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Does one atom of an element show the properties of a bulk of that element?
For example, does one atom of gold have a golden color?

IT depends on the property. Some properties of bulk materials result from interactions among the atoms making up the material (electrical conductivity and colour in metals for example). Other properties are common (atomic number for example).

There is an interesting caveat about gold however. The reason that gold has a different color than most other metals is in fact an atomic absorption effect, not metallic or electric. Silver has a strong absorption as well but it is in the UV whereas for gold it is in the visible part of the spectrum. See this and this for example. Gold atoms absorb blue light, but they would not have a metallic appearance, that requires that they are in bulk and there are conduction electrons.

Molecular properties such as melting point, conductivity, elasticity all require a multiplicity of atoms or molecules to be observed since each of those properties are defined interactions between the same particles. Color is different - when a photon hits an atom the atom's electrons are excited, the electron will fall back to it's unexcited (ground) state and when it does it emits a photon.  This photon will not always have the same energy as the photon absorbed, when this is the case the rest of the energy was converted into thermodynamic energy (heat). Since the color of a wave of light can be defined by its energy then atoms can have "colors". This is a bit of an oversimplification but I hope it helps!
Edit: as @alchimista pointed out the color of a single atom will not likely represent the color of a sample of that element large enough to see. Also the light reflected from most atoms will likely cover the entire visible spectrum which will result in the same color of light being reflected as was used to illuminate it.

Atoms can have colors all right, but not necessarily the same as the bulk material.

I suggest editing because colour, and specifically for a metal, dramatically depends on the number of atoms bound in the sample. Gold nanoparticles can probably span the entire visible and even the reflectivity should change. It could even be that colour is even more size dependant than, let us say, the melting point of a nanoscale sample. The answer is good just for pointing out the difference between collective and atomic properties but the colour example is misleading.

That is also the point where it becomes quite interesting to look at statistical thermodynamics, so the combination of larger amounts of particles with quantum mechanics leading to the classical macroscopic description of many fundamental thermodynamic equations.


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Exceptionally weak basicity of pyridine
It is often explained that the relatively lower basicity of pyridine, relative to most amines, is due to the fact that the $\ce {N}$ lone pair resides in an $\ce {sp^2}$ hybridised orbital, instead of an $\ce {sp^3}$ hybridised orbital. the higher s character of the lone pair-bearing orbital thus holds on to the electron pair more tightly as the lone pair experiences more of the nuclear charge, decreasing the availability of the lone pair for donation to an acid.
Consider the data provided by Clayden, Warren & Greeves (2012), on p. 726. Yes, we observe the lower basicity of pyridine, relative to piperidine. This is consistent with the above rationalisation. However, based on the explanation above, the relative basicities of a typical imine and pyridine should be similar since in both molecules, the lone pair resides in an $\ce {sp^2}$ hybrid. Based on the data, it seems that the basicity of pyridine is around 3 orders of magnitude lower than that of a typical imine. Why is that the case?

Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

I am not sure a long answer is required. The reason is the aromaticity of pyridine. Write down the two equilibria to see it.

@Alchimista Well... My educated guess would be aromaticity of the pyridine ring as well. But how exactly does aromaticity play a part in the acid-base behaviour?

Look at the difference between the two equilibria. Namely the relative stabilization of the charge in the two cases. In pyridine not mesomeric effect are at work without destroying the aromatic sextet. In imines "you can" stabilise the iminiun ion.

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What happens when a symmetry element is added or removed from a point group?
What new point group is formed if symmetry element i is added to C3 point group?

Then it's not a group.

@Zhe the book answer says that S6 point group is formed

A group plus one extra element is usually not a group at all. A group plus one extra element and the elements produced by it is usually another, bigger group (in this case $S_6$).

The set is not closed under composition, so it's not a group. If you take the closure with $i$, then yes, it's $S_{6}$. The easiest way to see is to write out the elements of the group.

@IvanNeretin how do we know what bigger group is formed let's say if we add $i$ to $C_{3v}$

If it isn't obvious right away, we may have to write down all elements of the new group, one by one.

@IvanNeretin the element added will combine with all the existing elements to generate new elements, how do we find what new elements are created in non trivial cases?

Like you said: we combine the new element with all the existing elements, thus getting some new elements, then we do the same with them and so on, until we stop getting any new elements.

I combine $C_{3}$ and $i$ by applying operations on point (x,y,z) and get a new point, how would I know to which symmetry element the new point corresponds?

Forget the points, work with matrices.

Inversion $i$ is equivalent to a two-fold improper rotation $S_2$. Introducing a $2n$-fold improper rotation axis to the existing $C_n$ point group will result in a new $S_{2n}$ point group, e.g. in your case $C_3 ⊗ i = S_6$.

Is there a way to solve for the new group

Can it be solved using a character table?

@rishabhx64 Yes, but that would be an overkill for this case.

I'd say character tables are not for this purpose at all.

@andselisk take this case add $i$ to $C_{3v}$

@IvanNeretin how do we solve such questions then? The thing is that the chapter in which this question is given in my textbook, has not introduced character tables, and I can't find a way to solve such questions

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Why does delocalization lead to stability?
If the electrons are relatively more stable when they are delocalized than when they are localized between two atoms, then why do we have bonds formed in the first place? Forming of bonds between two atoms means that electrons localized between the two atoms, if electrons tend to delocalize then why not all electrons (including the sigma bond electrons) are just delocalized? What is the driving force for delocalisation ? How does delocalisation the energy of the system?

The tendency to delocalize is offset by electrostatic attraction (to the nuclei).

An electron system delocalises automatically under certain circumstances, namely if it's a conjugated pi system. It's not a property of the electron, but of this specific bond system!

You have good answers - but I'd like to add: In general in chemistry, if you allow a "thing" to have more options on how to configure itself, it will have a lower energy state and that will present as stability.

Electrons "want" two thing: be close to a nucleus and have the freedom to move. Also, being fermions, they are "forbidden" by the Pauli exclusion principle to have exactly the same state (e.g. same set of quantum numbers in an atom) as another one.
The electron distribution in a molecule is a compromise of the two "wants" while not trying anything "forbidden". In the $\ce{H2}$ molecule, electrons get to enjoy the attraction of two nuclei while having more space available than when bound to an isolated hydrogen atom. Because of the Pauli exclusion principle, only two electrons get to enjoy that low energy.
Electrons associated with conjugated double bonds also have "more room" without being further away from the nuclei than electrons associated with isolated double bonds.
Sorry for the anthropomorphizing... for a more rigorous treatment, look up particle in a box in a quantum chemistry book.

Thanks, although I can't get rigorous way to understand that without QM, I got the intuition

If you think of an electron as a de Broglie wave, and confine it to a box, you can show that its energy $E$ is inversely proportional to the squared inverse of the length $l$ of one side of the box, that is, $$E \propto \frac 1{l^2} $$  Therefore, if you make the box larger (increase $l$) you lower $E$. This is a purely quantum-mechanical result due to the wave property of particles and it applies equally well to electrons confined by the attractive electrostatic potential generated by the positively charged nuclei in a molecule.
By the way, if you are wondering why the energy of the de Broglie wave would be higher under tighter confinement, think of the length of the box as equal to the wavelength of the particle. The result follows since a shorter wavelength means a higher frequency and a higher frequency implies a higher energy.


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Decrease in temperature of an aqueous salt solution decreases conductivity
Why does the conductivity of a water solution drop as the temperature decreases?
How are these two related?

I'm in the middle of understanding the relation since it's a question made by my professor at the university. There is no data to compare yet although I would be glad to provide a much more detailed question to help others who may read this.

I've asked a follow-up question: Is there an electronic component to water conductivity?

According to the Stokes-Einstein-Debye theory, and assuming the ionic composition remains constant (say for a fully dissociated salt), the main factor accounting for the response of the conductivity to temperature is the change in the viscosity of the solvent.
In the SED theory the frictional drag coefficient $f$ of a charged particle is proportional to the viscosity $\eta$:
$$ f \propto \eta$$
As a result the electrical mobility $\mu$ of an ion of charge $q$ is inversely proportional to the viscosity, since
$$ \mu =q/f\propto \eta^{-1}$$
and since the specific conductance $ \kappa$ depends linearly on the mobilities (approximately, at constant ionic strength), $$ \kappa \propto \eta^{-1}$$
Since the viscosity usually increases with decreasing temperature, the conductivity decreases.

Decrease in temperature has two effects, both attributing to lower electrolytic conductivity:

decreases the mobility of the charge carriers (e.g. $\ce{H3O+}$ and $\ce{OH-}$ for pure water);
suppresses auto-ionization of water (higher $\mathrm{p}K_\mathrm{w}$), reducing the total number of charge carriers.


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Combination of several unit symbols when finding new pressure from ideal gas law

As a car is driven, the tires heat up and the volume and the pressure change. Calculate the new pressure of a 35.0 L tire initially at 235 kPa that expands to 36.5 L when it heats up from 15.0 °C to 43.0 °C. (4 marks)

$P_1 = \pu{235 kPa}$
$P_2 = ?$
$V_1 = \pu{35.0 L}$
$V_2 = \pu{36.5 L}$
$T_1 = \pu{15.0 °C} = 15.0 + 273 = \pu{288 K}$
$T_2 = \pu{43.0 °C} = 43.0 + 273 = \pu{316 K}$
The number of moles of gas in the balloon is not changing, so $n_1 = n_2$.
$$P_1V_1/n_1T_1 = P_2V_2/n_2T_2$$
(since $n_1 = n_2$ we can cancel them from both sides)
$$P_1V_1/T_1 = P_2V_2/T_2$$
(cross multuple to allow for isolation of $P_2$)
$$P_1V_1T_2 = P_2V_2T_1$$
(divide both sides by $V_2T_1$ to isolate $P_2$)
$$P_2 = P_1V_1T_2/V_2T_1$$
(now plug in values)
$$\begin{align}
P_2 &= (\pu{235 kPa})(\pu{35.0 L})(\pu{316 K})/(\pu{36.5 L})(\pu{288 K})\\
  &= 2 599 100 / 10 512 \\
  &= \pu{247 kPa}
\end{align}$$
The new pressure of the tire would be 247 kPa.
I like keeping track of the units throughout solving, but not sure how to do that with three units...
Can I write $P_2 = 2599100\cdot\pu{kPa/L/K}/10512\cdot\pu{L/K}$ (L and K cancel out and leave me with kPa), or is it recommended not to record the units in that step and only write the final unit with the final answer?

Section 5.1 Unit symbols in SI Brochure [1, p. 130] has an answer:

In forming products and quotients of unit symbols the normal rules of algebraic multiplication or division apply. Multiplication must be indicated by a space or a half-high (centred) dot (⋅), since otherwise some prefixes could be misinterpreted as a unit symbol. Division is indicated by a horizontal line, by a solidus (oblique stroke, /) or by negative exponents. When several unit symbols are combined, care should be
  taken to avoid ambiguities, for example by using brackets or negative exponents. A solidus must not be used more than once in a given expression without brackets to remove ambiguities.
$\pu{m kg/(s3 A)}$,
  or $\pu{m kg s–3 A–1}$,
but not $\pu{m kg/s3/A}$,
nor $\pu{m kg/s3 A}$.

