How does NaCl maintain its crystalline structure? My understanding is that $\mathrm{NaCl}$ is an ionic compound, in which $\mathrm{Cl}$ becomes (effectively) $\mathrm{Cl^-}$ and $\mathrm{Na}$ becomes $\mathrm{Na^+}$. So I understand why I would get a "sea" of particles that would stick together.

But why does the above mean that it will have a face centered cubic structure with the ions held in place so rigidly?

Crystals have inspired a great many chemists because they are fascinating for a good reason. Not only are they aesthetically pleasing, but they serve as an excellent subject to tour a variety of theoretical subjects important for understanding high-level chemistry.

Crystalline materials are made up of periodic structures. We’re only going to primarily focus on binary compounds where there is not a high degree of covalency. There are several ways to think about this problem, but let’s start with the melting of a crystal.

We say that at some definite temperature a highly ordered crystal will melt into a liquid. Those of us familiar with the language of equilibrium thermodynamics might recognize that the change in free energy for this phase change can be written, at constant temperature, as,

$$ G_\text{liquid} - G_\text{crystal} = H_\text{liquid} - H_\text{crystal} - T ( S_\text{liquid} - S_\text{crystal} ) $$

$$ \Delta G = \Delta H - T \Delta S $$

If we suppose that this process is spontaneous then we would say that the change in Gibbs’ free energy is negative, i.e. $\Delta G < 0$. This is true if and only if,

$$\Delta H < T \Delta S$$

Traditionally we interpret this as saying that there is a thermally-driven increase in entropy when we melt a highly ordered crystal into a liquid which more than offsets the energy cost associated with the enthalpies of the interactions holding that crystal together. A chemist tends to learn early on that the reverse is not necessarily true: at some definite temperature a perfect crystal rarely forms from the liquid. This inability to just heat up any substance and always produce a perfect crystal by cooling illustrates how crystal formation is a case of kinetic- rather than thermodynamic- control. So the process by which you form your crystal could possibly result in a different crystal structure. Sometimes crystal structures change just by altering the temperature of the chamber you’re measuring the crystal structure in! Now neither of these cases apply to sodium chloride to the best of my knowledge.

The formation of an ionic crystal such as sodium chloride is a delicate balance between electrostatic attraction and Pauli repulsion. Electrostatic attraction says that between two different charges, $q_+$ and $q_-$, there is a Coulomb force given by, $$F= \frac{k q_+ q_-}{r^2}$$

where $r$ is the distance between the two charges. If one plays with the numbers then it’s easy to see that at short distances the force is strongest, but there is a limit to how close they may come together. Eventually a repulsive force due to a quantum mechanical principle called the Pauli Exclusion Principle overpowers the attraction. An equilibrium results in which the atoms sit a certain distance from one another so that, if you will humor me, the “forces” between them balance out. This is why we traditionally represent crystal packing using marbles with a unique radii. The radii of the hard marble represents where the Pauli repulsion overpowers the attraction.

You might say, “Sure, we have these kinetic, electrostatic and quantum mechanical factors to consider, but how do these help with the final crystal structure?” Hold your horses, we’re getting there.

A famous mathematician and scientist thought about the most efficient ways to pack spheres of the same size together. By most efficient I mean this in terms of what FedEx considers efficient, fitting things together into the smallest possible volume. This is also what electrostatics want. Kepler suggested that the best way to pack spheres with this in mind can be maximized in two ways: fcc and hcp.

This conjecture, however, rests on all the spheres being uniform. We can’t assume this is the same for the atoms in table salt because the cation, sodium, is said to be smaller than the chloride anion. A tool that is useful as a guide for helping predict the structure is based off of the relative size of the cation, $r_\text C$, to the anion, $r_\text A$. This radius ratio, $r_\text C / r_\text A$, is mostly only useful for simple binary species. These $r_\text C$ and $r_\text A$ are tabulated as ionic radii in many chemistry books; they are contrived by various rules proposed by researchers, such as Linus Pauling, applied to the experimentally determined interionic distances. In this system we find it useful to discuss how many neighbors an anion or a cation has of the opposite charge.

The coordination number ($\mathrm{CN}$) tells us how many neighbors a type of atom has around it. A certain $r_\text C/r_\text A$ gives us a feel for the probable CN. From one source I was able to calculate the ratio for $\ce{NaCl}$ as $r_{\ce{Na+}}/r_{\ce{Cl-}} = 0.564$. Most textbooks will say that if your ratio is $0.414{–}0.732$, then you have $\mathrm{CN}=6$. I’ll show you how to calculate the minimum value for $\mathrm{CN}=6$. The easiest way to obtain the maximum is to obtain the minimum for $\mathrm{CN}=8$.

Briefly, if we put a small cation on a $XY$ plane and surround it in the manner shown below then we could also place two circles in the $Z$-axis above and below the central cation for $\mathrm{CN}=6$. (I won’t show those two in the $Z$-axis for clarity.) We would say that a given $\mathrm{CN}$ is stable only if the spheres are all touching each other. We would say that for the ratios drawn that a $\mathrm{CN}=8$ is not stable because our cation’s sphere would be too small and wouldn’t touch all eight of its neighbors. A higher density is always preferred, so $\mathrm{CN}=4$ is not preferred when $\mathrm{CN}=6$ is stable.

Clearly, $\angle \mathrm{DAC} = 45^{\circ}$. Moreover, we can say that $\overline{\mathrm{DC}} = r_\text A$ and $\overline{\mathrm{AC}} = r_\text A + r_\text C$. Trigonometry will tells us that $$\sin 45^{\circ} = \frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}$$ Substituting in those values just determined,
$$\frac{\sqrt{2}}{2} = \frac{r_\text A}{\overline{r_\text A + r_\text C}}$$

Solving for $r_\text C/r_\text A$ gives,

$$r_\text C/r_\text A=\frac{2}{\sqrt{2}} -1 = 0.414$$

which is the lower bound desired.

So we can say that the $\mathrm{CN}=6$ for the cation. Similarly you can figure out that $\mathrm{CN}=6$ for the anion. This is why we say it has a 6:6 coordination pattern, which is indicative of a sodium chloride type structure. Cesium chloride, on the other hand, is 8:8 and indicative of its structure type. We tend to say that these patterns are consistent with a certain crystal structure because all the pieces fit together consistently. This geometric concept is based purely on sphere packing and is not without its limitations.

A more quantitative way to figure out which geometry is preferred is to calculate the electrostatic energies for the different geometries. A Russian by the name of Kapustinskii derived a formula to do this,

$$E_\text{lattice} (\mathrm{kJ/mol}) = \frac{\alpha}{r_\text C + r_\text A} $$

From here you need to put it in terms of the radius ratio with a constant $r_\text A$,

(I should note that this is evil algebra: $ \frac{a/b}{d/b + 1}= \frac{a}{b+d} $ where $a=\alpha$, $b=r_\text A$, $d=r_\text C$.)

$$E_\text{lattice} = \frac{\alpha/r_\text A}{\frac{r_\text C}{r_\text A} + 1} $$

Basically $\alpha$ is different for a change in the crystal structure type and from there you compare which has the preferred energy. You can reproduce the geometric trends using this more quantitative model. There are problems with this approach too, such as the neglected dipole and quadrupole and covalency. Yes, even in $\ce{NaCl}$ there is some covalent character. I think we say that $\ce{NaCl}$ $67\ \%$ ionic character. These become important when two different crystal structures are close in energy or when more exact calculations are required.

How does $\ce{NaCl}$ maintain its crystalline structure? It is a good question.

Because in crystalline structure of $\ce{NaCl}$  6 electrons of $\ce{Na+}$ ion are surrounded by 6 electrons of $\ce{Cl-}$ ion and 6 electrons of $\ce{Cl-}$ ion are surrounded by 6 electrons of $\ce{Na+}$. That's how $\ce{NaCl}$ maintains its crystalline structure.


Why is the chemical formula for ethanoic acid CH3COOH and not C2H4O2? Why does the hydrogen ion move to the front of the formula and not to the subscript of the H already in the formula?

The first one is preferred since its makes it easy to see that it has a carboxyl group and a methyl group, so its easier to draw the Lewis structure and figure out the molecule's properties. For example, from the first formula one can see that it is an acid. However, its more difficult to figure that out from the second one.


Unit of the equilibrium constant: contradiction of Bridgman's theorem? The following equation is standard in thermodynamics:

$$
\Delta G^\circ=-RT\log(K)
$$

where $K$ is the equilibrium constant.  In dimensional analysis, Bridgman's theorem tells us that the argument of a transcendental function (like $\log$) must always be dimensionless.  But $K$ may have dimensions (depending on the particular equilibrium).  Why is this OK?

Note: working out dimensions explicitly gives:

\begin{align}
[R]&=EN^{-1}\Theta^{-1}\\
[T]&=\Theta\\
[\Delta G^\circ]&=EN^{-1}
\end{align}

where $\Theta$ is the dimensions of temperature, $N$ is the dimensions of number and $E$ = dimensions of energy = $MLT^{-2}$.  From these, we see that the quantity $\log(K)$ ought to be dimensionless, implying that $K$ should be dimensionless as well.  But it isn't always.

