly from the surroundings when 1 mol of ice melts at 0°C, as illustrated in Figure \(\Page {6}\). The surroundings constitute a sample of low-density carbon foam that is thermally conductive, and the system is the ice cube that has been placed on it. The direction of heat flow along the resulting temperature gradient is indicated with an arrow. From Equation \(\ref{Eq2}\), we see that the entropy of fusion of ice can be written as follows: \[\Delta S_{\textrm{fus}}=\dfrac{q_{\textrm{rev}}}{T}=\dfrac{\Delta H_{\textrm{fus}}}{T} \label{Eq5}\] By convention, a thermogram shows cold regions in blue, warm regions in red, and thermally intermediate regions in green. When an ice cube (the system, dark blue) is placed on the corner of a square sample of low-density carbon foam with very high thermal conductivity, the temperature of the foam is lowered (going from red to green). As the ice melts, a temperature gradient appears, ranging from warm to very cold. An arrow indicates the direction of heat flow from the surroundings (red and green) to the ice cube. The amount of heat lost by the surroundings is the same as the amount gained by the ice, so the entropy of the universe does not change. In this case, \[\begin{align} ΔS_{fus} &= \dfrac{6.01\, kJ/mol}{273\, K} \\[4pt] &= 22.0\, J/(mol•K) \\[4pt] &= ΔS_{sys}.\end{align}\] The amount of heat lost by the surroundings is the same as the amount gained by the ice, so \[\begin{align} ΔS_{surr} &= \dfrac{q_{rev}}{T} \\[4pt] &= \dfrac{ −6.01\, kJ/mol}{273\, K} \\[4pt] &= −22.0\, J/(mol•K).\end{align}\] Once again, we see that the entropy of the universe does not change: \[\begin{align} ΔS_{univ} &= ΔS_{sys} + ΔS_{surr} \\[4pt] &= 22.0 \,J/(mol•K) − 22.0\, J/(mol•K) \\[4pt] &= 0 \end{align}\] In these two examples of reversible processes, the entropy of the universe is unchanged. This is true of all reversible processes and constitutes part of the second law of thermodynamics: the entropy of the universe remains constant in a reversible process, whereas the entropy of the universe increases in an irreversible (spontaneous) process. The entropy of the universe during a spontaneous process. It also during an observable non-spontaneous process. As an example of an irreversible process, consider the entropy changes that accompany the spontaneous and irreversible transfer of heat from a hot object to a cold one, as occurs when lava spewed from a volcano flows into cold ocean water. The cold substance, the water, gains heat (q > 0), so the change in the entropy of the water can be written as ΔS = q/T . Similarly, the hot substance, the lava, loses heat (q < 0), so its entropy change can be written as ΔS = −q/T , where T and T are the temperatures of the cold and hot substances, respectively. The total entropy change of the universe accompanying this process is therefore \[\begin{align} \Delta S_{\textrm{univ}} &= \Delta S_{\textrm{cold}}+\Delta S_{\textrm{hot}} \\[4pt] &= \dfrac{q}{T_{\textrm{cold}}}+\left(-\dfrac{q}{T_{\textrm{hot}}}\right) \label{Eq6}\end{align}\] The numerators on the right side of Equation \(\ref{Eq6}\) are the same in magnitude but opposite in sign. Whether ΔS is positive or negative depends on the relative magnitudes of the denominators. By definition, T > T , so −q/T must be less than q/T , and ΔS must be positive. As predicted by the second law of thermodynamics, the entropy of the universe increases during this irreversible process. Any process for which ΔS is positive is, by definition, a spontaneous one that will occur as written. Conversely, any process for which ΔS approaches zero will not occur spontaneously as written but will occur spontaneously in the reverse direction. We see, therefore, that heat is spontaneously transferred from a hot substance, the lava, to a cold substance, the ocean water. In fact, if the lava is hot enough (e.g., if it is molten), so much heat can be transferred that the water is converted to steam (Figure \(\Page {7}\)). When molten lava flows into cold ocean water, so much heat is spontaneously transferred to the water that steam is produced. Tin has two allotropes with different structures. Gray tin (α-tin) has a structure similar to that of diamond, whereas white tin (β-tin) is denser, with a unit cell structure that is based on a rectangular prism. At temperatures greater than 13.2°C, white tin is the more stable phase, but below that temperature, it slowly converts reversibly to the less dense, powdery gray phase. This phenomenon was argued to plagued Napoleon’s army during his ill-fated invasion of Russia in 1812: the buttons on his soldiers’ uniforms were made of tin and disintegrated during the Russian winter, adversely affecting the soldiers’ health (and morale). The conversion of white tin to gray tin is exothermic, with ΔH = −2.1 kJ/mol at 13.2°C. : ΔH and temperature ΔS and relative degree of order : Use Equation \(\ref{Eq2}\) to calculate the change in entropy for the reversible phase transition. From the calculated value of ΔS, predict which allotrope has the more highly ordered structure. Note: Whether failing buttons were indeed a contributing factor in the failure of the invasion remains disputed; critics of the theory point out that the tin used would have been quite impure and thus more tolerant of low temperatures. Laboratory tests provide evidence that the time required for unalloyed tin to develop significant tin pest damage at lowered temperatures is about 18 months, which is more than twice the length of Napoleon's Russian campaign. It is clear though that some of the regiments employed in the campaign had tin buttons and that the temperature reached sufficiently low values (at least -40 °C) to facilitate tin pest. Elemental sulfur exists in two forms: an orthorhombic form (S ), which is stable below 95.3°C, and a monoclinic form (S ), which is stable above 95.3°C. The conversion of orthorhombic sulfur to monoclinic sulfur is endothermic, with \(ΔH = 0.401\, kJ/mol\) at 1 atm. 1.09 J/(mol•K) S During a spontaneous process, the entropy of the universe increases. \[\Delta S=\dfrac{q_{\textrm{rev}}}{T}\nonumber\] A reversible process is one for which all intermediate states between extremes are ; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. The second law of thermodynamics states that in a reversible process, the entropy of the universe is constant, whereas in an irreversible process, such as the transfer of heat from a hot object to a cold object, the entropy of the universe increases.

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In the presence of dioxygen, iron(II) species are readily oxidized to iron(III) species. In the presence of water, iron(III) species frequently associate into \(\mu\)-oxodiiron(III) dimers. For iron(II)-porphyrin complexes this process may take only milliseconds at room temperature. The following mechanism was proposed in 1968 for the irreversible oxidation of iron(II)-porphyrinato species; subsequent work has largely confirmed it. \[Fe^{II} + O_{2} \rightleftharpoons Fe^{III}—O_{2}^{I-} \tag{4.29a}\] \[Fe^{III}—O_{2}^{I-} + Fe^{II} \rightleftharpoons Fe^{III}—O_{2}^{II-}—Fe^{III} \tag{4.29b}\] \[Fe^{III}—O_{2}^{II-}—Fe^{III} \rightarrow 2Fe^{IV}=O \tag{4.29c}\] \[Fe^{IV}=O + Fe^{II} \rightarrow Fe^{III}—O—Fe^{III} \tag{4.29d}\] In particular, the dimerization reaction (4.29b) may be rendered less favorable by low temperatures (< -40 °C) or by sterically preventing the bimolecular contact of an Fe -O moiety with an Fe moiety. In the latter case, sterically bulky substituents on the equatorial ligand surround the coordinated O ligand and the other axial position, trans to the coordinated dioxygen ligand, is protected with a nitrogenous base, such as imidazole, or with additional bulky substituents on the equatorial ligand (Figure 4.14). The protein effectively provides such protection and thus plays a key role in preventing the bimolecular contact of two hemes. The first observation of reversible binding of dioxygen to an iron(II)-porphyrin in the absence of protein was made in 1958. In that pioneering study, a heme group was immobilized on a polymer support specially modified to contain imidazole functions. The structurally characterizable hemoglobin or myoglobin species was replaced by a noncrystalline structurally uncharacterized polymer. Why does this irreversible oxidation not occur analogously for cobalt systems? Step (4.29c) involves cleavage of the O—O bond, which in H O has a bond energy of 34.3 kcal/mol or in Na O of 48.4 kcal/mol. By way of comparison, for O the bond energy is 117.2 and for HO • it is 55.5 kcal/mol. A simple molecular orbital picture gives insight into why an Fe =O species is stabilized relative to the analogous Co =O species. From Figure 4.15 we see that for metals with electronic configuration d , where n \(\leq\) 5, no electrons occupy the antibonding orbital \(\pi\)* for Fe -O or Fe =O moieties. For Co (d ) the extra electron goes into the antibonding orbital \(\pi\)*. As predicted by the model, Mn is observed indeed to behave like Fe . A second oxidation pathway does not require the bimolecular contact of two iron(II)-porphyrins. Coordinated dioxygen may be released not as O , as in normal dioxygen transport, but, as noted in Section I.C, as a superoxide radical anion O in a process called : \[Fe^{III}—O_{2}^{I-} \rightleftharpoons Fe^{III} + O_{2} \tag{4.30}\] This process is assisted by the presence of other nucleophiles that are stronger than the superoxide anion, such as chloride, and by protons that stabilize the O anion as HO : \[Fe_{III} + Cl^{-} \rightleftharpoons Fe^{III}—Cl^{-} \tag{4.31}\] The formation of methemoglobin occurs , probably by the above mechanism, at the rate of ~ 3 percent of total hemoglobin per day. If exogenous reductants are present, then further reduction of dioxygen can occur: \[2H^{+} + Fe^{III}—O_{2}^{I-} + e^{-} \rightarrow Fe^{V}=O + H_{2}O \tag{4.32}\] Such processes are important, for example, in the cytochrome P-450 system. With suitably small reductants, oxygenase activity also has been observed for hemoglobin A. This has led to the characterization of hemoglobin as a "frustrated oxidase." Note the formal similarity between this process (Equation 4.32) and the bimolecular irreversible oxidation of iron(II) porphyrins: the second Fe(II) complex in Reaction (4.29b) functions like the electron in Reaction (4.32). The end products of the irreversible bimolecular oxidation of Fe species contain the Fe —O—Fe fragment. Given the facile formation of \(\mu\)-oxodiiron(III) species, it is not surprising that the Fe—O—Fe motif is incorporated into a variety of metalloproteins, including the oxygen-carrier hemerythrin (Figure 4.10), the hydrolase purple acid phosphatase, the oxidoreductases ribonucleotide reductase and methane monoxygenase, an iron-sulfur protein rubrerythrin, and the iron-transport protein ferritin. In