text:
Are the bonds within bismuth pentafluoride (BiF5) covalent or ionic?
When googling $\ce{BiF5}$ most results call it bismuth pentafluoride such as Wikipedia.
I was under the impression that, since it is an ionic-bonded compound, Roman numerals should be used to identify the charge of bismuth, followed by fluorine with an ide ending such as the following:

Bismuth(V) Fluoride

So, I am wondering whether the bonds are either not ionic or that I've misidentified it?

For all nomenclature purposes, one should refer to the official IUPAC recommendations, which in this case are laid down in Nomenclature of Inorganic Chemistry — IUPAC Recommendations 2005, the ‘Red Book’. The relevant section for this question is section IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals
The introduction of this section (IR-5.1) states:

The simplest type of compositional name is a stoichiometric name, which is just a reflection of the empirical formula (Section IR-4.2.1) or the molecular formula (Section IR-4.2.2) of the compound. In stoichiometric names, proportions of constituent elements may be indicated in several ways, using multiplicative prefixes, oxidation numbers or charge numbers.

The following details on Stoichiometric Names of Elements and Binary Compounds (section IR-5.2) detail how to derive a stoichiometric name:

The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., given in Table IV). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. The final vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage). The two parts of the name are separated by a space in English.

Thus, the stoichiometric name of $\ce{BiF5}$ is bismuth pentafluoride.[1]
In the later section Generalized Stoichiometric Names, guidance is given on how to name compounds based on one constituent’s charge or oxidation number (IR-5.4.2.2).

It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. […]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Thus, according to this nomenclature, $\ce{BiF5}$ (determination of bismuth’s oxidation state is unambiguous) is bismuth(V) fluoride.

These two different styles of naming binary compounds are equivalent. In no way do they include any statement concerning the bonding situation. In fact, these names explicitly do not contain any information concerning structure or connectivity: dinitrogen tetraoxide is the correct name for both $\ce{O2N-O-N-O}$ and $\ce{O2N-NO2}$ — and the generalized name nitrogen(IV) oxide is also correct for $\ce{NO2}$.
Notably, usage of one or the other nomenclature cannot be used to distinguish between covalent or ionic bonding. As a case in point, $\ce{CCl4}$ — a rather clearly covalently bonded molecule — can be referred to as:

carbon tetrachloride
carbon(IV) chloride

And vice-versa, $\ce{BaCl2}$ — a very ionic compound — can be referred to as:

barium dichloride
barium(II) chloride
barium chloride[1]

Therefore, neither bismuth pentafluoride nor bismuth(V) fluoride make any reference to the type of bonds present in the compound; it is up to the chemist to use their previously acquired knowledge to realise whether bonds are predominantly covalent or predominantly ionic or something else.

Note:

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound […]. The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances […].

(Section IR-5.2)
Therefore, bismuth pentafluoride is permitted alongside monobismuth pentafluoride; and barium chloride is permitted, since only barium dichloride is reasonable.

But isn't the stoichiometric naming more commonly used in covalent compounds and the oxidation naming in ionic compounds?

@user34388 More commonly doesn’t say a thing.

What is the point of going against common use?

@user34388 The point is that there is no true distinction in common use. It is just that some compounds are commonly called one way and some others are commonly called another way. Common usage I am used to would say tin(II) chloride and tin(IV) chloride, although one is predominantly covalent and the other is predominantly ionic.

Nothing is completely ionic or completely covalent; it's just a label.
Convention is that electronegativity difference defines the nature of the bond, with 1.7 being the determining number.
However, if you look up a table, the electronegativity difference between bismuth and fluorine is 1.96, which although is conventially ionic, is really between the two.
Hence, it gets named as bismuth pentafluoride as well as bismuth (V) fluoride (this name is also on the same Wikipedia page).

so if nothing is ever completely either one does that mean either naming system(roman numeral or prefix) is appropriate for any bond? because my school has always stated that it is one or the other and is never both unless you have a polyatomic ion and a nonmetal bonded