Furthermore, suppose $K$ has dimensions $NL^{-3}$ (for an equilibrium of the form $A+B\leftrightarrow AB$, say).  Suppose now that we scale the units for number by a factor $a$.  Then we get new values for $K,R,\Delta G^\circ$, given by:

\begin{align}
\hat K&=K/a\\
\hat R&=aR\\
\hat{\Delta G^\circ}&=a\Delta G^\circ
\end{align}

From our original equation:

\begin{align}
\Delta G^\circ&=-RT\log(K)\\
a\Delta G^\circ&=-aRT\log(K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(a\hat K)\\
\hat{\Delta G^\circ}&=-\hat RT\log(\hat K)-\hat RT\log(a)
\end{align}

So the new quantities do not satisfy the old equation.  Rather, they satisfy the old equation, but with a constant factor of $-\hat RT\log(a)$ added on.  What is going on here?

The problem is that people are often sloppy with the definition of quantities. The equilibrium constant $K$ in your first equation is indeed a dimensionless quantity while the equilibrium constant $K_c$ that is usually used to describe an equilibrium in a solution is not. I will take some detour to show where they come from and how they are connected.

From thermodynamics it is known that the Gibbs free energy of reation is given by
\begin{equation}
 \Delta G = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum \nu_{i} \mu_{i} \ ,
\end{equation}

where $\xi$ is the extent of reaction and $\nu_{i}$ and $\mu_{i}$ are the stochiometric coefficient and the chemical potential of the $i^{\text{th}}$ component in the reaction, respectively.
Now, imagine the situation for a ideal system consisting of two phases, one purely consisting of component $i$ and the other being a mixed phase comprised of components $1, 2, \dots, k$, in equilibrium.
Since the system is in equilibrium and shows ideal behavior we know that the chemical potential of component $i$ in the mixed phase (having the temperature $T$ and the total pressure $p$), $\mu_{i}(p, T)$, must be equal to the chemical potential, $\mu^{*}_{i}(p_{i}, T)$, of the pure phase having the same temperature but a different pressure $p_{i}$, whereby $p_{i}$ is equal to the partial pressure of component $i$ in the mixed phase, namely
\begin{equation}
 \mu_{i}(p, T)= \mu^{*}_{i}(p_{i}, T).
\end{equation}
From Maxwell's relations it is known that
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = \left( \frac{\partial}{\partial p} \biggl(\frac{\partial G^{*}}{\partial n_{i}}\biggr) \right)_{T} = \Biggl( \frac{\partial}{\partial n_{i}} \underbrace{\biggl(\frac{\partial G^{*}_{i}}{\partial p}\biggr)}_{=\, V_{i}} \Biggr)_{T} = \left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}
\end{equation}

but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to
\begin{equation}
 \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i}
\end{equation}

and one gets
\begin{equation}
 \left( \frac{\partial \mu^{*}_{i}}{\partial p} \right)_{T} = v_{i} \ ,
\end{equation}

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase.
Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get
\begin{equation}
 \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p   \ ,
\end{equation}
so that, introducing the mole fraction $x_{i}$,
\begin{equation}
 \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ .
\end{equation}
Please, note that there is a dimensionless quantity inside the logarithm.
Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$
\begin{equation}
a_{i} = \frac{\varphi_{i} p_{i}}{p^0}
\end{equation}
where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition.
When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises
\begin{equation}
 \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ .
\end{equation}
Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets
\begin{equation}
 \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ ,
\end{equation}
where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure.
Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that
\begin{equation}
\ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}
So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant
\begin{equation}
 K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ .
\end{equation}
which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via
\begin{equation}
 K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
it is more or less proportional to $K$ and thus gives qualitatively the same information.

Edit: If the solution at hand behaves like an ideal solution then by definition it's activity/fugacity coefficient is equal to one. Furthermore the state of ideality is defined with respect to standard states: for an ideal solution this is $c^{\ominus} = 1 \, \text{mol}/\text{L}$. Using this together with the ideal gas law on the relation between $K$ and $K_{c}$
\begin{equation}
 K = \prod_i [\underbrace{\varphi_{i}}_{= \, 1}]^{\nu_{i}} \Bigl(\underbrace{\frac{RT}{p^{0}}}_{\substack{= \, 1/c^{\ominus} \, = \, 1 \, \text{L} / \text{mol} \\ \text{per definition}}}\Bigr)^{\sum_i \nu_{i}} K_{c} \qquad \Rightarrow \qquad K = \left(\frac{L}{\text{mol}} \right)^{\sum_i \nu_{i}} K_{c} \ .
\end{equation}
one sees that for an ideal solution $K_{c}$ is identical to $K$ scaled by a dimensional prefactor.

Edit: I forgot to mention there are also "versions" of the equilibrium constants that are defined in terms of partial pressures or mole fractions which provide a more suitable description for gas equlibria but all those "versions" can be traced back to the original equilibrium constant.

Yes, K is unitless, always, if defined by your equation:

$$\Delta G^\circ=-RT\log(K)$$

There is no way around this, unless you use some strange units for free energy, R or T.

Beyond the obvious mathematical answer, if you dig deeper you find that equilibrium constants are unitless because the expression relating $K$ to $ \Delta G^o$ requires the definition of reference states. This is because $\Delta G^o$ is defined as the free energy change for the transformation of reagents to products in the specified reference states. As a result, equilibrium constants contain ratios of activites (which can be expressed as functions of concentration) of reagents and products in their reference states and at equilibrium.

For instance, for a species i in the gas phase terms in the equilibrium constant involving that species consist of ratios of fugacities $(f_i/f^o)^\nu$ where $\nu$ is a stoichiometric factor, $f_i$ is the fugacity of the species, and $f^o$ is the fugacity in the reference state (fugacities become equal to pressures in the limit of zero pressure). The resulting equilibrium constant is a product of such unitless ratios, and therefore also unitless.

That equilibrium constants are sometimes reported with units is contrary to accepted conventions, but is convenient: specifying a set of units together with the equilibrium constant implicitly declares the reference state(s).


Why do chromyl bromide and chromyl iodide not exist? It is known that chromyl chloride exists (popularly known as the product of a test for presence of chloride ion) but chromyl bromide and chromyl iodide are unknown.

Why is it so? Why don't they exist? I'd guess it has something to do with the large size of bromine and iodine compared to chlorine, but I'm not sure.

What are the possible reasons for their non existence?

Chromyl Bromide does exist.  See Halogen Chemistry at page 246:

  Deep violet needles of chromyl bromide are obtained by treating chromium(VI) oxide with $\ce{CF3COBr}$ and HBr in the presence of phosphorus pentoxide. A less pure sample is made by treating chromyl chloride with a large excess of hydrogen bromide. The compound, which is hygroscopic, is purified by vacuum distillation at —135° (Zellner, 1949; Krauss and Starke, 1962).

The book continues:

  Chromyl iodide has not been prepared and may not exist because of the powerful oxidizing property of the chromium (VI) oxidation state.

See similarly The Chemistry Of Chromyl Compounds Chem. Rev., 1958, 58 (1), pp 1–61:

  Chromyl bromide has likewise been isolated within the past few years (273,
  274). The modern work corrects earlier reports of the compound (77).

  While there has been one report of the formation of chromyl iodide as a deep
  red liquid from a mixture of dichromate, iodide, and sulfuric acid (87), a majority of observers (77, 96, 128, 196, 255) deny this observation and the existence of the compound. There is little likelihood that the iodide can exist, in view of the large free-energy change attendant on the reaction:

  $\ce{2Cr^{+6} + I- -> I^{+5} + 2Cr^{+3}}$

This is because of the oxidation potentials of the halogens. Br- anion is oxidised to Br2 by Cr+6 but Cl- is not. Exactly what i am saying perhaps you would like to refer (E=RC-RA) E is electrode potential RC reduction potential at cathode and RA reduction potential at anode


Trouble with Converting from Structural Formulas into Newman Projections I have the following question from my Organic Chemistry textbook:

I'm asked to convert the following line diagram into a Newman Projection.

The arrow pointing to the carbon signifies "looking down the carbon".

My logic was the following:

At the carbon that is being pointed at by the arrow, there are 3 bonds to other carbons, all of which are in the same plane due to the absence of any 'dashed lines' or 'solid wedges' that would indicate otherwise.

I then concluded that the geometry of that carbon is trigonal planar and that it has a lone pair, since all the bonds to other carbons are in the same plane.

However, drawing a Newman Projection for an atom with trigonal planar geometry proves to be impossible.

I'm not sure how to go about converting this line diagram into a Newman projection.

How can I do so?

Thank you.

this is how you can convert the str.s into each other. It should have had some wedge/dash or sawhorse information... for the conversion.


How do we know the natural abundance of isotopes on Earth? Without knowing the Average Atomic Mass or the percent abundance, how do we know that Protium is the most prevalent hydrogen isotope? What methods did scientists use to come to this conclusion? How did they reach such a conclusion?
I'm still starting out in learning chemistry and I felt like I hit a boulder. I know the formulas and how to get the natural abundance, but I haven't heard a concrete explanation of the concept. I know that it's impossible to survey all the hydrogen elements on Earth to see which isotope is abundant, so what method did scientists use?