ferritin higher-order oligomers are formed. This \(\mu\)-oxodiiron(III) moiety has a distinctive fingerprint that has made it easy to identify this motif in proteins. Regardless of the number (4, 5, 6, or 7), geometry (tetrahedral, square pyramidal, tetragonally distorted octahedral, or pentagonal bipyramidal), and type of ligands (halide, RO , RCOO , aliphatic N, or aromatic N) around the iron center, and of the Fe—O—Fe angle, the magnetic susceptibility at room temperature lies in the range 1.5 to 2.0 Bohr magnetons per Fe —O—Fe group, equivalent to about one unpaired electron. In other words, the high-spin (S = \(\frac{5}{2}\)) iron centers are strongly antiferromagnetically coupled. Other bridging groups, such as OH , Cl , carboxylate, alkoxide, or phenoxide, give very weak coupling. The asymmetric Fe—O stretch, (Fe—O), lies in the range 730 to 880 cm ; in multiply bridged complexes this mode is weak in the infrared region. The symmetric vibration, (Fe—O), forbidden in the infrared region for linear, symmetric Fe—O—Fe groups, occurs in the range 360 to 545 cm . The symmetric mode is usually, but not always, observed by resonance Raman techniques upon irradiating on the low-energy side of the Fe—O chargetransfer band that occurs at about 350 nm. Few dinuclear iron(II) complexes are known where the ligands approximately resemble those believed or known to occur in the family of \(\mu\)-oxodiiron(III) proteins. The dioxygen-binding process in hemerythrin has no close nonbiological analogue. Although spectroscopically similar to oxyhemerythrin, the unstable monomeric purple peroxo complex formed by the addition of hydrogen peroxide to basic aqueous Fe (EDTA) solutions remains structurally uncharacterized. Iron porphyrins, the active sites of the hemoglobin family, have a rich magnetochemistry. Iron porphyrins may be octahedral (two axial ligands), square pyramidal (one axial ligand), or square planar (no axial ligand). The metal d orbitals, now having partial porphyrin \(\pi\)* character, are split, as shown in Figure 4.16. The radius of the metal atom is much greater when it is high spin (S = 2 for Fe , S = \(\frac{5}{2}\) for Fe ) than when it is low spin (S = 0 for Fe , S = \(\frac{1}{2}\) for Fe ). This difference influences Fe—N separations, porphyrin conformation, and the displacement of the iron center with respect to the porphyrin plane. For iron(II)-porphyrins, two strong-field axial ligands, such as a pair of imidazoles or an imidazole and carbon monoxide, lead to diamagnetic complexes (S = 0) with the six 3d electrons occupying those orbitals of approximate t symmetry. In a classic experiment in 1936, Pauling and Coryell proved that oxyhemoglobin and carbonmonoxyhemoglobin are diamagnetic. * * There was a considerable flurry of interest when an Italian group, using a SQUID (Superconducting Quantum Mechanical Interference Device), reported that at room temperature oxyhemoglobin was significantly paramagnetic. Not surprisingly, several theoretical papers followed that "proved" the existence of low-lying triplet and quintet excited states Subsequently, the residual paramagnetism was doubted and shown to arise from incomplete saturation of hemoglobin by O ; in other words, small amounts of deoxy hemoglobin remained Since oxygen affinity increases with decreased temperature, the concentration of paramagnetic impurity decreased with decreasing temperature. No axial ligands at all may lead to a spin state of S = 1, with unpaired electrons in the d and d orbitals. Five-coordinate iron(II)-porphyrinato complexes are commonly high spin, S = 2, although strong \(\sigma\)-donor \(\pi\)-acceptor ligands, such as phosphines, carbon monoxide, nitric oxide, and benzyl isocyanide, enforce a low-spin state. Five-coordinate iron(II)-porphyrinato complexes with aromatic nitrogenous axial ligands, such as pyridine or 1-methylimidazole, bind a second such axial ligand 10 to 30 times more avidly than the first to give the thermodynamically and kinetically (d , S = 0) stable hemochrome species, a process that is avoided by hemoglobins. That is, the equilibrium constant for the following disproportionation reaction is greater than unity, \[Fe—N + Fe—N \rightleftharpoons N—Fe—N +Fe \tag{4.33}\] except for bulky ligand N, such as 2-methylimidazole and 1,2-dimethylimidazole, for which the five-coordinate species predominates at room temperature even with a mild excess of ligand: \(\tag{4.34}\) For iron(III)-porphyrinato complexes, strong-field ligands lead to low-spin (S = \(\frac{1}{2}\)) complexes. A pair of identical weak-field ligands, such as tetrahydrofuran, leads to intermediate-spin (S = \(\frac{3}{2}\)) species. Five-coordinate species are, with few exceptions, high-spin (S = \(\frac{5}{2}\)), with all five 3d electrons in separate orbitals. Spin equilibria S = \(\frac{1}{2} \rightleftharpoons\) S = \(\frac{5}{2}\) and S = \(\frac{3}{2} \rightleftharpoons\) S = \(\frac{5}{2}\) are not unusual. Specific examples of these spin systems are given in Table 4.4. Higher oxidation states are found in some other hemoproteins. Fe(V)-porphyrin systems actually occur as Fe(IV)-porphyrin cation radical species, and Fe(I)-porphyrin systems exist as Fe(II)-porphyrin anion radical species. Substantial structural changes occur upon the addition of ligands and upon changes in spin state. In one mechanism of cooperativity these changes are the "trigger" (metrical details are deferred until the next section). Spectral changes in the UV-visible region are observed also (Figure 4.17) and may be monitored conveniently to evaluate the kinetic and thermodynamic parameters of ligand binding to hemoglobin. a) Could be placed in Fe column b) Could be placed in Fe column with spin = 0. c) Non-linear Fe—NCS moiety.

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All organic and biological compounds contain carbon and hydrogen, usually with various other elements as well. Hydrocarbons are compounds containing only carbon and hydrogen, but no other types of atoms. Since all organic compounds contain carbon and hydrogen, looking at hydrocarbon spectra will tell us what peaks are due to the basic C&H part of these molecules. It is sometimes useful to think of the C&H part of a molecule as the basic skeleton or scaffolding used to construct the molecule. The other atoms often form more interesting and active features, like the doors, windows and lights on a building. The simplest hydrocarbons contain only single bonds between their carbons, and no double or triple bonds. These hydrocarbons are variously referred to as saturated hydrocarbons, paraffins or alkanes. Examples of alkanes include hexane and nonane. (You can take a look at the to see what these names tell you about the structure.) Look at the IR spectrum of hexane. You should see: If you look at an IR spectrum of any other alkane, you will also see peaks at about 2900 and 1500 cm . The IR spectra of many organic compounds will show these peaks because the compound may contain paraffinic parts in addition to parts with other elements in them. ,

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The biological properties of molecules depends strongly on their polarity, because and play a large part in "noncovalent attractions" between molecules. Noncovalent attractions which are responsible for the DNA double helix, and antibody-antigen bonding can be understood in terms of alone, but sometimes the polarity of the whole molecule must be considered. For example, the suitability of water for biogenesis depends on the polarity of the whole molecule, and often involves the critically important "hydrophobic effect". The polarity of the molecule leads to extensive "hydrogen bonding" between molecules: The hydrophobic effect accounts for the insolubility of nonpolar molecules, like CH . Note that nonpolar molecules may contain polar bonds! The goal of this section is to differentiate between and . To do that, we start with important biological, chemical, and physical properties of a few small molecules: First, look at the melting points and boiling points of substances in the table below. Surprisingly, they are not related in any way to the molecular weights. This makes sense if you remember that molecular weight is a nuclear property. Melting and boiling involve breaking bonds between molecules, so like all bonding, they involve electronic structure. They depend on the size of atoms and molecules, but by size we mean , not . Second, notice that the small water molecule is exceptional. It has the largest dipole moment, and a boiling point that is over 100°C higher than molecules of similar size. Molecules of very low boiling points have zero polarity, indicated by the molecular dipole moment in debyes The molecular dipole moments determine all the biological properties mentioned above, so we must explain why water is so polar, and methane, CH , is a , even though it contains . First, let's look at the water molecule. The O atom in H O is surrounded by four electron pairs, two bonded to H atoms and two lone pairs. Oxygen has a higher electronegativity (3.34) than hydrogen (2.2), so we have two dipole bond vectors pointing from H to O. There are also two lone pairs on O, enhancing its negative charge, so there is a resultant dipole.   