@MohammadAli For example, CO2 clearly has more covalent character than ionic, so the covalent naming is used. In this case, since it is really around the middle, both namings are used. (Also, the polymer structure looks covalent to me.)

my teacher just emailed me so if i don't call 2 non metals covalent and a metal and a non metal ionic bonding i don't get the mark, but I'm sure you know what your doing so I'm going to mark this as the right answer and upvote

@MohammadAli I am quite sure your teacher would agree that BeH2 is covalent although it is metal + nonmetal... the reality is more complicated than that.

you would think that but all tests are marked using a checklist system with "0%" tolerance which means that their is no leniency. the school says that it isn't about whats right but instead about what is on the school board website and your ability to "demonstrate that you can learn stuff"

@MohammadAli I mean, BeH2 is taught as covalent even in school...

your school clearly isn't "highest rated" in my city. But yea I'm sure your school was better than mine

@MohammadAli is it taught in your school that there are some species that are octet-deficient, i.e. have less than 8 outermost electrons?

no idea what "species" is but if you mean an ion then yes. we are taught how to balance the charges to form a neutrally charged molecule.

for partial covalent character in ionic bonds see fajan' rule and for partial ionic character of covalent bonds see pauling equation or hannay and smith equation

Most of this answer is correct, but it fails to address the question’s misinterpretation.

@Jan I thought I addressed it in the first paragraph...

That wasn’t the question’s misinterpretation; see my answer.


text:
Work done in reversible isothermal expansion

A cylinder fitted with a frictionless piston is filled with $10~\mathrm{mol}$ of gaseous carbon tetrachloride and immersed in a large bath of water maintained at $300~\mathrm{K}$. The volume of the gas is initially at $V_1~\mathrm{L}$, and the external pressure on the piston is slowly decreased until the volume reaches $V_2~\mathrm{L}$. Calculate the work for this process, in $\mathrm{kJ}$, assuming the gas can be treated as ideal.

I think I have the general idea of how to solve this problem but I'm not sure how to find the external pressure. So far, I'm using the equation
$$\int \mathrm{d}w = \int_{V_2}^{V_1}-p_{\mathrm{ext}}\,\mathrm{d}V$$
since it is pressure-volume work and have tried calculating $p_{\mathrm{ext}}$ using $pV=nRT$ and the final volume of the system, but this doesn't seem to give the correct answer. How does it have to be solved?

If you could show us your work, then someone here might be able to point out where exactly you went wrong. Cheers!

I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.
However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.
This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).
Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)
$$\begin{align}
w &= -\int_1^2 p\,\mathrm{d}V \\
&= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V
\end{align}$$
and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)
$$\begin{align}
w = &= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\
&= -nRT\ln\left(\frac{V_2}{V_1}\right)
\end{align}$$
Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore
$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$

It's worth noting that carbon tetrachloride has a boiling point (at atmospheric pressure) of well over 300 K, so we can at least infer that both V1 and V2 correspond to starting and ending pressures that are significant below one atmosphere.

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text:
How does one experimentally determine virial coefficients?
The Virial expansion is a systematic correction to the ideal gas law which takes the form of an expansion around the number density, $\rho$, of the gas in question. It looks like this: $$\frac{p}{k_bT}=\rho+B_2(T)\rho^2+B_3(T)\rho^3+\cdots$$
The form of this equation is not surprising. An ideal gas is simply one for which the coefficients $B_2$, $B_3$, etc. are equal to zero. These are called the second and third Virial coefficients.
According to Wikipedia (and I have heard this elsewhere), the Virial coefficients systematically correct the ideal gas law by considering specific interactions. To quote the Wikipedia article,

The second virial coefficient $B_{2}$ depends only on the pair interaction between the particles, the third ($B_{3}$) depends on 2- and non-additive 3-body interactions, and so on.

This quote gives me quite a good idea of how I would determine these coefficients theoretically using $\textit{ab initio}$ calculations, but I can't imagine how I could calculate only the 2-body interactions experimentally...
Indeed, this paper seems to present one way to determine them by the so-called Burnett method, but I don't have a way of accessing this paper and have never heard of this method.
So, if anyone could give me a summary of how to determine these coefficients, I would appreciate that and am quite interested to hear the answers.