Because we can sample the environment with great precision
You seem to assume that we can't know the exact makeup of a sample unless we already know the components. This is wrong. Mass spectroscopy can reliably measure the exact mass of every isotope in a sample and the relative abundance of each. If, for example, a sample of sea-water is measured, we can reliably tell the exact proportion of all the hydrogen isotopes and all the oxygen isotopes in it.
Given that his is possible, estimating the true natural abundance is a matter of statistical sampling. Ocean and river water is fairly well mixed so this is easier than it sounds (at least when there are no processes that radically alter the isotopic composition). So we have a lot of samples, for example, from different oceans and different locations in those oceans. Given that, we can be fairly confident that the isotopic composition doesn't vary a lot and, therefore, we can get a good estimate of the "natural" abundance of each isotope.
But the techniques we use are precise enough to notice that there are some (small) variations in that composition. Some processes do (very slightly) alter the isotopic ratios. Evaporation tends to happen slightly faster for lighter isotopes, for example, as do some biological processes. Some of those processes are temperature dependent leading to the ability to identify the ambient temperature preset when some fossils were alive by measuring the ratio of oxygen isotopes in them.
The key point is that we do have tools that can measure isotope ratios very precisely. So we know the ratios in any sample very precisely. With a very large number of samples, we can generate a reliable picture of the typical abundance of each isotope which is what we refer to as the natural abundance.

These numerical values have been determined by mass spectrometry.
In brief, it starts form a box like a match box in the vacuum. No air in the room around the box. There are many plates and tubes in this box. First a small tube is introduced for delivering some hydrogen gas in the box. Second two thin metallic points have been introduced in advance one through the roof and one through the bottom of the box. Then a high tension is applied between the two points : a spark is produced that breaks the $\ce{H2}$ molecules into parts, producing $\ce{H}$ atoms, electrons, $\ce{H2^+}$ and $\ce{H+}$ ions. These fragments will soon recombined as soon as they get out of the spark. Except if they are prevented to do it by applying an outer tension. And indeed there are two vertical metallic plates on the lefthand and righthand sides of the box.
One of these vertical plates is charged positively and the other negatively. This last one attracts positive ions like $\ce{H+}$ and $\ce{H2+}$. They will move and fall on this plate. Not all ! Because this plate plate is pierced in its middle, and just on the other side there is another plate which is charged at a higher voltage. So the positive ions arriving in the middle of this hole are more strongly attracted by the second plate and they cross the hole.
Well ! This second plate is also pierced in its center, and it is followed by a third pierced plate, etc. The last plate is not pierced but it is far away from the last but one plate. This whole system of plates produces a thin beam of charged particules crossing all holes before being discharged when hitting the last plate. But between the last plate and the last but one plate, a big magnet is approached that is strong enough to deflect the trajectories of the ions in the beam. Small ions like $\ce{H+}$ are more deviated than heavier ions like $\ce{H2+}$. So by dividing the last plate into many independent small parts, you may know whether and how many ions have been slightly or strongly deviated. So you may know the proportion of isotopes in the beam.
OK ?


Can an atom have more than 8 valence electrons? If not, why is 8 the limit? According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained.

So, can an atom have more than 8 valence electrons?

If this is not possible, why can't an atom have more than 8 valence electrons?

2017-10-27 Update

[NOTE: My earlier notation-focused answer, unchanged, is below this update.]

Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.

However, the same bonding mechanisms that enable formation of greater-than-8 valence shells also enable alternative structural interpretions of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. Manishearth's excellent answer explores this issue in much greater detail than I do here.

Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:

(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or

(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.

While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.

The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic ("astronomical” might be a more apt word).

A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggests that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.

Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.

Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.

That is why my overall answer to this question is simply "yes".

Question: Why are octets special?

The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?

In a nutshell, the reason is that for any energy level  above the special case of the $n=1$ shell (helium), the "closed shell" orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.

It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.

The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus to avoid the larger energy penalties imposed by Pauli exclusion.

Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.

A final note

I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.

(A hint: If you have ever watched people create a two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)

Original 2014-ish Answer (Unchanged)

This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.

I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.

The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.

Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)

Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.

Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites.

Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.

2014-04-04 Update

Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.

The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.

The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.

The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.

Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."

Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.

Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologues, one of the consequences of which is that they have nearly identical physical chemistry properties.

Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:

Let's take a look at the periodic table: in the first row, there's only two elements: Hydrogen and Helium. They do not follow an octet rule. Hydrogen can only have a maximum of two electrons on the valence orbital. It turns out that the octet rule is not exclusive, meaning it is not the only rule that helps understand Lewis structure and electron configuration. Why do we use the octet rule, then?

Every period in the periodic table represents an energy shell of an atom. The first period represents the shell K, the first energy level, which has only the s orbital. Every orbital can only be filled with 2 electrons, both with a quantum spin towards opposite directions. Thus the maximum number of electrons possible for the first energy level shell, K, is 2. This is reflected in the fact that Helium is a noble gas, yet only contains 2. The second energy level shell, L, has the s orbital and the extra 3 p orbitals. Those add up to four orbitals or 8 electrons. Because the elements most commonly used are in the second and third period, the octet rule is in frequent use.

Elements of the third energy level are very similar. They still follow the octet rule, because even though now the have 5 d orbitals, no orbital needs to be filled. The electronic configuration shows that 4s is filled before 3d, so they don't need to fill the d orbital, thus they usually also obey the octet rule. However, third energy level shell elements, unlike second-row elements, (see Gavin's comment fir reference) are not limited to the octet rule. They can form hypervalent molecules in certain cases where the use that d orbital and fills — this is not the case with all apparent hypervalent molecules, SF6 is not hypervalent, it uses weak ionic bonds and polarity, but there still are hypervalent molecules out there. It will always depend on which state is more convenient in terms of electrostatics.

At the fourth energy level shell, there are f orbitals introduced, but we are not even close to filling them at that point because we first need to fill the d orbitals. The 5 d orbitals signify 10 electrons, plus the previous eight from the octet rule, sum up to 18. This is the reason why there are 18 columns in the periodic table. Now, a new rule superposes, and this is the well known 18-electron rule, which was mentioned above. Transition metals obey this rule with more frequency than not, though there are occasions in which they still obey the octet rule. At this point, with so many orbitals to fill, and with electrostatics playing a role in electronic configuration, we can obtain different cations from the same element with certain metals. That is also why they don't discuss oxidation state numbers with transition metals like they do with the first three rows of the table.


Total number of stereoisomers of 2,3,5,6-tetrachlorooctane Total number of stereoisomers for:

I tried the following combinations:
RRRR, SSSS, RRRS, SSSR, RRSS, RSRS, SRRS, RSSR, RSRR, SRSS, RRSR, SSRS, SSRR, SRSR.
So, total number of stereoisomers I got is 12, but my teacher says it's 16. I don't know how to figure out what other four isomers are.

If you have studied combinatorics, and if you know the fundamental principal of counting, then this is basically making a four letter word with two letters- R and S. Each place has two choices, and you have four spaces to fill, so you get 2×2×2×2= 16 optical isomers for this molecule. Since the terminal alkyl groups are different, so there is no chance of a meso-compound. And as Ivan rightly stated in the comment, to list the isomers, make all possible combinations of R and S(you have missed a few as pointed out by porphyrin in the comments)

There would be 16 stereoisomers as suggested  by other's answers.But there would be less no
of enantiomer as this molecule posses plane of symmetry in certain configurations


Electrolysis of concentrated sulphuric acid In my textbook it is given that for electrolysis of dilute sulfuric acid  at anode following reactions can occur:

At moderate concentrations

$\ce{2H2O ->  O2 + H+ +4 e-}$

And for high concentrations

$\ce{2SO4- ->  S2O8^2- +2 e-}$

SRP value for first reaction is less than second and hence the first reaction should take place. But why does that not hold for high concentrations?

This is because in concentrated sulfuric acid solution equilibrium condition will favour more water molecules than oxidation of water into protons. So there will be an over potential required (to go against the equilibrium) , that is extra potential beyond the theoretical reduction potential derived from thermodynamics to complete the reaction. So the total potential required will be theoretical Plus overpotential. In case of oxidation of sulphate reduction potential will be much less that for water ,thus oxidation of sulphate happens.


Calculating the EMF of cell Find the EMF of the following cell :
  $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$
  Given:  $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$
  and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$
And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$
Then using the Nernst equation,
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$
And substituting,
$n=2$,
$[\text{Products}]=2.53\times10^{-7}$,
$[\text{Reactants}]=0.004$,
$T=\pu{298K}$,
$R=\pu{8.314JK^{-1}mol^{-1}}$, and $F=\pu{96500C}$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i  misunderstood something?

Use RT/F = 0.059 or 0.06
It will simplify your calculations.