The polarity of water leads to hydrogen bond networks in ice or , which leads to destruction of plant tissue when water freezes. Snowflake crystals have shapes dictated by the angles between hydrogen bonded water molecules in Figure 1. Plants have several mechanisms to avoid ice damage. More important ramifications of water's polarity are mentioned later. If water were linear, the dipole moment vectors of the two bonds would cancel each other by vector addition, as shown in Figure 5(a) below: When more than one polar bond is present in the same molecule, the polarity of one bond may cancel that of another. Thus the presence of polar bonds in a polyatomic molecule does not that the molecule as a whole will have a dipole moment. In such a case it is necessary to treat each polar bond mathematically as a and represent it with an arrow. The of such an arrow shows how large the bond dipole moment is, while the of the arrow is a line drawn from the positive to the negative end of the bond. Adding the individual bond dipole moments as vectors will give the overall molecular dipole moment. Other arrangements of equivalent bonds that give zero dipole moment in this way are shown in Figure 5. The case of three equal bonds 120° apart, or a tetrahedral arrangement of equal bonds at 109° bond angles. Any combination of these arrangements will also be nonpolar. The molecule PF for example, is nonpolar since the bonds are arranged in a trigonal bipyramid, as discussed in "The Shapes of Molecules." Since three of the five bonds constitute a trigonal arrangement, they will have no resultant dipole moment. The remaining two bonds have equal but opposite dipoles which will likewise cancel. If we replace any of the bonds shown in Figure 5 with a different bond, or with a lone pair, the vectors will no longer cancel and the molecule will have a resultant dipole moment. In methane, there are four C - H bonds, all slightly polar (the electronegativities of C and H are 2.55 and 2.20 respectively), but they are arranged so that the vectors cancel one another because methane is tetrahedral as shown in the Jmol model below. You may click the Magnetic Dipole check box to see the dipole in place on the model of methane. Because methane is nonpolar, it does not dissolve in water. Other hydrocarbons and oils, like vegetable oil, can be made to dissolve by adding "amphiphilic" molecules (like soap), which have a polar ("hydrophilic", water loving) end that dissolves in water, and a nonpolar (hydrophobic, "water fearing") end that dissolves the oil. Amphiphilic molecules often make up cell walls, where the hydrophillic end of the molecule can face outward: As another example of this vector addition, consider the SO molecule. The dipole moments of the three S―O bonds are represented by the three arrows in Figure 6 below, at "1:00 o'clock", "5:00 o'clock", and "9:00 o'clock". The sum of vectors at "2:00 o'clock" and "5:00 o'clock" may be obtained by the parallelogram law—lines parallel to the original vectors intersect at the tip of their resultant, a horizontal line ending at "3:00 o'clock". The resultant is exactly equal in length and exactly opposite in direction to the bond dipole at "9:00 o'clock". Therefore the net result is zero dipole moment. Note, however, that the melting point and boiling point of SO are very high for a nonpolar substance. This is because it does not exist as a simple molecule in the solid state , but rather as a covalently bonded trimer (Figure 7), so covalent bonds need to be broken to melt or boil SO . At first glance, you might expect NH to have a zero molecular dipole moment, like SO . But in NH or NF , the N atom is surrounded by four electron pairs in an approximately tetrahedral arrangement. Since all four pairs are not equivalent, the molecule is polar. This option will not work correctly. Unfortunately, your browser does not support inline frames. In NF , the dipole moment, though, is surprisingly small because the lone pair on N cancels much of the polarity of the N―F bonds, which place electron density on the more electronegative (4.0) F atoms. Note that polar NH is soluble in water, because each has charges which can attract one another, but nonpolar CH is insoluble in water. The insolubility of CH is due to the fact that it has no charges for water to attract, but also because water bonds to itself so well. A full explanation requires the concept of entropy, because it's a rearrangement of the liquid water structure, shown in the first figure, which prevents nonpolar molecules from dissolving. This is called the "hydrophobic effect", and it has a lot to do with formation of cell walls and other biological structures like the lipid bilayers shown above, as well as forcing proteins into specific conformations. The "structure creating" results of the hydrophobic effect can be seen with the "Magic Sand" model shown in the second Figure, and in several YouTube videos. Which of the following molecules are polar? About how large a dipole moment would you expect for each? (a) CF ; (b) CHF ; (c) CH F; (d) CH F .; VSEPR theory predicts a tetrahedral geometry for CF . Since all four bonds are the same, this molecule corresponds to Figure 2 . It has zero dipole moment. This option will not work correctly. Unfortunately, your browser does not support inline frames. For CHF , VSEPR theory again predicts a tetrahedral geometry. However, all the bonds are not the same, and so there must be a resultant dipole moment. The C―H bond is essentially nonpolar, but the three C―F bonds are very polar and negative on the F side. Thus the molecule should have quite a large dipole moment: This option will not work correctly. Unfortunately, your browser does not support inline frames. The resultant dipole is shown in color. This second jmol demonstrates the dipole when one hydrogen atom in methane is replaced by a highly electronegative fluorine. Fluorine pulls the electron strongly away from the carbon atom, creating a dipole moment pointing from carbon to fluorine. The third jmol depicts the dipole moment when two hydrogen atoms in methane have been replaced by fluorine atoms. Each fluorine pulls the electrons in the carbon-fluorine bonds away from the carbon, creating a net dipole pointing between the two fluorine atoms.

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To understand ionic reactions, we need to be able to recognize whether a particular reagent will act to or to . Reagents that acquire an electron pair in chemical reactions are said to be ("electron-loving"). We can picture this in a general way as a heterolytic bond breaking of compound \(X:Y\) by an electrophile \(E\) such that \(E\) becomes bonded to \(Y\) by the electron pair of the \(XY\) bond. Thus Reagents that donate an electron pair in chemical reactions are said to be ("nucleus loving"). Thus the \(X:Y\) bond also can be considered to be broken by the nucleophile \(Nu:\), which donates its electron pair to \(X\) while \(Y\) leaves as \(Y:^\ominus\) with the electrons of the \(X:Y\) bond: Thus, by definition, electrophiles are electron-pair acceptors and nucleophiles are electron-pair donors. These definitions correspond closely to definitions used in the generalized theory of acids and bases proposed by G. N. Lewis (1923). According to Lewis, an acid is any substance that can accept an electron pair, and a base is any substance that can donated an electron pair to form a covalent bond. Therefore acids must be electrophiles and bases must be nucleophiles. For example, the methyl cation may be regarded as a , or an electrophile, because it electrons from reagents such as chloride ion or methanol. In turn, because chloride ion and methanol donate electrons to the methyl cation they are classified as , or nucleophiles: The generalized Lewis concept of acids and bases also includes common proton-transfer reactions.\(^1\) Thus water acts as a base because one of the electron pairs on oxygen can abstract a proton from a reagent such as hydrogen fluoride: Alternatively, the hydronium ion (\(H_3O^\oplus\)) is an acid because it can accept electrons from another reagent (e.g., fluoride ion) by donating a proton. A proton donor can be classified as an electrophile and a proton acceptor as a nucleophile. For example, hydrogen chloride can transfer a proton to ethene to form the ethyl cation. Therefore hydrogen chloride functions as the electrophile, or acid, and ethene functions as the nucleophile, or base: What then is the difference between an acid and an electrophile, or between a base and nucleophile? No great difference until we try to use the terms in a sense. For example, if we refer to acid strength, or acidity, this means the position of in an acid-base reaction. The equilibrium constant \(K_a\) for the dissociation of an acid \(HA\), or the \(pK_a\), is a quantitative measure of acid strength. The larger the value of \(K_a\) or the smaller the \(pK_a\), the stronger the acid. A summary of the relationships between \(K_a\) and \(pK_a\) follow, where the quantities in brackets are concentrations: or By definition, \(-\text{log} \: K_a = pK_a\) and \(-\text{log} \: \left[ H_3O^\oplus \right] = pH\); hence or However, in referring to the strength of reagents as electrophiles or nucleophiles we usually are not referring to chemical equilibria but to . A good nucleophile is a reagent that reacts rapidly with a particular electrophile. In contrast, a poor nucleophile reacts only slowly with the same electrophile. Consequently, it should not then be taken for granted that there is a parallel between the acidity or basicity of a reagent and its reactivity as an electrophile or nucleophile. For instance, it is incorrect to assume that the strengths of a series of bases, \(B:\), in aqueous solution will parallel their nucleophilicities toward a carbon electrophile, such as methyl chloride: In what follows we will be concerned with the rates of ionic reactions under conditions. We shall use the term repeatedly and we want you to understand that a nucleophile is any neutral or charged reagent that supplies a pair of electrons, either bonding or nonbonding, to form a new covalent bond. In substitution reactions the nucleophile usually is an anion, \(Y:^\ominus\), or a neutral molecule, \(Y:\) or \(HY:\). The operation of each of these is illustrated in the following equations for reactions of the general compound \(RX\) and some specific examples: An is any neutral or charged reagent that accepts an electron pair (from a nucleophile) to form a new bond. In the preceding substitution reactions, the electrophile is \(RX\). The electrophile in other reactions may be a carbon cation or a proton donor, as in the following examples: \(^1\)The concept of an acid as a and a base as a is due to Bronsted and Lowry (1923). Previous to this time, acids and bases generally were defined as substances that functioned by forming \(H^\oplus\) or \(OH^\ominus\) in water solutions. The Bronsted-Lowry concept was important because it liberated acid-base phenomena from the confines of water-containing solvents by focusing attention on rather than the formation of \(H^\oplus\) or \(OH^\ominus\). The Lewis concept of generalized acids and bases further broadened the picture by showing the relationship between proton transfers and reactions where an electron-pair acceptor is transferred from one electron-pair donor to another. and (1977)