I personally prefer to use powers of the units — subjectively, this makes the expressions clearer and also reserves a slash "/" for separating the symbol of the quantity from the units on plots and in the table headers. However, in your case you don't need any of that. According to the ideal gas law
$$pV = nRT$$
and
$$n_iR = \text{const} \quad \implies \quad \frac{p_iV_i}{T_i} = \text{const},$$
so you have a simple proportion
$$\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \quad \implies \quad p_2 = \frac{V_1T_2}{V_2T_1}p_1.$$
Obviously, there is no need to juggle around with the units as all of them except for the pressure units are cancelled out (note how I also grouped homogeneous quantities within the fraction one under another, and wrote pressure separately for clarity):
$$p_2 = \frac{V_1T_2}{V_2T_1}p_1 = \frac{\pu{35.0 L}\cdot\pu{316 K}}{\pu{36.5 L}\cdot\pu{288 K}}\cdot\pu{235 kPa} = \pu{247 kPa}.$$
And yes, generally it is considered a good practice not to do any intermediate calculations unless you are explicitly asked so, and solve the problem algebraically in the first place.
References

BIPM. Le Système International d’unitès / The International System of Units (“The SI Brochure”), 8th ed.; Bureau international des poids et mesures: Sèvres, 2006.


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Complementary base pairing
In the molecular structure of a DNA it is seen that purines (adenine and guanine) bind with pyrimidines (cytosine, uracil and thymine) through hydrogen bonds, and this is always between a purine and a pyrimidine. Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines?

There can, but essentially it doesn’t fit into the double helix so well. That’s arguably better suited for Biology.

https://en.wikipedia.org/wiki/Hoogsteen_base_pair

The reason that you predominantly find A:T and G:C base pairs in double-stranded DNA is because the DNA polymerase make a reverse-complementary strand when they copy one strand of DNA. You could make this question more relevant to the chemistry site by asking about the intermolecular interactions during synthesis that make mis-incorporation highly unlikely, even though in double-helical RNA segments, there is a large variety of pairings including purine:purine or pyrimidine:pyrimidine (so they "fit" into a double helix just fine).

Can't there ever be hydrogen bonds between 2 purines or between 2 pyrimidines?

Yes, it is possible to have hydrogen-bonded base pairs (and triplets and quadruplexes) beyond the canonical Watson-Crick base pairs A:T and G:C. You can explore these in 3D on the DSSR-Jmol site for example. If you look at the 1ehz structure (a tRNA from yeast), which loads automatically on DSSR-Jmol, and click on non-canonical base pairs, you will see some examples. Below is an A:A pair with two symmetric hydrogen bonds from that structure:


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MO Diagram "greatest stabilisation when orbitals are close in energy"
I was studying MO diagrams in Chemistry and we were given a diagram explaining this:

I understood the diagram furthest to the left describing the situation where the Electrophile (E) and Nucleophile (Nu) result in the largest energy gain when both are equal in energy.
However, the lecturer later said that when there is a large difference in energy between E and Nu (far right), the energy gain is negligible. He didn't go further to explaining why this was true.
I tried looking on the internet for an answer but was led to s-p mixing and content I am not familiar with nor find relevant to the question.
I'm confused here. What explains this phenomena? Or, is it just an effect observed without explanation - like a Postulate?

Fundamentally it's mathematics. The energies of the MOs are the eigenvalues of the matrix $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix},$$ where $(\alpha_1,\alpha_2)$ are the energies of the two constituent orbitals and $\beta$ is (loosely speaking) the overlap between them.
For the same size of $\beta$, if $\alpha_1 \approx \alpha_2$, then the eigenvalues can differ quite a lot from $(\alpha_1,\alpha_2)$. On the other hand, if $\alpha_1$ is very different from $\alpha_2$, then the eigenvalues will be closer to $(\alpha_1,\alpha_2)$.
As an illustration consider the matrix
$$\begin{pmatrix}1.1 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$
This has eigenvalues of $1.55$ and $0.55$, which are relatively distant from the "original energies" of $1.1$ and $1$. On the other hand, the matrix
$$\begin{pmatrix}3 & 0.5 \\ 0.5 & 1 \end{pmatrix}$$
has eigenvalues $3.12$ and $0.88$, which are closer to $3$ and $1$.
Taking a more abstract perspective, the eigenvalues of $$\begin{pmatrix}\alpha_1 & \beta \\ \beta & \alpha_2 \end{pmatrix}$$ are
$$\lambda_\pm = \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\sqrt{(\alpha_1 - \alpha_2)^2 + 4\beta^2}}{2}$$
and you can play around with this expression to gain some insight. For example, in the limit where  $(\alpha_1 - \alpha_2)^2 \gg 4\beta^2$ (corresponding to a large energy difference between the original interacting orbitals), the eigenvalues reduce to
$$\lambda_\pm \to \frac{\alpha_1 + \alpha_2}{2} \pm \frac{\alpha_1 - \alpha_2}{2}$$
which are simply $\alpha_1$ and $\alpha_2$.

Perhaps this is overly complicated. The math is clear for advanced level chemistry, but if @BobSmith doesn't understand what happens when orbitals mix, maybe he needs to first grasp the concept underneath the math. He will probably be lost wondering why you do the calculations that you do.

These formulae do not seem to show that the antibonding MO is more antibonding than the bonding MO is bonding. I suppose it is because you are not working with the rigorous physics here and you are just using simpler mathematics to illustrate the point that is relevant to this question?

@chemicalromance I agree with your point, but I make no apologies for writing what I have written. The nature of SE allows multiple people to write multiple answers and the asker may accept anything they deem to be most useful. This may not be directly helpful to OP, but it may be useful to somebody else with a different background. Unfortunately, I don't really have the time or desire to explain Hückel theory from the ground up. The point is: if you think you have a more helpful approach, please post it (indeed, I see you already have!).

@TanYongBoon In essence, yes, it is a simplification. It is explained here in greater detail https://chemistry.stackexchange.com/a/52160/16683

@orthocresol So the reason is that you are employing simple Huckel theory which does not take into account the overlap integral S?

@TanYongBoon yes, but it suffices to illustrate the point in the question.

You can think (in a gross simplification) of an orbital as an orbit, a circular path where electrons go to feel the attraction of the nucleus. If the attraction is big, then the orbit is small and the energy is low. If the attraction is low, the orbit is larger and the energy rises.
Now, when two atomic orbitals (AO from now on) mix, the electrons switch from feeling the attraction of just one nucleus to being attracted by two. That is, they "jump" to new orbits called molecular orbitals (MO from now on), one bonding (lower energy) and one antibonding (higher energy). The bond will be stronger when the overall energy of all the electrons is lower.
When two AOs mix:

If both AOs have the same energy, then they go from the same initial state to the same final state and the electrons are shared equally between the atoms. You can see that in the leftmost diagram.
If one AO is lower in energy (Nu in this example), then the attraction from the nucleus of this atom will be greater even when combined to form MOs; the electrons in the bonding MO will be shared but they will "spend more time" closer to the Nu atom. Thus we draw the bonding MO slightly closer to the Nu AO, and the antibonding MO closer to the E AO, as evidenced by the diagram in the middle. As the difference in energy increases, we move the MOs even closer in energy to the AOs because the Nu atom attracts the electrons even more strongly, like the diagram to the right.

Now, in the special case of a nucleophile bonding with an electrophile, the electrophile's AO is empty while the nucleophile's is full (again an oversimplification1). As all the electrons are there, the initial state has an overall energy determined by the energy of the nucleophile's AO.
When the bond forms, all the electrons occupy the only bonding MO. Then, the final state has an energy determined by the energy of the bonding MO.
Hence, the bond strength is determined by the overall difference in the energy of the electrons in the initial and final states, which in this diagram1 is directly proportional to the energy difference between the Nu AO and the bonding MO (confusingly labeled "energy gain" in the graphic, even though the overall energy is lower). So, for equal energies of the starting AOs, this so-called "energy gain" is biggest.

1 Often there are more than one AO and more electrons involved.
2 This varies wildly from molecule to molecule, because not every atom has the same electronic configuration.


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Why do double bonds cause kinks in fatty acid chains?
Fatty acids can either be saturated or unsaturated. If they are saturated, they stack easily and form solids at room temperature. If they are unsaturated, however, double bonds are introduced, and these double bonds introduce kinks into the structure of the fatty acid molecule. In a different question on this site, it was suggested that the reason why double bonds cause kinks to form is that they interrupt the zig zag shape of the molecule.
However, what was left unexplained is how double bonds interrupt the zig zag shape. If we were dealing with a cis fatty acid, and both hydrogens were located on top, why would this lead to a kink?

Instead of zig-zag, zig-zag, zig-zag etc. it now goes zig-zag, zig, zig-zag, zig- etc (the bold zig is due to the cis double bond). So the direction of the alkyl chain on one side of the cis double bond is different from that of the alkyl chain on the other. You could partially correct for that "wrong turn" by rotating a single bond by $180^\circ$, but that changes a staggered conformation into an eclipsed one.
Here is a link to a 3D model: http://www.biotopics.co.uk/jsmol/fattyacids.html and a snapshot in case the link goes away:

Is there an extra zig because of the hydrogen on the next carbon occupying the region where the next carbon would have needed to go to form the next zag? (and if so, can you add it to your answer?)

A cis-bond itself is a zig-zig (or U-shaped) while a trans-bond is zig-zag (just like a single bond in staggered antiperiplanar conformation). So yes, compared to the trans-bond you are swapping the single bonds to hydrogen and to the carbon on one side of the double bond.


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Reaction between carboxylic acids and amines
Would some be able to explain why when I mix a carboxylic acid, like ethanoic acid, and an amine, say ethylamine, I get ethylammonium ethanoate. But when I mix a bi-functional group carboxylic acid (say ethanedioic acid) and a bi-functional group amine (say 1,2-diaminoethane) I get a condensation reaction and an amide bond is formed.
This also leads to why does ethanoic and ammonia yield ammonium ethanoate and not ethanamide. I understand that the ammonia is acting as a base in this reaction, but why doesn't it favour being a base rather than a nucleophile?
To me both of the reaction seen the same. Can someone elaborate this for me, or am I missing a key piece of knowledge.

Did you actually do this? Yes, acetic acid and ethylamine form salts and I would expect the same result with oxalic acid and ethylene diamine at ambient temperature. At elevated temperature, the first example would yield n-ethylacetamide and the latter a polymeric amide since both components are bifunctional. Mixing 1,6-ethylenediamine and adipic acid at ambient temperature you do not get nylon. By the way, real chemists, at least of my generation, say acetic acid and not ethanoic acid.

So is it a temperature that stops the anime acting as a base and starts acting as a nucleophole? But why?  I only use ethanoic acid as that what I have to use for my exams as it’s IUPAC standards

At ambient temperature the amine reacts as a base forming a salt. At elevated temperature the amine will act as a nucleophile. In your first example a sealed tube would be required to avoid sublimation. This is not a normal non-industrial way to make amides. Usually the acyl chloride or anhydride of the acid is reacted with the amine.

Unrelated to the question, but the preferred IUPAC name is actually “acetic acid”, not the systematic name “ethanoic acid”.

I should add that oxalic acid and ethylenediamine can also form the 6-membered monocycle piperazine-2,3-dione.

