Why is the molecular structure of water bent? Water has the formula $\ce{H2O}$ and we can draw a Lewis structure with two lone pairs on the central oxygen. As a physics student and not a chemist, I think to myself, "Okay, there are two lone pairs, they will repel each other" and so we arrive at the conclusion of bent shape.

The issue in my mind is that I don't see why the lone electron pairs have to exist on the same side of the atom. Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)? If that were true, then there would be a resonance structure between the two states and we would get a linear geometry. Clearly I'm running around in circles here, please someone enlighten me!

I mean, there is a time and place for VSEPR, and this is probably as good a time as any, because all beginning chemistry students go through it. The actual model has already been explained multiple times, so I will only briefly say that according to this theory, there are four pairs of electrons around the central oxygen. In order to minimise electron-electron repulsions, these pairs adopt a tetrahedral arrangement around the oxygen. It does not matter which two are lone pairs and which two are connected to hydrogen atoms; the resulting shape is always bent.

What's worth bearing in mind (and hasn't been explained very carefully so far) is that VSEPR is a model that chemists use to predict the shape of a molecule. The truth is that there is no real way to predict the shape of a molecule, apart from solving the Schrodinger equation, which is not analytically possible for water. Everything else is an approximation to the truth. Some of these approximations are pretty accurate, such as the use of density functional theory. Some of them are extremely crude, and VSEPR falls into this category: it essentially treats electrons as classical point charges, and seeks to minimise the electrostatic repulsion between these point charges. As a physics student you should know better than to do this. Thus, while it predicts the correct result in this case, it is more in spite of the model rather than because of the model. And you should not be surprised to hear that in some slightly more complicated cases, VSEPR can predict entirely wrong outcomes.

As you learn more chemistry you will find that there are increasingly sophisticated ways of explaining molecular geometry. Most revolve around molecular orbital theory. For a qualitative method, you have Walsh diagrams which have been explained at Why does bond angle decrease in the order H2O, H2S, H2Se?. For a more rigorous method you would likely have to run some quantum chemical computations, e.g. Are the lone pairs in water equivalent?. Of course, the drawback of this is that it becomes more and more difficult to extract true chemical understanding from the numbers. Although it should also be said that you cannot extract any true chemical understanding from the VSEPR model.

What interests me more is the followup question:

  Also, wouldn't the Schrödinger equation provide an equally plausible structure for water with the lone pairs on the opposite side of the oxygen from what we assume (imaging the electrons on the top or on the bottom of the oxygen in the Lewis structure)?

Because the Hamiltonian of the water molecule is invariant upon rotation, this means that indeed, any orientation of the water molecule is equally likely. However, this only refers to the orientation of the water molecule as a whole. It does not say anything about the internal degrees of freedom, such as the bond angle.

In the absence of any external force, the molecule is free to bend in whichever direction it likes, and most water molecules indeed do do this as they float through space or swim in a lake. But it will always be bent.

  If that were true, then there would be a resonance structure between the two states and we would get a linear geometry.

If you were to think of a single particle in a double-well potential, say something with

$$V = \begin{cases}
\infty & x < -b \\
0      & -b \le x \le -a \\
\infty & -a < x < a \\
0      & a \le x \le b \\
\infty & x > b
\end{cases}$$

then because of the symmetry of your system, in every eigenstate of your system, the expectation value of $x$ would be $\langle x \rangle = 0$. This is quite similar to your argument. In the case of water, let's set the oxygen nucleus to be at the origin. Because it can point either up or down, the expectation value of the hydrogen nucleus position along the up-down axis would be exactly level with the oxygen atom, i.e. 0. In fact, don't stop there: it can point to the left or the right, and to the front or the back. So the hydrogen nucleus has a position expectation value of exactly $(0, 0, 0)$, i.e. right inside the oxygen nucleus.

Does that mean it's actually there, though? In our contrived double-well system, it's patently impossible for the particle to be at $x = 0$, because $V = \infty$ there. If you were to measure its position, you would never find it at $x = 0$; you would only find it in the left-hand side $[-b, -a]$, or the right-hand side $[a,b]$. Just because the particle has an expectation value of $\langle x \rangle = 0$ does not mean that it is physically there, or that $x = 0$ is somehow its equilibrium state. You're confusing an expectation value with a genuine eigenstate (which is what a resonance structure is).

In exactly the same way, if you ever were to measure the properties of water (and bear in mind that practically every interaction with a water molecule is, in effect, a measurement), we would find that it is indeed always bent.

There are eight valence electrons on the oxygen that have similar energies then there is a gap and two electrons very close to oxygen core follow. These eight electrons form four spin-pairs and those point in directions that more or less minimise the repulsion. These directions are called "tetrahedral" in Chemistry language, since they point from the center (= oxygen atom core) to the vertices of an tetrahedron. Now the two protons are connected with two of those "electron pairs" and two are "alone". Hence you obtain the bent shape of H$_2$O.

(This is kind of trivialized version of the VSEPR model.)


Why does the dissolution of calcium hydroxide have a negative entropy? We had an experiment on the dissolution of calcium hydroxide and we obtained the experimental entropy value $\pu{-203 J mol^-1 K^-1}$ with a percent error of $26.7\%.$ Doesn't dissolving a solid in a solvent increase entropy?

When ionic compounds dissolve in water, water molecules form a complex ion around the calcium cation.  The presence of the hydroxide anion has a similar effect, though the polarities of the water ligands is reversed.
Thus, dissolution of Ca(OH)2 in water is actually a chemical reaction.  In some cases, complex ions may be more structured than the crystals of undissolved ions.       In such a case, the entropy change of solution will be negative.

Entropy can be measured and aproached differently, (i) from an equation that relates entropy and Enthalpy to measure the heat of reaction, the remainder of the work done per degree of Temperature (°) is
or if the change in enthalpy is measured then simply.
In this case the Vector of entropy has a negative direction hence an endothermic process and the magnitude is 203.
(ii) Considering the amount of microstates, Boltzmann constant would relate the energy and the temperature, between solid and liquid dissolution; As a function of Omega (W).
For Omega (W) evaluate the microstates in your solid sample and the liquid volume using Boltzmann Relation.
The change in Entropy between solid and liquid state is measure by Math approach, and is proven by Covention I and II (Physical Chemistry; Ira N. Levine Pg.295 6th Ed) for the Activity Coefficients on Gibbs free energy of standard solid state (S* solid 83.36 J/mol*K) and ideal model for gas state (S°gas,1 bar 285.62).
An increase in Entropy would be, then proven, as a solid State entropy is of approx. 83 J/molK and raise to 203 J/molK, through an endothermic process.
NIST Ca(OH)2 https://webbook.nist.gov/cgi/cbook.cgi?ID=C1305620&Mask=FFF


Structure of the compound C9H7O2Cl? A compound having formula $\ce{C9H7O2Cl}$   (O = oxygen) exists predominantly in enol form (A) and also in keto form (B). On oxidation with $\ce{KMnO4}$ it gives m-chlorobenzoic acid as one of the products. Identify (A) and (B).
(subjective question; no options were given)

I tried it and found the double bond equivalent as 6, 4 of which were utilized in benzene ring. Also, the benzylic carbon must have at least one hydrogen. Therefore I determined the following structure as (A):

But my book suggests (A) should be:

Please explain whether both are correct or only one product will exist.
Source: Joint Entrance Exam (JEE) 2003 India

Both the answers are correct but the most appropriate will be B as the chelation occurs because of hydrogen bonding which makes it more stable.
H-bonding is  possible in your structure but the ring is less stable
.


Is every chemical reaction in equilibrium? I read that every chemical reaction is theoretically in equilibrium in an old textbook. If this is true how can a reaction be one way?

Yes, every chemical reaction can theoretically be in equilibrium.  Every reaction is reversible.  See my answer to chem.SE question 43258 for more details.
This includes even precipitation reactions and reactions that release gases.  Equilibrium isn't just for liquids!  Multiphase equilibria exist.
The only thing that stops chemical reactions from being "in equilibrium" is the lack of the proper number of molecules.  For a reaction to be in equilibrium, the concentrations of reactants and products must be related by the equilibrium constant.
$$ \ce{ A <=> B} $$
$$ K = \frac{[B]}{[A]} $$

When equilibrium constants are extremely large or small, then extremely large numbers of molecules are required to satisfy this equation.  If $K = 10^{30}$, then at equilibrium there will be $10^{30}$ molecules of B for every molecule of A.  Another way to look at this is that for equilibrium to happen, there need to be at least $10^{30}$ molecules of B, i.e. more than one million moles of B, in the system for there to be "enough" B to guarantee an equilibrium, i.e. to guarantee that there will be a well-defined "equilibrium" concentration of A.

When this many molecules are not present, then there is no meaningful equilibrium.  For very large (or very small) equilibrium constants, it will be very difficult to obtain an equilibrium.  In addition to needing a megamole-sized system (or bigger), the system will have to be well-mixed, isothermal, and isobaric.  That's not easy to achieve on such large scales!

Update Commenters suggest that "irreversible" reactions do not have an equilibrium.  This is true, but tautological.  In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree.  To say otherwise would violate microscopic reversibility.  A reaction that was 100% irreverible would have an equilibrium constant of infinity.  But if $K= \infty$, then $\Delta G^{\circ} = -RT \ln{K}$ would turn into $\Delta G^{\circ} = -\infty$. So to get infinite energy we would just have to use 100% irreversible reactions!  Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

No, every reaction isn't in equilibrium with its products. Consider the following irreversible reaction: $$\ce{BaCl2(aq) + H2SO4(aq) -> BaSO4(ppt) + 2HCl(aq)}$$.

By definition if the reaction is irreversible then there is no equilibrium for that reaction.

If there were an "equilibrium" for the reaction then the equation
would be something like: $$\ce{K_{eq}} = \dfrac{\ce{[BaSO4][HCl]^2}}{
   \ce{[BaCl2][H2SO4]}}$$ and such an equilibrium just doesn't exist
since when the barium sulfate precipitates there could be a microgram
or a kilogram as the product.
Think of it another way - adding $\ce{HCl}$ (in dilute solution) or $\ce{BaSO4}$ won't shift the reaction to the left. (Adding more HCl would shift $\ce{HSO4^{-} <-> H+ + SO4^{2-}}$ in concentrated solutions, which is besides the point I'm trying to make.)
There is a solubility product for barium sulfate, but the solubility product doesn't depend on the amount of barium sulfate precipitate, nor the concentration of HCl. So the solubility product isn't for the overall reaction but rather for part of the system:

$$\ce{[Ba][SO4^{2-}] = K_{sp}}$$

(Full disclosure - Theoretically the barium sulfate solubility product wouldn't depend on the HCl concentration, but really that isn't quite true. The barium sulfate solubility product really depends on the activity of the barium and sulfate ions, so the ionic strength of the solution matters.)