The Burnett method is described in this paper. http://nvlpubs.nist.gov/nistpubs/jres/75C/jresv75Cn3-4p165_A1b.pdf Essentially a near perfect gas is used to calculate an apparatus constant. The apparatus can then be used to measure a gas which does not have near ideal gas behavior. The apparatus has two chambers and a pressure gauge and kept and some temperature. The second chamber is isolated from the first and evacuated. The first chamber is filled with the gas at some measured pressure. The valve between the two chambers is opened and the pressure measured again.

re: ab initio calculations I think this would be a lot harder than you imagine. This would in essence be calculating the critical temperature, pressure and volume for a compound.

Well I think it would require quite a lot of computational time and I would almost certainly end up using MD due to the necessity of a large number of particles, but what I really meant is that conceptually, it is quite easy to imagine to what would need to be done. In practice it'd probably be tedious like anything else. And thanks for the link.

These coefficients can be calculated based on 2 ways:
1) Based on experimental data
In this method, from pressure-volume-temperature (PVT) data these coefficients will be calculated. Consider the following equation:
$$\vartheta\left(\frac{P\vartheta}{RT} - 1\right) = B(T) + \frac{C(T)}{\vartheta} +\ldots$$
$$\vartheta = \frac{1}{\rho} =\mathrm{Molar\ volume}$$
This equation is a re arange form of Virial equation. Now accroding to this equation, if you plot the left hand side of aforementioned equation versus molar density, the intercept of that line represents second virial coefficient and the slope of that line is the third virial coefficient. If the curve was not a straight line, we should consider more virial coefficients for them.
2) From potential functions.
In this method, the virial coefficients are calculated by potential functions such as L-J model, Kihara model, Exp-6 model and etc. The basis of This method is statistical thermodynamics.

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text:
How to find the volume of ice given its density, and the volume and density of water used?
Just trying to understand density calculations, for example,
Water has a density of $\pu{0.997 g cm^-3}$ at $\pu{25 ^\circ C}$; ice has a density of $\pu{0.917 g cm^-3}$ at $\pu{-10 ^\circ C}$.
If a soft-drink bottle whose volume is $\pu{2.00 L}$ is completely filled with water and then frozen to $\pu{-10 ^\circ C}$, what volume does the ice occupy?

See also What happens if you cool water in a container too small for it to freeze?

It depends.... under steady state conditions

If the bottle is glass, then the glass shatters, and the frozen water sublimes so there is no ice left.
If the container can withstand the pressure without distorting and the container volume changes less than $\pu{0.005 L}$ with a temperature change from $\pu{25 ^\circ C}$ to $\pu{-10 ^\circ C}$, then the final volume will be $\pu{2.00 L}$.
If container was so flimsy so that it distorted at a slight change in atmospheric pressure, and it was initially filled completely ($\pu{2.00 L}$) at $\pu{25 ^\circ C}$; then the ice would occupy a volume of:
$$\pu{2.00 L} \cdot \frac{\pu{0.997 g cm^3}}{\pu{0.917 g cm^3}}= \pu{2.17 L}$$
If the container distorts, but is pressurized, then the volume will be
$$\pu{2.00 L} \le V \le \pu{2.17 L}$$

Thank you so much!  I realize you are probably at Phd level chem but your answer was way over my head, there was only one simple answer and it was 2.17 so thank you.

I was trying to be a smart ass. I was not really trying to do your homework.

Damn man coulda fooled me but its not hw. Im practicing dimensional analysis.

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text:
How to calculate the electromotive force of a silver chloride/ silver bromide cell?