You have a false dichotomy.
Diamines and diacids can form salts with doubly charged anions and cations. Sadly, simply searching SigmaAldrich’s catalogue for the phrase diammonium did not turn up any example; however, I assume this is because these salts are unlikely to be used and thus wouldn’t sell. Chemically, proton transfer is one of the fastest reactions possible and thus simply mixing a diacid and a diamine in a suitable solvent would result in the diammonium diate salt as it would for an ammonium monoate salt.
Amide bond formation is a different reaction that typically requires activation or harsher conditions (e.g. heating). It is a condensation that results in the release of water. All that is required is the presence of an amino group and a carboxylic acid. So aminoethane and ethanoic acid can be heated in solution to produce N-ethylacetamide.
The one difference between the mono and the di systems is that the mono system has no of polymerising or cyclising. So the amide bond will always consist of one acid fragment and one amine fragment while in the case of a diamine and a diacid polymerisation or cyclisation can occur.

So what is it exactly about the “harsher conditions” that make the amine act as a nucleophile and not a base?

@H.Linkhorn In introductory chemistry, you would probably use heat. The salt forms immediately at any temperature at which the suitable solution is liquid. For the condensation reaction to occur, the activation barrier must be overcome which is easily achieved by heating up. Another one is removal of water e.g. using molecular sieves. In preparative chemistry, you would usually not use the actual acid but an activating ester (e.g. phenyl thiocarboxylate) or in situ activator such as DCC.

So, does that mean that when you are forming polyamides, a salt if formed first and then you have to heat the mixture to drive off the water leading to polymerisation?


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Is there any electronic component to water conductivity?
Answers to Decrease in temperature of a aqueous salt solution decreases conductivity indicate that the electrical conductivity of salt solutions arises from the mobility of ionic species and therefore the temperature dependence of conductivity is related to viscosity.
Question: Is there any measured or predicted electronic component to water conductivity as well, where the charge carriers are electrons rather than ions?
This could take place via charge exchange (migration of bound electrons) or even perhaps via solvated electrons, or some other mechanism.
In this case I'm interested in both pure water, and salt solutions.

Well, while sodium dissolved in ammonia may get you solvated electrons, same metal leads to explosion in water, so it's still no.

There might be other ways for the solvated electrons to occur. But regardless of their origins, all solvated electrons will react with water, and that quite vigorously.

@IvanNeretin freeing up extra electrons to water (say by ionizing radiation) may result in some vigorous recombination of course, but that's not the same thing as claiming there is zero ionization and zero free electrons in equilibrium. If there is a source that says so, it would be great to mention it or even include it in an answer. I'm trying to move towards an answer and away from continuing to justify why it's okay to ask the question. Thanks!

The question Is there any electronic component to water conductivity? is clear and narrowly defined. It's a yes or no question! Votes to close as "too broad" without any helpful comments or suggestions seem spurious. These may be helpful  1, 2 though they are at extreme temperature and/or pressure; unfortunately they're paywalled and the libraries here are closed for lunar new year:

To support IvanNeretin's claim, note that the ionization energy of water is 1218 kJ/mol. This means that the relative distribution at room temperature between water and ionized water is roughly $3.2\times 10^{213} : 1$, so in effect, it does not ionize.

@Zhe if you can support that for both pure water and salt solutions (last sentence of question) then you can rule out one possible mechanism for electronic conductivity. That would be a helpful answer to post!

I agree with the commenters that electrical conduction is very unlikely, but it's worth going through some possible mechanisms:

actual solvated electrons: As others have noted, free electrons would be expected to react rapidly with protons, even in a basic solution, so this changes quickly to a scenario of sequential electron transfer between protons, so let's do that next.
Sequential electron transfer between $\ce{H.}$ and $\ce{H+}$: Let's assume the solution is strongly acidic, so protons are abundant, and a proton gets reduced at the cathode to a hydrogen atom radical. Based on the bond dissociation energies, abstraction of $\ce{H.}$ from water to form $\ce{H2}$ and $\ce{HO.}$ is slightly unfavorable, the hydrogen radical would be preferred over the hydroxyl radical. (The H-H bond formed has a BDE of ~ $\pu{105 kcal/mol}$, while the $\ce{H-OH}$ bond broken has a BDE of ~$\pu{120 kcal/mol}$.) The problem is the rate of quenching by reaction of two hydrogen atom radicals to form $\ce{H2}$ (which is water electrolysis produces hydrogen gas). I couldn't find a rate constant for that, but there is a published rate constant for recombination of hydroxyl radicals in water that is around $\pu{10^10 M-1 s-1}$. As you might expect, that's essentially diffusion limited, so the rate constant of hydrogen atom recombination is going to be at least as high. If we optimistically assume that transfer of the electron from $\ce{H.}$ to $\ce{H+}$ has a comparable rate constant, you would still have to have a very low concentration of radical and very short path to travel in order for an electron to make it from a cathode to an anode, but it doesn't seem theoretically impossible.
The third possibility would be sequential transfer of electrons from $\ce{HO-}$ to $\ce{HO.}$. In a strongly basic solution, this also seems like a theoretical possibility given a very short path and a very low concentration of radical, assuming there are no other molecules in solution that can quench the radical.

I'm not suggesting that either of these theoretical possibilities actually ever occurs, just that these are the mechanisms that seem most likely to me.

Actually electronic conduction can occur in water.  You'll probably find it in your daily rounds, if you know where to look.
Water, like all condensed matter, has a band structure.  This is discussed with some references below, as the presence of band structures in liquids is not as intuitive as that in crystalline solids.  Its valence band is filled with the electrons from the bonding molecular orbitals of the water molecules, while the conduction band that lies higher in energy, derived from antibonding orbitals, is empty.  As with insulators generally, transport of an electron into the conduction band requires a much larger amount of energy input than is available to an average molecule from thermal energy, so we expect hardly any electrons make to this transition and be electrically conducted.
But in the presence of an electric field, which is necessary to make electrical conduction meaningful anyway, the energy levels of the bands are not uniform; they depend on the position in the electric field as well as the molecular orbital structure.  An energy level corresponding to the valence band at one place may correspond to the conduction band some distance away.  And that means an electron can become a conduction electron by quantum tunneling through that distance.  This picture (source[1])

shows what is happening to the band structure given such a strong electrical field, in this case at an electrode surface.  Which gives a hint as to what's coming...
There's a catch.  The chapter in our quantum mechanics textbook that describes tunneling also tells us that tunneling is most probable if the particle/wave has to move only a short distance.  With the electric fields we ordinarily experience macroscopically, the distance required to get the valence and conduction bands to line up energetically is too long; the tunneling probability is too small to notice.  But if there were a field of at least hundreds of millions of volts per meter, then the required distance becomes short enough to make significant tunneling a realistic scenario.  We have broken the insulator down.
What sort of monstrous device is needed to create such a field?  How about the battery in your car or smoke detector? As illustrated here in a basic form, electrochemical reactions at solid electrodes depend on forcing electron transfer into the electrolyte; the surface automatically builds up enough of an electric field to enable electrons to tunnel and react.

From https://electricalacademia.com/batteries/voltaic-cell-working-construction/.
This conduction occurs over only a few atomic layers, so we don't see the electronic conduction in the watery phase directly.  We can, however, see the resulting reaction and, of course its energy input or output as voltage.

The band structure of water has been widely studied.  For example, this paper[2] uses calculations on water clusters to identify a gap of 6.9 electron volts in water, exceeding even the electronic band gap in diamond.  Another reference from this answer gives an even higher band gap.  Quoting from that answer:

do Couto et al have published an article[3] characterizing the density of states (band structure) of water using DFT theory combined with MC sampling. Extrapolating from clusters to bulk water, the HOMO-LUMO gap is 8.55 eV (Table 3).

(Additional discussion in the referenced answer.)

References:

Saraf, S.; Giraldo, M.; Paudel, H. P.; Sakthivel, T. S.; Shepard, C.; Gupta, A.; Leuenberger, M. N.; Seal, S. Photoelectrochemical analysis of band gap modulated TiO2 for photocatalytic water splitting. Int. J. Hydrogen Energy 2017, 42 (15), 9938–9944 DOI: 10.1016/j.ijhydene.2017.01.232.
Coe, J. V.; Earhart, A. D.; Cohen, M. H.; Hoffman, G. J.; Sarkas, H. W.; Bowen, K. H. Using cluster studies to approach the electronic structure of bulk water: Reassessing the vacuum level, conduction band edge, and band gap of water. The Journal of Chemical Physics 1997, 107 (16), 6023–6031 DOI: 10.1063/1.474271.
Cabral Do Couto, P.; Estácio, S. G.; Costa Cabral, B. J. The Kohn-Sham density of states and band gap of water: From small clusters to liquid water. The Journal of Chemical Physics 2005, 123 (5), 054510 DOI: 10.1063/1.1979487.

Never say anything is impossible unless dictated so by a law of physics.

Some models have electrons tunneling from one site (transition metal ion) in an enzyme complex to another over distances of a couple of nanometers. I'm not sure that this behavior could be excluded for a solution of a transition metal complex in water. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4213181/

Could this current be continued over some period, or would it in practice degenerate into a spark?

another water question: Does water really have strong EM absorption at 3 kHz in solid and 2 GHz in liquid? Why the huge shift?

Besides the useful answers above, search the keyword "aquated electron" or better still "solvated electron". People have been studying theses systems.
For example: Barnett, R. N.; Landman, U.; Nitzan, A. Dynamics and spectra of a solvated electron in water clusters. The Journal of Chemical Physics 1988, 89 (4), 2242–2256 DOI: 10.1063/1.455067.
Of course a free electron cannot exist in water before that will reduce water. It would an interesting thought experiment to irradiate water with gamma rays and simultaneously measure conductivity. You can search along those lines. On the other hand, Grotthus mechanism takes a "proton" movement in solution. For example people still do not fully understand as to why H+ ion has the highest conductivity in aqueous solution. The Grotthus mechanism was proposed a century ago and it resurfaced again.

From here "Other ions with low surface charge density (such as Cl-, ClO-, HO2- and O2·-) will also favor the gas/liquid interfaces [928a] as probably do hydrated electrons 1841, 1874." I don't have the background to interpret these; can't even access them until the lunar new year is over and the libraries here open up. I understand that there is a continuum of theories about water, from widely accepted, to... not so much.

Once a professor joking said that a physical chemist cannot differentiate between methane and a cow! Sometimes the pure simulations are far from reality. The two suggested papers are highly quantum mechanical in nature (pure simulations). I don't have the background either. I can send those papers if I have your email. Can you contact via Researchgate?

Okay I understand what you mean, some might say all models are wrong; some are useful. I can just wait until Monday to have a look, thank you!

Any conduction by non-ionic electrons in water would require some concentration of these electrons. Mithoron pointed out the extreme reactivity of sodium as an example of how it couldn't work. Using a less-reactive metal, such as zinc, might put electrons into the water, but they would be quickly attracted to aquated H+ ions that exist at 10 e-7 moles/liter and turned into molecular hydrogen. The zinc would go in as Zn++.
So the prediction of electronic conduction in water has to be negative, because even pure water contains an electron-grabbing, electron-neutralizating component that wouldn't allow free electrons to travel across any significant distance between electrodes.

This is a helpful answer, thank you! I mentioned two possibilities I could think of; I wonder if you can also rule out charge exchange (migration of bound electrons) with a similar argument?

Charge-transfer complexes can exhibit high conductivity, apparently even in solution (https://en.wikipedia.org/wiki/Charge-transfer_complex). But this apparently thru the complex, not completely thru the solvent. I think passage of electrons past the H+ ions in water is not likely.