Calculate the standard oxidation potential from given potentials The standard electrode potentials for the following redox couples are given as: >$E_0(\ce{Fe^{2+}/Fe}) = x~\mathrm{V}$ and $ E_0(\ce{Fe^{3+}/Fe^{2+}}) = y~\mathrm{V}$
  What will be potential $E_0(\ce{Fe^{3+}/Fe})$ (in $\mathrm{V}$)?

My approach: $\ce{Fe^{3+} + e- + Fe^{2+} + 2e- -> Fe^{2+} + Fe}$

So $E_0(\ce{Fe^{3+}/Fe})=(x+y)~\mathrm{V}$

You can write down the three half reactions:

\begin{aligned}
 \ce{Fe^{2+} + 2e- &-> Fe}      & E_1&= x\,\mathrm{V}\\
 \ce{Fe^{3+} + e-  &-> Fe^{2+}} & E_2&= y\,\mathrm{V}\\
 \ce{Fe^{3+} + 3e- &-> Fe}      & E_3&= z\,\mathrm{V}
\end{aligned}

The third half-reaction is simply the sum of the first two. However, we cannot just sum $E_1 + E_2$ in order to obtain $E_3$. The problem is with the meaning of the potentials, and their relation to the stoichiometry.

An half reaction with potential $E$ in volts means that the elecrons have some energy assigned to them. The meaning of volts is joules per coulomb. The charge of one mole of electrons is 96500 C (this is Faraday's constant: 1 F = 95600C/mol).

Take a look at the first half reaction. It transfers two moles of electron per mole of reaction. This means the electrons carry

$x$ joules per coulomb, or
$96500x$ joules per mole of electrons, or
$2\times 96500x$ joules per mole of reaction.

In the same fashion, the second half reaction has the electrons carrying

$y$ joules per coulomb,
$96500y$ joules per mole of electrons,
the same $96500y$ joules per mole of reaction.

The third half reaction is the sum of the first two, so its electrons carry

$2\times 96500x + 96500y$ joules per mole of reaction, which is
$3\times 96500 z$ joules per mole of reaction, or
$96500z$ joules per mole of electrons, or
$z$ joules per coulomb.

Now you can simplify this to $2x + y = 3z$ and calculate the potential you want.

Your approach is a correct one.

Explanation: (I call the given two reduction half cell reactions as 1 and 2 respectively)

The required half cell reaction $\ce{Fe^{3+} -> Fe}$ could be obtained by adding reaction 1 and 2 .Thus the net $\ce{E0}$ value would be the sum of $\ce{E0}$ values of reaction 1 and 2 (as both are reduction half cell reactions.)

Hence net $\ce{E0(Fe^{3+}/Fe^{2+})} = (x + y) \, \text{V}$


What does pO2 of blood mean and why do we use it? I understand the basic Dalton's law of partial pressures in gases. Also, Henry's law of diffusion, says, the concentration of gas dissolved in a fluid is proportional to the partial pressure above it.

So if we say that the $p(\ce{O2})$ of oxygenated blood is $\pu{100 mmHg}$, where is the free gas existing in equilibrium with dissolved gas? Does it mean that the blood has a concentration of oxygen equal to that when placed in a surrounding of $p(\ce{O2}) = \pu{100 mmHg}$? If yes, why don't we directly report in concentrations instead? Is it easier to measure?

Wikipedia also says that the Henry's law doesn't stand if the gas is reacting. But isn't oxygen reacting with the Haemoglobin?

There is a good explanation in Relating oxygen partial pressure,
saturation and content: the
haemoglobin–oxygen dissociation
curve Breathe 2015; 11: 194–201

  The partial pressure of oxygen (also known
  as the oxygen tension) is a concept which often
  causes confusion. In a mixture of gases, the
  total pressure is the sum of the contributions
  of each constituent, with the partial pressure of
  each individual gas representing the pressure
  which that gas would exert if it alone occupied
  the volume. In a liquid (such as blood), the partial
  pressure of a gas is equivalent to the partial
  pressure which would prevail in a gas phase in
  equilibrium with the liquid at the same temperature.
  With a mixture
  of gases in either the gas or
  liquid phase, the rate of diffusion of an individual
  gas is determined by the relevant gradient of its
  partial pressure, rather than by its concentration.
  While in a gas mixture, the partial pressure
  and concentration of each gas are directly proportional,
  with oxygen in blood the relationship
  is more complex because of its chemical combination
  with haemoglobin. This allows blood
  to carry an enormously greater concentration
  (content) of oxygen than, for example, water
  (or blood plasma). Measurement of $p_\ce{O_2}$, therefore,
  does not give direct information about the
  amount of oxygen carried by blood.

So blood $p_\ce{O_2}$ does not correspond to a particular concentration of oxygen, because the concentration of haemoglobin can vary, and most of the oxygen is bound to the heme iron.

$P_\ce{O_2}$ is the partial pressure of oxygen in a hypothetical gas phase which would make the blood oxygen and gas phase oxygen be in equilibrium.

I believe medical oximeters read % saturation relative to the partial pressure of oxygen in the atmosphere. Thus a reading of 94 - 95% (normal) implies that the amount of oxygen in the blood is 94 - 95% of that which would be found were the blood in equilibrium with air. I think the reason it's done that way is for convenience. If the oximeter reads 50% the gas passer knows right away to increase the partial pressure of oxygen he is feeding the patient by 45% to get saturation back to around 95%.


Why don't free radicals undergo rearrangement like carbocation/carbonium ions do? Why don't free radicals and carbanions undergo rearrangement like carbocation/carbonium ions undergo in a reaction?

First off, your observation is correct, rearrangements involving free radicals and carbanions are noticeably less common than rearrangements involving carbocations.

Many common carbocation rearrangements involve 1,2 shifts.  If we look at the MO diagram for a 1,2 shift and compare shifting a substituent to an empty p orbital (carbocation, case A in diagram) or to a filled p orbital (carbanion, case B in diagram), we see that the carbocation process involves a 2-electron system whereas the carbanion process is a 4-electron system.  The 2-electron process is allowed to proceed themally according to orbital symmetry, but the 4-electron process would need to proceed with inversion of configuration (bonding to the other lobe of the p orbital as we transform from reactant to product) in order to be thermally allowed - and this would be geometrically difficult.

(Source - V8rik at the English Wikipedia via Wikimedia Commons)

If we consider hydrogen shifts and replace the p orbital in our digram with an s orbital, we see that a 1,2 hydrogen shift in a carbocation is thermally allowed, but the corresponding shift in a carbanion is thermally forbidden (with an s orbital, there is no "other" lobe and proceeding with "inversion" is not possible)

Since thermal 1,2 shifts with retention at the migrating center are not allowed in carbanions by orbital symmetry, and a 1,2 shift with inversion at the migrating center is geometrically difficult, such processes would involve higher energy transition states.  As a consequence of the higher energy, 1,2 shifts in carbanion systems are less common than orbital symmetry allowed, thermal 1,2  shifts involving carbocations. One might reasonably expect that a radical process would be somewhere in between these two cases; and again the transition state would be of higher energy (since it is not orbital symmetry allowed) making such rearrangements less common.

While radical and carbanion shifts are not as common as carbocation shifts, they do occur.  In the case of radicals it requires high temperatures (hydrocarbon cracking) or weak bonds (peroxides for example). Carbanionic rearangements are more common when the carbanion is stabilized.  This can be achieved by having an electronegative heteroatom attached to the carbanionic center (as in the Wittig rearrangement) or by placing the anion on an $\ce{sp^2}$ or $\ce{sp}$ hybridized carbon, rather than an $\ce{sp^3}$ hybidized carbon (the Fritsch rearrangement provides an example here).

Free radicals cannot go into rearrangements because they are electrically neutral having only one unpaired electron .
But free radical rearrangements is possible in bi radicals and halogens because they both have loan pairs .
As halogens have loan pairs so they are electrically charged and and can attract single electron present in free radicals.


Reaction between halogens and concentrated nitric acid I have read that the reaction between $\ce{I2}$ and conc. $\ce{HNO3}$ gives $\ce{HIO3}$ and $\ce{NO2},$ but there is no mention of the other halogens' products. Do they all give halous acids and nitrogen dioxide? Specifically with chlorine, how does the reaction proceed?

This question can easily be addressed considering reduction potential of each compound. Concentrated nitric acid is a powerful oxidizing acid, which has a high reduction potential:

$$
\begin{align}
\ce{NO3- + 4H+ + 3e- &<=> NO + 2H2O} &\quad E^\circ &= \pu{0.957 V} \tag1\\
\ce{NO3- + 3H+ + 2e- &<=> HNO2 + H2O} &\quad E^\circ &= \pu{0.934 V} \tag2\\
\ce{2NO3- + 4H+ + 2e- &<=> N2O4 + 2H2O} &\quad E^\circ &= \pu{0.803 V} \tag3
\end{align}
$$

Now, let's look at reduction potential of halides:

$$
\begin{align}
\ce{F2 + 2e- &<=> 2F-} &\quad E^\circ &= \pu{2.866 V} \tag4\\
\ce{Cl2 + 2e- &<=> 2Cl-} &\quad E^\circ &= \pu{1.3583 V} \tag5\\
\ce{Br2 + 2e- &<=> 2Br-} &\quad E^\circ &= \pu{1.066 V} \tag6\\
\ce{I2 + 2e- &<=> 2I-} &\quad E^\circ &= \pu{0.5355 V} \tag7\\
\end{align}
$$
Accordingly, we can list oxidizing power at standard conditions in decreasing order:

$$\ce{F2} > \ce{Cl2} > \ce{Br2} > \ce{NO3-} > \ce{I2}$$

Therefore, at standard condition, concentrated $\ce{HNO3}$ acid (meaning $\ce{NO3-}$ in acidic medium) can easily oxidize only $\ce{I2}.$ One such example is documented in Ref.1. According to the authors, the oxidation of iodine by hyperazeotropic nitric acid ($\pu{16–21 M}$) was shown to proceed in at least two steps:

$$
\begin{align}
\ce{I2 + 3HNO3 &<=>[$k_1$][$k_2$] 2I+ + 2NO3- + 2HNO2 + H2O}\\
\ce{I+ + 6HNO3 &<=>[$k_3$][$k_1$] I^5+ + 4NO3- + 2HNO2 + 2H2O}
\end{align}
$$

Since reduction potentials of $\ce{HNO3}$ and $\ce{Br2}$ are in such close range that each can oxidize the other. In fact, the kinetic studies of this phenomenon has been performed in literature (Ref.2).

In different aspect, there is a publication (Ref.3) claiming that the presence of bromine in fuming nitric acid makes it a brominating reagent (electrophilic aromatic substitution):

This bromination has worked well with deactivated aromatic nucleus. In absence of bromine, the reagent nitrate the ring. Thus, it is safe to suggest that $\ce{HNO3}$ might promote formation of '$\ce{Br+}$', meaning oxidation.