Note:- I have converted ln into log.
So formula becomes:-
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{0.059}{n}\log\frac{[\text{Products}]}{[\text{Reactants}]}$$

Just try this in problems. Calculations would become easier.


Why can enolizable aldehydes undergo Cannizzaro reactions? Why does 2-methylpropanal undergo a Cannizzaro reaction even though it has an $\alpha$-hydrogen?

There is never an either–or in chemistry; it is typically a both instead. In the case of a hydroxide ion approaching an aldehyde with an α-hydrogen, it can both deprotonate the hydrogen creating the corresponding enolate and attack the carbonyl nucleophilicly to generate the required intermediate for a Cannizzaro reaction.

Most of the time, if there are two (or more) possible pathways and neither of them is intrinsically less hindered due to a strong reason, both pathways will be taken.

Referring to Jans answer one might ask for a reason that in the case of 2-methyl-propanal the canizzaro reaction gets favoured over the aldol reaction in comparison to other $\alpha$-deprotonizable aldehydes.

Take a look at the intermediate of the aldol reaction: You will get a trifold substituted carbo-anion, which is quite unfavourable in terms of stability. (Talking basic conditions as these are also required for canizzaro reactions)


Lifetime of a (reversible) hydrogen fuel cell system I am trying to determine how long a system that constantly converts water into H and O and then back will last, as in roughly how many cycles. Now this system can either be made of say a reversible fuel cell or a non-reversible fuel cell combined with a dedicated electrolysis machine (if there is a difference in lifetime). As far as I can tell, platinum is used (almost) everywhere as catalyst and as far I can determine it seems to corrode somewhat throughout use, especially in terms of splitting the oxygen in the fuel cell. I am no professional though and I can only find limited information so I would appreciate it if someone could please tell me roughly how long a system like this will last and how.

Fuel cell manufacturers quote a range of different lifetimes which depend on many factors. Backup cells tend to have lower lifetimes as the start-stop process degrades them compared to continuous operation. On the other hand, use as engines in vehicles requires a reasonably long life.

Impure fuel will damage the platinum (CO for example will poison the catalytic surface). But platinum is recyclable so either poisoning or other factors won't necessarily be a fundamental barrier to use (some companies take used cells back at the end of life and reuse the platinum). Membranes and electrodes also degrade mechanically depending on duty cycle and any mechanical stabilisation in the design.

Nedstack quotes lifetimes ranging from 2,000 hours for backup cells (which presumably means years of actual operation as they only get used when other power is cut which isn't a routine happening) and 8,000-16,000 hours for continuous operation (which is two years). Tests of fuel cells in vehicles have reported lifetimes of 7,500 hours (which presumably relates to several years of use in practice), though they seem to need more maintenance than diesel engines.

I have actually purchased and run three fuel cell stacks. One was dead on arrival and never converted any hydrogen to electricity. If you ever buy one, you will notice no warranty, no return policy and no expected useful life. This led us to actually invent a way to probe PEMFC health without putting hydrogen in, and we got a patent on that. The second fuel cell ran for about 40 hours, but the performance degraded quickly over that time. The third one also degraded and died within 200 hours. These were commercial stacks used with extreme care in a research lab facility by expert engineers. Given that demonstration cars and buses also don't last very long, I think we know why fuel cells are not going anywhere. The nature of the degradation is such that there isn't much that can be done to make the PEMFC more robust. See publication by S. Page in Journal of IEEE.


What exactly is the 'shuttle effect' in sulfur batteries? Ref :

https://www.electropages.com/blog/2020/01/why-lithium-sulphur-batteries-are-taking-so-long-be-used-commercially

  Because they continually shuttle between the electrodes, they never change back into a neutral state (which the surface polysulphide ions do)

I don't think this explains the 'shuttle effect' clearly. I still don't understand what it is.

During discharging, lithium ions and sulfur form Li2S8 at the cathode, then it becomes Li2S4, Li2S2, Li2S with continued discharging. During charging, the Li2S move towards the cathode, now, why can't it be reversely reduced to Li2S4, Li2S8. Why can't they ever change back into a neutral state?

OP's title indicated OP is looking for the real mechanism of  the Shuttle Effect in $\ce{Li-S}$ batteries. Yet, OP is asking different questions in the text body. I'd try to address as much as possible in this answer.

A most recent review article (Ref.1) states that:

  Lithium-sulfur ($\ce{Li-S}$) batteries are one of the most promising batteries in the future due to its high theoretical specific capacity ($\pu{1675 mAh g−1}$) and energy density ($\pu{2600 Wh kg−1}$). However, the severe capacity fading caused by shuttle effect of polysulfide needs to be addressed before the practical application of $\ce{Li-S}$ batteries.

To understand what is the Shuttle Effect, OP need to first understand the working chemical mechanism of $\ce{Li-S}$ batteries, and can take a head stat in that direction with this article about Lithium–Sulfur Battery, which gives a fair description of the chemistry involve in the battery. The same Wikipedia article simply described the Shuttle Effect as follows:

  The lithium polysulfide $\ce{Li2S}_x$ ($6 < x < 8$) is highly soluble in the common electrolytes used for $\ce{Li-S}$  batteries. They are formed and leaked from the cathode and they diffuse to the anode, where they are reduced to short-chain polysulfide and diffuse back to the cathode where long-chain polysulfide are formed again. This process results in the continuous leakage of active material from the cathode, lithium corrosion, low coulombic efficiency, and low battery life.

The formation of polysulfide shuttle effect is classified into five steps in Ref.1:

  Formation of long-chain polysulfide,
  Detaching of polysulfide from sulfur host,
  Dissolution of polysulfide into electrolyte,
  Migration of polysulfide toward lithium anode side, and
  Reaction between polysulfide and lithium anode.

If interested, OP may read Ref.1 and Ref.2, which thoroughly discussed the mechanism and possibilities of how to prevent the Shuttle Effect in depth. For example, one such techniques is modifying the separator, which has been done successfully recently (Ref.3):

References:

Wenchen Ren, Wei Ma, Shufen Zhang, Bingtao Tang, "Recent advances in shuttle effect inhibition for lithium sulfur batteries," Energy Storage Materials 2019, 23, 707-732 (https://doi.org/10.1016/j.ensm.2019.02.022).
Chao Deng, Zhuowen Wang, Shengping Wang, Jingxian Yu, "Inhibition of polysulfide diffusion in lithium–sulfur batteries: mechanism and improvement strategies," J. Mater. Chem. A 2019, 7(20), 12381-12413 (https://doi.org/10.1039/C9TA00535H).
Yanpeng Fan, Zhihui Niu, Fei Zhang, Rui Zhang, Yu Zhao, Guang Lu, "Suppressing the Shuttle Effect in Lithium–Sulfur Batteries by a UiO-66-Modified Polypropylene Separator," ACS Omega 2019, 4(6), 10328-10335 (https://doi.org/10.1021/acsomega.9b00884).

Apparently, the main drawback of the Lithium-Sulfur battery is that it should work so : At the anode made of lithium, $Li^+$ ions are emitted and go to the cathode made of sulfur. This sulfur should accept the electrons, to make $S^{2-}$ ions which neutralize the arriving $Li^+$ ions. And it should produce a deposit of insoluble $Li_2S$ (or $Li_2S_2$, or $Li_2S_4$) on the cathode. The trouble is that sometimes, the sulfur does accept electrons, but the $S$ atom is part of a big molecule $S_6$ or $S_8$, and that this molecule fixes an electron producing $S_6^-$ or $S_8^-$ ions, which do not neutralize $Li^+$ ions. On the contrary, they cross the electrolyte in the "wrong" sense, and get deposited on the anode. This destroys the cell. Nobody has yet found the way of getting rid of this stupid problem.


Converting Lactic Acid to Potassium Lactate I am looking to try an experiment incorporating potassium lactate into soap making. This seems to be new territory for soap makers in that sodium lactate is usually used. I found a bottle of "Lactic Acid 88%" at a beer making supplier and already have 90% potassium hydroxide but do not know how much of one is needed to neutralize the other. Can anyone point me to the right tables or resources to do the calculations?

OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$.  Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid.  Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

\begin{align}
m_{\ce{KOH}}
  &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}
 = \pu{311.7 g}\\
m_{\mathrm{LA}}
  &= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88}
 = \pu{511.8 g}
\end{align}

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.

90 g pure lactic acid reacts with 56 g pure potassium hydroxyde. As your substances are not pure, you must use 102 g of your impure lactic acid (88%), and make it react with 62.2 g of your impure (90%) potassium hydroxyde.


What would happen if the reaction of reduction of carboxylic acids by LiAlH4 is carried out in alcohol? Which is the intermediate compound in the reduction of carboxylic acids by $\ce{LiAlH4}$ and what would happen if the reduction of carboxylic acids is carried out in alcohol?

From my knowledge I know that a geminal diol is formed and then elimination takes place to form $\ce{R-CH=O}$. If this is reacted with alcohol instead of $\ce{LiAlH4}$ again what are the expected products?

Alcohols are not an ideal solvent for LiAlH4, but the main issue is that there is no carboxylic center to where the hydride transfer from LiAlH4 can take place.