When you transform water into a metal, it becomes an excellent conductor even in the absence of ions:

[from the abstract] Insulating materials can in principle be made metallic by applying pressure. In the case of pure water, this is estimated to require a pressure of 48 megabar, which is beyond current experimental capabilities and may only exist in the interior of large planets or stars. Indeed, recent estimates and experiments indicate that water at pressures accessible in the laboratory will at best be superionic with high protonic conductivity5, but not metallic with conductive electrons1. Here we show that a metallic water solution can be prepared by massive doping with electrons upon reacting water with alkali metals. [...]

Interesting! While they haven't exactly measured electronic conductivity directly, the've presented a lot of optical data consistent with (and not easily explained without) the presence of metallic quantities of conduction electrons. Unfortunately with the metallic NaK directly below they can't (at the moment) directly demonstrate a transition of the thin water layer above it from transparent to opaque-reflective, hopefully a demonstration of the sudden appearance of a "water mirror" will come in the near future - via displacement of a reflected laser beam or interferometry for example.

In pure water, the electrical conductivity primarily arises from the presence of ions generated through the process of self-ionization, where water molecules dissociate into hydronium ions (H3O+) and hydroxide ions (OH-). These ions can carry electric charge and contribute to the conductivity of water. However, the concentration of these ions in pure water is relatively low, resulting in low electrical conductivity.
On the other hand, in the presence of dissolved salts or other ionic solutes, the conductivity of water can increase significantly. In such cases, the charge carriers responsible for conductivity are primarily the ions derived from the dissolved salts. These ions, being charged particles, can move and carry electric charge through the water, leading to increased conductivity.
However, it is important to note that in the absence of dissolved ions, the contribution of free electrons to the electrical conductivity of water is generally considered to be negligible. Solvated electrons, which are free electrons in the solution, can exist under certain extreme conditions such as in the presence of strong reducing agents. However, solvated electrons are highly reactive and short-lived species.
In summary, while ions play a significant role in the electrical conductivity of water, the contribution of free electrons to the conductivity is minimal in typical conditions of pure water or salt solutions. The conductivity of water is predominantly governed by the movement of ions and their ionic conductivity.

Welcome to Stack Exchange! I'm sorry for the down vote, but I'm not seeing anything that adds to what's already been covered in the other answers. It feels like auto-generated text or copy/pasted from from some general source, and in no way directly addresses "Question: Is there any measured or predicted electronic component to water conductivity as well, where the charge carriers are electrons rather than ions?"

I recognize patterns generated by ChatGPT which is banned to use for answers.


text:
Adhesion of polymerized fats to cast iron
I'm trying to make an informed decision as to which oil is best to use for cast iron seasoning. I've read all kinds of conflicting information online and the only consistency seems to be that seasoning involves polymerization of fatty chains.
This explains the formation of the coating itself, but I'm interesting in choosing the oil that is least likely to separate from the cast iron surface. So my question is this:
What is the chemical (or physical) reaction (or process) which is responsible for the adhesion between the polymerized oil and the surface of a cast iron pan? Is it different from that of unpolymerized oil?

Giving it some thought, pre-polymerization, we're probably talking about the process of wetting. So what happens when the fluid polymerizes into a semi or total solid? I can't help but notice similarities between this process and glue.

Thank you for asking this question. As a French, I am 200+% obsessed with food, cooking and every dish my grandma was cooking for the family. I am also an organic chemist so I don't buy the "everyday bullshit" (if that makes sense).
My first (personal) remark would be that if there were a way to quickly season a cast iron pan, every manufacture would use it and build its reputation on that.
This brings me to a more physical/chemical remark: (1) seasoning cast iron takes time ; (2) many different fats are recommended, from pork fat (low in unsaturated fat) to olive oil (high in unsaturated fat). But only unsaturated fat can polymerize.
Overall, that points to something else, which is that burnt fat would play a major role in seasoning. Burning fat will create carbon (which iron cast contains) and aromatic compounds (from Maillard's reactions) which have affinity for carbon.
As a whole, I would suggest you to cook in this/those pan(s) as often as you can and deglaze it to make some sauce as we do in French cuisine.


text:
Question about the cyclic hemiacetal form of compounds

Why is D not a valid answer for this question?

D is formed from 5-hydroxypentanal, a primary alcohol. Your compound is 4-hydroxypentanal. B is the answer.

Short and simple, if you count atoms from $\ce{O}$ (of $\ce{OH}$ group) to carbonyl $\ce{C}$, you'd find the number is 5. Thus, it makes 5-membered cycle. So, it can't be D.

D would not be a valid answer if you think about what forms or what happens that leads to formation of cyclic hemiacetal. In the course of formation of cyclic hemiacetal, the -OH oxygen has 2 lone pair and the C=O carbon has partial positive charge on it and so, in a nutshell the -OH oxygen attacks the carbon with partial positive charge and this leads to formation of hemiacetal. Now,the -OH is attached to 4th C atom and so during the course of ring formation, the 5th carbon will always lie outside the ring and this the reason why D cannot be a valid answer.
A scheme of this critical mechanism step is shown below. The carbon atoms are numbered for clarity. Additionally, the methyl group on carbon 5 is circled. Since the mechanism of the ring closure does not involve carbons 2-5, the structure cannot change at those positions.

Thanks for additional editing to the answer to make it more easier to understand. Actually I'm quiet new to this website so I'm figuring out how to do things here.So thank you.

Hi and welcome to the site! Don't doubt to take the tour for a quick intro and use the help center for any doubts you might have.


text:
Can a part of a compound be a reducing agent?
I have the equation:
$$\ce{2 Cu2O + Cu2S -> 6 Cu + SO2}$$
My doubt is that since copper is reduced (it goes from $+1$ to a $0$ O.S.), then which of the two reactants is reduced? Or why would it be fine to say that only copper is reduced and not an entire reactant? Because from my experience, I've seen that an entire reactant is considered to be reduced or oxidized. So is it right to tell only a part of the compound is oxidized?

Compounds are not reduced; elements are. To say that a compound is reduced is an abuse of language. Sometimes it is forgivable, sometimes it isn't.

So is it more correct to refer to a reactant as a reducing/oxidizing agent rather than saying "a compound is reduced/oxidized"?

Exactly. But then again, is can so happen that one element in a compound is oxidized and another reduced; what would you call that?

So that would then be referred to as a disproportionation reaction, I believe? And to determine which type of reagent it is, we would have to look at the particulars of the mole ratio and the individual changes in oxidation number of that particular reactant, right?

No and no. Disproportionation is a reaction where the same element reduces and oxidizes at the same time. As for the type of reagent, this is not something worth caring about. Call it a reducing and oxidizing agent at once, if you wish, or don't call it anything special at all.

,@Ivan Neretin. "is can so happen that one element in a compound is oxidized and another reduced; what would you call that?". How that is possible out of a disproportionation? I cannot see which third species can do so (be oxidized and reduced by two different elements originally in the same compound).

For accounting purposes, we can assign oxidation numbers to all the atoms. Oxygen could be -II in all species here, then sulfur would be +IV in $\ce{SO2}$ and -II in $\ce{Cu2S}$ (oxygen and sulfur are in the same group and both are more electronegative than copper, so assigning the same oxidation number in $\ce{Cu2S}$ and $\ce{Cu2O}$ makes sense). Oxidation numbers of atoms in elements are zero, and the copper in both compounds would be +I to balance out the -II of the chalcogens.
Once you are in agreement with your discussion partner on these oxidation states, you can make a statement about which atoms undergo oxidation or reduction: Sulfur gets oxidized, and copper in both copper compounds gets reduced.
In cases where only one atom in a compound changes oxidation states, or when all changes are in the same direction, we do talk about the compound being reduced or oxidized. For example, it is common to say that when you oxidize an aldehyde, you get a carboxylic acid (and when you reduce it, you get an alcohol or an alkane).
On the other hand, if you hydrate an alkene, some folks don't consider that a redox reaction.

I am restating the question below (to slightly adjust the formatting and make it easier to address):

I have the equation:

$$2\ce{{Cu_2O} + {Cu_2S}⟶{6Cu} + {SO_2}}$$
Here we have compounds $\ce{{Cu}_2O}$, $\ce{{Cu}_2S}$ for the reactants, and the products of metallic $\ce{Cu}$ and gas $\ce{S{O_2}}$; compounds share electrons and so it difficult to say that they are "reduced" or "reduce" since there is a shared bond in any case to start with.
Getting beyond that, however, we see that for $\ce{Cu_2O}$ that just like water that is $\ce{H_2O}$, Copper shares two of its electrons with Oxygen.  In the case of $\ce{Cu}$ Copper, these electrons are not shared anymore, so in this case $\ce{Cu}$ is reduced (the opposite of oxidation where electrons are shared), and the $\ce{O}$ is oxidized, since the copper is no longer sharing negative charge, increasing its positive charge.

My doubt is that since copper is reduced (it goes from +1
to a 0 O.S.), then which of the two reactants is reduced? Or why would it be fine to say that only copper is reduced and not an entire reactant? Because from my experience, I've seen that an entire reactant is considered to be reduced or oxidized. So is it right to tell only a part of the compound is oxidized?

Oxidation and reduction is usually formally defined in the sense as in this link https://chemed.chem.purdue.edu/genchem/topicreview/bp/ch19/oxred_3.php but I am trying to address the problem from a conceptual perception to make progress in answering the question at hand.
Also, for $\ce{{Cu}_2S}$, we have that the reactant $\ce{Cu}$ Copper is sharing two of its electrons with $\ce{S}$ Sulphur and in the product that metallic $\ce{Cu}$ Copper is no longer sharing these electrons, receives them back in the reaction, and therefore is also reduced in the forward reaction.
For Copper, it is a a metal at the end, and individual atoms are "covalent reduced" here, where we are taking oxidation and reduction to be defined in a sense of negative charge being moved (reduction) to the Copper atom in the case of this metal $\ce{Cu}$; but something to be careful about is that this charge that moves back to the Copper atom is not charge of the Oxygen atom or charge of the Sulfur atom.  Rather, this is shared charge that belongs originally to the neutral Copper atoms.
Now let us consider the compounds.  All the reactant compounds are neutral to start with.  But in the course of the reaction, the shared charge that belonged to the Copper atoms is returned to those Copper metal atoms.  And the movement of the charge away from the Oxygen $\ce{2 O}$ and from $\ce{2{Cu}_2O}$ results in the lowering of negative charge shared there on the $\ce{2 O}$ and is therefore oxidation, though not in an ionic sense since the charge was only shared to begin with.
Similarly, the reactant compound $\ce{{Cu}_2S}$ has a movement of shared charge to the Copper $\ce{{Cu}_2}$ and away from the $\ce{S}$ Sulfur, a covalent oxidation of the Sulfur.  The usual application of oxidation and reduction is for ionic compounds and all compounds of the reactants and the products are covalent or metallic bonds.
But to conceptually address the question, the shared metallic electrons are returned to metallic copper $\ce{Cu}$ (a non-ionic reduction characteristic) and in the course of the reaction the shared electrons are taken back from the final Sulfur Dioxide Compound product $\ce{SO_2}$ (a non-ionic oxidation characteristic).
If it were not enough to consider the movement of negative charge for Copper reactants towards the metal in the course of the reaction -- there is an additional bonding process!
Interesting enough also, is that the Copper atoms have completely received their covalent-bonded charge back - but that means that it had been taken from somewhere.  And the only product left is $\ce{SO_2}$; so that implies that the move of negative charge in the reaction towards the Copper metal product (considered here like a covalent reduction) must result in a covalent-like oxidation (movement of negative charge away) of the $\ce{SO_2}$ and this effect probably influences the resulting strength of the resulting neutral metallic and neutral covalent bonds at the end of the reaction.
The chemical Sulfur Dioxide $\ce{SO_2}$ is covalent, meaning that the Oxygen atoms $\ce{O_2}$ newly share their charge with the Sulfur atom $\ce{S}$ in the course of the reaction. Although this bond is covalent, charge still moves away from the Oxygen atoms towards the Sulfur atoms, having a reduction effect on the Sulfur and an Oxidation effect on the Oxygen atoms (within the covalent compound)!
I appreciate an upvote if this answer addresses the question as well as comments from any angle!