References:

J. C. Mailen, T. O. Tiffany, "The reaction of iodine with concentrated nitric acid," Journal of Inorganic and Nuclear Chemistry 1975, 37(1), 127-132 (https://doi.org/10.1016/0022-1902(75)80139-4).
Istvan Lengyel, Istvan Nagy, Gyorgy Bazsa, "Kinetic study of the autocatalytic nitric acid-bromide reaction and its reverse, the nitrous acid-bromine reaction," J. Phys. Chem. 1989, 93(7), 2801–2807 (https://doi.org/10.1021/j100344a021).
Alexander M. Andrievsky, Vera I. Lomzakova, Mikhail K. Grachev, Mikhail V. Gorelik, "Aromatic Bromination in Concentrated Nitric Acid," Open Journal of Synthesis Theory and Applications 2014, 3(2), Article ID:44941, 5 pages (doi: 10.4236/ojsta.2014.32003).

$\ce{HNO_3}$ does not react with $\ce{Cl_2}$, as $\ce{Cl_2}$ is produced in the famous "aqua regia", which is a mixture of nitric and chlorhydric acid, having reacted according to :$$\ce{HNO_3 + 3 HCl -> NOCl + Cl_2 + 2 H_2O}$$
This equation is given in Cotton and Wilkinson, Chapter 5-13. In Chapter 12-8, it is stated that NOCl is never pure. It is decomposed to $\ce{Cl_2 + NO}$ to the extent of $0.5$% at room temperature.

NOBr is produced in a mixture $\ce{NO + Br_2}$ and it is an equilibrium, where ~ $7$ % of NOBr is decomposed at room temperature and $1$ atm.

Ref.: F. Albert Cotton, Geoffrey Wilkinson, Advanced Inorganic Chemistry, A Comprehensive Text, 3rd Ed., 1972, Interscience Pub., John Wiley & Sons, Inc. New York, London.


Mercury assisted deprotection of dithiane 1,3 dithianes are used in umpolung reactions and can be deprotected by mercury containing compounds - but could anyone help explain what the exact mechanism is? Many texts seem to skim over this by explaining it is similar to acetal hydrolysis, but I'd really like a more detailed explanation as to the role of the compounds. A reaction I have seen involves $\ce{HgO}$, $\ce{HgCl2}$, $\ce{CH3CN}$ and $\ce{H2O}$. If it is similar to acetal hydrolysis, the $\ce{H2O}$ is probably to complete the final step of attacking the $\ce{S=H}$ bond, but what about the initial step?

I am assuming one sulfur atom forms a bond to the $\ce{Hg}$, but which Hg compound would it be in this case, the $\ce{HgO}$ or the $\ce{HgCl2}$? What is the role of the $\ce{CH3CN}$ in this case? Thanks!

$\ce{Hg^2+}$ has a high affintity to thiols and other sulfur functional groups, and therefore $\ce{HgCl2}$ forms a Lewis acid-base adduct with the dithiane. The cyclic sulfonium cation undergoes ring opening, and after nucleophilic attack by water, a second Hg-S bond under ring closure is formed. Deprotonation finally yields the deprotected carbonyl compund and the cyclic mercury dithiolate.

Ok, I'm speculating a bit here: The mixture of acetonitrile and water is probably just used to keep all components in solution. HgO serves as an oxidant; alone it might be able to convert the 1,3-dithiane to the corresponding 1-oxide, which doesn't necessarily results in its cleavage. $\ce{ Hg^{2+}}$ on the other hand has a high affinity to (di)thiols.


Will the sulfur-iodine cycle and Haber Method work at lower pressures and temperatures? I have a year-long research project next year in my senior year of high school, in which we have to make some product or prototype to solve some issue.

My interest was in the Liquid Fluoride Thorium Reactor but considering that I do not have the access I need, I opted to invest in the heat it produces.

One use as proposed by Kirk Sorensen proposed using the heat from the reactor to fuel the sulfur-iodine cycle to split water into $\ce{H2}$ and $\ce{O2}$.  Then use those combined with carbon collected from the waste of coal and fossil fuel plants and nitrogen from the atmosphere to make methanol dimethyl-ether and ammonia.

My teacher and I have ruled out the first two, but decided that we might be able to make ammonia.  My question is whether it would be feasible and practical to use the sulfur-iodine cycle in combination with the Haber process to split water and make ammonia at a much lower temperature and pressure to ensure safety.

Yes this is possible but you must remember it is a reversible reaction so if more pressure creates ammonia, applying less will cause the reaction to reverse creating some of the reactants again.


A metal, M, forms two fluorides containing 40.43% F and 50.44% F by mass. Identify the metal A metal, $\ce{M}$, forms two fluorides containing 40.43 % $\ce{F}$ and 50.44 % $\ce{F}$ by mass. Identify the metal.

I get that you can find out how much of the metal reacts with 1 mole of fluorine atoms in the 40.43 % one. This can be calculated to be 27.99 g.

Then from that you can work out that 28.49 g of fluorine reacts with the metal in the 50.44 % one and that that is about 1.5 moles of $\ce{F}$.

I'm a bit lost after that. How might you continue from here?

So the law of multiple proportions conveys that the fluorides contain fluorine in a 1.5:1 or 3:2 ratio. This means the possibilities are

$\ce{MF2}$ and $\ce{MF3}$, or
$\ce{MF4}$ and $\ce{MF6}$

Since there is twice as much fluorine, the mass of $\ce{M}$ in $\ce{MF2}$ must be 56 g. The next one is 112 g. If there are no errors in what was said, the metal must be iron or cadmium. What makes one a better candidate over the other?

You need to use the general concepts of the law of definite proportions and the law of multiple proportions to solve this:
Sample 1 is 59.57% $\ce{M}$ and 40.43% $\ce{F}$.
Sample 2 is 49.56% $\ce{M}$ and 50.44% $\ce{F}$.
Assume $\ce{M_{x}F_{y}}$ for sample 1 and $\ce{M_{a}F_{b}}$ for compound 2.
Let $m$ by the molar mass of $\ce{M}$. We have
$$\frac{19.00y}{mx + 19.00y}=\frac{40.43}{100}$$
and
$$\frac{19.00b}{ma + 19.00b}=\frac{50.44}{100}$$
We can rearrange the equations to:
$$m\frac{x}{y}=28.0$$
and
$$m\frac{a}{b}=18.7$$
Note that $$\frac{28.0}{18.7}\approx\frac{3}{2}$$
so we can start by trying:
$$m\frac{3}{1} = 28.0$$
and
$$m\frac{2}{1} = 18.7$$
This is weird because the compounds would be $\ce{M3F}$ and $\ce{M2F}$.
Fortunately, as long as the proportions work, it's fine, so divide both fractions by 6:
$$m\frac{1}{2} = 28.0$$
and
$$m\frac{1}{3} = 18.7$$
This gives $m=56.0$, fairly close to the molar mass of iron which is $55.9$.
Cadmium is a bad candidate, since that would be $\ce{CdF4}$ and $\ce{CdF6}$.

Another approach would be to assume that one of the compounds is of the type $\ce{MF_n}$ with $n$ a small integer. Using a spreadsheet (or just building a table) you use the small integer multiples of $18.9984$ (say across the top 6 of the columns $n=1,2,3,4,5,6$ — hexavalent compounds are fairly common, higher valences are possible but much less likely. For the rows' first column you choose the values of the % of F. This allows you to calculate the mass (which we can (but don't have to) assume is the atomic weight (atomic mass)) of the metal). That calculation is that the mass of the F atoms is the respective % of that row, so the total mass is the Fs mass ÷ %F. Subtract the mass of Fs from this and you get the mass of the M. (Yes, you could have done the same thing using the $100-\%\ \ce{F}$ values, maybe even simpler, IDK).

Then, you look at the entries, and see two (and only two) pairs that are "interesting" - meaning they're close to one another. Under 2F ($37.9968$) is $55.984$ and under 3F is $56.001$ and under 4F is $111.970$ and under 6F is $112.002$. These suggest that Fe at $55.845$ or Cd at $112.414$ are good matches. In fact if you compute the difference between their theoretical %s and the measured values, the sum of the square of the differences are $\ce{Fe} = 0.41$ and $\ce{Cd} = 0.00$, which indicates that statistically Cd is the "better" answer. You then should compare the values and determine if the difference between them is meaningful given the measured results. Since the %s were given to 4 places, I think that Cd is the "more correct" answer, the error for Fe is simply too large, IMHO.

This approach has another advantage: you can inspect the table to see if there are any values which are small multiples of an actual atomic weight of a metal (atomic mass). This would allow you to evaluate the assumption that the formula is $\ce{MF_n}$ and not $\ce{M2F_n}$ or $\ce{M3F_n}$, depending on how good your mental arithmetic is.


How do I call the functional group of this alkane? Given the following chain of carbon atoms (imagine they are all saturated with the correct number of hydrogen atoms), how do I call it? 4-(dimethyl-ethyl)-octane? I'm having trouble naming the short carbon chain extending from the 4th carbon atom of the octane. Can you help me out please?

Under the 2004 provisional recommendations by IUPAC, one possible systematic name is 4-(2-methylpropan-2-yl)octane, which indicates that the longest chain in the substituent is three carbons long, is connected to the main carbon chain via the second carbon (-2-yl), and has a methyl group on the second carbon (2-methyl).

4-(1,1-dimethylethyl)octane is also possible, and indicates that there is an ethyl substituent on the fourth carbon in octane, which is further substituted with two methyl groups on the first carbon where it branches off (see section P-29.4).

However, the preferred IUPAC name is 4-tert-butyloctane; the tert-butyl group is used to describe a $\ce{-C(CH3)3}$ group and is preferred over either of the two above systematic names (see section P-56.2.2.2).

I don't have access to the 2013 recommendations but I believe it should not be different.

2,2-dimethyl-3-propylheptane

Choose the 7 carbon chain instead. Not necessary to be the longest chain in this case


Formation of zwitterion from amino acids Why it is easier for the carbonyl group to lose a proton to become negatively charged and the amino group to accept a proton to become positively charged?

I know it has to do with the very polar bond due to the oxygen. And the $\ce{H+}$ ions attracted to the lone pair of electrons on $\ce{N}$ in the amino group.

But I don't understand why would an electronegative atom such as oxygen wants to lose a proton.

I was told that they will attract more electrons towards towards them. But that does not help in my understanding of why protons will leave.

The reason amino acids exist largely in their zwitterionic form at biological $\mathrm{pH}$ (usually around 7) is due to the $\mathrm{p}K_\mathrm{a}$ of the constituent groups. It's not that the oxygen 'wants' to lose a proton, but more that at that $\mathrm{pH}$ the equilibrium lies towards the deprotonated state (things are rarely 100% protonated/deprotonated).

In general, carboxylic acids easily lose their proton to form carboxylates (the anionic charge is delocalised over the three atoms making the proton loss 'favourable'), and amines act as bases, to pick up a proton.