For the galvanic cell: $\ce{Ag|AgCl(s)|KCl(0.2M)||KBr(0.001M)|AgBr(s)|Ag}$, calculate the electromotive force (EMF) generated.
  $K_\mathrm{sp}(\ce{AgCl}) = 2.8\times10^{-10}$;
  $K_\mathrm{sp}(\ce{AgBr}) = 3.3\times10^{-13}$

I have tried this forming the cell reaction which is coming out to be
$$\ce{AgCl + Br- <=> AgBr + Cl-}$$
then using the $K_\mathrm{sp}$ of both salts I found the equilibrium constant which should be
$$K_\mathrm{eq} = \sqrt{\frac{K_\mathrm{sp}(\ce{AgCl})}{K_\mathrm{sp}(\ce{AgBr})}}$$
then I put it in the formula
$$E_\mathrm{cell} = -0.059 \log(K_\mathrm{eq})$$ but my answer is coming wrong.

Have a look at the Nernst equation

Consider them as 2 separate Ag/AgX/X-  half cells and do no forget to involve X- concentrations.

$$\begin{align}
E_{\ce{Ag/AgBr}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgBr}}{[\ce{Br-}]} \\
E_{\ce{Ag/AgCl}}&=E_{\ce{Ag/Ag+}}^{\circ}+0.059\log\frac{K_{\rm sp, AgCl}}{[\ce{Cl-}]} \\
EMF&=0.059 \cdot \left| \log \left( {\frac
{ K_{\rm sp, AgCl} \cdot [\ce{Br-}]}
{ K_{\rm sp, AgBr} \cdot [\ce{Cl-}]}
} \right) \right| \\
EMF&=0.059 \cdot \left| \log \left( {\frac
{ 2.8\times10^{-10} \cdot 0.001}
{ 3.3\times10^{-13} \cdot 0.2}
} \right) \right| \\
EMF&=0.059 \cdot \left| \log \left( {\frac { 2.8}{ 0.66} } \right) \right| \\
EMF&=0.037 \rm \ V\\
\end{align}$$


text:
Molecular weight of Polymers
Addition polymerisation is defined in my textbook as the process in which a molecule or compound is formed by combination of small molecules (monomers) implicitly having greater molecular mass than either of the monomers.
"A molecule is the smallest part of a substance which can exist independtly."
By this definition, one should find THAT part (Smallest freely existing part) of any polymer and then extract a molecular formula for the polymer. But it does not happen in actual. For example, polythene is assigned a molecular formula $(C_2H_4)_n$.
I just want to know what kind of molecule is determined by this kind of molecular formula as far as the definition of molecule is concerned?
And by what means I can find out its (polythene's) molecular mass?

related http://chemistry.stackexchange.com/questions/34446/what-are-the-units-of-85000-molecular-weight-of-a-polymer

Mithoron gave a good link. // The short answer is that polyethylene doesn't have a particular molecular mass, so there is no exact value for $n$. So no matter what the process the resulting polymer will be a mixture of chain lengths.  The polymerization process can be "tuned" to give short lengths, medium lengths or long lengths so that the resulting plastic has different properties.


text:
Reaction of glucose with 2,4-dinitrophenylhydrazine (DNPH)
In its open-chain form, glucose possesses an aldehyde group; however, glucose does not test positive with 2,4-dinitrophenylhydrazine (which typically forms a yellow/orange/red precipitate with carbonyl compounds).
Why is this so? Is this a reflection of the cyclic hemiacetal form (where the aldehyde group is not present)?
Both forms are in equilibrium so I'd expect it to give a positive test.

Consider also that glucose gives a positive test in both the Tollens' and Benedict's test. While it is true that less than one in hundred exists in the aqueous solution as the cyclic form, this is not sufficient to explain why it does not give a positive test with 2,4-dinitrophenylhydrazine but yet gives a positive test in Tollens' test and also in Benedict's test.

Interestingly, aqueous glucose reacts to give a positive test in both the Tollens' test and Benedict's test. It is indeed peculiar why it gives a negative test with 2,4-DNPH.