Cu++ ion is reduced to Cu]0]. The reducing agent is sulfide Whether the mechanism is stepwise decomposition of CuS then reduction of CuO by elemental S or some concerted mechanism requires some chemical study. Since oxidation numbers are assigned by fiat there is little need to be concerned about bond covalency.

@jimchmst For oxidation and reduction I am going from starting reactants to final products. (What is Copper(I) Oxide?) https://byjus.com/chemistry/copper-oxide/ notes that "Copper(I) Oxide is also called as cuprous oxide, an inorganic compound with the chemical formula $\ce{Cu2O}$. It is covalent in nature. "  The statement  "Cu++ ion is reduced to Cu]0]" is in the opposite direction of the question:  (the product Copper metal to a reactant covalent bond),  The forward reaction is the question $\ce{2Cu2O + Cu2S⟶6Cu + SO2}$; that oxidized in a reverse reaction is reduced in a forward reaction.


text:
Free energy of solvation and Henry's law constant
I am trying to calculate the free energy of solvation of $\ce{CO2}$ from its Henry's law constant. As given on Wikipedia, the dimensionless Henry's law constant is $0.83$. If I try to calculate $\Delta G$ from this, I got
$$\begin{align}
\Delta G &= -RT\ln K \\
  &= -\pu{8.3145 J mol-1 K-1}\times\pu{298.15 K}\times\ln 0.83 \\
  &= \pu{0.46 kJ mol-1} \\
  &= \pu{0.11 kcal mol-1}.
\end{align}$$
However, I saw an article saying that the experimental value is $\pu{0.24 kcal mol-1}$. The article cited from the CRC Handbook of Chemistry and Physics, while Wikipedia cited from this compilation, which I think is also a reliable source.
So what did I do wrong in my calculation of $\Delta G$ from the Henry's law constant?

It seems your calculation is fine. I find it a bit suspicious that the article from 2018 cites very old edition of CRC Handbook from 1995. Newer 97th edition from 2017 doesn't include free energies of solvation at all, only enthalpies of hydration. The difference in numbers might stem from different approaches of determination of the Gibbs free energy.

Also that article contains interesting passage on p. 17: "Thus, at QM/MM level of theory the value of hydration free energy of $\ce{CO2}$ amounts to 0.01±0.92 kcal/mol. This value is in good agreement with the experimental value of 0.24 kcal/mol". This confidence interval looks suspiciously broad and I don't see how it's in a "good agreement" at all. Probably I'm wrong and theoretical chemists who know better could comment on that.

I think solvation free energies having an uncertainty of 1 kcal/mol is not uncommon in theoretical calculations. The solvation free energy happens to be close to zero so that the uncertainty appears huge.

The CRC Handbook of Chemistry and Physics [1] provides data for the solubility of $CO_2$ in $H_2O$ originally from reference [2]. The reported data are mole ratio solubilities $\chi$ of the gas at a partial pressure of one atmosphere (101.325 kPa), but can readily be converted into Henry's law solubility in other units:
$$H_{CP} = 1000 \delta_{H2O} \chi /M_{H2O} = 997 \times 6.15 \times 10^{-4}/18 = 0.034 mol/L atm$$
or in molal and bar units
$$H_{mP} = 1000\chi /1.013 (bar/atm)M_{H2O}=1000\times 6.15 \times 10^{-4}/(18 \times 1.013) = 0.034 mol/kg bar$$
Reference [2] is one of the original sources of the data tabulated in the Wikipedia, compiled by Rolf Sander in reference [3]. Other references in [3] provide similar values of $H_{CP} \approx 0.034$ mol/Latm. The value of $H_{CP}$ in the wikipedia is therefore consistent with the solubility reported in the CRC Handbook.
The article by Prasetyo and Hofer [4], referred to in the OP, cites the CRC Handbook as the source of the following reported experimental value (see Table 4): $$\Delta_{solv} G^o = 0.24 kcal/mol$$
However, from $H_{mP} = 0.034$ mol/kgbar (or $H_{CP} =  0.034$ mol/Latm) we obtain $$\Delta_{solv} G^o = -RTlog(H_{CP} \frac{P^o}{m^o})= 2.0 kcal/mol$$ which is far larger than the value reported in the article.
To attempt further progress, note that comparing standard free energies of solvation requires awareness of the associated reference states.
Prasetyo and Hofer indicate (pg 6474) that their choice of standard state is P° = 1 bar, m° = 1 mol/kg, and T = 298.15 K. The experimental values Prasetyo and Hofer compare to simulation results (see Table 4) refer presumably to this standard state. But this is approximately the same as the molar standard state associated with the data reported in the Wikipedia (P° = 1 atm, c° = 1 M, 298.15 K). Certainly the energies computed in the two scales don't differ significantly, as shown above. I therefore suspect that a different standard state was used in their calculation, or free energies for different (chemical) processes are being reported.
It is worth adding that the value $\Delta_{solv} G^o = 0.24 kcal/mol$ is encountered in other articles that also reference the 75th Edition of the CRC Handbook.
References
[1] The CRC Handbook of Chemistry and Physics, 85th Edition, section 8-87, SOLUBILITY OF SELECTED GASES IN WATER, L. H. Gevantman.
[2] R. Crovetto, Evaluation of Solubility Data for the System CO2-H2O, J. Phys. Chem. Ref. Data, 20, 575, 1991.
[3] R. Sander: Compilation of Henry's law constants (version 4.0) for water as solvent, Atmos. Chem. Phys., 15, 4399-4981 (2015).
[4] N. Prasetyo and T.S. Hofer J. Chem. Theory Comput. 2018,  14, 12, 6472-6483

I wonder how you obtained 2.0 kcal/mol. I guess you calculated it by -RT ln H? But your H has units and it should not represent an equilibrium constant. That's why I used a dimensionless H in my calculation.

@mck H is reported with units so people understand the associated reference state, not because you are supposed to take the logarithm of the units. H is not unitless, but $HP^o/m^o$ or $HP^o/c^o$ are, depending on the definition of H. Since $P^o = 1, c^o=1$ and $m^o=1$ in the appropriate units, it all works out. See my changes to the answer.


text:
Solubility of lead cyanide
I read that lead(II) iodide is soluble in excess KI solution due to complexation, but lead(II) cyanide does not dissolve in excess cyanide solution. This seems wrong since cyanide is a much better ligand that iodide, so shouldn't it be able to form a complex more easily?

Cyanide is not always a much better ligand than iodide.  It preferably complexes some things while iodide preferably complexes others.  Lead(II) is one of those things that prefers iodide.

Cyanide is generally better than iodide in transition metal complexes because it acts as a pi-acceptor. It has empty orbitals that overlap with the partially filled d-orbitals of the metal, lowering the energy of these electrons.
Iodide is a sigma-donor, meaning that it donates electron density to empty metallic orbitals. The effect is small compared to a pi-acceptor in a transition metal, because the major contribution to stability is the energy of d-orbital electrons.
However in Pb(II) complexes, the d-orbitals are full already, so here the sigma-bonding mechanism for stabilisation is key. As cyanide isn't as good of a donor as iodide, then iodide forms a complex whereas cyanide doesn't.

I can only guess here, but main-group complexes and transition-metal complexes are sometimes a little different. Once you leave the d-orbital chemistry you have to ask yourself where you get low laying, empty orbitals. If you take $\ce{BiI3}$ for example. This is a $\ce{Bi^3+}$ so there is still one lone-pair. Hence you will get a tetrahedron with three $\ce{I^-}$ and one lone-pair. If you add for example $\ce{NaI}$ it will dissolve, but how? The next empty d-orbitals would be the 6d-orbitals, here the 7s would be much more favourable but they are still quite high. What's actually happening is that you shift electron density from the extra iodide-ligand into an anti-bonding $\ce{Bi-I}$-bond of the $\ce{BiI3}$ molecule.
The same thing goes for $\ce{PbI2}$. You have the two iodide-ligands and one lone-pair. So there is already a free spot on the metal. That means first you get $\ce{[PbI3]^-}$ and then, by a similar mechanism a $\ce{[PbI4]^2-}$.
We call this a hypervalent compound.
That is at least how Olaf Kühl describes it in his book.
At this point I can only guess. If the approaching Iodide / Cyanide acts as a Lewis-base this means that it will donate electron density into empty orbitals. If those empty orbitals are the anti-bonding σ*-orbitals of a $\ce{Pb-L}$-bond, we require those orbitals to be low in energy.
The problem is that I don't really know how to compare an iodide with a cyanide in terms of covalency and electronegativity here. The bond energies in $\ce{Pb-X}$ decrease from fluoride to iodide. Hence I expect the $\ce{Pb-I}$-bond to be quite weak. The problem is that cyanide is a soft base as well and only a little smaller.


text:
Rate law equation - orders?

Consider the following reaction:
$$\ce{2 H2 (g) + 2 NO (g) -> N2 (g) + 2 H2O (g)}$$
The elementary steps of this reaction are:
$$
\begin{align}
&1^\text{st}~\text{step:} &\ce{H2 (g) + 2 NO (g) &-> N2O (g) + H2O (g)} &&\text{(slow)} \\
&2^\text{nd}~\text{step:} &\ce{N2O (g) + H2 (g) &-> N2 (g) + H2O (g)} &&\text{(fast)} \\
\end{align}
$$
The reaction is exothermic overall.
(a) What is the rate law for this reaction? Justify your answer.
$$\text{rate} = k[\ce{NO}]^2[\ce{H2}]$$
The 1st step of the reaction is the rate limiting step.

How did they know the order of the reaction with respect to $\ce{NO}$ was $2$, and $1$ for $\ce{H2}$? With other questions the orders are something they state in the question and are unrelated to the coefficients.

You are correct, the reaction order cannot be deduced from the brutto-equation and is usually determined experimentally.
However, here you are presented with the elementary reactions and their rates. Obviously, the slowest step is the bottleneck and controls the overall kinetics, hence the total rate matches the rate of the slowest step. And as this is an elementary reaction, you just apply the law of mass action and use the stoichiometric coefficients for the rate equation.

With other questions the orders are something they state in the question and are unrelated to the coefficients.

a- If the reaction is multi- step," the orders are may or may be not related to the coefficients" that the order and the rate equation of a given reaction cannot be reliably deduced from the stoichiometry and must be determined experimentally.
b- If the reaction is one step," the orders are related to the coefficients". In this case, the exponents in the rate law equation (or order of the reaction) are equal to the number of colliding molecules or to the stoichiometric coefficients of those reactants in the balanced equation that represents that collision. There is, of course, the possibility of bimolecular collisions, and the probability of termolecular collision is weak, especially for fast reactions.  Four-molecule collisions are unlikely to occur in most cases.