At a $\mathrm{pH}$ of 7, the medium is basic enough to cause the carboxylic acid to be deprotonated, and the amine to be protonated. If you look at the diagram below, you can see that increasing or decreasing the pecan shift the equilibrium towards fully protonated (at low $\mathrm{pH}$) or fully deprotonated (at high $\mathrm{pH}$) amino acid.

Ionisation state of amino acids over 0-14 $\mathrm{pH}$ range, taken from Biochemistry, Berg (7 ed.)

You can see as, by the cleavage of proton the carboxylate ion becomes resonance stabilized which provides high stability and also due to the character of amino group as a lewis base it will be easier to attract the proton loosen by the carboxylic group.
So this type of reaction is proposed to explain the aspects such as high M.P.(Melting Point), B.P., etc. Because now it behaves as a salt( in the form of zwitter ion).


Whats is the difference between Tris and Tris-acetate? I would like to run the following protocol to detect plasmid presence in bacteria.
Would be possible to replace 'Tris- acetate' with just Tris?

We don't have 'Tris- acetate' in the lab.
What is the role of acetate here?

Hot alkaline method for all plasmid sizes and bacteria (Kado & Liu,1981), modified

Centrifuge 2-3 ml of culture, resuspend pellet in 1 ml of solution containing 0.04 M Tris- acetate, pH 8.0 (adjust pH with glacial acetic acid) and 2 mM EDTA
Add 2 ml of lysis buffer (0.05 M Tris, 3% SDS, pH 12.50, adjusted with 2 N NaOH) and mix
Incubate at 60-68°C for 30-45 min (strain dependent)
Add to hot samples 6 ml of phenol/chloroform (1:1) and mix gently to complete emulsification
Separate phases by centrifugation at 10.000 x g for 15-20 min at RT and transfer the upper aqueous phase carefully (avoid interphase which contains debris) to new tube containing 1 volume of chloroform. Mix and centrifuge again for separation of phases

"Tris" simply means "three" and is used in a number of different contexts. However, in biochemistry it's normally used to refer to the common buffering agent tris(hydroxymethyl)aminomethane. You have to be a little careful, though, as there are other uses of "Tris" in biology, which may not refer to tris(hydroxymethyl)aminomethane. (For example, there's two other buffers called "bis-tris methane" and "bis-tris propane", each of which have very different properties from tris(hydroxymethyl)aminomethane.)

Your situation is referring to the standard tris(hydroxymethyl)aminomethane, though. However, when specifying a buffering system, it's often insufficient to specify just the buffering agent used. A buffer consists of a weak acid or weak base in near-equal concentration to its conjugate base or acid. Tris is the weak base being used, but to have a decent buffer you need to also have the conjugate acid form.

As this conjugate acid is charged, there's an obligate counter ion. Different buffering systems may use different counter ions. That's what the acetate is doing here - acting as a counter ion for the conjugate acid of tris(hydroxymethyl)aminomethane.

That's what "Tris-acetate" is specifying. It's using a Tris buffering system with an acetate counter ion. This buffering system is normally obtained by using enough Tris base to get the desired concentration, and then slowly adding enough acetic acid to obtain the desired pH.

Alternatively, you could use something like the Henderson–Hasselbalch equation to calculate the appropriate proportion of base and conjugate acid for your desired pH, and then specifically mix those two forms together. This isn't typically done with Tris-acetate, as the acetate counter ion form of Tris isn't commonly sold. However, Tris hydochloride is, so if you had a system which used a chloride counter ion, you could use this approach. (Though in practice you'll want to aim a little on the basic side, and adjust to the desired pH by titration - the numbers which come out of the Henderson–Hasselbalch equation are theoretically pure, and don't account for non-ideal deviations in behavior and things like weighing errors.)

By Tris you most likely mean Tris base. Make sure that the formula is C4H11NO3 and the cas number is 77-86-1. Both should be on a label. If this is the case - you can use glical acetic acid. Slowly mix 121.14 g of Tris in 350 mL of water with 60 g of glical acetic acid. Bring volume to 500 mL. This will give you 2M solution of Tris-Acetate.


Forming the resonance structures for the nitrite anion Yesterday I learned about resonance structures and came across this problem.

The resonance structures are formed when there is a delocalized π electron cloud. I understand that for benzene, there are two resonance structures. For making π bonds I know that there are three main reasons:

The two orbitals should overlap each other.
There should only be one electron for each orbital.
The two orbitals should be in one level.

In benzene there are six $\mathrm{p}_z$ unhybridized orbitals, therefore they can share their electrons with each other (delocalize the π bond electrons). So benzene has 2 resonance structures.

But I don't understand how this works for the nitrite anion, $\ce{NO2-}$.

In $\ce{NO2-}$ there is also π electron delocalization, so there are 2 resonance structures. Now I am going to say how I think the delocalization is happening.

When the first structure changes into the second structure, nitrogen has to make a π bond with the right side oxygen with each of the $\mathrm{p}_z$ orbitals before, but now that π bond's two electrons take to the right side oxygen, and then there are 2 electrons in the right side $\mathrm{p}_z$ orbital and no electrons in the nitrogen $\mathrm{p}_z$ orbital. But the left side oxygen (in the first structure, $\ce{O-}$) has 2 electrons in its $\mathrm{p}_z$ orbital. Then the nitrogen $\mathrm{p}_z$ and oxygen $\mathrm{p}_z$ overlap each other and make a π bond.

I have said before "There should only be one electron for each orbital." But, according to my understanding above, the π bond is made by $\mathrm{p}_z$ with 2 electrons (oxygen) and $\mathrm{p}_z$ with no electrons (nitrogen) overlapping each other.

So, is there any mistake in my imagination?

Yes. I think that there is a mistake in your imagination. Nitrogen is evidently $sp^2$-hybridised in this case. Nitrogen has 5 valence electrons. Two electrons are used to make 2 $\sigma$ bonds with two hybrid orbitals. Another two electrons become the lone pair and reside also in a hybrid orbital. This leaves us with one electron in one unhybridised $p$ orbital. This $p$ orbital is then used to make the $\pi$ bond to either oxygen atom, contributing one electron to the $\pi$ bond while the oxygen atom also contributes one electron from its $p$ orbital.


Stability of Bohr Orbits In class we had been taught that Rutherford's model was unsuccessful because it failed to show that the orbits are stable because the electrons would lose energy because of electromagnetic radiation.

Bohr had suggested that electrons would move in orbits having well defined energy levels, but my question is that what would prevent these electrons from emitting radiation and collapsing into the nucleus.

Not every model is perfect. Rutherford's model suffered from the problem of electromagnetic radiation but answered important questions about the structure of atoms by showing the existence of a positively charged nucleus.

Bohr improved on this by realising that if electrons could only have certain energies then he could explain other things such as the spectra of hydrogen. As far as I am aware he did not attempt to explain why electrons are fixed in these orbits - this was only explained later by the advent of quantum mechanics. However, this does not mean that Bohr's model was useless since it accurately predicts useful things.

We do not have to know everything about why a model works in order for it to be useful. Einstein did not know why the speed of light was constant but he assumed that it was (based on all the available experimental evidence) and derived special relativity, which is an extremely useful theory. Nowadays we have quantum mechanics and general relativity which we know are incompatible but they are both exceptionally useful theories so we do not ignore them.

As you might have studied Bohr's model includes various orbits. An orbit is a Defined as a particular region in space where the electron moves and has constant energy.
A particle that has constant energy will not collapse in the nucleus. He even went forward to calculate these specific distances from the nucleus he came to the conclusion equation mvr=nh/2pi where the terms stand for their general meaning. By this equation we get to know about specific distances from the nucleus where the electron can be present without losing any energy.
See the thing is Bohr's model was only created for hydrogen, what you are actually seeing is the adaptations made to the Bohr's model that help us calculate values like velocities, energies, radii, etc.
The Logic behind the discarding of the Rutherford model was that the fact that classical electromagnetism states that ACCELERATING  charges product EM waves and hence an electron would lose energy.(Circular Motion is a motion where constant centripetal acceleration is needed hence the electron actually experiences the constant acceleration). This constant energy loss would collapse the structure proposed in almost 10^(-8) Seconds as you might have studied. But this does not happen.


Most probable point for finding an electron in the 1s orbital of a Hydrogen atom My professor says that the most probable point for finding an electron
in a 1s orbital of a hydrogen atom is at its origin.

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

I can understand the reasoning he gave about the origin being the highest probable point for finding an electron, but then why are the graphs of the radial distribution function so?

It makes a difference if you look in a point or in a volume element.

For the radial distribution we sum up the probabilities within (very thin) shells at different distances from the core.
If the radius get's bigger the volume in this shell gets bigger.
Remember that the surface of a sphere goes by $4 \cdot \pi \cdot r^2$ so the volume of the shell is essentially for all practical purposes $4 \cdot \pi \cdot r^2 \cdot \Delta r$.
At the core the radius is zero which means the volume is zero and thus the probability is zero too.

At the same time the probability of finding the electron in a point at a certain distance goes down, but the volume goes up.
Because of the scaling of the volume and the "decay" of the probability to find it in a point further away it turns out that there's a maximum of finding the electron at a certain distance away from the core (but not a single point at that distance).

He explains this by citing the fact that the square of the wave function which gives the probability density is maximum at the origin.

Not exactly.

The answer by @dsva explains why this is wrong and I'll just expand that in a minute, but first note that it's easy to see why this is wrong.

If the electron was most likely to be at the origin it would be likely to interact with the nucleus, something we expect not to happen.  In fact there is a small chance of this because the nucleus is a non-zero size.

Essentially he's missing the need to sum over the shell, and the shell's volume is zero at the origin because of the volume element $dV = 4\pi r^2dr$ with $r=0$.

The probability of finding the electron in a region of zero radial size is zero, but we can evaluate the relative probability for two radii.  For 1-s orbitals that is :

$$\frac {r_1^2 e^{-{r_1}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$

and

$$\frac {r_2^2 e^{-{r_2}/a}}{r_1^2 e^{-{r_1}/a}+r_2^2 e^{-{r_2}/a}}$$

For $r_1=0$ we clearly get a relative probability of zero compared to any $r_2\neq 0$.

  At the same time, we all agree that the Bohr radius is the distance at which probability of finding the electron is maximum for 1s orbital.

This is because of what an expectation value is and how it is calculated in QM.

An expectation value is an average over the entire space of the effect of an operator on a wavefunction.  It is not simply the operator multiplied by the probability density.

In short an expectation value can be considered an average over all space, whereas the probability density is not an average over all space.


Why does IR spectroscopy specifically require a change in dipole moment, and why doesn't this apply to other types of spectroscopy? I would really appreciate any amount of detail/depth. I do not have a background in quantum, which is probably why I don't know the answer to this. However, any answer referencing quantum concepts is welcome and appreciated; I will follow up on any resources provided.

My question is basically as follows: Why does IR absorption require a change in dipole moment, and why is it that other types of spectroscopy (UV Vis, microwave) don't have this requirement? What causes this requirement specifically for IR?

Thank you!