So in this case, the side-reaction between alcohol and LiAlH4 can be regarded to be minimal and the major product will be the alcohol produced from the carboxylic acid.

P.S. : A diagrammatic representation of the reactions for LiAlH4


Why are covalent bonds directional? It is said that covalent bonds are directional, while ionic bonds are not. Why? Is it because of the orientation/directional properties of the overlapping orbitals?

The meaning of covalent bonds being directional is that atoms bonded covalently prefer specific orientations in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes.

The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you said, by overlapping the atomic orbitals of participant atoms. And usually only few patterns of overlap are possible, consequently, only few spatial arrangements of atoms are possible.

Ionic bonds are different: there is no electron sharing (or atomic orbital overlaps) and the number of anions surrounding a cation is limited by the charges of the ions, their sizes, and the efficiency of the lattice packing.

In covalent bond mutually sharing present between more or less electronegative element.
Eg; if we consider two elements A and B.electronegative of A is more than B.then electron density more on A and less on B.
As a result B easily lose electron means that their is direction of losing electron and it's direction depends upon electronegative of element hence covalent bond is directional bond


Are the breaking of bonds simultaneously endothermic and exothermic? I understand that energy is NEEDED to BREAK bonds. But don't chemical bonds already have some form of potential energy in them already? So wouldn't it make sense that even though energy is required to break them, their potential energy is also RELEASED once they are broken?

In other words, shouldn't the bond breaking process be considered simultaneously endothermic AND exothermic?

Side note: I'm studying grade 12 level chemistry right now, so I don't have too much knowledge on the nitty gritty details of this concept. That being said, It would be greatly appreciated if anyone could explain this in very simple terms.

Bond breaking does NOT release energy!!

In the course of a chemical reaction, you typically break some bonds and form some new bonds. The bond breaking is always endothermic. The formation of new bonds is exothermic, so depending on whether the old bonds or the new bonds were stronger, the reaction overall can be either endothermic or exothermic.

When somebody says chemical bonds “contain potential energy” it is a HIGHLY misleading statement. You justifiably interpreted this as: when you break bonds, they release the potential energy that was stored within. This is, unfortunately, wrong. That phrase should honestly be banned, but somehow it has always persisted.

What that phrase really means is more nuanced. It means that some chemical bonds can be broken in a reaction which subsequently produces more energy via bond formation.

Bond breaking is always endothermic!

For further reading see: Is Bond Formation "Strictly" Exothermic?

A bond is always some local energy minimum. You definitely need to add some energy to get out of there, but when you separate the remaining parts to infinity, you might get more energy back than you put into it. Certainly not if you split water or sodium chloride.

So you're principally correct, but the two (or more, breaking a bond will usually induce further rearrangement in the parts) processes cannot be separated. Can't break a bond but not remove the parts from each other.


Is ΔS of a system related to temperature and change in enthalpy? I often see the equation $\Delta S_\text{universe}$ = $\Delta S_\text{surroundings}$ + $\Delta S_\text{system}$, where $\Delta S_\text{surroundings}$ can be re-expressed in terms of the enthalpy change of the system as

$$\Delta S_\text{surroundings} = \dfrac{-\Delta H_\text{system}}{T}$$.

This leaves me wondering why $\Delta S_\text{system}$ cannot also be re-expressed (or at least is never written) as a term with some combination of $\Delta H$ and $T$ (Maybe as $$\Delta S_\text{system} \overset{?}{=} \frac{- \Delta H_\text{surroundings}}{T}$$ although this would make $\Delta S_\text{universe}$ equal $0$).

Instead, $\Delta S_\text{system}$ is usually given as a standard value, seemingly independent of temperature. Is there a reason why we can only express the change in entropy of the surroundings, but not the change in entropy of the system, as related to the change in enthalpy and the temperature of a reaction?

If the system experiences a change at constant pressure, the change in enthalpy of the system $\Delta H_{system}$ is equal to the heat Q transferred from the surroundings to the system.  So the heat received by the surroundings is $-\Delta H_{system}$.  Typically, the surroundings is regarded as an ideal constant temperature reservoir, and the reservoir is considered as being "internally reversible."  So, if T is the temperature of the reservoir, the change in entropy of the surroundings is $\Delta S = -\frac{\Delta H_{system}}{T}$.  The system is typically not regarded as experiencing an internally reversible change, so the same kind of approach can not be applied to the system.

I don't know if you have done some statistical mechanics, but maybe this is related to counting of states. First of all, we express the entropy change of the surroundings in terms of enthalpy change and temperature of the system because we simply can define our system in most cases and so can circumvent and measures quantities that defines a system (enthalpy change, for example, can be measured in a calorimetry expt.) then the entropy change of the surrounding can be inferred because the that of the universe is zero (only for reversible processes).

Second, for the counting I mentioned earlier, Boltzmann defines entropy as
$$S = k\ln W$$
$W$ represents the count of possible states of the system/surrounding. But the  number of states of the system multiplied by those of the surroundings equal the states of the universe. So if you know two quantities (here, system and universe) through knowing their entropies, you can get the remaining. Sure, this doesn't prevent you from doing what you suggested, which is expressing system entropy change as surroundings' enthalpy change and temperature, but you can't define them specifically for the surroundings and don't have a readily available tool to measure them.


Why does radium have a higher first ionisation energy than barium? I'm wondering why radium appears to buck the general trend that first ionisation energies decrease as you move down a group in the periodic table: barium (the group 2 element preceding it) has a first ionisation energy of $\pu{502.9 kJ/mol}$, whereas radium has a slightly higher first I.E. of $\pu{509.3 kJ/mol}$ (from the Wikipedia, although my textbook agrees).

Is there any explanation for this at present? (I imagine that quantum mechanics may be involved in some way, but I'm not entirely sure how)

There are most likely other examples of the trend being broken, but this is the only one I've come across so far and I'm curious to know why this is the case.

I think it's also important to mention relativistic effects here.  They already start becoming quite visible after $Z=70$, and $\ce{Ra}$ lies a good bit after that.

In very heavy atoms, the electrons of the $\ce{1s}$ orbital (actually, all orbitals with some electron density close to the nucleus, but the $\ce{1s}$ orbital happens to be the closest and therefore most affected) are subjected to very high effective nuclear charges, compressing the orbitals into a very small region of space. This in turn forces the innermost electrons' momenta to be very high, via the uncertainty principle (or in a classical picture, the electrons need to orbit the nucleus very quickly in order to avoid falling in). The momenta are so high, in fact, that special relativity corrections become appreciable, so that the actual, relativistically corrected momenta, ($p_{\text{relativistic}}=\gamma p_{\text{classical}}$) are somewhat higher than the approximate classical momenta. Again via the uncertainty principle, this causes a relativistic contraction of the $\ce{1s}$ orbital (and other orbitals with electron density close to the nucleus, especially $\ce{ns}$ and $\ce{np}$ orbitals).

The relativistic contraction of the innermost orbitals creates a cascade of electron shielding changes among the rest of the orbitals. The final result is that all $\ce{ns}$ orbitals are contracted, getting closer to the nucleus and becoming shifted down in energy. This is relevant to the question because the $\ce{7s}$ valence electrons in $\ce{Ra}$ are more attracted to the nucleus than one would expect from a simple trend analysis, since they rarely take into account the increase of relativistic effects as one goes down the periodic table.

Thus, the first (and second) ionization energy of $\ce{Ra}$ becomes higher than expected, to the point that there's actually a upward blip in the downward trend. Eka-radium ($Z=120$) would have far stronger relativistic effects, and can be expected to have a significantly higher ionization energy compared to $\ce{Ra}$. In fact, relativistic effects will conspire to make the group 2 metals slightly more noble! Though the periodic table becomes such a mess near the super heavy elements that it's hard to say whether it'll be a clearly visible trend, or just one effect to be combined with several others.

Well it's all about the shielding effect,f subshell have many holes and thus the nuclear influence on the outer s electrons increases,so valence electrons are much tightly bounded to the nucleas.hope you got the answer


Finding concentration and moles given final and initial pH Questions:

How many moles of $\ce{NaOH}$ are needed to change $500~\mathrm{L}$ of solution with $\mathrm{pH}\ 2$ to $\mathrm{pH}\ 11$?
A swimming pool contains 2 million litres of water at $\mathrm{pH}\ 7.80$ and the $\mathrm{pH}$ needs to be changed to $6.80$. How many moles of $\ce{HCl}$ are needed?

Not quite sure how to tackle these types of questions.
Would I have to find the moles of hydroxide/hydrogen ions to neutralise the solution first (ie. moles of hydroxide ions needed to reach $\mathrm{pH}\ 7$) and then from there calculate the amount of hydrogen/hydroxide ions needed to get to final $\mathrm{pH}$?
Or do I just calculate the difference between the moles of ions in final $\mathrm{pH}$ with moles of ions in initial $\mathrm{pH}$?
I have tried both methods but I don't have the answer to the questions so I'm just trying to understand the concept.