How did they know the order of the reaction with respect to $\ce{NO}$ was $ 2$, and $1$ for$\ce{ H2}$?

The above-mentioned reaction mechanism has two steps. The first step is the slow step followed by a fast step. Since the total rate of the reaction is determined by the rate of the slow step, which is described as the Rate-Determining Step, so the rate of the reaction is equal to the rate of the reaction of the first slow step.
$$\ce{\mathrm{r}= \mathrm{k}[{NO}]^2[{H_2}]}$$
$$\color{red}{\text{Another proposal for the mechanism of reaction}}$$
The first  slow step of the above-mentioned mechanism  in the question indicates that three molecules ( two molecules of $\ce{NO} $, one molecule of $\ce{H2}$) collide with each other at the same space and time (Termolecular Collisions), but the probability of termolecular collision is weak, so, there is another mechanism excludes termolecular collision as the following:
\begin{array}{ }\ce{
\text{First step:}~~~2NO_\mathrm{(g)} <=>[k_\mathrm{1}][k_\mathrm{-1}]N2O2  &\quad\left (\mbox{Fast }\right)\\
\text{Second step:}~~~N2O2\mathrm{(g)}+H_2\mathrm{(g)} -> N2O\mathrm{(g)}+H2O \mathrm{(g)} &\quad\left(\mbox{Slow }\right)\\
\text{Third step:}~~~N2O\mathrm{(g)}+H2 \mathrm{(g)}-> N2\mathrm{(g)}+H2O \mathrm{(g)} &\quad\left(\mbox{Fast }\right) \\
}\end{array}
$$\color{red}{\text{ Describing the proposed  reaction mechanism }}$$
The reaction mechanism consists of three steps where the first step is not the slow step, but the slow step precedes by a fast step followed by a fast step. In this case, where the slow step is preceded by a fast step, the first step reaches a state of equilibrium, beginning with the reverse of the reaction of the first step.
This is because the first fast step produces a reactive intermediate material $\ce{ N2O4}$ at a fast rate, but the second slow step cannot consume these $\ce{ N2O4}$  at a similar rate, leading to the accumulation of $\ce{ N2O4}$, so, the probability of collisions of the $\ce{ N2O4}$ molecules increases,  this leads to reverse the reaction in the first step, thus the reaction in the first step reaches  at equilibrium.
The forward  and backward reactions are fast in the first step, but the reaction of the third step  is fast in the forward direction
$$\color{red}{\text{ Deducing the order or rate equation from the reaction mechanism }}$$
Where $K_1$ is the rate constant of the forward reaction, as $K_{-1}$ is the rate constant for the backward reaction of the first step, and $K_2 $ is the rate constant for the reaction of the slow step.
If the concentration of  $\ce{ N2O2}$ which  produced from the first step increased, the reaction rate increased in the opposite direction, leading to the arrival of the concentration intermediate material $\ce{ N2O4}$   to a fixed value. This is called Steady State, where the concentration of $\ce{N2O2}$ is constant,
The rate of appearance of the intermediate material $\ce{N2O2}$  equal the rate of its disappearance, this material produced from the first step in the forward direction and consumed from the first step in the opposite direction, and in the second step also.
$$\text{ Production rate of}~ \ce{N2O2}  = \text{ Consumption rate of}~ \ce{N2O2}$$
$$k_\mathrm {1}\ce{[NO]^ 2}= k_\mathrm{-1}[\ce{N2O2}] + k_\mathrm {2} \ce{[NO] [H2]}$$
But $k_2 << k_{-1}$ because the second step is the slow step, so the term $k_2\ce{ [NO] [H2] }$ can be neglected  at the far right of the last equation.
$$k_1\ce{[NO]^2}= k_\mathrm{-1}\ce{ [N2O2] }$$
$$[\ce{N2O2}] =\frac{ k_1 [NO] ^ 2 }{ k_\mathrm{-1}}$$
The value of$ [\ce{N2O2}]$ from the last equation can now be compensated in the reaction rate equation which represents the slow step.
$$\text{Rate} = \ce{k_2[N2O2][H2] = \frac{k_1k_2}{k_\mathrm{-1}}\times{[NO]^2 [H2]} = k[[NO]^2[H2]}$$
In it we see that k expresses the value of the fractional $\frac{k_1k_2}{k_\mathrm{-1}}$


text:
Looking for experimental data for titration curves
I am looking for real, open source and free experimental dataset (not image, scan of table) of acid/base titration curves: pH versus titrant volume for many combination (strong/weak acid/base, polyprotic acids, etc.).
I could not find something satisfactory over the internet. Where could I find this kind of dataset?

Sorry. My answer did really seem unattuned to the question. But I liked looking at the curves!

The only thing I can recommend is papers. Look up titrations in Google Scholar and then use open databases to download the paper itself, such as Library Genesis.


text:
Calling compounds electron deficient
I believe this is a conceptual question and might sound silly, but please help me with it:
I can understand when ions like $\text{Co}^{3+},\text{Mg}^{2+}$ are called electron deficient, because they have no valence electrons, but how do I call a compound like $\text{AlCl}_3$ electron deficient? What's the basis for calling such a compound electron deficient?

Is aluminium's octet complete?

Yeah, I believe aluminium's octet is complete.

@DhrubajyotiGhosh No, it's not. Count the valence electrons again.

Being electron-deficient would imply that the molecule possesses vacant MO which can feasibly accept electrons from an effective Lewis base. And yes, AlCl3 has an incomplete octect(6 electons).

An interesting question is whether $\ce{AlCl3}$ really exists under the given conditions. Perhaps we rather have $\ce{Al2Cl6}$. Then counting might be different.

@andselisk Sorry, so silly of me. Yeah, aluminium does have an incomplete octet...

@DhrubajyotiGhosh No, it's not silly, it's fine. Only the one who does nothing is never mistaken. You can post an answer to your question if you solved the issue yourself:)


text:
ionisation isomers

With ionisation isomers, how do you know which ligand (out of $\ce{Br}$ and $\ce{NH3}$ for example) exchanges with the compound outside the bracket?

Well, as net charge is always the same, you are free to exchange any. I do remember water being involved in Cobalt chloride complexes, with change in lattice water. The same must apply for ammonia, too.

Your question is not very clear. Do you mean how to make the difference between the 2 compounds if you have them in 2 different jars? Or do you mean how to study that exchange in a solution?


text:
Do the salt bridge ions have to match the ions in the half-cells solutions?
Does the salt bridge in a galvanic cell have to have the same ions as the ones in the solutions? E.g. does it need to have nitrate ions if the half-cells contain copper nitrate and zinc nitrate, or could nitrate ions still be used if the solutions had sulfate instead?

Any ions will move in the electric field. It is not that the ions in the cell would build a border wall against the ions of different kind and make them pay for it. It is just that the ions from the salt bridge would slowly get into the half-cells and vice versa. As long as this doesn't cause any unwanted reactions, it is not a big deal.

In short: a priori no, it need not.
The purpose of the salt brigde is to close the electric circuit (transport of charges) between the two half-cells.  To work as intended, all you need to do is to fill the bridge with any salt whose ions, both cation and anion do not react with either half cell components, nor under the external potential applied, and ideally have the same mobility in the electrolyte of the bridge, expressed by the transfer number.  This is why aqueous solutions of KCl are used so often for this purpose, but you are free to choose differently.
And not to forget, the tube normally ends by a diaphragm at each side, to slow down considerbly the mechanical exchange of matter (ions) between the bridge's content and the half cells.


text:
Composition difference between #2 and #2D fuel oil
I have read that #2 and #2D fuel oil are essentially the same, but the #2D might have less sulfur. I have read others that say that #2 has a higher cloud point, so if that is true, then there are some differences.
I tried burning #2D in my furnace and it generated a noticeable oil fume in the house compared to #2 which generates no fumes. Why would #2D be generating oily fumes if it is cleaner?

For the US (your mileage may vary), ASTM defines three grades, 2S15, 2S500, and 2S5000. There is no 2D defined by ASTM standards.

? See: Diesel Fuel Grades


text:
Order of Nucleophilicity
I know this question has been asked many times but I need to confirm this doubt which has been pestering me for the past few days.
Suppose in an $\mathrm{S_N2}$ reaction (in polar aprotic solvent) we have two nucleophiles:

ethoxide $(\ce{EtO-})$ ion;
ethyl thiolate $(\ce{EtS-})$ ion.

Now, since we know fluoride $(\ce{F-})$ is stronger nucleophile than iodide $(\ce{I-})$ in polar aprotic solvent, i.e. nucleophilicity parallels basicity.
We should expect the alkoxide ion to be more nucleophilic than thiolate since alcohols are weaker acids than thiols.
But my book says the opposite. Is this a exceptional case? If so, could it be predicted in advance that thiolates are better than alkoxide even though fluoride is stronger than iodide?
Any help would be much appreciated.

There is not a direct correlation between pKa and nucleophilic strength; pKa is mostly used for leaving group ability. Since the nucleophilic attack occurs via overlap of the nucleophile's HOMO with the p orbital of a carbon atom in the alkyl halide, the major factor in determining the nucleophilicity is how high the HOMO is (higher HOMO --> more nucleophilic). Since S has lower nuclear charge than O, the HOMO in the EtS- ion is higher than that in the EtO- ion (electrons less bound by nuclear charge). Thus, EtS- is more nucleophilic.

Predicting nucleophilicity is often a non-trivial task due to the large number of factors to be considered (e.g. thermodynamics of the interaction, orbital interactions, solvent interactions, etc.).

It has been established here that basicity and nucleophilicity are distinct concepts. Why then is basicity often used as a measure of nucleophilicity?
To answer this question, we have to consider the several factors that affect nucleophilicity, which is a kinetic concept (i.e. it is related to the activation energy required for the reaction to take place). My answer here has very clearly put out the factors that affect nucleophilicity. Chiefly, they are the thermodynamics of the interaction (think of the bond energies of bonds being formed and broken), orbital interactions (primarily, a kinetic consideration) and of course, there are the usual steric and solvent considerations.
Being a thermodynamic concept, basicity is related to how strong are the bonds being formed between the nucleophile and the substrate carbon atom. It is essentially the thermodynamic factor which I have mentioned above. The generalisation "more basic means more nucleophilic" has been made as it has been observed that basicity is a good measure of nucleophilicity in most cases. However, it is important to note that consideration of basicity neglects other factors I have highlighted, particularly orbital interactions.
Although thiolates fall behind alkoxides on the thermodynamics factor, they win on the kinetics factor and that seems to be the reason for their higher relative nucleophilicity. The more highest occupied MO of the sulfur atom seems to be closer in energy to the lowest unoccupied MO of the substrate. This results in rather favourable orbital interactions. For more information on this, you can take a look at the portion on Klopman-Salem equation here.

So, are you saying that the 'thermodynamic' factor (strength of carbon-halide) bond is responsible for the halides trend and the 'kinetic' factor is responsible for the thiolate and alkoxide trend?...By the way in one of your references it is written that, methoxide is better nucleophile than methyl thiolate..Why is that?

The two factors (thermodynamics and kinetics) are always at play. However, in some cases, one outweighs the other. Yes, that particular reference cites that based on intrinsic nucleophilicities (i.e. solvent-free nucleophilicity). You are considering a polar aprotic solvent, thus solvent factors would come into play, and this could change the relative nucleophilicity order.