It was proposed by my classmate that it is perhaps due to the differences in reaction conditions. The conditions for the Tollens' test and Benedict's test are alkaline. However, in the reaction with 2,4-DNPH, the reaction medium is supposedly neutral. This would mean that the rate of conversion between the cyclic and linear forms is rather low in the neutral medium (i.e. the conversion reaction is uncatalysed). In the alkaline medium, the $\ce {OH^-}$ acts as a catalyst. This can allow for the  rapid conversion from the linear form to the cyclic form when the linear form reacts with DNPH.

What I can say about this is that the isomerisation of the cyclic hemiacetal form of Glucose in the open chain aldehyde may explain the positive Tollens' and Benedict's test, but may not be the correct reason, because if it has to be the isomerization only, hemiacetals will break more easily in acidic medium than basic conditions and therefore glucose should have been more prone towards positive 2,4-DNPH test. I believe that there should be some other explanation for this kind of observation, other than isomerisation to aldehyde.

I think this answer will help you: https://chemistry.stackexchange.com/questions/22269/does-glucose-react-with-bradys-reagent

That is because when glucose reacts with 2,4-dinitrophenylhydrazine, instead of corresponding imine, amine analogue of acetal is preferred - ring gets closed and forms colourless DNPH substituted glucose.
Due to the two strongly electron withdrawing nitro groups on the aromatic ring, the electron cloud moves forward to the chain of hydrazine, so it prefers the ring closure. There is no imine that can be detected, and it will not give a positive test.

The test is based on imine formation. The formation of imine from carbonyl compound occurs in equilibrium ergo the principle of Le Chatelier is applicable to the test. According to this principle the increase of the concentration of at least one of the products results in the decrease of degree of conversion of the reactant.
I guess that you are trying to perform the test in aqueous solution, because glucose has poor solubility in ethanol. The second product of the conversion of carbonyl to imine is water. So, performing the reaction in the tremendous excess of water, you are in fact pushing the degree of conversion toward zero in accordance with the principle of Le Chatelier.
That's why the expected precipitate is not formed.

Imine and eanmine formation is reversible but the formation of DNP's, phenyl hydrazones, osazones, oximes and semicarbazides are not reversible under the mild conditions applied to imines and enamines. The latter group is used as derivatizing agents and are stable. These derivatizing agents all have N-N or N-O bonds. Was the DNP derivatization done under the same conditions applied to phenylhydrazone formation of glucose?


text:
Where does the label ‘Dq’ to denote the field split in coordination compounds come from?
As established in a previous question, coordination compounds typically have a field split between the $\mathrm{t_{2g}}$ and the $\mathrm{e_g}$ d-orbitals.[1] This energy difference can be explained by the crystal field theory which assumes negative point-charges approaching a complex and destabilising orbitals pointing towards said point charges.
Commonly, the energy difference between the lower and higher orbitals is referred to by the relative value of $\Delta$ or the equally relative value of $10~\mathrm{Dq}$.
The latter is of interest to me. It is clearly used as a (relative) unit, because ligand field stabilisation enthalpies (LFSE’s) are typically given in values of $\mathrm{Dq}$ — e.g. $\ce{[Cr(H2O)6]^3+}$ (hexaaquachromium(III)) has an LFSE of $-12~\mathrm{Dq}$. But what does $\mathrm{Dq}$ mean and what does it derive from?

Note:
[1]: This is explicitly considering the octahedral case. Tetrahedral compounds split the d-orbitals into $\mathrm{t_2}$ and $\mathrm{e}$. This distinction is, however, not relevant to the question.

an answer here https://www.researchgate.net/post/Where_does_the_expression_Dq_for_the_splitting_of_the_ligand_field_comes_from implies that Introduction to Ligand Field Theory By B.N.Figgis: Chap-2 discusses this is detail

One of the earliest papers is by R. Schlapp & W. Penney Phys. Rev. v42 p 666, 1932 (see also  v41 p194 1932). They calculate the susceptibility of ions in cubic fields. In the hamiltonian D is just one of several  terms and is a constant scaling  as $D(x^4+y^4+z^4)$  and  q is a ratio of matrix elements.