Both the cases are similar, so I'll show you the calculation for the fist case only:

The $[\ce{H+}]$ ion concentration changes from $\pu{10^-2~M}$ to $\pu{10^-11~M}$. Assuming water at $\pu{25^\circ C}$, this implies the change in $[\ce{OH-}]$ ion  concentration from $\pu{10^-12~M}$ to $\pu{10^-3~M}$. Since the volume of the solution is fixed at $\pu{500L}$, the amount of $\ce{OH-}$ ions has changed from $500\times\pu{10^-12 moles}$ to $500\times\pu{10^-3 moles}$. This change is exactly the number of moles of $\ce{NaOH}$ that need to be added to this solution, since $\ce{NaOH}$ is a strong base and dissociates with $100\%$ yield.


electronegativity and how it relates to bond strength my question is, if the strength of the bond between two atoms increases as the difference in electronegativity increases as well.

so say, a H-F bond would be stronger than a O-H bond.

I'm a little confused because I know that acidity increases from left to right across the periodic table. So H-F is a stronger acid than H-CH3. Wouldn't this mean that H-F has weaker bonds than H-CH3?

Bond strength usually refers to the energy required to break a bond homolytically--with one electron going to one of the atoms and the other electron going to the other atom. In the case of HF this requires more energy than in CH4. Acidity is the tendency for the bond to break heterolytically--with two electrons going to one atom and none going to the other atom. In the case of HF, we get the two ions H+ and F-. This still requires a lot of energy, but in water solution, the ions are stabilized by hydrogen bonding, so ionization is much more favorable. Methane on the other hand would ionize to give a CH3- anion, which is a lot worse than F- due to carbon being less electronegative.

First off, you're comparing an ionically bonded $\ce{HF}$ with a covalently bonded $\ce{CH4}$.
HF is a stronger acid than $\ce{CH4}$ because its able to furnish out $\ce{H+}$ ions due to its ionic bonding.

So I think your question boils down to: Why is ionic bonding in $\ce{HX}$  (X=Halogen) stronger than covalent bonding?

This is because in covalent bonding there is only a mutual agreement to share electrons whereas electrostatic attraction in ionic bonding.


Basicity of substituted pyridines Which of the following pyridines are most basic?

I would've considered D, since chlorine is the most electronegative, and would therefore have the largest partial negative charge, making it more basic than the nitrogen substituent. The answer given, however, is B. How can I reason this for myself?

Basicity has everything to do with electron density on the electron-pair donor.  Electronegative subsituents actually decrease basicity because they draw electrons towards themselves, though nitrogen and chlorine have comparable electronegativities (3.04 and 3.16 respectively), so I'm not sure you can use that argument here. Instead of inductive effects, let's consider the resonance structures of our molecules:

$\hspace{5.6cm}$

A positive charge on a nitrogen is always more favorable than a positive charge on a chlorine. From this we know nitrogen is donating its electrons into the ring to a much greater extent than chlorine. Because the electron density is much greater on the heterocyclic nitrogen in B, it will therefore be the more basic of the two.

I think N,N-Dimethylpyridine is the most basic compound due to + I efects of two methyl groups .


Is there an easy way to make PVC from PVA? Say, I took Elmer's glue (or a similar Polyvinyl Acetate) and I inserted some Chlorine Tabs and warm it up while stirring.

Would the electronegative Chlorine turn some of it it into Polyvinyl Chloride (PVC)?

Would I get a Dioxin Compound too?

Would I get any other toxic Side-Product?

Choring Tablets have positivly charge cholrine atoms you want negative, so chlorine tablets will not work.  And no you won't get a dioxin compound unless you burn it, then you may ge some.  If you want to make PVC from PVA you can use Hydrochloric acid but you may not get a good yield.


Energy of the n-th level for an atom Going through the Bohr's model and his assumptions, I came across with this formula to find the energy of the n-th level of any atom:

$$E = - \frac{Z k_e e^2}{2r_n} = -\frac{Z^2(k_e e^2)^2m_e}{2\hbar n^2} \approx \frac{-13.6Z^2}{n^2}~\pu{eV}$$

Now, let's say we take as an example the atom of hydrogen, the energy of the $n = 1$ and only level is ($\pu{-13.6 eV}$). Now, according to Wikipedia (not really a reliable source) this can be interpreted as

  An electron in the lowest energy level of hydrogen ($n = 1$) therefore has about $\pu{13.6 eV}$ less energy than a motionless electron infinitely far from the nucleus.

I don't really understand why there is a "minus" in the result. Why does it say that it has less energy than a motionless electron when he, himself, suggested that the electron's acceleration does not result in radiation and energy loss?

I also understand this formula only worked for the atom of hydrogen (that's why I took it as an example) and due to this problem and that's why the Sommerfeld model was postulated.

It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is.  The total energy is still zero. The "falling" from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.

For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy.  When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until "crashes" - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.)  The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6.  This is  called binding energy, and is negative by the convention physicists and chemists have followed since long ago.

BTW, that formula is only for any atom with exactly one electron, and Z not too big.  With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.

Oh!!! Let me tell you what actually happens.
Hopefully you might have been studying physics along with chemistry because the concept used is physics.
..........................................
The thing is this question is very important and often a point of confusion. First lets prove it mathematically then i will reason it out.
Potential energy is calculated by the formula kq1q2/r
So the charge on the nucleus as a whole is +ive and that on the electron is -ive so when you take the product you get a -ive. Hence the potential energy of this system is -ive with respect to the standard infinite distance condition which means that the electron is at an infinite distance from the  nucleus (unbound electron). Hence the energy is negative in comparison to an unbound electron.

Now in the just above formula if you imagine an electron to be a positively charged particle then you would get a positive potential energy which actually means that the electron will be repelled until it retches infinite separation.
From this we actually need to interpret that the -ive total energy implies a bound system, like that of an electron in an orbit. On the other hand +ive or zero energies represent an unbound system where particles stay or prefer to stay at infinite separation.
Remember The potential energy of any bound system is -ive!!!


Why is meta-nitrophenol less acidic than para-nitrophenol? My professor said while explaining the acidic nature of phenol that meta-nitrophenol is less acidic than para-nitrophenol, and gave the reason that resonance doesn't play on meta, but I didn't get it.

The acidity of nitrophenols (or any acid for that matter) is determined by the stability of the conjugate base. In the case of m-nitrophenol and p-nitrophenol, the relative stability can be determined by looking at the resonance structures.

You can see that p-nitrophenol has an additional resonance structure where the negative charge is delocalised onto the oxygens of the nitro group, which are highly electronegative and therefore stabilise the charge effectively. This stabilisation is not possible in the case of m-nitrophenol because there is no opportunity via resonance for delocalisation onto the nitro group.

Refer to the structures the negative charge never gets over META position and hense resonance does not affect the  electron density at META.

and hense nitro on META will not be able to pull electron from ring via (-M) and hense there would be negligible efffect on polarity of O-H bond.


Why do carbonyl compounds undergo nucleophilic and not electrophilic addition? Why aldehydes and ketones undergo nucleophilic addition reactions and not electrophilic addition?

Alkenes show electrophilic addition because of pi bond, which is also present in aldehydes and ketones. But there is nucleophilic addition only. Why?

They sure do. Except that the oxygen has lone pairs, which are more nucleophilic than the $\pi$ bond. Once those electrons attack an electrophile, it further deactivates the $\pi$ bond as a nucleophile.

Aldehydes and ketones prefer nucleophilic addition reactions Instead of electrophilic addition reactions this is because in case of nucleophilic addition reaction a stable intermediate is formed as compared to the electrophilic addition reactions where the intermediate formed is carbocation which is highly unstable.


Why is the boiling point of heavy water more than that of normal water? In class we learnt that the London forces become stronger as the relative molecular mass increases. Not just as in organic chemistry but in things like the halogens.

However, as I understand, the London forces are just temporary dipoles forming indicating that this would have nothing to do with the molecular mass but instead the number of electrons because there will be more electrons to form the dipole. This would mean that the trend would be stronger London Forces with more electrons instead of increasing molecular mass.

How can that explain the slightly higher boiling points of heavy water (deuterium oxide ~101˚C) than normal water?

The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed.

You can find a good explanation of the physics behind these effects on this page.

I think due to the greater size of deuterium atom, positive charge would be more dispersed, thus more stable. So deuterium's electron can come out with comparatively more ease, thereby reducing deuterium's electronegativity.


How is mean square displacement defined in Brownian motion? I am trying to calculate mean square displacement (MSD) of water in a simulation to calculate its diffusion coefficient. However I am little confused on its exact form. Can anyone please explain it in little detail?
My confusion is: from wikipedia first paragraph on MSD page

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

My understanding of it was that:

$$MSD =(x(0)-x_\text{0})^2/1 + \{(x(0)-x_\text{0})^2 + (x(1)-x_\text{0})^2\}/2 + \{(x(0)-x_\text{0})^2+(x(1)-x_\text{0})^2+(x(2)-x_\text{0})^2\}/3$$
... and so on

That is that each term in MSD is cumulative sum of all previous terms devided by time. However most of online programms that i saw (for example matlab programme at the last page of this link)
tries to calculate MSD as  just simple square of distance between initial position and position at time T. Can some one please explain if it is a cumulative sum of previous displacements or not.