Are the solvent factors responsible for changed trend in methoxide and thiolate? If yes, how so? I thought that polar aprotic solvent didn't affect the nucleophiles and they were relatively free as opposed to polar protic...By the way, your previous answers were very helpful..

I am not entirely sure myself because solvent effects are often not straightforward. The category "polar aprotic solvents" itself is a very broad one with solvents with quite different properties. I am uncertain of how polar aprotic solvents can influence the nucleophilicity order in this case but clearly, there is some sort of influence. That's all I can help you with.

Please remember that nucleophilic substitution at a saturated carbon atom does not directly correlate with pKa.
The pKa of a substance (in the Bronsted-Lowry definition) relates to the substance's ability to have strong interactions with $\ce{H^+}$. Thus, this interaction is mainly electrostatic in nature; since $\ce{H^+}$ is a very small, weakly polarizable ion, according to hard-soft acid base theory, strong interactions with this ion are largely electrostatic in nature, i.e. proportional to charge density.
However, nucleophilic substitution at a saturated carbon atom occurs mainly through orbital interactions, not through electrostatic interactions. Namely, the highest occupied molecular orbital (HOMO) of the nucleophile interacts with the LUMO (sigma antibonding orbital) of the substrate (usually alkyl halide).

Since higher energy HOMOs interact better with LUMOs, molecules or ions with higher energy HOMOs are more nucleophilic.
With the specific case that you gave ($\ce{EtO-}$ and $\ce{EtS-}$), consider that sulfur is a larger atom, and the electrons in HOMO in $\ce{EtS-}$ are going to be less tighly held by nuclear charge than electrons in the HOMO in $\ce{EtO-}$. Alternatively, you can consider that the highest energy level orbital in S is 3p while that of oxygen is 2p --> electrons are more tighly held and lower in energy in $\ce{EtO-}$ --> $\ce{EtO-}$ is less nucleophilic.

Thus, if you consider the relative energy levels HOMOs in related molecules and ions, you get the general trend below:

References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

So why us the halide trend reversed in polar aprotic solvent?

This is likely due to the absence of hydrogen-ion interactions in aprotic solvent. In protic solvents, the high charge density fluoride ion is solvated while the lower halides are not. This solvation is absent in aprotic solvents. In this case, the solvent effects blur the general trend that we observe just considering the energy level of the HOMO. Just remember that there are several factors other than HOMO energy (such as solvent effects and substrate character) that influence nucleophilicty. Given a reaction, you need to consider these effects together.


text:
Bond cleavage of 3 membered cyclic ring on heating
I was told to propose a mechanism for the following conversion:

One obvious step is the conversion of cyclic ketal to ketone because of acidic conditions. But I am unable to understand how the three membered ring opened to give an alkene? (However, I can understand  that the three membered ring is unstable)
I feel that somehow, a carbocation is being generated and E1 type elimination is taking place.
What is happening exactly? And does this type of thing take place with any three membered ring under given conditions?

Break the three mbered ring to form an enol. Rearrange the carbocation and heat it to form an alkene in the mean while your enol tautomerizes to a ketone in acidic medium. Havent the solutions been released along with the paper though?

@AvnishKabaj No solution was given with the paper.

Since you get hydrolysis of the ketal, you should list the hydronium ion over the arrow. With just a proton as a catalyst, the ring cleavage can occur with the retention of the ketal. Temporary opening of the ketal to the enol ether and protonation of the ketone causes fragmentation. The ketal can then reform.

Possible mechanism for cyclopropyl ring opening of given compound in acid medium:

Is there any name to the step in which the alkene attacks to open the cyclopropyl ring? Is it related to Neighbouring Group Participation?

Not really to my understanding. But, you see when the first enol formed, it was conjugated all the way to the other ketone (you can write few resonance to see how stabilizing it is).


text:
H2O and Le Châtelier’s principle
According to Le Châtelier’s principle, the adding of a reactant causes a shift in the direction of products, to minimize the modification. But what about $\ce{H2O}$? When can it be considered as a reactant? Consider that it’s concentration doesn’t appear in the equilibrium constant when the solution is very diluted.
For example, in the following reaction:
$$\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)}$$
does the adding of water cause the system to shift in the direction of products?

Well, the solvent is always considered an active species in the reaction. When you write "CO2(aq)" it really means CO2 surrounded by H2O molecules !

Related, if not a duplicate: Does adding water to a reaction mixture shift equilibrium?

Equilibrium expressions in the thermodinamically strict way should be written with activities, not concentrations:
$$
a_i = c_i \times \gamma_i
$$
Where $a_i$ is activity, $c_i$ is the chosen concentration unit (molar fraction for water), and $\gamma_i$ is the activity coefficient, a number that measures how much the species i differs from ideal conditions in that particular reaction system. You can think of it as an "effective" concentration. In the ideal case, $\gamma_i = 1$ and $a_i = c_i$, so generally that is what we use to calculate, concentrations.
In a diluted aqueous solution, all that a water molecule "sees" is more water molecules, with an ocassional solute species passing by. Not only is $c_{H_2O} \approxeq 1$ (molar fraction), but also the interaction is practically zero, and each species behaves like it would by itself, so $\gamma_{H_2O} \approxeq 1$ as well. $a_{H_2O} = 1$, and adding more water only decreases interaction further; we can safely omit $a_{H_2O}$ from the equilibrium expression.
However, in a very concentrated solution water is closer to the solute, and there are significant interactions between them (e.g. attraction to anions by hydrogen bridging) so $\gamma_i \neq 1$ for all the species. If you add water (or remove it, in a mental experiment) in such a system, the interactions significantly change, and so does $\gamma_{H_2O}$ (and $c_{H_2O}$). The equilibrium shifts! We make big mistakes in calculations if we omit $a_{H_2O}$ from the equilibrium expression.
In the picture below (extracted from here) you can see a plot derived of experimental measures of water activity [$a_{H_2O}$] vs. molar fraction of the solute [x] in a perchloric acid solution.

It is up to you to decide when to start considering water in the equilibrium calculations, but normally we work with small concentrations so it's OK to ignore water activity (and the activity coefficients of the other species) altogether.  Note that a molar fraction of 0.1 means a solution about 6 M in the acid, whereas a $\mathbf{10^{-3} M}$ solution represents a molar fraction of $\mathbf{2 \times 10^{-5}}$.
Additionally, as $\gamma_i$ for every species depends on the concentration of all the species in the system, the calculations can get really tricky if you switch to activities.


text:
pH of 0.05 M NaH2PO3
What is the pH of a $\pu{0.050 M}$ solution of $\ce{NaH2PO3}$? I am given that $\mathrm{p}K_\mathrm{a1} = 2.0$ and $\mathrm{p}K_\mathrm{a2} = 6.58$.
I am getting two different answers, $3.94$ and $4.30$, using two different methods that I have learned:
Method 1
$\ce{H2PO3-}$ can either be protonated or deprotonated. The protonation reaction has
$$K = \frac{10^{-14}}{10^{-2}} = 10^{-12},$$
while the deprotonation reaction has
$$K = 10^{-6.58} = 2.63\times 10^{-7}.$$
Since the protonation reaction is significantly less favorable than the deprotonation ($K$ ratio is less than $1000$), we can ignore the protonation and only focus on the deprotonation. This yields:
$$10^{-6.58} = \frac{x^2}{0.05-x} \to x = [\ce{H+}] = \pu{1.15e-4 M},$$
$$\mathrm{pH} = -\log(x) = 3.94.$$
Method 2
With a diprotic acid that can either be protonated or deprotonated, the $\mathrm{pH}$ is given by the following approximation:
$$pH = 1/2(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a1}),$$
$$\mathrm{pH} = 1/2 \times (2.0 + 6.58) = 4.29.$$
This is the correct answer as presented in the text. Why is the first method invalid?

The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.
The criteria of $K_1 > 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H2A produces only a small amount of HA- and K2 is small as well, so ionisation of HA- would be negligible and we can ignore it.
However with an ampholytic species, while K2 is smaller than K1, the initial reactant is the same (HA-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.

I think this question has some discrepancies. Asking is the pH of a $\pu{0.050 M}$ $\ce{NaH2PO3}$ solution, and given that $\mathrm{p}K_{a1}=2.0$ and $\mathrm{p}K_{a2}=6.58$ for the salt. Keep in mind that $\ce{H3PO3}$ is a diprotic acid even though it has three $\ce{H}$s, according to Wikipedia-1. If the question really meant sodium salt of $\ce{H3PO3}$, then $\mathrm{p}K_{a1}$ should have been in the range of $1.26–1.3$ at $\pu{25 ^oC}$ (reported as $1.257$ in Ref.1), and $\mathrm{p}K_{a2}=6.7$ (Wikipedia-1). See the illstration of atom arrangement of phosphorus acid:

On the other hand, if the question meant sodium salt of $\ce{H3PO4}$ instead, then $\mathrm{p}K_{a1}$ should have been $2.147$ at $\pu{25 ^oC}$ (Ref.2), and $\mathrm{p}K_{a2}=7.20$ (Wikipedia-2 & Ref.2).
Based on those literature values, $\mathrm{p}K_{a1}$ in the question is more closed to that of $\ce{H3PO4}$ ($2.0$ vs $2.15$) while given $\mathrm{p}K_{a2}$ is more closed to that of $\ce{H3PO3}$ ($6.58$ vs $6.7$). Therefore, one may determine that this question cannot resolve. Yet, I decided to use both values individually in routing calculations usually does by a biochemist to see what would happen:
If If you used $\mathrm{p}K_{a2}=6.58$,
$$K_{a2} = \frac {\alpha^2}{0.05-\alpha} ~~~ \left(\text{from the equation:} ~  \ce{H2PO3- + H2O <=> HPO3^2- + H3O+} \right)$$
$$\text{when solve for} ~\alpha ~\text{gives:}~~ \alpha = \left(0.05 \cdot K_{a2} \right)^{\frac 12}~~~ \text{since}~~0.05 \gt \gt \gt K_{a2} ~~\text{and hence}~~0.05 \gt \gt \gt \alpha  $$
$$\text{Thus,}~~~\mathrm{p}\ce{H} = \mathrm{p}\alpha = \frac{\mathrm{p}K_{a2}- \log (0.05)}{2}$$
$$\text{If}~\mathrm{p}K_{a2}=6.58~\text{then,}~~~\mathrm{p}\ce{H} = \frac{6.58- \log (0.05)}{2} = 3.94$$
$$\text{If}~\mathrm{p}K_{a2}=7.20~\text{then,}~~~\mathrm{p}\ce{H} = \frac{7.20- \log (0.05)}{2} = 4.25$$
Thus, I can only conclude that you have done nothing wrong on assuming neglegible protonation in first calculation. But, I assume real $\mathrm{p}K_{a2}$ is $7.20$, and calculation should be done accordingly.
References:

J. W. Larson, M. Pippin, “Thermodynamics of ionization of hypophosphorous and phosphorous acids. Substituent effects on second row oxy acids,” Polyhedron 1989, 8(4), 527-530 (https://doi.org/10.1016/S0277-5387(00)80751-2).
J. W. Larson, K. G. Zeeb, L. G. Hepler, “Heat capacities and volumes of dissociation of phosphoric acid (1st, 2nd, and 3rd), bicarbonate ion, and bisulfate ion in aqueous solution,” * Canadian Journal of Chemistry* 1982, 60(16), 2141-2150 (https://doi.org/10.1139/v82-306).