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Symmetrical reagents do not differentiate between the members of a pair of enantiomers for the same reason that an ordinary sock fits equally well on a right foot as on a left foot. However, asymmetric or chiral reagents can differentiate between enantiomers, especially by having at least some difference in reactivity toward them. A good analogy is the comparison between the ease of putting a left shoe on a left foot and a left shoe on a right foot. The difference may not be very pronounced for simple compounds with only one or two chiral centers, but generally the larger and more complex that chiral reagent becomes, the greater is its selectivity or power to discriminate between enantiomers and diastereomers as well. The property of being able to discriminate between diastereomers is called , and this is an especially important characteristic of biological systems. For example, our ability to taste and smell is regulated by chiral molecules in our mouths and noses that act as receptors to "sense" foreign substances. We can anticipate, then, that enantiomers may interact differently with the receptor molecules and induce different sensations. This appears to be the case. The two enantiomers of the amino acid, leucine, for example, have different tastes - one is bitter, whereas the other is sweet. Enantiomers also can smell different, as is known from the odors of the two carvones. One has the odor of caraway and the other of spearmint. Some animals, and especially insects, rely on what amounts to a "sense-of-smell" for communication with others of their species. Substances synthesized by a particular species, and used to send messages in this way, are called . Many of these substances have rather simple molecular structures because they must be reasonably volatile and yet they are remarkably specific in the response they induce. When stereoisomerism is possible, usually only one isomer is effective. The sex attractant of the silkworm moth has been identified as -10- -12-hexadecaden-1-ol, \(30\), familiarly known as "bombykol," and that of the gypsy moth is 2-methyl- -7-epoxy-octadecane, \(31\), or "disparlure": There is hope that insect sex lures can be used to disrupt the mating pattern of insects and thereby control insect population. This approach to pest control has important advantages over conventional insecticides in that the chemical lures are specific for a particular species; also they are effective in remarkably low concentrations and are relatively nontoxic. There are problems, however, not the least of which is the isolation and identification of the sex attractant that is produced by the insects only in minute quantities. Also, synergistic effects are known to operate in several insect species such that not one but several pheromones act in concert to attract the opposite sex. Two notable pests, the European corn borer and the red-banded leaf roller, both use -11-tetradecenyl ethanoate, \(32\), as the primary sex attractant, but the pure cis isomer is ineffective unless a small amount of trans isomer also is present. The optimum amount appears to be between \(4\%\) and \(7\%\) of the trans isomer. We shall discuss many other examples of biological stereospecificity in later chapters. and (1977)

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Recall that the entropy of a system, \(S\), can be calculated if the partition function, \(Q\), is known: \[S=\frac{U}{T}+k_B \ln{Q} \nonumber \] where \(Q\) is: \[Q=\sum_j{e^{-\frac{E_j}{kT}}} \nonumber \] The internal energy of the system can also be calculated from the partition function: \[U=k_BT^2\left(\frac{\delta \ln{Q}}{\delta T}\right) \nonumber \] Combining these equations, we obtain: \[S=k_B\ln{Q}+k_BT\left(\frac{\delta \ln{Q}}{\delta T}\right) \nonumber \] The third law of thermodynamics states that entropy of a perfect crystal at absolute zero is zero. We can show that our equation for entropy in terms of the partition function is consistent with the third law. Plugging in our equation for the partition function into our equation for entropy, we obtain: \[S=k_B\ln{\sum_j{e^{-\frac{E_j}{kT}}}}+k_BT\left(\frac{\delta \ln{\sum_j{e^{-\frac{E_j}{kT}}}}}{\delta T}\right) \nonumber \] As \(T\) goes to zero, all the states will be in their lowest energy configuration. If the ground state is \(n\) degenerate, then \(n\) states will be in the ground state: \[E_1=E_2=...=E_m \nonumber \] This is true for all the next states. For example, if the next state is \(m\) degenerate, then: \[E_{n+1}=E_{n+2}=...=E_{n+m} \nonumber \] This gives us the result: \[\sum_j{e^{-\frac{E_j}{kT}}} = ne^{-\frac{E_1}{kT}}+me^{-\frac{E_{n+1}}{kT}}+... \nonumber \]

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In a previous section on we looked at essential minerals that are sometimes called , but are really only usable to our bodies when they are supplied as . Both potassium and iodine are essential nutrients, but each must be supplied in a compound if it is to be of any use (let alone nontoxic) to the body. Iodine deficiency leads to thyroid dysfunction and goiter, but very little iodine is required, so goiter is rare. Potassium deficiency (hypokalemia, after kallium, the German name for potassium and source of the symbol, K) leads to muscle weakness, cramps, and constipation . The Recommended Daily Allowances for iodine and potassium are 150 µg and 4700 mg respectively, so we need about 30,000 times as much potassium as iodine. Luckily, most foods (especially oranges, potatos, and bananas) supply potassium, and KI is only necessary as therapy.   As we saw previously, the element iodine I is a toxic, purple crystalline solid, while the nutrient must be supplied as part of a compound, like potassium iodide (KI). Iodine has a melting point of just 113 C, where it also vaporizes significantly, and it boils at just 184.3 C. Iodine has a high density of 4.93 g/cc. We'll see below that KI has entirely different properties. Similarly, the element potassium (K) is a metal with the lowest density (0.89 g/cc) of any metal except lithium. It cuts like butter, has a melting point of only 63.38 C and a boiling point of 759 C. In the solid, K atoms are arranged in the lattice shown below. Potassium reacts explosively with water as shown in the video below or on : Body centered cubic jmol: SID10534500 Because of the toxicity and reactivity of the element potassium, nutrient sources must contain the potassium in a compound like potassium iodide, KI, where it has entirely different properties. Potassium iodide is a white crystalline solid that looks and tastes like table salt (NaCl), dissolves in water, has a melting point of 631 C, and a boiling point of 1330 C. It's used, along with sodium iodide (NaI) to "iodize" salt. Violet, solid iodine (I ) White crystalline potassium iodide Here we have three views of the same reaction: At the top, the macroscopic appearance of the reactants and products; below them, the microscopic or atomic level representation of the atoms or molecules; and finally, a symbolic represenation in the form of a chemical equation. The solid state of the reactants is indicated by the "s" in parentheses, and the crystalline state of the product is indicated by "(c)". Liquids are designated by (l), gases by (g), and aqueous (water) solutions as (aq). This equation may be interpreted microscopically to mean that 1 potassium atom and 1 iodine molecule react to form 2 potassium iodide units, but there is no such thing as an isolated KI unit. Rather, there's an extended lattice of alternating K ions and I . When the potassium and iodine are heated together, they react even more energetically than potassium and water, to give the KI according to the equation above. This illustrates , which states that atoms are the units of chemical changes. Notice that there are just as many potassium atoms after the reaction as there were before the reaction. The same applies to iodine atoms. Atoms were neither created, destroyed, divided into parts, or changed into other kinds of atoms during the chemical reaction. The balanced chemical equation reinforces this idea. The view of solid KI shown above is our first microscopic example of a . A compound is made up of two (or more) different of atoms. Since these atoms may be rearranged during a chemical reaction, the compound can be decomposed into two (or more) different elements. The 1:1 ratio of K atoms to I atoms implied by the formula KI (subscript "1"s are assumed, so KI = K I ) agrees with Dalton’s fourth postulate that atoms combine in the ratio of small whole numbers.

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Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio. Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10 m), while radio waves can be several hundred meters in length. The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of , called , rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as: \[E = \dfrac{hc}{\lambda} \tag{4.1.1}\] where E is energy in kJ/mol, (the Greek letter ) is wavelength in meters, is 3.00 x 10 m/s (the speed of li mber known as . Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s . When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression: \[\lambda \nu = c \tag{4.1.2}\] where (the Greek letter ‘ ) is frequency in s . Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 10 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the . (Image from ) Notice that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest. Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kJ/mol of photons? In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed. Here is the key to molecular spectroscopy: Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of E, the molecule will specifically absorb radiation with wavelength that corresponds to E, while allowing other wavelengths to pass through unabsorbed. By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure. These generalized ideas may all sound quite confusing at this point, but things will become much clearer as we begin to discuss specific examples.

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The establishes very important relationships between the statistics for two distributions that are related in a particular way. It enables us to understand some important features of physical systems. The central limit theorem concerns the distribution of averages. If we have some original distribution and sample it three times, we can calculate the average of these three data points. Call this average \(A_{3,1}\). We could repeat this activity and obtain a second average of three values, \(A_{3,2}\). We can do this repeatedly, generating averages \(A_{3,3}\), , \(A_{3,n}\). Several things will be true about these averages: There is nothing unique about averaging three values. We could sample the original distribution seven times and compute the average of these seven values, calling the result \(A_{7,1}\). Repeating, we could generate averages \(A_{7,2}\), , \(A_{7,m}\). All of the things we say about the averages-of-three are also true of these averages-of-seven. However, we can now say something more: The distribution of the \({\mathrm{A}}_{\mathrm{7,i}}\) will be less spread out than the distribution of the \({\mathrm{A}}_{\mathrm{3,i}}\). The corresponding probability density functions are sketched in Figure 13. The central limit theorem relates the mean and variance of the distribution of averages to the mean and variance of the original distribution: \(\boldsymbol{N}\) \({\boldsymbol{\sigma }}^{\boldsymbol{2}}\) \(\boldsymbol{N}\) \({\boldsymbol{A}}_{\boldsymbol{N}}\) \(\boldsymbol{\sigma }^{\boldsymbol{2}}/\boldsymbol{N}\) \(\boldsymbol{N}\) It turns out that the number, \(N\), of trials that is needed to get a good estimate of the variance is substantially larger than the number required to get a good estimate of the mean.