see also page 91 here https://books.google.com/books?id=w5AZBQAAQBAJ&pg=PA92&dq=10dq+x4%2By4%2Bz4&hl=en&sa=X&ved=0ahUKEwjD_8rHqKDPAhUD4YMKHXjJC5gQ6AEIJzAA#v=onepage&q=10dq%20x4%2By4%2Bz4&f=false

I currently happen to have the book by Figgis that Max mentioned. Chapter 2 is devoted to a mathematical formulation of crystal field theory, which I did not bother reading in detail because I do not understand any of it. (The corollary is: If you have a better answer, please post it!) It seems that both $D$ and $q$ are collections of constants defined in such a way so as to make the octahedral splitting equal to $10Dq$. In particular it is defined that
$$D = \frac{35ze}{4a^5}$$
and
$$q = \frac{2e\langle r^4\rangle}{105}$$
where $ze$ is the charge on the ligands (in crystal field theory, the ligands are treated as point charges), $a$ is the distance of the ligands from the central ion, and $\langle r^4 \rangle$ is the mean fourth power radius of the d electrons in the central ion.
These quantities can be derived from first principles (that is precisely what crystal field theory is about) and lead to the expression of the crystal field electrostatic potential in Cartesian coordinates:
$$\mathbf{V}_{\mathrm{oct}} = D\left(x^4 + y^4 + z^4 - \frac{3r^4}{5}\right)$$
I followed some references given in the book which ultimately ended up in one of the papers in porphyrin's comment: Phys. Rev. 1932, 31, 194. There's more discussion there.
In particular, as far as I can tell, it seems that $D$ is called that simply because it is a quartic term in the potential (the first three terms having coefficients of $A,B,C$), and $q$ is called that because it is related to a series of constants $p_J$ which depend on the total spin quantum number $J$ (consistent with $p$ and $q$ being commonly used letters for mathematical constants).

Just to clear up any potential confusion with porphyrin's comment which at first glance seems to be missing one $r^4$ term in the electrostatic potential, the above article writes:

It is sufficient to write the potential $V = D(x^4+y^4+z^4)$ since this can be made to satisfy Laplace's equation by adding a function of $r^2$ (viz. $-3Dr^4$). [The addition of this term] corresponds to superposing a spherically symmetrical field, and merely shifts all levels equally.

There's still a factor of $1/5$ missing compared to the expression above, but I'm not particularly inclined to figure out how it arises, and I don't think it is of much importance anyway since the only difference is effectively to shift the zero of energy.

This is a good answer, but two improvements would be: 1. pointing out that the "q" portion as defined in the answer is specific to hydrogen-like d-orbitals, and 2. there are other higher and lower order terms in the $V_{oct}$ expression, but they do not contribute to splitting for d-orbitals (but the 6th order terms do contribute for f-orbitals) as explained in slides 9 and 10 here http://neel.cnrs.fr/UserFiles/file/MCBT/Equipe25/Elsa/ChampCristallinGiens2009.pdf

Damnit. I was just reading Figgis & Hitchman for unrelated reasons and I stumbled across the section this section. This question immediately came to mind, but it looks like orthocresol got to it! On p. 36, there is a good explanation that shows mathematically how the energies -4Dq and 6Dq correspond to the wavefunctions for eg and t2g d orbitals. I hope that some day I will understand this in greater depth, but it's time to leave the library and get some rest.

That's a really old article you are working from, using Gaussian units. More contemporary books would use SI units, arriving at $D=\frac{35ze}{16\pi\epsilon_0a^5}$. You can see the factor of 1/5 by just plugging in to Laplace's equation: $\nabla^2D\left(x^4+y^4+z^4+cr^4\right)=0$ and solving for $c$. Or you could expand the field near the origin due to $6$ charges $ze$ each at distance $a$ from the origin in $O_h$ configuration to fourth order in $\frac ra$ or $4$ in $T_d$ configuration and get the quantitative results.