All the types of spectroscopy you mention have a similar condition for a spectrum to be observed.  This is that the molecule has to be able to interact with the radiation in the first place and then the radiation has to be of the correct energy to change the energy of the molecule from one level to another.

A plain (classical) explanation is as follows. (We implicitly assume that the radiation has the correct energy). As radiation is electric dipole in nature (it consists of oscillating electric and magnetic fields) if the molecule also generates an oscillating electric field it can interact with the radiation; energy is taken from the radiation field and enters the molecule. Thus a molecule may gain a quantum of vibrational or rotational energy or an excited state may be produced.

In the case of a vibrating diatomic molecule only if it is hetero-nuclear e.g. HCl, will it have a dipole and thus an oscillating dipole and oscillating electric field as it vibrates. Thus hetero-nuclear diatomics show an IR spectrum, homo-nuclear diatomics do not. In a polyatomic molecule vibrations can distort the molecule, e.g. benzene is distorted so that it is no longer 6 fold symmetric, so that a  dipole is produced and an IR spectrum may result. What form this takes depends on the molecule's symmetry and group theory point groups are used to determine this.

In a rotating diatomic molecule we usually use a rigid-rotor model. If the molecule has a dipole then this clearly changes as the molecule rotates, produced an oscillating electric field, and so a microwave (rotational) spectrum is produced. A homo-nuclear diatomic e.g. $\ce{O2}$ has no microwave spectrum.

In a UV/visible transition an excited state may be produced but only if the electron distribution in the excited state is different from that of the ground state, i.e. a dipole is produced that can interact with the radiation. A molecule such a benzene has a forbidden first electronically excited state due to its symmetry (which is however observed weakly due to other effects) and so absorbs very little compared to aniline (amino-benzene) or many dye molecules where symmetry does not limit any electronic dipole formation.

Thus although the details vary, the underlying ideas are the same. You should understand now that the statements about spectra in your question are not all correct. The rules used to determine what transitions are allowed or not allowed (called forbidden) are called 'selection rules'.

You can read about this in more detail, as a quantum version, in most undergraduate phys. chem. textbooks.

You may realise that the vibrational frequencies of homo-nuclear molecules are known even though they have no ir spectrum and this is because they can be measured by Raman spectroscopy. This depends on the change in shape/size of the 'electron cloud' (polarisability) a molecule possesses. Electro-magnetic radiation also possesses an oscillating magnetic field but this interacts far more weakly with molecules than its electric counterpart and can often but not always be ignored, the exception being NMR spectroscopy.

As per my knowledge there are two reasons one is: we use infrared radiation as a source such that it helps to induce vibration in the molecule that results in the difference in dipole moment . Second : due to the reduced mass(mu) in hookes law equation given below,


Effect of increasing product concentration on time of completion of reaction This was the question asked in our chemistry exam:

In the experiment of determination of equilibrium constant of esterification of ethanol, if , instead of using
clean dry bottle, if a bottle already containing slight amounts of ethanoate and water is used to carry out the
reaction under identical (remaining) conditions, what would be the consequence?

According to our lecturer it's answer is "The reaction will take lesser time to reach equilibrium". But according to Le Chatelier's principle, increase in concentration of products shifts the reaction backwards. So it should take more time to reach equilibrium.
Please help!!

In equilibrium reactions the equilibrium is approached exponentially in time. The lifetime of the exponential is the reciprocal of sum of the forwards and back rate constants multiplied by the equilibrium concentration of the species, thus for reaction $\ce{A + B \leftrightharpoons C + D }$ the lifetime of the reaction is $\tau = 1/[k_1 (A_e+B_e) + k_{-1} (C_e+D_e) ] $ where $k_1$ is the forward rate constant and $k_{-1}$ the backwards one, and subscript $e$ means equilibrium amount of a species. Thus if you add more material you change the equilibrium concentrations  and so the lifetime gets shorter as its a reciprocal in the concentrations

The lifetime is the time taken to change to $1/e$ of the initial amount, approx to 38%, whereas the half-life is the time taken to halve the initial amount. The lifetime is the reciprocal of the total rate constant for reaction which is why it is used instead of the half-life.

(The exception is the simple $\ce{A\leftrightharpoons B}$ reaction when the lifetime is independent of the equilibrium amounts ). In equilibrium reactions the lifetime is also called the relaxation time.


Why is law of mass action only applicable for elementary reaction? Why can't we apply law of mass action for non-elementary reaction? Why does is vary with the rate law in case of non-elementary reactions?

The Law of Mass action states:

The rate of a reaction is directly proportional to the concentration of reactants raised to powers equal to their stoichometric coefficients

This is only valid for elementary reactions since they happen in a single step. As a result, such simple relations hold good.

But what about the more complicated ones. Certain reactions, like the oxidation of ethylbenzene to benzoic acid using hot $\ce{KMnO4}$ don't happen in a single magical step. They involve lots of reactive intermediates and transition states, so it's basically a bunch of several elementary reactions. You can't consider the whole reaction as an elementary reaction.

But you can formulate a rate equation for non elementary reactions by considering them as a bunch of elementary ones. However such formulations involve knowing the mechanism of the reaction in question, along with other data.

If I ever knew why the tLoMA held for elementary reactions, I've forgotten. I can, however, give an example of a situation applicable to your question illustrating why it doesn't necessarily hold for complex reaction systems. Consider the two elementary reactions A+B→ C+D and A+C→E  the first thing I'd ask you to do is express the equation for the total equilibrium constant. The second thing I'd ask you to consider is if the first reaction had an extremely high rate and the second had an extremely low one (on the order of years-1). The overall reaction will be dependent almost exclusively (except for the very very start) on the rate of the second reaction, tLoMA will (and must) fail.


Why does HNO2 not have resonance? A question on the 1996 AP Chemistry Free Response asks:

  The $\ce{N-O}$ bonds in the $\ce{NO2-}$ ion are equal in length, whereas they are unequal in $\ce{HNO2}$. Explain.

Since the nitrogen dioxide ion has resonance, the $\ce{N-O}$ bonds are equal as resonance is in reality a hybrid of all of the possible structures for a certain molecule. However, the answer then assumes that $\ce{HNO2}$ has no resonance but rather one single and one double bond ($\ce{N-O}$ and $\ce{N=O}$), and therefore not all of the $\ce{N-O}$ bonds are equal.

How is this possible? Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

Resonance is a remnant of valence bond theory, which is necessary because it is impossible to describe delocalised bonding within a localised bonding scheme. Resonance structures are not real. One Lewis structure describes one electronic configuration with fully localised electrons. For many molecules it is completely sufficient to describe the bonding situation with one such structure in a first order approximation. For many other molecules this is completely insufficient. For $\ce{NO2-}$ this is the case, hence you need a superposition of more than one structure to get closer to a qualitative description. This caveat is not necessary in molecular orbital theory, since this is designed to treat delocalisation.

Nitrite anion

One question therefore arises from the above: Are the $\ce{N-O}$ bond lengths equal because there is resonance, or is there resonance because the $\ce{N-O}$ bond lengths are equal?

This is obviously more of a philosophical question, but it shows the dilemma of the whole concept. The bonding of any given molecule is a result of many different factors; to name but a few: repulsion of the nuclei, repulsion between the electrons, attraction between electrons and nuclei.

For $\ce{NO2-}$ there is one key question, that gives us a very important clue as to why the bond lengths are equal: Why should one oxygen atom be different from the other? In fact the molecule is symmetric (C2v), therefore the bond lengths must be equal.
If you try to write a Lewis structure that respects this symmetry, you will end up with an electron sextet on nitrogen (and a formal charge of +2), which certainly can only be an insufficient approximation. If you write a Lewis structure with fewer formal charges, keeping symmetry, you will notice that nitrogen will have 10 electrons. That would mean that it would have to use d-orbitals, which simply are not available for nitrogen (more on that another time). There is another way to write a Lewis structure, however, this one has to break the symmetry, and you'll notice that there is a second one, which is equal. In order to keep the symmetry you introduce resonance, to state, that the "true" electronic structure is a superposition of (at least) those two configurations.

TL;DR: The $\ce{N-O}$ bonds in the nitrite anion are equal, because the oxygens are equal.

Nitrous acid

It is false to say that in $\ce{HONO}$ is no resonance. And it is also wrong to say that a lack of resonance makes the bonds unequal. If the solution manual claims this, it is wrong.

  Why can $\ce{HNO2}$ not have 2 possible structures where the single and double bonds switch between the two oxygens?

These are valid resonance structures of nitrous acid. You have to keep in mind though, that resonance does not change the geometry of a molecule (or molecular structure). The contribution of the Lewis structure with fewer formal charges to the total wave function will be much higher than the contribution of the other structure in this case. This is not a general statement, there are molecules where the charge separated Lewis structure has a higher contribution to the wave function.

By analogy to the anion case, there is one key question, that gives us a clue about the molecular structure: If you bind a proton to one oxygen, are both oxygen really equal anymore? In this case the symmetry of the molecule is reduced to Cs. Therefore the bonds, cannot be equal.
For this molecule you will have no problem writing a single Lewis structure that satisfies every imposed criteria.
To answer the initial question: The bonds in $\ce{HONO}$ are unequal, because the proton is bound to one of the oxygens. It is a very polar, but covalent bond. The electron density of the $\ce{N-O}$ bond is therefore shifted towards the $\ce{H-O}$ bond, and simultaneously reduced for the $\ce{N-O}$ bond when compared to the anion. Therefore the $\ce{HO-N}$ bond is longer than the $\ce{N=O}$ bond (I implied the most contributing Lewis structure here).

TL;DR: The $\ce{N-O}$ bonds in the nitrous acid molecule are unequal, because the oxygens are unequal.

There frequently is the notion of a most stable resonance structure. While this is very popular, it is completely wrong. I again refer to the question What is resonance, and are resonance structures real? Resonance is used to describe the electronic structure of a molecule, there is no change in the molecular structure allowed. It is necessary to understand, that only the description in terms of all possible resonance structures is in itself correct, i.e. the theoretical limit of valence bond theory, everything else is an approximation. However, not all resonance structures will have the same contribution to the wave function. Whenever someone speaks of the most stable resonance structure, it is likely that they mean the resonance structure with the highest contribution to the wave function. But even if you know what is meant, it is still wrong to speak of a most stable resonance structure, it must be corrected.

(I currently don't have the time to provide a full fledged answer, so instead I was trying to give some pointers in the right direction. I might come back to amend it whenever I have more time.)

The idea is that in a resonant structure, all the structures are equally stable -- for instance, in CO3:

, you can see that the formal charges on the Os initially are -1 on 2 of them and 0 on the last; these charges are consistent in each structure. However, in your example, the structures are:

The formal charges on the Os in this case are 0 and 0 on the left, as compared to 1 and -1 on the right. These charges aren't the same, so this is a case of more stable structure rather than resonant structure.