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The formation of hydrogen ions (hydroxonium ions) and hydroxide ions from water is an endothermic process. Using the simpler version of the equilibrium: \[ H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}\] Hence, the forward reaction, as written, "absorbs heat". According to , if you make a change to the conditions of a reaction in dynamic equilibrium, the position of equilibrium moves to counter the change you have made. Hence, if you increase the temperature of the water, the equilibrium will move to lower the temperature again. It will do that by absorbing the extra heat. That means that the forward reaction will be favored, and more hydrogen ions and hydroxide ions will be formed. The effect of that is to increase the value of \(K_w\) as temperature increases. The table below shows the effect of temperature on \(K_w\). For each value of \(K_w\), a new pH has been calculated. It might be useful if you were to check these pH values yourself. You can see that the pH of pure water decreases as the temperature increases. Similarly, the pOH also decreases. If the pH falls as temperature increases, this does mean that water becomes more acidic at higher temperatures. A solution is acidic if there is an of hydrogen ions over hydroxide ions (i.e., pH < pOH). In the case of pure water, there are always the same concentration of hydrogen ions and hydroxide ions and hence, the water is still neutral (pH = pOH) - even if its pH changes. The problem is that we are all familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that to calculate the neutral value of pH from \(K_w\). If that changes, then the neutral value for pH changes as well. At 100°C, the pH of pure water is 6.14, which is "neutral" on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14. Similarly, you can argue that a solution with a pH of 7 at 0°C is slightly acidic, because its pH is a bit lower than the neutral value of 7.47 at this temperature. Hence, there is an excess of \(H^+\) ions vs. \(OH^-\) ions. Jim Clark ( )

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The economies of the highly industrialized nations of the world are based in large part on energy and chemicals produced from petroleum. Although the most important and versatile intermediates for conversion of petroleum to chemicals are compounds with double or triple bonds, it also is possible to prepare many valuable substances by . In such substitutions, a hydrogen is removed from a carbon chain and another atom or group of atoms becomes attached in its place. A simple example of a substitution reaction is the formation of chloromethane and chlorine: \[ \ce{CH_4 + Cl_2 \rightarrow CH_3Cl + HCl}\] The equation for the reaction is simple, the ingredients are cheap, and the product is useful. However, if we want to decide in advance whether such a reaction is actually feasible, we have to know more. Particularly, we have to know whether the reaction proceeds in the direction it is written and, if so, whether conditions can be found under which it proceeds at a convenient rate. Obviously, if one were to mix methane and chlorine and find that, at most, only \(1 \%\) conversion to the desired product occurred and that the \(1 \%\) conversion could be achieved only after a day or so of strong heating, this reaction would be both too unfavorable and too slow for an industrial process. One way of visualizing the problems involved is with energy diagrams, which show the energy in terms of some arbitrary that is a measure of progress between the initial and final states (Figure 4-4). Diagrams such as Figure 4-4 may not be familiar to you, and a mechanical analogy may be helpful to provide better understanding of the very important ideas involved. Consider a two-level box containing a number of tennis balls. An analog to an energetically favorable reaction would be to have all of the balls on the upper level where any disturbance would cause them to roll down to the lower level under the influence of gravity, thereby losing energy. If the upper level is modified and a low fence added to hold the balls in place, it will be just as energetically favorable as when the fence is not there for the balls to be at the lower level. The difference is that the process will not occur without some major disturbance. We can say there is an to occurrence of the favorable process. Now, if we shake the box hard enough, the balls on the upper level can acquire enough energy to bounce over the barrier and drop to the lower level. The balls then can be said to acquire enough to surmount the barrier. At the molecular level, the activation energy must be acquired either by collisions between molecules as the result of their thermal motions, or from some external agency, to permit the reactants to get over the barrier and be transformed into products. We shortly will discuss this more, but first we wish to illustrate another important concept with our mechanical analogy, that of and . With gentle shaking of our two-level box, all of the balls on the upper level are expected to wind up on the lower level. There will not be enough activation to have them go from the lower to the upper level. In this circumstance, we can say that the balls are not equilibrated between the lower and upper levels. However, if we shake the box and , no matter whether we start with all of the balls on the lower or upper level, an will be set up with, on the average, most of the balls in the energetically more favorable lower level, but some in the upper level as well. To maintain a constant average fraction of the balls at each level with vigorous and continued shaking, the at which balls go from the upper to the lower level must be equal to the that they go in the opposite direction. The balls now will be between the two levels. At equilibrium, the fraction of the balls on each of the two levels is wholly independent of the height of the barrier, just as long as the activation (shaking) is sufficient to permit the balls to go ways. The diagrams of Figure 4-4 are to be interpreted in the same general way. If thermal agitation of the molecules is sufficient, then equilibrium can be expected to be established between the reactants and the products, whether the overall reaction is energetically favorable (left side of Figure 4-4) or energetically unfavorable (right side of Figure 4-4). But as with our analogy, when equilibrium is established we expect the major portion of the molecules to be in the more favorable energy state. What happens when methane is mixed with chlorine? No measurable reaction occurs when the gases are mixed and kept in the dark at room temperature. Clearly, either the reaction is energetically unfavorable or the energy barrier is high. The answer as to which becomes clear when the mixture is heated to temperatures in excess of \(300^\text{o}\) or when exposed to strong violet or ultraviolet light, whereby a rapid or even explosive reaction takes place. Therefore the reaction is energetically favorable, but the activation energy is greater than can be attained by thermal agitation alone at room temperature. Heat or light therefore must initiate a pathway for the reactants to be converted to products that has a low barrier or activation energy. Could we have predicted the results of this experiment ahead of time? First, we must recognize that there really are several questions here. Could we have decided whether the reaction was energetically favorable? That the dark reaction would be slow at room temperature? That light would cause the reaction to be fast? We consider these and some related questions in detail because they are questions and the answers to them are relevant in one way or another to the study of reactions in organic chemistry. Presumably, methane could react with chlorine to give chloromethane and hydrogen chloride, or chloromethane could react with hydrogen chloride to give methane and chlorine. If conditions were found for which both reactions proceeded at a finite rate, equilibrium finally would be established when the rates of the reactions in each direction became equal: \[ \ce{CH_4 + Cl_2 \rightleftharpoons CH_3Cl + HCl}\] At equilibrium, the relationship among the amounts of reactants and products is given by the equilibrium constant expression \[K_{eq} = \dfrac{[CH_3Cl,HCl]}{[CH_4,Cl_2]} \label{4-1}\] in which \(K_\text{eq}\) is the equilibrium constant. The quantities within the brackets of Equation \(\ref{4-1}\) denote either concentrations for liquid reactants or partial pressures for gaseous substances. If the equilibrium constant \(K_\text{eq}\) is \(1\), then on mixing equal volumes of each of the participant substances (all are gases above \(-24^\text{o}\)), reaction to the will be initially faster than reaction to the left, until equilibrium is established; at this point there will be more chloromethane and hydrogen chloride present than methane and chlorine. However, if the equilibrium constant were \(1\), the reaction initially would proceed faster to the and, at equilibrium, there would be more methane and chlorine present than chloromethane and hydrogen chloride.\(^4\) For methane chlorination, we know from experiment that the reaction goes to the right and that \(K_\text{eq}\) is much greater than unity. Naturally, it would be helpful in planning other organic preparations to be able to estimate \(K_\text{eq}\) in advance. It is a common experience to associate chemical reactions with equilibrium constants greater than one with the evolution of heat, in other words, with negative \(\Delta H^\text{0}\) values. There are, in fact, many striking examples. Formation of chloromethane and hydrogen chloride from methane and chlorine has a \(K_\text{eq}\) of \(10^{18}\) and \(\Delta H^\text{0}\) of \(-24 \: \text{kcal}\) per mole of \(CH_3Cl\) formed at \(25^\text{o}\). Combustion of hydrogen with oxygen to give water has a \(K_\text{eq}\) of \(10^{40}\) and \(\Delta H^\text{0} = -57 \: \text{kcal}\) per mole of water formed at \(25^\text{o}\). However, this correlation between \(K_\text{eq}\) and \(\Delta H^\text{0}\) is neither universal nor rigorous. Reactions are known that absorb heat (are endothermic) and yet have \(K_\text{eq} > 1\). Other reactions have large \(\Delta H^\text{0}\) values and equilibrium constants much less than \(1\). The problem is that the energy change that correlates with \(K_\text{eq}\) is not \(\Delta H^\text{0}\) but \(\Delta G^\text{0}\) (the so-called change of " ")\(^5\), and if we know \(\Delta G^\text{0}\), we can calculate \(K_\text{eq}\) by the equation \[ \Delta G^o =-2.303 RT \log_{10} K_{eq} \label{4-2}\] in which \(R\) is the gas constant and \(T\) is the absolute temperature in degrees Kelvin. For our calculations, we shall use \(R\) as \(1.987 \: \text{cal} \: \text{deg}^{-1} \: \text{mol}^{-1}\) and you should not forget to convert \(\Delta G^\text{0}\) to \(\text{cal}\). Tables of \(\Delta G^\text{0}\) values for formation of particular compounds (at various temperatures and states) from the elements are available in handbooks and the literature. With these, we can calculate equilibrium constants quite accurately. For example, handbooks give the following data, which are useful for methane chlorination: Combining these with proper regard for sign gives and \(\text{log} \: K_\text{eq} = -\left( -24.7 \times 1000 \right)/ \left(2.303 \times 1.987 \times 298.2 \right)\), so \(K_\text{eq} = 1.3 \times 10^{18}\). Unfortunately, insufficient \(\Delta G^\text{0}\) values for formation reactions are available to make this a widely applicable method for calculating \(K_\text{eq}\) values. The situation is not wholly hopeless, because there is a relationship between \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) that