$\color{red}{\text{PartA.Preface:}}$
The salt ($\ce{NaH2PO3}$)  is an ionic salt, and most sodium salts fully dissociate in aqueous solution; so you can consider all ($\ce{NaH2PO3}$)  to be converted to ($\ce{H2PO3-}$)   ions and $\ce{Na+}$ ions into the water. $\ce{Na+}$  is weak acid(a spectator ion) it won't undergo hydrolysis or take part in any reactions, so the initial amount of sodium in solution will remain constant. We can thus ignore it.
($\ce{H2PO3-}$)   is amphoteric that behaves both: as an acid(releasing $\ce{H+}$ in aqueous solution forming $\ce{HPO3^{-2}}$ ions which cannot produce  $\ce{PO3^{-3}}$ ions) in water and as a base(can undergo hydrolysis forming ($\ce{H3PO3}$) acid. The autoionization of water produces very few hydrogens and oxygen ions.
The main reactions which are possible to take place are :
\begin{array}{ }\ce{
a)Dissociation :NaH2PO3 &~->H2PO3-  +  Na+&\quad\left (\mbox{Full Dissociation}\right)\\
b)Ionization:H2PO3- &~<=> HPO3^{-2} + H+ &\quad\left(K_\mathrm{2}\approx{2.63\times10^{-7}}\right) \\
c)Hydrolysis:H2PO3-  +  H2O &~<=> H3PO3 + OH- &\quad\left(K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{1}}=\frac{10^{-14}}{1\times10^{-2}}=1\times10^{-12}\right) \\
d)Autoionization:H2O &~<=> H+ + OH- &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
}\end{array}
so :
The ($\ce{NaH2PO3}$) solution contains the following $5$ species:
the ions$\ce{ Na+ ,H+ ,OH−, H2PO3-, HPO3^{-2} }$ and undissociated acid $\ce{H3PO3}$.
$\color{red}{\text{PartB : Forming "Equilibria ,Mass balance and Proton balance " , equations :}}$
We need $6$ independent equations in order to specify the unknown concentrations of the $6$species present in an aqueous solution of ($\ce{NaH2PO4}$) . these equations are:
1- Equilibria. We have $3$ equilibrium equations:The equilibrium constants of the two stages of ionization of phosphorous acid(This acid is diprotic (readily ionizes two protons), not triprotic Wikipedia) and the dissociation equilibrium of water :
\begin{align}
K_\mathrm{1} & = \frac{[\ce{H+}][\ce{H2PO3-}]}{[\ce{H3PO3}]} \approx{1 \times 10^{−2}} \tag{1}\\
K_\mathrm{2} & = \frac{[\ce{H+}][\ce{HPO3^2-}]}{[\ce{H2PO3-}]} \approx{ 2.63 \times 10^{−7}} \tag{2}\\
K_\mathrm{w} & = {[\ce{H+}][\ce{OH-}]} = 1 \times 10^{−14} \tag{3}
\end{align}
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $C_\ce{P}$  .
For the solution of ($\ce{NaH2PO43}$), these equations would be
a- The mass balance equation for the cation ($\ce{Na+}$), but it is trivial:
$$C_\ce{P} =[\ce{Na+}]= [\ce{NaH2PO3}]_0 =\pu{0.05}\tag{4}$$
b- The mass balance equation for the anion of ($\ce{H2PO3-}$) ;  this concentration is the sum of all concentrations of species that are related to or derived from ($\ce{H2PO3-}$). In this solution the relevant species are:  $\ce{H2PO3-, HPO3^{-2} , H3PO3}$  , so we have the  following mass balance equation :
$$C_\ce{P}  =[\ce{H3PO3}]+[\ce{H2PO3–}] +  [\ce{HPO3^2–}]     \tag{5}$$
3- Proton Balance Equation: In the proton balance equation, the sum of the concentration terms for species that form by proton consumption is equated to the sum of the concentration terms for species that are formed by the release of a proton.
When ($\ce{NaH2PO3}$ is added to water, $\ce{H2PO3-}$, does react with water and is considered to be the starting species, so:
The species $\ce{ H3PO3}$  results with the consumption of one proton(by $\ce{H2PO3-}$ acting as a base).
The species of $\ce{ H+}$  can form with the consumption of one proton (by water acting as a base).
The species of $\ce{HPO_3^{-2}}$  can form with the release of one proton (by $\ce{H2PO3-}$acting as acid).
The species of $\ce{ OH-}$  can form with the release of one proton (by water acting as acid).
Thus, we have the  following proton balance equation :
$$\ce{[H3PO3] + [H+] = [HPO3^{-2}] + [OH-]}$$
$$\ce{[H3PO3] + [H+] = [HPO3^{-2}]  +\frac{ K_\mathrm{w}}{[\ce{H+}]}}\tag{6}$$
$\color{red}{\text{Part C}}$ : Derive quartic equation to calculate $[\ce{H+}]$ in the solution:
The sequence of derivation:
Step one :
Using equations (1,2) to  eliminate  $\ce{ [H2PO3-], [HPO3^{-2}] }$  successively:
1- Rearrange equation($1$) to get $[\ce{ H2PO3-}]$ :
$$ [\ce{H2PO3-}]  =\frac{K_\mathrm{1} [\ce{H3PO3}]}{[\ce{H+}]}\tag{7}$$
,and substitute$ [\ce{H2PO3-}]$ of equation($7$) in equation($2$) to get $[\ce{ HPO3^{-2}}]$ :
$$[\ce{HPO3^{-2}}] = \frac{K_\mathrm{1} K_\mathrm{2}  [\ce{H3PO3}]}{[\ce{H+}]^2} \tag{8}$$
Step two:
1- Substitute the species  $\ce{ [H2PO3-]}\text{of equation(7),and} [HPO_3^{-2}]\text{of equation(8) }\text {in equation}(5)$. This lead to
$$C_\mathrm{P}= [\ce{H3PO3}] + \frac{K_\mathrm{1} [\ce{H3PO3}]}{[\ce{H+}]} +\frac{K_\mathrm{1} K_\mathrm{2}  [\ce{H3PO3}]}{[\ce{H+}]^2 }$$
2- Rearrange the above equation to get $[\ce{H3PO3}]$( to express as a function of $ C_\pu{P}\text{and}[\ce{H+}]$ ):
$$[\ce{H3PO3}]=\frac{C_\ce{p} [ H^+]^2}{[ H^+]^2+K_\mathrm{1} [ H^+]+K_\mathrm{1} K_\mathrm{2} }\tag{9}$$
3-Substitiute $[\ce{H3PO3}]$ of equation($9$) in the  equations ($8$),  in order to express $ [\ce{HPO3^{-2}}]$ as a functions of $C_\pu{p}$  and $[\ce{H+}]$:
$$[\ce{HPO3^{-2}}]=\frac{K_\mathrm{1}K_\mathrm{2}C_\pu{p}}{[ H^+]^2+K_\mathrm{1} [ H^+]+K_\mathrm{1} K_\mathrm{2} }\tag{10}$$
Step three:
1- Now, replace these expressions$(9),(10)$ in proton balance equation ($6$), to get :
$\ce{\frac{C_\mathrm{p} [ H^+]^2}{[ H^+]^2 +K_\mathrm{1} [ H^+] +K_\mathrm{1} K_\mathrm{2} }+ [{H^+}]=\frac{K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} }{[ H^+]^2 +K_\mathrm{1} [ H^+] +K_\mathrm{1} K_\mathrm{2}}+\frac{ K_\mathrm{w}}{[\ce{H+}]}\tag{11}}$
2- Reduce to same denominator to get  :
$\ce{ C_\mathrm{p} [ H^+]^3 + [ H^+]^4 +K_\mathrm{1} [ H^+]^3 + K_\mathrm{1} K_\mathrm{2} [ H^+]^2 = K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} [ H+] +  K_\mathrm{w}[ H^+]^2 +K_\mathrm{1}K_\mathrm{w} [ H^+] +K_\mathrm{1} K_\mathrm{2}K_\mathrm{w} \tag{12}}$
3- Put everything on the same side:
$\ce{ [ H^+]^4 + C_\mathrm{p} [ H^+]^3 +K_\mathrm{1} [ H^+]^3 + K_\mathrm{1} K_\mathrm{2} [ H^+]^2 -   K_\mathrm{w}[ H^+]^2 - K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} [ H+]  -K_\mathrm{1}K_\mathrm{w} [ H^+] -K_\mathrm{1} K_\mathrm{2}K_\mathrm{w}  =    0 \tag{13}}$
4- Grouping and rearranging  the terms to get the final formula of quartic equation:
$$\color{red}{\ce{ [ H^+]^4 + (C_\ce{p}  +K_\mathrm{1}) [ H^+]^3 + (K_\mathrm{1} K_\mathrm{2}  -   K_\mathrm{w})[ H^+]^2   -(K_\mathrm{1}K_\mathrm{2}C_\mathrm{p} + K_\mathrm{1} K_\mathrm{w}) [ H^+]-K_\mathrm{1} K_\mathrm{2} K_\mathrm{w} =    0 \tag{14}}}$$
$\color{red}{\text{Part D:}}$  Calculations  and answers :
1- Substitute the values of $k_\mathrm{1} , k_\mathrm{2}, \text{and} C_\ce{P}$ , in quartic equation ($14$) , letting $X$ denote[$\ce{H+}$] as the following
:

2-  Solve for (X) from the quintic equation ($18$) by usingwolframalpha

There exists two positive real solution to the quartic equation( $14$) which is$ X=\color{red}{[\ce{H+}]\approx{8.07\times{10^{-5}}}\pu{M}}
 ~\text{and}[\ce{H+}]\approx{3.2\times\mathrm{10^{-4}}\pu{M}}$
$$\color{red}{\pu{pH}=4.1}~\text{and}~\pu{pH}=3.5$$


text:
Why does phosphine have a dipole moment and a higher boiling point than carbon tetrafluoride?
Phosphine, PH3, and carbon tetrafluoride, CF4, are small molecules of a similar size and the same mass of 88 au. CF4 has a dipole moment of 0, which is unsurprising given its tetrahedral shape. However, I would expect P-H bonds to be non-polar (P and H have electronegativities of 2.19 and 2.2) and therefore the molecule to also have a dipole moment of about 0, despite being trigonal pyramidal and therefore asymmetrical. So I would expect it to exhibit similar intermolecular forces and a similar boiling point to CF4. Yet PH3's dipole moment is quoted as 0.58 D. And PH3 has a boiling point of -88 °C and CF4's is -128 °C.
Why does phosphine have such a higher dipole moment and boiling point than CF4 despite it having non-polar bonds?

Mass of phosphine is nowhere near 88. Not that it matters much, though.

Damn it, I confused myself because it has a boiling point of -88C. It's PF3 that has a mass of 88 as well.
So actually my question should be: why does phosphine have a higher boiling point than phosphorus trifluoride?

The dipole moment of $PH_3$ can be attributed to the lone pair on P, which is directed away from all three P-H bonds. You can imagine that there is a higher electron density near the lone pair, while the electron density is more or less uniform near the nonpolar P-H bonds. This does not happen for $CF_4$, so it has no dipole moment.

Thank you. Actually I screwed up my question completely. I meant to ask: why does PH3 have a higher b.p. than PF3 when it is less massive and has virtually non-polar bonds, whereas PF3 has polar bonds and is asymmetrical? They both have the same shape.