Image sources: http://www.chem.ucla.edu/~harding/tutorials/resonance/draw_res_str.html
https://s3mn.mnimgs.com/img/shared/content_ck_images/images/HNO2.png


Displacement of leaving groups in organic reactions When I initially learned about substitution reactions, there would be times where a poor leaving group had to be changed into a better one in order to proceed with a substitution reaction. For instance, changing a poor leaving group (-OH) to a better one, such as water:

After the "good" leaving group is made, it is displaced via an SN2 reaction. The displacement of the water required a nucleophile. However, I recently came across this mechanism (dehydration):

In this case, the water leaves by itself. My question is:

If I ever come across an instance where water (a good leaving group), must leave, how should I assess whether it can leave BY ITSELF or not?

In that kind of reactions, the full explanation would require advanced quantic calculation to determine the energy of all the present entities.

In chemistry, we usually prefer giving a half-quantitative explanation to explain the formation of the observed products.

In your example A —> B + C + D, A is a secondary alcohol and it CAN be transformed to a carbocation leading to B... but you can see it is far from perfect since B represents only 50% of the yield. The result would be much more clear-cut with a primary alcohol (unstable carbocation) or a tertiary one (very stable carbocation). I think you are expected to explain all that.


Forces of attraction of hydrated potassium ion in aqueous solution I want to study the intermolecular forces (IMFs) in hydrated potassium ion, $\ce{K+ (aq)}$ in an aqueous solution of $\ce{KCl}$. According to my thoughts, among the 4 IMFs I know:

London dispersion forces (LDFs);
Dipole-Dipole interaction;
Hydrogen bonding;
Ion-dipole forces

LDFs are definitely included, but my question is whether dipole-dipole is included as well (because KCl is a polar substance)?

Dipole-dipole interaction, specifically, will probably not happen, or they at least won't contribute significantly to IMFs. The thing with these different chemical bonds is that they are mostly based in identical features, and sometimes calculated differently, taking different things in account.
An ionic bond will be stronger than a polar bond, so that is why dipole-dipole interaction seems irrelevant, in comparison to ion-dipole forces.

All the kinds of interaction mentioned are derivatives of electrostatic interaction, also know as Coulomb interaction. LDFs are a direct derivative of Van der Waals bonds, but if you look more physically at all those bonds you can think of them as electrostatic interaction.

There are no intermolecular forces in KCl because it is an ionic compound. IMFs only exist in covalent compounds between molecules.


Mass of produced carbon dioxide I am studying for a big test and I ran across something that I couldn't solve no matter how hard I tried. (reading through old tests)

Burning ethanol produces carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$).

a) Write down a balanced chemical equation that illustrates the complete burning of ethanol.

b) What is the mass of produced carbon dioxide ($\ce{CO2}$), when completely burning 575 liters of ethanol?

NTP conditions apply. Ethanols density is 0.8 g/cm³.

MC = 12 g/mol, MH = 1 g/mol, MO = 16 g/mol

For the first part I came up with $\ce{C_2H_6O (l) + 3O2(g) --> 2CO_2 (g) + 3H2O (g)}$

Then I calculated that $M_\ce{CO_2}$ = 44 g/mol and $M_\ce{H_2O}$ = 18 g/mol

I also calculated that 575 liters of ethanol weighs 460 kg

I know that m = nM

However I am not sure what to do next.
I may also be using some weird terms here as this is my translation and I've never dealt with math/chemistry in English.

Ethanol's molecular mass $M(\ce{C2H6O})=46\ \mathrm{g/mol}$

$$n=\frac mM,$$
so
$$\frac{460\,000\ \mathrm g}{46\ \mathrm{g/mol}} = 10\,000\ \mathrm{mol}$$

So having 10 000 moles we can multiply that by the molecular mass of carbon dioxide and multiply again by two because there are two carbon dioxides.

$$10\,000\ \mathrm{mol}\times44\ \mathrm{g\ mol^{-1}}\times2=880\,000\ \mathrm g=880\ \mathrm{kg}$$

1.u can easily analyse that 1mole of etanol produces 2 moles co2.
2.so first calculate M for ethanol=46g/mol.
3.m calculated by u for C2H5OH is 460000gm
4.so no of moles=m/M=460000/46=10000 moles.
5.now 1mole C2H5OH produces 2mole co2 so
using unitary method 10000 moles of C2H5OH will produce=10000×2 Co2 moles.
6.molecular mass of co2=44g/mol,20000 moles=20000×44=880000gm=880kg of co2 is ur ans.


Consequences of ortho effect on resonance Consider any carboxylic acid with any group at ortho position. According to ortho effect, the resonance of the $\ce{C=O}$  with the ring is inhibited.

Is this resonance entirely lost? Or some little amount still exists?

Can resonance be present when the participating orbitals are not in the same plane?

If they are off the plane due to steric effect,  then resonance wouldn't occur, hence the (-)R effect of - COOH group vanishes and hence "generally" acidity of the respective acid increases.


What exactly is formed when K2Cr2O7 is reduced? I read on Wikipedia and in many books that when $\ce{K2Cr2O7}$ is reduced, it forms $\ce{Cr^3+}$ and a change is noticed from orange color to green. I want to know in what exact complex/salt/compound $\ce{Cr^3+}$ exists that it shows green color.

I know $\ce{[CrCl3].3H2O}$ shows green color but in many of these reactions, there is no $\ce{Cl-}$ ion, so I guess that is not formed. Also, $\ce{Cr^3+}$ can exist as $\ce{[Cr(H2O)6]^3+}$ but since it is violet in color, that possibility is also ruled out. So basically, how does the solution becomes green in color?

The dichromate ($\ce{Cr2O7^2-}$) ions are strong oxidizing agents at low pH. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. In redox reactions in aqueous acid solution, the aquated $\ce{Cr^3+}$ ion is produced according to following half-reaction (Electrochemical Series):
$$\ce{Cr2O7^2- + 14H3O+ + 6e- <=> 2Cr^3+ + 21H2O} \space \space \space \pu{E^0 = 1.36 V}$$

Assume that the $\ce{Cr^3+}$ ion in aqueous solutions is in its simplest ion form: the hexaaquachromium(III) or $\ce{[Cr(H2O)6]^3+}$. This ion is violet-blue-grey in color. However, when $\ce{Cr^3+}$ ion is formed during a reaction in aquous acid solution, it is often appeared in green color. Thus, we always describe the green color due to $\ce{Cr^3+ (aq)}$, implying it as the hexaaquachromium(III) ion ($\ce{[Cr(H2O)6]^3+}$), but that's actually not the case.

What's really happening in the solution is that one or more of the water molecules in each of $\ce{[Cr(H2O)6]^3+}$ ions get replaced by other negative ions in the solution, typically by $\ce{SO4^2-}$ (if the acid used is $\ce{H2SO4}$) or $\ce{Cl-}$ (if the acid is $\ce{HCl}$). For example, if one of the water molecules is replaced by a $\ce{SO4^2-}$ ion, $\ce{[Cr(H2O)6]^3+}$  becomes $\ce{[Cr(H2O)5SO4]+}$, which is green (notice also the change in the charge on the ion). You can check this by yourself: warm freshly prepared aqueous  chromium(III) sulphate hexahydrate solution, which would change its color of purple-blue or violet ($\ce{[Cr(H2O)6]^3+}$) to green ($\ce{[Cr(H2O)5SO4]+}$) due to the ligand exchange discussed above [Chem Guide].

In $\ce{CrCl3.6H2O}$, on the other hands, the green color of the solid itself or its aqueous solution is due to the $\ce{[Cr(H2O)4Cl2]+}$ ion. Hence, actually more proper way of writing the formula is $\ce{CrCl2(H2O)4Cl.2H2O}$. In contrast to previous case, the green $\ce{CrCl2(H2O)4Cl.2H2O}$ solution would slowly becomes violet in color by standing at room temperature overnight, by slow ligand exchange with solvent (water)[Chem Guide].

Interesting fact: Chromium was discovered by Louis Nicholas Vauquelin in 1797 who named his new element after the Greek word chroma — which means color.

Late Additions:

Although $\ce{Cr^3+}$ ions complexes with most ligands to give green color, it gives vivid line of colors with other ligands (note it names after chroma!). Some interesting examples are illustrated in the picture below:

Also see:

Chromium(III) nitrate: Blue-violet crystals (anhydrous), and purple crystals (nonahydrate), according to Wikipedia (see the picture below).

$$\ce{4K2Cr2O7 ->[\Delta] 4K2CrO4 + 2Cr2O3 + 3O2}$$

Chromium(III)oxide is formed in a normal decomposition reaction and the colour changes from orange to green.


Why isn't my galvanic cell making enough voltage? I'm in kind of a pickle here. I made a galvanic cell of Lead and aluminium for a chemistry internal assessment but unfortunately, it didn't turn out so well and neither me nor my teacher has any clue why.
The E0 value of the cell is 1.53v. I got a maximum reading of 0.29v

If you don't have time to read here is what I think the sources of error could be:
1. I used filter paper as a salt bridge and I don't think it was thick enough to allow ions to flow through it.
2. I used around a 1M concentration of KCl to dip the salt bridge in.
3. I suspect either of the metals was not correct but this seems improbable as I got a voltage just not the correct one.
4. The solutions of AlNO3 and PbNO3 were kept in round-bottomed flasks with a glass cork in them for a few days before I decided to do the experiment.
5. I got a negative reading for the voltage which was weird.
6. I got a higher voltage reading only when I touched the aluminium rod to the salt bridge in the solution

I used the following procedure:

I made 0.01M, 0.1M, 0.25M and 0.5M solutions of aluminium nitrate and a 0.1M solution of lead nitrate (both well dissolved) from lab equipment.
I poured each of them into separate 50ml beakers. I then placed an aluminium rod (that was 50% submerged) into the AlNO3 beaker and a thin piece of lead (that was well sanded before each reading) that was fully submerged in PbNO3.
I connected each of the metals to a multimeter via alligator clips
I then prepared a salt bridge using 5x25cm strips of filter paper which were dipped in a solution of KCL of about 1M concentration. The filter strips were folded and stapled.
I then placed the salt bridge in both the beakers and conducted the experiments.

I'm not in school right now and I don't have time to conduct the experiment again as the assessment is due very soon.

Just keep in mind that Eo is determined under highly controlled conditions and concentrations. I sense two key problems here. The major problem is already stated above that the resistance of a wet filter paper must be high, and this is giving you lower voltage. Evaporation of water from the filter paper would make the situation worse. Since you are not repeating the experiment anyway, make sure to identify the causes of lower voltage. Secondly, both lead and aluminum nitrate tend to form hydroxides in water if you don't add little bit of acid in them. Did you notice that lead solutions were cloudy or they were clear and transparent?

As a side note, negative voltage reading of the battery just means that your connections of the lead are the other way round on the voltmeter. Test your voltmeter with a battery keeping the signs on the battery in your mind. Once you get a positive reading of a simple AAA cell you have correctly made the connections.