also involves \(T\) and another quantity, \(\Delta S^\text{0}\), the standard of the process: \[ \Delta G^o = \Delta H^o -T \Delta S^o \label{4-3}\] This equation shows that \(\Delta G^\text{0}\) and \(\Delta H^\text{0}\) are equal when \(\Delta S^\text{0}\) is zero. Therefore the sign and magnitude of \(T \Delta S^\text{0}\) determine how well \(K_\text{eq}\) correlates with \(\Delta H^\text{0}\). Now, we have to give attention to whether we can estimate \(T \Delta S^\text{0}\) values well enough to decide whether the \(\Delta H^\text{0}\) of a given reaction (calculated from bond energies or other information) will give a good or poor measure of \(\Delta G^\text{0}\). To decide whether we need to worry about \(\Delta S^\text{0}\) with regard to any particular reaction, we have to have some idea what physical meaning entropy has. To be very detailed about this subject is beyond the scope of this book, but you should try to understand the physical basis of entropy, because if you do, then you will be able to predict at least qualitatively whether \(\Delta H^\text{0}\) will be about the same or very different from \(\Delta G^\text{0}\). Essentially, the entropy of a chemical system is a measure of its or . Other things being the same, the more random the system is, the more favorable the system is. Different kinds of molecules have different degrees of translational, vibrational, and rotational freedom and, hence, different average degrees of molecular disorder or randomness. Now, if for a chemical reaction the degree of molecular disorder is different for the products than for the reactants, there will be a change in entropy and \(\Delta S^\text{0} \neq 0\). A spectacular example of the effect of molecular disorder in contributing to the difference between \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\) is afforded by the formation of liquid nonane, \(C_9H_{20}\), from solid carbon and hydrogen gas at \(25^\text{o}\): \[\ce{9C(s) + 10H_2(g) \rightarrow C_910_{20}(l)}\] with \(\Delta H^o = -54.7 \, kcal\) and \(\Delta S^o = 5.0 \, kcal\). Equations \(\ref{4-2}\) and \(\ref{4-3}\) can be rearranged to calculate \(\Delta S^\text{0}\) and \(K_\text{eq}\) from \(\Delta H^\text{0}\) and \(\Delta G^\text{0}\):   and \[K_{eq} = 10^{-\Delta G^o/2.303 \,RT} = 10^{-5.900/(2.303 \times 1.987 \times 298.2)} = 4.7 \times 10^{-5}\] These \(\Delta H^\text{0}\), \(\Delta S^\text{0}\), and \(K_\text{eq}\) values can be compared to those for \(H_2 + \frac{1}{2} O_2 \longrightarrow H_2O\), for which \(\Delta H^\text{0}\) is \(-57 \: \text{kcal}\), \(\Delta S^\text{0}\) is \(8.6 \: \text{e.u.}\), and \(K_\text{eq}\) is \(10^{40}\). Obviously, there is something about the entropy change from carbon and hydrogen to nonane. The important thing is that there is a great in the constraints on the atoms on each side of the equation. In particular, hydrogen molecules in the gaseous state have great translational freedom and a high degree of disorder, the greater part of which is lost when the hydrogen atoms become attached to a chain of carbons. This makes for a large \(\Delta S^\text{0}\), which corresponds to a in \(K_\text{eq}\). The differences in constraints of the carbons are less important. Solid carbon has an ordered, rigid structure with little freedom of motion of the individual carbon atoms. These carbons are less constrained in nonane, and this would tend to make \(\Delta S^\text{0}\) more positive and \(\Delta G^\text{0}\) more negative, corresponding to an increase in \(K_\text{eq}\) (Equations \(\ref{4-2}\) and \(\ref{4-3}\)). However, this is a effect on \(\Delta S^\text{0}\) compared to the enormous difference in the degree of disorder of hydrogen between hydrogen gas and hydrogen bound to carbon in nonane. Negative entropy effects usually are observed in ring-closure reactions such as the formation of cyclohexane from 1-hexene, which occur with substantial loss of rotational freedom (disorder) about the \(C-C\) bonds: There is an even greater loss in entropy on forming cyclohexane from ethene because substantially more freedom is lost in orienting three ethene molecules to form a ring: For simple reactions, with the same number of molecules on each side of the equation, with no ring formation or other unusual changes in the constraints between the products and reactants, \(\Delta S^\text{0}\) usually is relatively small. In general, for such processes, we know from experience that \(K_\text{eq}\) \(\Delta H^\text{0}\) \(-15 \: \text{kcal}\) \(\Delta H^\text{0}\) \(+15 \: \text{kcal}\). We can use this as a "rule of thumb" to predict whether \(K_\text{eq}\) should be greater or less than unity for vapor-phase reactions involving simple molecules. Some idea of the degree of success to be expected from this rule may be inferred from the examples in Table 4-5, which also contains a further comparison of some experimental \(\Delta H^\text{0}\) values with those calculated from bond energies. Suppose \(\Delta G^\text{0}\) is positive, what hope do we have of obtaining a useful conversion to a desired product? There is no simple straightforward and general answer to this question. When the reaction is reversible the classic procedure of removing one or more of the products to prevent equilibrium from being established has many applications in organic chemistry, as will be seen later. When this approach is inapplicable, a change in reagents is necessary. Thus, iodine does not give a useful conversion with 2,2-dimethylpropane, \(1\), to give 1-iodo-2,2-dimethylpropane, \(2\), because the position of equilibrium is too far to the left (\(K_\text{eq} \cong 10^{-5}\)): Alternative routes with favorable \(\Delta G^\text{0}\) values are required. Development of ways to make indirectly, by efficient processes, what cannot be made directly is one of the most interesting and challenging activities of organic chemists. To reach an understanding of why methane and chlorine do not react in the dark, we must consider the details of the reaction occurs - that is, the . The simplest mechanism would be for a chlorine molecule to collide with a methane molecule in such a way as to have chloromethane and hydrogen chloride formed directly as a result of a breaking of the \(Cl-Cl\) and \(C-H\) bonds and making of the \(C-Cl\) and \(H-Cl\) bonds (see Figure 4-5). The failure to react indicates that there must be an energy barrier too high for this mechanism to operate. Why should this be so? First, this mechanism involves a very precisely oriented "four-center" collision between chlorine and methane that would have a low probability of occurrence (i.e., a large decrease in entropy because a precise orientation means high molecular ordering). Second, it requires pushing a chlorine molecule sufficiently deeply into a methane molecule so one of the chlorine atoms comes close enough to the carbon to form a bond and yield chloromethane. Generally, to bring nonbonded atoms to near-bonding distances (\(1.2 \: \text{A}\) to \(1.8 \: \text{A}\)) requires a large expenditure of energy, as can be seen in Figure 4-6. Interatomic repulsive forces increase rapidly at short distances, and pushing a chlorine molecule into a methane molecule to attain distances similar to the \(1.77\)-\(\text{A}\) carbon-chlorine bond distance in chloromethane would require a considerable amount of compression (see Figure 4-7). Valuable information about interatomic repulsions can be obtained with space-filling models of the CPK type ( ), which have radii scaled to correspond to actual atomic interference radii, that is, the interatomic distance at the point where curves of the type of Figure 4-6 start to rise steeply. With such models, the degree of atomic compression required to bring the nonbonded atoms to within near-bonding distance is more evident than with the ball-and-stick models. It may be noted that four-center reactions of the type postulated in Figure 4-5 are encountered only rarely. If the concerted four-center mechanism for formation of chloromethane and hydrogen chloride from chlorine and methane is discarded, all the remaining possibilities are . A slow stepwise reaction is dynamically analogous to the flow of sand through a succession of funnels with different stem diameters. The funnel with the smallest stem will be the most important bottleneck and, if its stem diameter is much smaller than the others, it alone will determine the flow rate. Generally, a multistep chemical reaction will have a slow (analogous to the funnel with the small stem) and other relatively , which may occur either before or after the slow step. A possible set of steps for the chlorination of methane follows: Reactions (1) and (2) involve dissociation of chlorine into chlorine atoms and the breaking of a \(C-H\) bond of methane to give a methyl radical and a hydrogen atom. The methyl radical, like chlorine and hydrogen atoms, has one election not involved in bonding. Atoms and radicals usually are highly reactive, so formation of chloromethane and hydrogen chloride should proceed readily by Reactions (3) and (4). The crux then will be whether Steps (1) and (2) are reasonable under the reaction conditions. In the absence of some , only collisions due to the usual thermal motions of the molecules can provide the energy needed to break the bonds. At temperatures below \(100^\text{o}\), it is very rare indeed that the thermal agitation alone can supply sufficient energy to break any significant number of bonds stronger than \(30\) to \(35 \: \text{kcal mol}^{-1}\). The \(Cl-Cl\) bond energy from Table 4-3 is \(58.1 \: \text{kcal}\), which is much too great to allow bond breaking from thermal agitation at \(25^\text{o}\) in accord with Reaction (1). For Reaction (2) it is not advisable to use the \(98.7 \: \text{kcal} \: C-H\) bond energy from Table 4-3 because this is one fourth of the energy required to break all four \(C-H\) bonds ( ). More specific are given in Table 4-5, and it will be seen that to break one \(C-H\) bond of methane requires \(104 \: \text{kcal}\) at \(25^\text{o}\), which again is too much to be gained by thermal agitation. Therefore we can conclude that Reactions (1)-(4) can not be an important mechanism for chlorination of methane at room temperature. One might ask whether dissociation into ions would provide viable mechanisms for methane chlorination. Part of the answer certainly is: Not in the vapor phase, as the following thermochemical data show: Ionic dissociation simply does not occur at ordinarily accessible temperatures by collisions between molecules in the vapor state. What is needed for formation of ions is either a highly energetic external stimulus, such as bombardment with fast-moving electrons, or an ionizing solvent that will assist ionization. Both of these processes will be discussed later. The point here is that ionic dissociation is not a viable step for the vapor-phase chlorination of methane. First, we should make clear that the light does more than provide energy merely to lift the molecules of methane and chlorine over the barrier of Figure 4-4. This is evident from the fact that very little light is needed, far less than one light photon per molecule of chloromethane produced. The light could activate either methane or chlorine, or both. However, methane is colorless and chlorine is yellow-green. This indicates that chlorine, not methane, interacts with visible light. A photon of near-ultraviolet light, such as is absorbed by chlorine gas, provides more than enough energy to split the molecule into two chlorine atoms: Once produced, a chlorine atom can remove a hydrogen atom from a methane molecule and form a methyl radical and a hydrogen chloride molecule. The bond-dissociation energies of \(CH_4\) (\(104 \: \text{kcal}\)) and \(HCl\) (\(103.1 \: \text{kcal}\)) suggest that this reaction is endothermic by about \(1 \: \text{kcal}\): Use of bond-dissociation energies gives a calculated \(\Delta H^\text{0}\) of \(-26 \: \text{kcal}\) for this reaction, which is certainly large enough, by our rule of thumb, to predict that \(K_\text{eq}\) will be greater than 1. Attack of a methyl radical on molecular chlorine is expected to require somewhat more oriented collision than for a chlorine atom reacting with methane (the chlorine molecule probably should be endwise, not sidewise, to the radical) but the interatomic repulsion probably should not be much different. The net result of \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) and \(CH_3 \cdot + Cl_2 \longrightarrow CH_3Cl + Cl \cdot\) is formation of chloromethane and hydrogen chloride from methane and chlorine. Notice that the chlorine atom consumed in the first step is replaced by another one in the second step. This kind of sequence of reactions is called a because, in principle, one atom can induce the reaction of an infinite number of molecules through operation of a "chain" or cycle of reactions. In our example, chlorine atoms formed by the action of light on \(Cl_2\) can induce the chlorination of methane by the : In practice, chain reactions are limited by so-called processes. In our example, chlorine atoms or methyl radicals are destroyed by reacting with one another, as shown in the following equations: Chain reactions may be considered to involve three phases: First, must occur, which for methane chlorination is activation and conversion of chlorine molecules to chlorine atoms by light. Second, steps convert reactants to products with no net consumption of atoms or radicals. The propagation reactions occur in competition with steps, which result in destruction of atoms or radicals. Putting everything together, we can write: The chain-termination reactions are expected to be exceedingly fast because atoms and radicals have electrons in unfilled shells that normally are bonding. As a result, bond formation can begin as soon as the atoms or radicals approach one another closely, without need for other bonds to begin to break. The evidence is strong that bond-forming reactions between atoms and radicals usually are , that there is almost no barrier or activation energy required, and the rates of combination are simply the rates at which encounters between radicals or atoms occur. If the rates of combination of radicals or atoms are so fast, you might well wonder how chain propagation ever could compete. Of course, competition will be possible if the propagation reactions themselves are fast, but another important consideration is the fact that the . Suppose that the concentration of \(Cl \cdot\) is \(10^{-11} \: \text{M}\) and the \(CH_4\) concentration \(1 \: \text{M}\). The probability of encounters between two \(Cl \cdot\) atoms will be proportional to \(10^{-11} \times 10^{-11}\), and between \(CH_4\) and \(Cl \cdot\) atoms it will be \(10^{-11} \times 1\). Thus, other things being the same, \(CH_4 + Cl \cdot \longrightarrow CH_3 \cdot + HCl\) (propagation) would be favored over \(2Cl \cdot \longrightarrow Cl_2\) (termination) by a factor of \(10^{11}\). Under favorable conditions, the methane-chlorination chain may go through 100 to 10,000 cycles before termination occurs by radical or atom combination. Consequently the efficiency (or ) of the reaction is very high in terms of the amount of chlorination that occurs relative to the amount of the light absorbed. The overall rates of chain reactions usually are slowed very much by substances that can combine with atoms or radicals and convert them into species incapable of participating in the chain-propagation steps. Such substances are called , or . Oxygen acts as an inhibitor in the chlorination of methane by rapidly combining with a methyl radical to form the comparatively stable (less reactive) peroxymethyl radical, \(CH_3OO \cdot\). This effectively terminates the chain: To a considerable degree, we can predict reactivities, provided we use common sense to limit our efforts to reasonable situations. In the preceding section, we argued that reactions in which atoms or radicals combine can well be expected to be extremely fast because each entity has a potentially bonding electron in an outer unfilled shell, and bringing these together to form a bond does not require that other bonds be broken: The difference between the average energy of the reactants and the energy of the transition state is called the (Figure 4-4). We expect this energy to be smaller (lower barrier) if a weak bond is being broken and a strong bond is being made. The perceptive reader will notice that we are suggesting a parallel between reaction rate and \(\Delta H^\text{0}\) because \(\Delta H^\text{0}\) depends on the difference in strengths of the bonds being broken and formed. Yet previously ( ), we pointed out that the energy barrier for a reaction need bear no relationship to how energetically feasible the reaction is, and this is indeed true for complex reactions involving many steps. But our intuitive parallel between rate and \(\Delta H^\text{0}\) usually works quite well for the rates of steps. This is borne out by experimental data on rates of removal of a hydrogen atom from methane by atoms or radicals (\(X \cdot\)), such as \(F \cdot\), \(Cl \cdot\), \(Br \cdot\), \(HO \cdot\), \(H_2N \cdot\), which generally parallel the strength of the new bond formed: Similarly, if we look at the \(H-C\) bond-dissociation energies of the hydrocarbons shown in Table 4-6, we would infer that \(Cl \cdot\) would remove a hydrogen most rapidly from the carbon forming the weakest \(C-H\) bond and, again, this is very much in accord with experience. For example, the chlorination of methylbenzene (toluene) in sunlight leads to the substitution of a methyl hydrogen rather than a ring hydrogen for the reason that the methyl \(C-H\) bonds are weaker and are attacked more rapidly than the ring \(C-H\) bonds. This can be seen explicitly in the \(\Delta H^\text{0}\) values for the chain-propagation steps calculated from the bond-dissociation energies of Table 4-6. The \(\Delta H^\text{0}\) of ring-hydrogen abstraction is unfavorable by \(+7 \: \text{kcal}\) because of the high \(C-H\) bond energy (\(110 \: \text{kcal}\)). Thus this step is not observed. It is too slow in comparison with the more favorable reaction at the methyl group even though the second propagation step is energetically favorable by \(-37 \: \text{kcal}\) and presumably would occur very rapidly. Use of bond-dissociation energies to predict relative reaction rates becomes much less valid when we try to compare different kinds of reactions. To illustrate, ethane might react with \(F \cdot\) to give fluoromethane or hydrogen fluoride: It is not a good idea to try to predict the relative rates of these two reactions on the basis of their overall \(\Delta H^\text{0}\) values because the nature of the bonds made and broken is too different. Faced with proposing a mechanism for a reaction that involves overall making or breaking of more than two bonds, the beginner almost invariably tries to concoct a process wherein, with a step, all of the right bonds break and all of the right bonds form. Such mechanisms, called , have three disadvantages. First, they are almost impossible to prove correct. Second, prediction of the relative rates of reactions involving concerted mechanisms is especially difficult. Third, concerted mechanisms have a certain sterility in that one has no control over what happens while they are taking place, except an overall control of rate by regulating concentrations, temperature, pressure, choice of solvents, and so on. To illustrate, suppose that methane chlorination appeared to proceed by way of a one-step concerted mechanism: At the instant of reaction, the reactant molecules in effect would disappear into a dark closet and later emerge as product molecules. There is no way to prove experimentally that all of the bonds were made and formed simultaneously. All one could do would be to use the most searching possible tests to probe for the existence of discrete steps. If these tests fail, the reaction still would not be concerted because other, still more searching tests might be developed later that would give a different answer. The fact is, once you accept that a particular reaction is concerted, you, in effect, accept the proposition that further work on its is futile, no matter how important you might feel that other studies would be regarding the factors affecting the reaction rate. The experienced practitioner in reaction mechanisms accepts a concerted mechanism for a reaction involving the breaking and making of more than two bonds as a last resort. He first will try to analyze the overall transformation in terms of discrete steps that are individually simple enough surely to be concerted and that also involves energetically reasonable intermediates. Such an analysis of a reaction in terms of discrete mechanistic steps offers many possibilities for experimental studies, especially in development of procedures for detecting the existence, even if highly transitory, of the proposed intermediates. We shall give many examples of the fruitfulness of this kind of approach in subsequent discussions. \(^4\)If calculations based on chemical equilibrium constants are unfamiliar to you, we suggest you study one of the general chemistry texts listed for supplemental reading at the end of Chapter 1. \(^5\)Many books and references use \(\Delta F^\text{0}\) instead of \(\Delta G^\text{0}\). The difference between standard Gibbs energy \(\Delta G^\text{0}\) and the Gibbs energy \(\Delta G\) is that \(\Delta G^\text{0}\) is defined as the value of the free energy when all of the participants are in standard states. The free energy for \(\Delta G\) for a reaction \(\text{A} + \text{B} + \cdots \longrightarrow \text{X} + \text{Y} + \cdots\) is equal to \(\Delta G^\text{0} - 2.303 RT \: \text{log} \: \frac{\left[ \text{X} \right] \left[ \text{Y} \right] \cdots}{\left[ \text{A} \right] \left[ \text{B} \right] \cdots}\) where the products, \(\left[ \text{X} \right], \left[ \text{Y} \right] \cdots\), and the reactants, \(\left[ \text{A} \right], \left[ \text{B} \right] \cdots\), do not have to be in standard states. We shall use only \(\Delta G^\text{0}\) in this book. \(^6\)The entropy unit \(\text{e.u.}\) has the dimensions calorie per degree or \(\text{cal deg}^{-1}\). and (1977)