The mean squared distance is a measure of deviation: how far away two things are.

Let's say we want to find the MSD of a particle from it's origin. What we do is find the square distance at specific time intervals, say every second. We then make a table of data:

$$\begin{array}{c|c|}
 \text{time} & \text{square distance} \\ \hline
 1 & 3.6 \\ \hline
  2 & 3.8 \\ \hline
 3 & 3.4 \\ \hline
 4 & 3.2 \\ \hline
\end{array}$$

The formula for the average is adding everything up and dividing by the number of things, so in this example:

$$MSD = \frac{3.6+3.8+3.4+3.2}{4}$$

What I've done above is stated more generally, and much more elegantly, as the formula you stated,

$$MSD = 1/T * \sum_\text{t=1}^\text{T} (x(t)-x_\text{0})^2$$

Your cumulative sum formula doesn't make intuitive sense, it instead finds the sum of the mean square distance at every time increment (there is no reason why you'd want to do that)

At least in the software I use, MSD is based only on the displacement vector from the initial position. It's computationally cheap and very good at differentiating between diffusion and Brownian motion. In the former you'll get almost always a smooth positive gradient whereas the latter would give random gradients.


Formula of the compound based on occupancy of lattice voids Atom of element $\ce{B}$ forms HCP lattice and those of the element $\ce{A}$ occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements $\ce{A}$ and $\ce{B}$?

My doubt is that since the element $\ce{B}$ is in HCP, the number of octahedral and tetrahedral voids will be 6 and 12, respectively, which means that $2/3 \times 12 = 8$ tetrahedral voids occupied by $\ce{A}$.

Since $\ce{B}$ is in HCP lattice, there must be 12 voids, i.e. the ratio will be $8 : 12 \implies  2 : 3$, so $\ce{A2B3}$ must be the answer.

But according to NCERT textbook (example 1.2) the answer is $\ce{A4B3}$.

Since B forms an HCP lattice, it's z=6 (z is the number of atoms power unit cell of the lattice) then the ratio comes out to be 8:6 that gives us A4B3.

Let no. of atoms of B used in hcp lattice =n
Number of tetrahedralvoids=2n
Number of atoms of A=(2/3)2n=(4/3)n

A:B=(4/3)n(n)

formulaof the compound is A4B3


Can glutamic acid and arginine form H-bond at physiological pH? I was wondering if, say, glutamic acid and arginine can form H-bonds at physiological pH?

According to the figure arginine has a $\ce{NH3+}$ group and glutamic acid a $\ce{COO-}$ group at physiological pH.
I know hydrogen bonding is possible amongst $\ce{NH2}$, $\ce{HF}$ and $\ce{OH}$. But $\ce{-NH2}$ is in the protonated form and the $\ce{-COOH}$ in the deprotonated form.

What will happen:

Will $\ce{-NH3+}$ donate its $\ce{H+}$ to $\ce{-COO^-}$ to form $\ce{-NH2}$ and $\ce{-COOH}$ and then form hydrogen bonds?
Will they attract each other because of the opposite charge, with resulting H-bonding
No H-bonding will occur.

(maybe the first and second option are combined, I didn't really know how to formulate it)

UPDATE
How is the H-bond possible in the following figure (I would think this is just an electrostatic interaction, this figure suggest it would a combination of both):

source of picture: Wikipedia

Yes, glutamate and arginine can form a hydrogen bond at physiological pH. Remember of course, that as soon as you enter aquaeous media, everything will be saturated by hydrogen bonds from the surrounding water molecules. But inside a protein an arginine—glutamate hydrogen bond is perfectly viable. Arginine will donate the hydrogen bond using an $\ce{N-H}$ bond and glutamate will receive towards a carboxylate.

Note, by the way, that your picture gives a wrong structure of protonated arginine. The correct protonation of the guanidine group is given in the image below (taken from Wikipedia).

As none of the chemists have picked up on this one, I will venture an answer.

First, I am answering on the assumption that this question relates to proteins as it was originally posted on Biology SE and you have previously asked chemical questions about proteins.

Second, I would have thought that you would be aware that weak interactions in proteins are usually categorized in three different types. The section on non-covalent bonds in Berg et al. states:

  The three fundamental noncovalent bonds are electrostatic interactions, hydrogen bonds, and van der Waals interactions. They differ in geometry, strength, and specificity. Furthermore, these bonds are greatly affected in different ways by the presence of water.

So I find it strange that you wish to make an electrostatic interaction (aka an ionic bond) involving integral positive and negative charges into a hydrogen bond which can be regarded as a sort of poor man’s ionic bond as it involves only partial positive and negative charges. To quote Berg et al. again:

  The relatively electronegative atom to which the hydrogen atom is covalently bonded pulls electron density away from the hydrogen atom so that it develops a partial positive charge (δ+). Thus, it can interact with an atom having a partial negative charge (δ-) through an electrostatic interaction.

The glutamic acid and arginine side-chains will be ionized at physiological pH because of the values of their ionization constants — the position of the equilibrium between the ionized and unionized forms. One extra hydrogen ion in the sea of hydrogen ions in the aqueous environment is not going to affect this one iota.

You should also be aware that the aqueous environment (in this case the surface of a globular soluble protein) is most unfavourable to the formation of hydrogen bonds between residues because of the competition of polar water molecules. To quote once more:

  A hydrogen atom of water can replace the amide hydrogen atom as a hydrogen-bond donor, whereas the oxygen atom of water can replace the carbonyl oxygen atom as a hydrogen-bond acceptor. Hence, a strong hydrogen bond between a CO group and an NH group forms only if water is excluded.

You should read the whole of this section (which also applies to ionic bonds) to appreciate how these bonds are most important in the hydrophobic interior of proteins.

In summary, the answer to your question is “No hydrogen bonding will occur — if anything there will be ionic bonding, but in an aqueous environment this will be disrupted.”

You clearly have a reason for wanting glu and arg to form hydrogen bonds. Perhaps you could explain why.


Does carbon monoiodide (CI) exist? Many of my students get chlorine $\ce{Cl}$, confused with carbon monoiodide $\ce{CI}$, despite my promises to use serif fonts whenever possible and to never give them carbon monoiodide in a question.

Everything I know about chemical bonding tells me carbon monoiodide should not exist ($\ce{CI4}$, sure), but weird things happen sometimes. My only results from internet searches for carbon monoiodide bring up a set of Quizlet flash cards.

Will my students prove me wrong if I tell them that carbon monoiodide does not exist (except possibly in extreme laboratory conditions).

Provided that mellitic anhydride ($\ce{C12O9}$, empirical formula $\ce{C4O3}$) is considered one of carbon oxides, then e.g. hexaiodobenzene ($\ce{C6I6}$, empirical formula $\ce{CI}$) could be considered a form of carbon(I) iodide, formally. But please rather don't teach that in regular lectures.

Sure

Formula:    CI
Element System: C-I
Element Names:  Carbon, Iodine
Molecular Weight:   138.915 g/mol
Name(s):    CI <g>
Carbon Monoiodide gas

See:
http://lb.chemie.uni-hamburg.de/static/ES/2_C_C-I.php?content=527/pz0cDmNo


How to identify zero order reactions? I have come across many reactions that are zero order reactions but at one glance I'm unable to tell if they are zero order or not.  Is there any criteria that can be used to identify these reactions or are they purely experimentally determined?

Determining the rate of the kinetics of a reaction isn't done with only one measurement, or looking at only one graph.

What zero order kinetics means, is that the rate of the reaction is independent of the concentration of chemical species. To determine the kinetics of a process, you need to take a series of measurements while altering one variable, namely the concentration of one of the chemical species under investigation.

In zero order kinetics, you should observe that as you change the concentration of reactant, for example, the rate of the reaction remains constant.

To be honest, we could only use scientific data to determine the order of an reaction , Say an example of one species determining reaction. You can only plot graphs to find out the shape of the curve and thus discovering whether it is zero first or second order reaction
If you would like , it is in a chapter called determination of reaction order


Why does solvent particles move towards solution when semi permeable membrane is added? When a SPM is added in between solute and solution the water molecules move towards salt water even though the concentration is same both sides. Why does this happen? According to SPM's perspective, concentration (of water) at salt water side is lower than that at fresh water side but how does that change the fact that the overall concentration is same both sides? My book says it depends on vapour pressure. If that's the case then osmosis should take place without SPM because it depends only on vapour pressure?

Here what happens is that the fresh water is a pure solvent and a pure solvent always has a greater vapor pressure than that of a solution. So due to the greater vapor pressure on water's side, the molecules of water will experience a force in salt water's direction as the pressure is less in salt water side. Therefore the water will flow towards the salt water.


Why does rearrangement not take place in Wurtz reaction? The Wurtz reaction is $$\ce{R-X + Na ->[ether] R-R}$$ where $\ce{R}$ is alkyl. But if we take butylchloride with $\ce{Na}$, the product is n-octane and not 3,4-dimethylhexane. I am unable to understand the fact that rearrangement does not happen even when it's a free radical mechanism.