I think, there is still something wrong with the salt bridge. It works like a resistor I guess. You loose voltage according the law of Ohm. U = R*I.  Did you try for a short time to use a wire between your two cells. If yes how is the voltage reading. Other possibilities are oxidising of the lead and aluminium as well AND KCl means chloride ions , which can form precipitation of PbCl2. Blocks the saltbridge.
Did you immerse both eletrodes in KCl solution for test? Exchange the saltbridge to potassium nitrate. KNO3.


Electrolysis of diluted sulfuric acid My teacher told me that electrolysis of dil. $\ce{H2SO4}$ (concentration below $50\%$) using inert electrodes results in gradual increase of the concentration of $\ce{H2SO4}$.

I can't seem to figure out why. I know that the sulfate ions will remain in the solution, but aren't the $\ce{H+}$ ions of $\ce{H2SO4}$ reduced at the cathode as well?

Another doubt I have is that I've seen different equations for the oxidation at the anode for this electrolysis in different books. Some of them show that $\ce{OH-}$ will be reduced, and some show that $\ce{H2O}$ will be reduced! I'm really confused about all this.

In a dilute solution of sulfuric acid, there are the following species present: $\ce {H_2O}$, $\ce {H^+}$, $\ce {OH^-}$, $ \ce {HSO_4^-}$, $\ce {SO_4^2-}$. Of course, the water molecules are present in the highest concentration, much higher than the other species, since it is a dilute solution. For simplicity, we would also represent the proton and hydroxide ion as unsolvated. However, do take note that they are in fact heavily solvated in aqueous solution.

In some textbooks, it may be said that for the electrolysis of dilute $ \ce H_{2}\ce S\ce O_{4}$, the oxidation half-equation is written as $\ce {2H^+ + 2e^- -> H_2}$ and the reduction half-equation is written as $\ce {4OH^- -> 2H_2O + O_2 + 4e^-}$. However, the concentrations of these ions in solution are rather small compared to the concentration of the water molecule in solution. Albeit these reactions may be taking place, they should not be the main reactions and are unlikely to be representative of what is actually happening in solution.

In fact, since water molecules are the dominant species in solution, it is more likely that they undergo the redox reactions. The following half-equations represent the reactions happening in the solution more accurately:

Reduction: $\ce {2H_2O + 2e^- -> H_2 + 2OH^-}$

Oxidation: $\ce {2H_2O -> 4H^+ + O_2 + 4e^-}$

Since water molecules are being used up in the electrolysis process, the concentration of the remaining ions increase as the solution is electrolysed. Thus, the concentration of the sulfuric acid solution should increase.

$\ce{OH-}$ can't be reduced. It may only be oxidized by giving an electron at the anode, while $\ce{H+}$ is reduced instead of $\ce{OH-}$.
Secondly, the reducing property of $\ce{H2O}$ is greater than that of $\ce{H+}$ ion. Moreover, it is greater in number than other ions like $\ce{H+}$ ions etc. So it will be preferentially discharged although it is already discharged but preferentially reduced at the cathode.


How do you identify reducing / non-reducing sugar by looking at structure? Identifying reducing / non-reducing sugar been confusing me for a while now , I know that reducing sugar contain aldehyde or ketone group . It's easy to identify them in monosaccharides but this becomes confusing in case of disaccharides!

I fail to see carbonyl group here, How can I identify reducing / non-reducing sugar by looking at structure in picture? can anyone help me out?

Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a hemiacetal.

Once you realize that a hemiacetal can equilibrate with a carbonyl (e.g. it is a carbonyl in disguise), identification of reducing sugars becomes easier.

Here is an example from Wikipedia, it is Maltose, the same as your third compound

See the aldehyde in the open isomer drawn on the right; see how it can close to form a hemiacetal on the left.

The right most c of maltose contains one OH bond and one OR bond and one H and one R. So it is a reducing sugar. R is any alkyl group. And this is the condition for reducing sugar.


Which orbitals of the hydrogen atom are degenerate for n=3? Which of the following orbitals are degenerate in the hydrogen atom with $n = 3$?

  A. II and III only;
  B. I and IV only;
  C. I, II, and IV only;
  D. II, III, and IV only;
  E. all.

The answer says its E. All of them.

First of all isn't there only 1 electron in hydrogen? And how could the $\mathrm{s}$ orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

First of all isn't there only 1 electron in hydrogen?

yes

  And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

"degenerate" means having the same energy.  "Degenerate" refers to a set of orbitals.  It doesn't make sense to say one orbital is degenerate.

Solving the non-relativistic Schrodinger equation, all the orbitals for a given "n" are degenerate.  Energy only depends upon n.

More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.

Ok so the best way to think of this is to compare two atoms. Take for example {Ne} which has 10 electrons so this atom would have a number of n=3. The problem asks what orbitals would be degenerate for n=3 and we know that H has a number of n=1 so this would be just 1s^1 the question is a trick question to make you think that all orbitals above the n=1 orbital are degenerate so really even though it gives you hydrogen its just to throw you off. Picture a n=3 atom for this problem that has no element attached to it in that case all orbitals would be equal energy therefore it would be option E because all orbitals s,p,and d would be degenerate. Hope that helps took my a while to conceptualize this.


Vapor Pressure Paradox This is quite an interesting situation I had thought of. Consider two beakers, each containing $\pu{180 g}$ of water. Also, suppose one of the two has $\pu{1.8 g}$ of glucose dissolved in it. Let's call the one without glucose as A, and the one with glucose as B. I keep a small bell-jar that is large enough to just cover both of the two vessels. Now I will try to measure the vapor pressure inside the jar. Here's a little diagram to show what I mean:

I do not have access to bell-jars or manometers so I'll try to use Raoult's law to find the vapor pressure inside the vessel. The law in a simplified way would be:

$$P_{\text{solution}}=P_{\text{solvent}}x_{\text{solvent}}$$

Where $x$ denotes mole fraction, and solute is considered non-volatile. Keeping vessel A in mind, I get the vapor pressure as 23.8 mmHg, while applying Raoult's law for vessel B, I get the vapor pressure as:

$$P=23.8\left(\frac{10}{10.01}\right)=23.7 \text{mmHg}$$

Now, both of the values can't be true at the same time, and that's what I believe is the paradox.

How can this paradox be resolved?

My thoughts are that perhaps water from vessel A condenses back onto B, thus slowly increasing water content in B, increasing the vapor pressure. However, since theres solute already in B, the vapor pressure can never reach the value provided by the pure water.

What are your thoughts? Does the paradox ever resolve or does it remain in an unbalanced situation?

Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.

Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.

An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.

Your post suggests (in fact, requires) you believe that Raoult's Law is accurate to 0.5%. It is not. Quantum Mechanics, the foundation of modern chemical theory, was created/discovered in the 1920s, Raoult's Law in the 19th Century. That's pretty good evidence that the law will be, at best, an approximation to the actual state of the system. As greg commented, you assume equilibrium - or at least, you should have understood that Raoult's Law requires equilibrium or near equilibrium. You are correct that since the two liquids have different vapor pressures, then equilibrium can not (yet) exist. The outcome would be (given an infinite amount of time, given a isolated system, given no bacteria nor oxidants present to consume the glucose, that the pure water (pure, right? no dissolved salts, but what about the dissolved air?) that the glucose solution would increase in mass & volume eventually overflowing its container. There are other considerations, which we ignore: surface tension and gravitational potential energy (adding more volume raises the height). One thing I learned fairly recently is that if you picture a cylinder of gas, the thermodynamics I learned in chemistry classes (including graduate level) and the thermodynamics I learned in freshman Physics, says that that cylinder (as a well insulated and closed system) will eventually attain thermal equilibrium. That is wrong. The temperature and density of the gas in the cylinder will depend on height. Thermal equilibrium will "never" be achieved.(although for a sufficiently "short" cylinder, the temperature and density gradient might be so small as to be impossible to measure.) ("never" meaning as long as the planet and cylinder exist. In the long run, the planet will most likely (we don't know for sure) be pulled into the Sun... but I digress...). Chemistry talks about many things, it is much more of a pragmatic science that Physics - which is more theoretical. (Although there may be more pragmatic physics used day-to-day than theoretical chemistry...) One way to divide chemical theory (there are many ways) is between Thermodynamics (equilibrium or near equilibrium) and Kinetics. I advise you always keep in mind whether a particular relationship or process or theory is thermodynamic (usually about energy, about path-independent initial and final states) or kinetics (about speed, shape, and specific process (path)).


How tetraamminecopper(II) is formed? I'm trying to get a sense of how $\ce{Cu^2+}$ and $\ce{NH3}$ are bonded together because it doesn't make any!
I understand that $\ce{NH3}$ has a lone pair of electrons that can be donated (right?), but since there are four molecules of $\ce{NH3}$, then there are 4 pairs, and $\ce{Cu^2+}$ only needs two!
the reaction in my textbook was written like this:
$$\ce{CuSO4.4NH3 <=> [Cu(NH3)4]^2+ + SO4^2-}$$
I would appreciate any clarification.

There are many ways to describe chemical bonds. Copper II ion binds to some anions with ionic bonds in solids, sometimes in covalent bonds (Ref 1).
In aqueous solution, copper II ions are surrounded (hydrated) with six water molecules, with the oxygen ends pointing in, toward the positive copper ion. Water is acting as a Lewis base. Ammonia is a stronger Lewis base - after all, it is basic in water. So an ammonia molecule will replace a water molecule in a hexa-aquated copper ion. Eventually, you get $Cu(NH_3)_4)^{2+}$.$(H_2O)_2$. Oh, I forgot, sometimes we leave out the water molecules, since we just assume that they are there - everybody knows that, since the teacher has explained that (unless the teacher hasn't explained that).
In the equation of the question, only four of the actual six water molecules are shown, because they are important for describing the interaction with $NH_3$, and the other two are just a complication, because I know you would ask, why don't they get replaced too, to give $Cu(NH_3)_6^{2+}$? After all, nickel, cobalt and chromium III do! Well, zinc doesn't; it forms a tetra-ammonia complex. And iron, manganese and aluminum don't even dissolve to form soluble ammonia complexes in alkaline solution. Reference 3 suggests that it is an equilibrium situation, but there are deeper explanations which you can explore later.
Ref 1. https://aip.scitation.org/doi/10.1063/1.1677079
Ref 2. https://www.chemguide.co.uk/inorganic/transition/copper.html
Ref 3. https://www.chemguide.co.uk/inorganic/complexions/aquanh3.html#top

Yes, you are right. This is not a covalent bond it is a coordination bond. The copper has a electron pair deficit (Lewis Base) and the Nitrogen (Lewis Acid) donates the elctrons pair which makes the molecule more stable in the end.


Formation enthalpy vs. bond enthalpy I have been working with thermochemistry at school for a while. I have been introduced to the two terms: formation enthalpy and bond enthalpy. However, I have not quite understood the difference between the two and would be very glad if someone could explain it!