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The boron family contains the semi-metal (B) and metals (Al), (Ga), (In), and (Tl). These elements are found in Group 13 (XIII) of the p block in the . Aluminum, gallium, indium, and thallium are metallic. They each have three electrons in their outermost shell (a full s orbital and one electron in the p orbital) with the valence electron configuration ns np . The boron family adopts oxidation states +3 or +1. The +3 oxidation states are favorable except for the heavier elements, such as Tl, which prefer the +1 oxidation state due to its stability; this is known as the . The elements generally follow except for certain Tl deviations: is the first element of Group 13 and is the only metalloid of the group. Its chemical symbol is B, and it has an atomic number of 5. Boron has the electron configuration [He] 2s 2p and prefers an oxidation state of +3. Boron has no natural elemental form; it forms compounds which are abundant in the Earth's crust. Boron is an essential nutrient for plants. There are a few locations where boron ores, known as borax, are found in great concentrations. Due to its lack of a complete octet, boron is a . It tends to forms hydrides, the simplest of which is diborane, \(B_2H_6\). Boron hydrides are used to synthesize organic compounds. One of the main compounds used to form other boron compounds is boric acid, which is a weak acid and is formed in the following two-step reaction: \[B_2O_{3 \;(s)} + 3 H_2O _{(l)} \rightarrow 2 B(OH)_{3 (aq)}\] \[B(OH)_{3 \;(aq)} + 2 H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + B(OH)^-_{4\; (aq)}\] Boron can be crystallized from a solution of hydrogen peroxide and borax to produce sodium perborate, a bleach alternative. The bleaching ability of perborate is due to the two peroxo groups bound to the boron atoms. is the most important metal in the boron family, with the chemical symbol Al and atomic number 13. It is used in lightweight alloys and is an active metal. It has the electron configuration [Ne] 2s 2p , and usually adopts a +3 oxidation state. This element is the most abundant metal in the Earth's crust (7.5-8.4%). Even though it is very abundant, before 1886 aluminum was considered a semiprecious metal; it was difficult to isolate due to its high melting point. Aluminum is very expensive to produce, because the electrolysis of one mole of aluminum requires three moles of electrons: \[Al^{3+} + 3e^- \rightarrow Al(l)\] Aluminum is a soft, malleable metal that is silver or gray in color. It is highly reactive, and therefore found in nature in compounds. Aluminum does not appear to react with water because it is aluminum is protected by a layer of Al O ; this effect is known as anodizing. The thickness of the layer varies based on galvanic reactions, but it prevents the metal from oxidizing further. Aluminum is used in many alloys to prevent corrosion. Aluminum can dissolve in both acids and bases—it is amphoteric. In an aqueous OH solution it produce Al(OH) , and in an aqueous H O solution it produce [Al(H O) ] Another important feature of aluminum is that it is a good reducing agent due to its +3 oxidation state. It can therefore react with acids to reduce H (aq) to H (g). For example: \[2Al (s) + 6H^+(aq) \rightarrow 2Al^{3+}(aq) + 3H_2(g)\] Aluminum can also extract oxygen from any metal oxide. The following reaction, which is known as the thermite reaction, is very exothermic: \[Fe_2O_3(s) + 2 Al(s) \rightarrow Al_2O_3(s) +2 Fe(l)\] has the chemical symbol Ga and the atomic number 31. It has the electron configuration [Ar] 2s 2p and a +3 oxidation state. The melting point is 29.8º C, slightly above room temperature. Gallium has the second lowest melting point (after mercury) and can remain in the liquid phase at a larger range of temperatures than any other substance. Gallium is industrially important because it forms gallium arsenide (GaAs), which converts light directly into electricity. Gallium is also used in conjunction with aluminum to generate hydrogen. In a process similar to the thermite reaction, aluminum extracts oxygen from water and releases hydrogen gas. However, as mentioned above, aluminum forms a protective coat in the presence of water. Combining gallium and aluminum prevents the formation of this protective layer, allowing aluminum to reduce water to hydrogen. has the chemical symbol In and the atomic number 49. It has the electron configuration [Kr] 2s 2p1 and may adopt the +1 or +3 oxidation state; however, the +3 state is more common. It is a soft, malleable metal that is similar to gallium. Indium forms InAs, which is found in photoconductors in optical instruments. The physical properties of indium include its silver-white color and the "tin cry" it makes when bent. Indium is soluble in acids, but does not react with oxygen at room temperature. It is obtained by separation from zinc ores. Indium is mainly used to make alloys, and only a small amount is required to enhance the metal strength. For example, indium is added to gold or platinum to make the metals more useful industrial tools. Thallium has the chemical symbol Tl and atomic number 81. It has the electron configuration [Xe] 2s 2p and has a +3 or +1 oxidation state. As stated above, because thallium is heavy, it has a greater stability in the +1 oxidation state (inert pair effect). Therefore, it is found more commonly in its +1 oxidation state. Thallium is soft and malleable. It is poisonous, but used in high-temperature superconductors. Because of its toxicity, thallium was widely used in insecticide and rat poison until this usage was prohibited in 1975 in the U.S. Both Be and Al are hydrated to produce [Be(H O) ] and Al(H O) , respectively. When reacted with water, both compounds produce hydronium ions, making them slightly acidic. Another similarity between aluminum and beryllium is that they are amphoteric, and their hydroxides are very basic. Both metals also react with oxygen to produce oxide coatings capable of protecting other metals from corrosion. Both metals also react with halides that can act as Lewis acids. ( highlight the blue areas to find the answers) 1. T/F In reality, aluminum forms a protective layer and does not react with water. 2. Which statement about Gallium is false? 3. T/F Thallium is highly toxic and therefore it is commonly used for rat poisons and insecticides in the United States. 4. Boron: 5. Aluminum Oxide: 6. T/F Aluminum is amphoteric 7. Which element is the only metalloid in the boron family? 8. When beryllium reacts with a halide, which of the following is true? 9. What is the electron configuration of thallium? 10. Which statement is False?

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This page looks at some aspects of manganese chemistry. It includes: two simple reactions of manganese(II) ions in solution (summarized from elsewhere on the site), and the use of potassium manganate(VII) (potassium permanganate) as an oxidizing agent - including its use in titrations. The simplest ion that manganese forms in solution is the hexaaquamanganese(II) ion - \([Mn(H_2O)_6]^{2+}\). Hydroxide ions (from, say, sodium hydroxide solution) remove hydrogen ions from the water ligands attached to the manganese ion. Once a hydrogen ion has been removed from two of the water molecules, you are left with a complex with no charge - a neutral complex. This is insoluble in water and a precipitate is formed. \[[Mn(H_2O)_6]^{2+} + 2OH^- \rightarrow [Mn(H_2O)_4(OH)_2] + 2H2O\] In the test-tube, the color changes are: I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colorless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. Ammonia can act as both a base and a ligand. In this case, at usual lab concentrations, it simply acts as a base - removing hydrogen ions from the aqua complex. \[[Mn(H_2O)_6]^{2+} + 2NH_3 \rightarrow [Mn(H_2O)_4(OH)_2] + 2NH_4^+\] Again, I have shown the original solution as the palest pink I can produce, but in fact it is virtually colorless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. There is no observable difference in appearance between this reaction and the last one. Potassium manganate(VII) (potassium permanganate) is a powerful oxidizing agent. Potassium manganate(VII) is usually used in neutral or alkaline solution in organic chemistry. Acidified potassium manganate(VII) tends to be a rather destructively strong oxidizing agent, breaking carbon-carbon bonds. The potassium manganate(VII) solution is usually made mildly alkaline with sodium carbonate solution, and the typical colour changes are: Potassium manganate(VII) oxidizes carbon-carbon double bonds, and so goes through the colour changes above. Ethene, for example, is oxidized to ethane-1,2-diol. The oxygen in square brackets is taken to mean "oxygen from an oxidizing agent". This abbreviated form of the equation is most commonly used in organic chemistry. You are very unlikely to have to write the complete ionic equation for this reaction at this level. This isn't a good test for a carbon-carbon double bond, because anything which is even mildly reducing would have the same effect on the potassium manganate(VII) solution. You could, however, use this reaction simply as a means of making the diol. Alkaline potassium manganate(VII) solution oxidizes any hydrocarbon side chain attached to a benzene ring back to a single -COOH group. Prolonged heating is necessary. For example: In the case of the ethyl side chain, you will also get carbon dioxide. With longer side chains, you will get all sorts of mixtures of other products - but in each case, the main product will be benzoic acid. Potassium manganate(VII) solution is used to find the concentration of all sorts of reducing agents. It is always used in acidic solution. For example, it oxidizes \[Fe^{2+} \rightarrow Fe_{2+} + e-\] \[ H_2O_2 \rightarrow O_2 + 2H^+ 2e^-\] \[ SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-\] In each case, the half-equation for the manganate(VII) ions in acidic solution is: \[ MnO_4^- + 8H^+ +5e^- \rightarrow Mn^{2+} + 4H_2O\] These equations can be combined to give you an overall ionic equation for each possible reaction. That, of course, also gives you the reacting proportions. For example, when the equations are combined, you find that 1 mole of MnO ions react with 5 moles of Fe ions. Having got that information, the titration calculations are just like any other ones. The potassium manganate(VII) solution always goes into the burette, and the other solution in the flask is acidified with dilute sulfuric acid. As the potassium manganate(VII) solution is run into the flask it becomes colourless. The end point is the first faint trace of permanent pink in the solution showing that there is a tiny excess of manganate(VII) ions present. There are two things you need to be aware of: Jim Clark ( )

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Henry's law is one of the gas laws formulated by William Henry in 1803 and states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid: \[ C =k P_{gas} \] where Raoult's and Henry's laws are limiting laws, generally applicable when the solute concentration goes to zero. In this limit the vapor pressure of any component in the solution depends linearly on its mole fraction, implying the absence of solute-solute interactions. Raoult's law describes the dependence of the vapor pressure of a solvent as a function of its mole fraction \(\chi_1\): \[\lim_{\chi_1\rightarrow1}\left( \frac{p}{\chi_1}\right) =p^\ast \nonumber\] where \(p^{\ast}\) is the vapor pressure of the pure solvent. Henry's law describes the dependence of the vapor pressure of a solute as a function of its concentration. In terms of mole fraction \(\chi_2\): For a binary mixture of pure substances it can be shown that the laws are complementary: if one law holds for one component then the other law holds for the other component. It can also be shown that the laws imply that the solution satisfies other criteria for ideality, including zero enthalpy and volume of mixing. What is Henry's constant for neon dissolved in water given: \(C_{Ne}=23.5 \,mL/L\) solution and STP (22,414 mL/mole gas) and pressure (1 atm)? Now we can rearrange our equation from above to solve for the constant: \[ C =k P_{Ne} \nonumber\] To use C we must convert 23.5 mL/L solution to Molarity. Since Ne is a gas, we can use our standard molar volume. Thus giving us: \[\dfrac{\text{23.5 mL/L soln}}{\text{1 mole Ne/22,414 mL}}= 0.00105\,M.\] Now we have solved for the solubility of Ne in the solution. C= 0.00105M and we know the pressure at STP is 1 atm, so we can now use our rearranged equation: \[k= \dfrac{C}{P_{Ne}} \nonumber\] Where C= 0.00105M, P = 1 atm, thus giving us k=0.00105 M/atm Compute the molar solubility in water that is saturated with air. This time, we need to use constant (k) that we just calculated and our \(P_{Ne}\) in air. \[ \begin{align*} C &=k P_{gas} \\[4pt] &=(0.00105\; M/atm)(0.0341\; atm) \\[4pt] &= 3.58 \times 10^{-5}\; M \end{align*}\]