The Wurtz Reaction mechanism involves radical anions. Neither radicals nor carbanions rearrange the same way that carbocations do.
The first step of the mechanism is a single electron transfer from $\ce{Na}$ to $\ce{R-X}$:
$$\ce{Na + R-X -> Na+ +[R-X].^{-}}$$
The alkyl halide radical anion fragments to a carbon radical and a halide anion:
$$\ce{[R-X].^{-}->R. + X-}$$
The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...
$$\ce{R-CH2-CH2. +.CH2-CH2-R->R-CH=CH2 + CH3-CH2-R}$$
... or coupling:
$$\ce{R-CH2-CH2. + .CH2-CH2-R -> R-CH2-CH2-CH2-CH2-R}$$
The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. $\ce{Na}$ is a good reducing agent, and that primary radical is readily reduced to an anion:
$$\ce{R. + Na -> R:^{-} + Na+}$$
The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:
$$\ce{R:^{-} + X-R -> R-R + X-}$$
The formation of the anion is actually problematic, especially if $\ce{R-X}$ is secondary or tertiary, since E2 elimination will compete.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$

You are making it too complicated, but it's really simple: $$\ce{Bu-Cl + 2Na + Cl-Bu -> Bu-Bu + 2NaCl}$$ where $\ce{Bu-{}}$ is the n-butyl group and $\ce{Bu-Bu}$ is n-octane.

You can understand wurtz's reaction much easier this way.


What kind of system is required for dynamic equilibrium to be set up? Most websites and books mention a closed system, but technically isn't an isolated system required? If energy can be transferred, then the position of of the equilibrium will continuously change and no equilibrium will be set up, right?

If energy can be transferred, it does not mean it is transferred.

If there is e.g. zero net thermal energy transfer, a system is in thermal equilibrium with the neighborhood, even if it is not isolated.

As Ivan says, an equilibrium is possible even in an open system, if net transfer of particular matter kinds is zero.

The thermodynamic equilibrium consists of thermal, radiative, chemical and mechanical equilibrium.

Note that thermodynamic equilibrium happens at following conditions:

  For a completely isolated system, the entropy $S$ is maximum at thermodynamic equilibrium.
  For a system with controlled constant temperature and volume, the Helmholtz free energy $A$ is minimum at thermodynamic equilibrium.
  For a system with controlled constant temperature and pressure, the Gibbs free energy $G$ is minimum at thermodynamic equilibrium.

$$\begin{align}
G&=H-TS \\
\Delta G&=\Delta H-T\Delta S \\
\end{align}$$

Total entropy of the not isolated system and it's surrounding increases in time.

Entropy of the system alone may decrease with time, if the Gibbs energy of system decreases, as this latter decrease also means
$$\Delta S_{\rm surrounding} +  \Delta S_{\rm system}  \gt 0$$

A closed system is required for the dynamic equilibrium to set up.


Does salt on the snow and ice covered roads and sidewalks make the nearby air temperature cooler? I  noticed when it snows and salt is melting the snow on the roads in the city, that the air temperature feels a lot cooler than the temperature indicated on the thermometer.
A 40 degree fahrenheit temperature feels like 30 degrees. Is it the moisture from melting snow or rain that makes the air temp feel ten degrees cooler or is it the salt brine (melted snow or ice  and salt?  Is it similar to adding salt  to the ice in an Hand powered ice cream churn to make the cream container colder to make ice cream?

As water melts due to adding a salt (AKA melting point depression), it absorbs energy from the environment and the temperature of the salt/ice/water mixture does, indeed, drop. Whether that is sufficient to cool the air above the ice perceptibly, though, is a matter of opinion.

Another effect that might make the air feel colder is increased humidity, particularly as the salt slush is splashed into the air by vehicles.

You could perform an experiment yourself to determine the air temperature above two containers, one with fresh snow and the other with snow and salt. If you do, please post your results as an answer!

No. It merely reduces the melting temperature of the ice. If you've ever noticed, salt water doesn't freeze until a much lower temperature than regular water. It is because of the same reason.


"Crystal radius" vs "Ionic radius" What is the difference between crystal radius and ionic radius? For $\ce{NaCl}$ for example the ions combine to a crystal, and I would have guessed they are equal. But I then found an example of data, where they are not.

I am aware that ionic radius is the radius of an atom's ion. Quite simple. So what is crystal radius?

To quote Erik Rakovský · 30.73  · 54.96 · Comenius University in Bratislava (from this discussion on researchgate)

Shannon published simultaneously two sets of ionic radii. So called "effective ionic radii" were made consistent with Pauling's r($\mathrm{O_2^-}$)=140 pm. "Crystal radii" are using Shannon's value r($\mathrm{O_2^-}$)=126 pm.

https://dx.doi.org/10.1107%2FS0567739476001551


IUPAC Nomenclature of organic compounds Can the prefixes like iso,neo be used in the IUPAC nomenclature of organic compounds and can all the common names of aromatic compounds be used in their IUPAC name?

Prefixes -  I think the prefixes neo, iso etc. are used only in common names. There's a different system in IUPAC for naming things having a branched chain.
Aromatic Compounds - The only aromatic compound that I know of having same common and IUPAC name is benzene.


Why does Nitrogen Oxide NO exist? The nitrogen oxide NO exists and has an uneven number of valence electrons. This is "impossible". Why does it exist ? Why is it not dimerized ?

It is possible due to the fact in NO oxygen has an electron configuration of a noble gas which makes it less energetic. And N shares 2 electrons and it gets close to a noble gas electron configuration.


How is seawater alkaline? If when water and carbon dioxide react they form carbonic acid, carbonate, and bicarbonate, how does seawater still have a pH of around 8? Doesn't a compound need an hydroxide ion to be a base?

The short answer to the title question is the ubiquitous presence of geologic and biogenic calcium carbonate $(\mathrm{p}K_\mathrm{a} = 9)$.  The oceans as a whole can largely be thought of as residing on beds of calcium carbonate. The following are some excerpts from this Wikipedia page:

Eggshells, snail shells and  most seashells are predominantly calcium carbonate.
Carbonate is found frequently in geologic settings and constitute an enormous carbon reservoir. Calcium carbonate occurs as aragonite, calcite and dolomite. The carbonate minerals form the rock types: limestone, chalk, marble, travertine, tufa, and others.
Calcium carbonate contributors, including plankton (such as coccoliths and planktic foraminifera), coralline algae, sponges, brachiopods, echinoderms, bryozoa and mollusks, are typically found in shallow water environments where sunlight and filterable food are more abundant. Cold-water carbonates do exist at higher latitudes but have a very slow growth rate.

Regarding your question

Doesn't a compound need an hydroxide ion to be a base?

Here you go:
$$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH-(aq)}$$
Although calcium carbonate is only sparingly soluble $(K_\mathrm{sp} = 3.3×10^{−9})$, as stated above the oceans are simply rife with it.

to advance the discussion:
consider the pH of water is temperature dependent

for pure H2O (without any minerals):
1.[H+]=[OH-] (always neutral) and  2.from 0 degrees to 60 degrees [H+] and [OH-] increase simultaneously. The pH of pure and neutral water drops from 7.47 to 6.14 from 0 to 100 degrees. For discussion purpose lets assume the standard temperature of 25 degrees (pH=7.0).

Doesn't a compound need an hydroxide ion to be a base?
First realize pH (and pOH) concerns a compound in water; second, every aquatic/H2O solution has both [OH-] and [H+], therefore
Stewart (stewart textbook of acid-base, 2.2 definitions) simply defines:
a base as a solution wherein [OH-]>[H+].

Due to the salts in solution:
[H+] > [OH-] (acidic) or [OH-] > [H+] (alkaline).

Salts quantatively determine [H+] and [OH-] because 1. electroneutrality is a general principle ([cations]=[anions]) and 2. concentrations of salts (order of 600 meq/l = 6e102) far outweigh [H+] and [OH]- (order of neq/l =10 -9) and 3. [H+]*[OH-] is Kw.

CO2 pushes an equilibrium (CO2 <-> H2CO3 <-> H+ and HCO3-) already determined by cations (Na+, Mg2+, Ca2+, K+, etc) and anions (Cl-, SO4 2-, HCO3-, etc).

the reason seawater is basic is because HCO3- is pushed into equilibrilium with CO2 (aq) and OH- by the cations.

for example a solution with 600 mmol NaCl and 2 mmol NaHCO3 (calculated with Aqion online Ph calculator). Mainly the dissociation of HCO3- into CO2 and OH- contributes to pH 7.71. If the same composition is analysed at 4 degrees the pH is 8.07 (calculated with phreeq online pH calculator).

the aquous species are:

In seawater the strong cations sodium, magnesium, calcium and potassium are all contributing to [OH-] being larger than [H+]

An example of cation and anion composition in mmol/l with equilibrium of monovalent and bivalent ions. (mM concentration is calculated for 20 degrees)
Calculated miliequivalents are: total cations 620,56meq/l and total anions 620,305 meq/l.