OK, here is the difference: formation enthalpy refers to the formation of a compound from elementary compounds, while bond energy is applicable when considering the formation of a molecule from isolated atoms. The relation between the two is pretty much explained by the Hess's law: the enthalpy of a reaction does not depend on its pathway, but only on the initial and final states.

Now consider the following two pathways for your example.

Pathway 1: we burn graphite in oxygen, just like $\ce{C + O2 -> CO2}$. There is some change of enthalpy associated with this reaction.

Pathway 2: we vaporize graphite to get atomic carbon (see how many C-C bonds are broken, take that into account), atomize $\ce{O2}$ to get elementary oxygen (again, some bonds are broken), and finally produce $\ce{CO2}$ (this time some bonds are formed). Now if you'll look up the energies of all bonds involved and carefully add/subtract them all together, you'll have a decent approximation of the reaction enthalpy obtained in the pathway 1.

(Normally it works the other way around, that is, we use reaction enthalpies to calculate bond energies, but... whatever.)

Formation Enthalpy: energy it takes to form a compound from elements

Bond Enthalpy: an approximation of the energy it takes to form a compound from anything by calculating the bond energies(bond energies of the reactants - bond energies of the products)


The thermal decomposition of NH4Cl When you heat $\ce{NH4Cl}$ it decomposes into $\ce{NH3 + HCl}$ gases. My question is why does it decompose into ammonia and hydrogen chloride gases on the molecular level. Is it because in the $\ce{N-(HCl)}$ bond, The $\ce{N-H}$ bonds breaks easily since it requires only $\pu{390 kJ/mol}$ and $\ce{H-Cl}$ requires far more heat energy? Thus the $\ce{N-H}$ bonds breaks first?

In the solid state of ammonium chloride, the nitrogen and hydrogen atoms are connected by covalent bonds, forming the ammonium ion. The bond length is about 98 pm. Each ammonium ion is surrounded by eight chloride ions (and in turn, each chloride ion is surrounded by eight ammonium ions). The distance between a chloride and the closest hydrogen atoms is about 237 pm. In contrast, the bond length in HCl is 127 pm.

In the reaction $$\ce{NH4Cl(s) -> NH3(g) + HCl(g)},$$

you lose the ionic interactions, and ammonium gets deprotonated while chloride gets protonated (lose an N-H bond and gain a H-Cl bond). The change in entropy for the reaction is positive, so it is favored at high temperature.

We might think that high temperature makes the H-Cl bond stronger and the N-H bond weaker. The bond dissociation energy for N-H is 390 kJ/mol while that of H-Cl is 432 kJ/mol. However, bond dissociation energies are not directly relevant because we are not transferring a hydrogen atom (making $\ce{NH3+ + HCl-}$). Instead, we are transferring a proton, so the pKa of ammonium and hydrogen chloride are more relevant but not directly either because this does not happen in aqueous solution.

The key is that the products are in the gas phase. If the products of the decomposition were ions and had opposite charges, they would not remain in the gas phase. In aqueous solution, you get $$\ce{NH4Cl(s) -> NH4+(aq) + Cl-(aq)}$$ because water is good at solvating ions. In the gas phase, the most stable combination of the atoms at hand is $\ce{HCl + NH3}$.

Here we have a protolytic reaction like

H2O + HCl <=> H3O+ + Cl-

NH3 + HCl <=> NH4+ + Cl-

There no other products possible.


Bond stability of C-C in ethane, ethene, and ethyne Which one of the three compounds has the highest $\ce{C-C}$ bond stability?

Ethane has $\mathrm{sp^3}$ hybridisation so less $\mathrm{s}$-character and lesser electronegativity compared to ethyne. Ethyne has high $\mathrm{s}$-character and high electronegativity. Therefore the covalent character of $\mathrm{sp^3-sp^3}$ in Ethyne should be stronger and have more stability right? the order would therefore be $\mathrm{sp-sp}$ >  $\mathrm{sp^2-sp^2}$ >  $\mathrm{sp^3-sp^3}$
Some online sites tell so, but others say since the $\ce{C}$ atoms are so close in ethyne that the nuclear repulsion would make it less stable. So ethane's $\ce{C-C}$ would be stronger and that therefore the order should be reversed
Edit: nutshell- sigma bond of which hybridisation is strongest among $\mathrm{sp}$, $\mathrm{sp^2}$ and $\mathrm{sp^3}$?
I took ethane, ethene, and ethyne just as a example

The truth is: in reality, sigma bonds, bond orders and hybridization do not exist (not even bonds or orbitals do). They are just concepts to explain the reactivity, stability and geometry of a molecule (and in the case of orbitals to build approximate many-particle wave functions). There is thus no way to compare a sigma-bond strength without choosing some model which introduces bias.
The least-biased way to get there is probably starting from the experimental dissociation energy already mentioned. These can be divided by the bond order (which, again, is not a thing, but quite well established):

ethane: 350 kJ/mol -> 350 kJ/mol per bond
ethene: 610 kJ/mol -> 305 kJ/mol per bond
ethyne: 835 kJ/mol -> 278 kJ/mol per bond

In general: for any chosen approach, the sigma bond in ethane will be the strongest even though the bond distance of ethyne is the smallest. The additional electrons of the pi bond(s) destabilize the bond.


Is it known for sure that bases feel slippery because of the production of soap/surfactant? Discussion around the question Why does bleach feel slippery? has started me thinking about the saponification explanation for the slippery feeling of basic solutions.
According to Wikipedia:

Alkaline drain openers can dissolve hair (containing proteins) and fats inside pipes via alkaline hydrolysis of amide and ester respectively:
$$\ce{RCONH2(amide or proteins)+ OH− → NH3 + RCOO−}$$
$$\ce{RCO2R’(ester or fats)+ OH− → R'OH + RCOO−}$$

and just above that:

Essentially, the hydroxide ions from the basic lye attack the carbonyl carbons of the fat, which eventually kicks off the hydrophobic tails of the triglyceride (tristearin/fat) to isolate glycerol.

So I am not sure if the "saponification" explanation for why a small amount of a fairly strong base feels slippery between thumb and forefinger really says that the reaction has produced a true soapy surfactant or that it should really say that it is lubrication by glycerol-like compounds that generate the intense slippery feeling.
Wikipedia also says:

The Stratum corneum (Latin for 'horny layer') is the outermost layer of the epidermis, consisting of dead cells (corneocytes). This layer is composed of 15–20 layers of flattened cells with no nuclei and cell organelles. Their cytoplasm shows filamentous keratin. These corneocytes are embedded in a lipid matrix composed of ceramides, cholesterol, and fatty acids. (emphasis added)

Corneocytes are keratinocytes and named as such because of the abundance of keratin protein filaments they contain.
I seem to remember my very smart high school chemistry teacher telling us that the breakdown of protein was also important, but I don't remember the production of soap as being so central. So I'd like to ask if there is any actual empirical evidence - ideally with a scientific reference, that demonstrates which molecules present between the fingers in this situation is primarily responsible for the slippery feeling. Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing inter-cellular lipid layers that would be exposed by the removal of successive layers of cells within the stratum corneum.

below: Epidermal layers, from here.

Potential candidates might include:

Soap/surfactant,
Glycerol produced from hydrolysis of "finger lipids",
Broken down protein products, or
the pre-existing [inter-cellular lipid layers][5] that would be exposed by the removal of successive layers of cells within the stratum corneum.

Very good guesses but none of the above actually (i think, what does "finger lipids" mean?). But the reason bases feel slippery to bare skin no matter how much you scrub first, is bases attack the phospholipid bilayer of your cells quite readily not only saponifying them, but breaking the inter-molecular bonds of molecules that already resemble a soap. Take for example phosphatidylcholine:

The phosphochloline already has all of the elements to make a soap and when base is added, the molecule does not need to interact with other molecules to achieve local charge balance, but now have the ions from the base to balance, which breaks intermolecular bonding creating a small layer of soap, making the base feel slippery.

As a chemist, I have no idea what "glycerol-like" means. I guess erythritol, sugars, sugar alcohols, sugar acids, amino sugars, and other polyols qualify. I just dissolved some sucrose on my fingers, and didn't notice any significant slip. Even more significant US Patent Application US 5709849 A claims that polyhydrics are "sticky", not slippery. So, your first task is justifying (to yourself, at least) your claim that polyhydrics like glycerol are slippery.  Also, your first Wikipedia link claims OH- "reoxidizes" Al2O3. That is ludicrous. Coming across some nonsense like that is a good reason (imho) to stop reading. Or, at best, take anything else claimed as dubious. Now your question is imho an interesting question but isn't clear enough. You seem to acknowledge that having a slippery feel is different than actually being slippery. We can measure the coefficient of friction between two surfaces, we can't objectively measure the feeling we get when pressing and rubbing thumb to forefinger. Broadly, I'd mechanistically reduce your question to:1. does a strong base change the physical structure of the skin? 2. does a strong base change the composition of the skin? and 3. does a strong base change the lubricants present between the two (skin) surfaces? (Assuming that we've established that the coefficient of friction does change in the presence of a solution of a strong base). There are some experiments you could do to determine whether it was a permanent change in structure and/or composition causing it. Water makes things slippery (in general). Soaps are well known lubricants. The use of soaps, surfactants, and oils is well known in the cosmetics formulator's repertoire, but it is mostly an area of empirical science. Feel and tactility aren't easily reduced to molecular effects. Distinguishing between mechanical (physical, structural) effects and chemical effects isn't easy when both occur together. It's certainly the case that hydrogen bonding is important for the adhesion/slip of skin. While soaps might orient at the surface, it's possible that the reduction in the number of 'available' h-bonding sites is what reduces friction/adhesion between skin and another polar surface. A strong base will provide more opportunity for chemical groups on the skin's surface to h-bond to it instead of to polar groups on the other surface. Based on the slippery feel my old finger tips have, another aspect to this is the ability of base to extract (solubilize) oils present on the skin surface which may be there to improve friction/tactility. My best guess, and it is just a guess, is that since surfactants at neutral pH feel slippery, that the main tactile effect of base on skin is the creation of soaps. But a definitive answer this is not.


Spin quantum number of electron Why is the spin quantum number of an electron +1\2 or -1\2 in an atom? Is this an experimental value or theoretical value?

It is both experimental and theoretical. Electrons have angular momentum, and sometimes we want to discuss whether it lines up with another direction (ie typically the direction an electromagnetic field, since electrons are charged their angular momentum makes them magnetism and that interacts with electromagnetic field).

How should we talk about the direction of the angular momentum? We can say "up" and "down" or "forwards" and "backwards", and sometimes we do that. But if we want to use it in calculations, it is easier to give them numbers. So what numbers should we give? They are opposite directions so really it makes sense to use opposite numbers. So what opposite numbers should we use? We can use any, so why not use a gap of "1" between them? So what opposite numbers have a gap of 1 between them? 1/2 and -1/2.

There are some other detailed reasons why 1/2 is a convenient number, but it is in my view fair to say we just want to represent two opposite directions in calculations, and the numbers 1/2 and -1/2 suit this well.