@user5713492, I appreciate the comment, but if I may, I would like to suggest you share this in the form of another answer. To be honest, I barely understand what you're saying (and I don't even remember what I read to come up with this answer!) Obviously, that is not your fault; it's just that this topic is too far removed from my usual interests.

I will expand my previous comment into an answer.
In the $O_h$ case the metal ion is surrounded by $6$ ligand ions, each with charge $q=-Ze$ located at $\langle\pm a,0,0\rangle$, $\langle0, \pm a,0\rangle$, and $\langle0,0,\pm a\rangle$.
The electrostatic potential of a point charge $q$ is
$$V=\frac{kq}R$$
where
$$k=\frac1{4\pi\epsilon_0}$$
is the Coulomb's Law constant and $\epsilon_0$ is the permittivity of free space and $R=\left\|\vec{r_f}-\vec{r_s}\right\|$ is the distance from the source point $\vec{r_s}$ to the field point $\vec{r_f}$.
Working for the moment with the potential $V_1$ due to the charge at $\langle a,0,0\rangle$,
$$R=\left\|\langle x,y,z\rangle-\langle a,0,0\rangle\right\|=\left\|\langle x-a,y,z\rangle\right\|=\sqrt{(x-a)^2+y^2+z^2}=\sqrt{a^2-2ax+x^2+y^2+z^2}=a\sqrt{1-\frac{2x}a+\frac{r^2}{a^2}}$$
Where $r^2=x^2+y^2+z^2$ is the squared distance from the origin. Then
$$V_1=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2x}a+\frac{r^2}{a^2}\right)^{-1/2}$$
Now we need to expand $(1+x)^{-1/2}$ to fourth order in $x$. The Taylor series yields
$$(1+x)^{-1/2}=1-\frac12x+\frac38x^2-\frac5{16}x^3+\frac{35}{128}x^4+O\left(x^5\right)$$
We could have obtained this result via algebra alone by assuming that $(1+x)^{-1/2}=1+ax+bx^2+cx^3+dx^4+O\left(x^5\right)$, then squaring and multiplying by $(1+x)$ so that
$$\left(1+ax+bx^2+cx^3+dx^4\right)^2(1+x)=1+O\left(x^5\right)$$
and then finding the unknowns $a$, $b$, $c$, and $d$ from the conditions that the coefficients of $x$, $x^2$, $x^3$, and $x^4$ should be zero. Then we have
$$\begin{align}V_1= & -\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2x}a+\frac{r^2}{a^2}\right]+\frac38\left[\frac{4x^2}{a^2}-\frac{4xr^2}{a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac{8x^3}{a^3}+\frac{12x^2r^2}{a^4}\right]+\frac{35}{128}\frac{16x^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This is complicated, but now we simplify by combining the potentials due to all $6$ ligands. We want to find group operations from $O_h$ that map the first ligand to the others, and fortunately the powers of $S_6$, the rotoreflection by angle $\frac{\pi}3$ about the axis through the origin and $\langle1,1,1\rangle$ will do the trick.
Its powers cycle the original ligand point through all the others. We just need to find the mappings of the powers of $x$ by the powers of $S_6$ and sum over all powers of $S_6$:
$$\begin{array}{c|cccccc|c}
f(x,y,z)&Ef&S_6f&S_6^2f&S_6^3f&S_6^4f&S_6^5f&\sum\limits_{i=0}^5S_6^if \\
\hline r^2&r^2&r^2&r^2&r^2&r^2&r^2&6r^2 \\
x&x&-z&y&-x&z&-y&0 \\
x^2&x^2&z^2&y^2&x^2&z^2&y^2&2r^2 \\
x^3&x^3&-z^3&y^3&-x^3&z^3&-y^3&0 \\
x^4&x^4&z^4&y^4&x^4&z^4&y^4&2\left(x^4+y^4+z^4\right)
\end{array}$$
So now
$$\begin{align}V_{O_h}= & \sum\limits_{i=0}^5S_6^iV_1 \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{6-\frac12\left[0+\frac{6r^2}{a^2}\right]+\frac38\left[\frac{8r^2}{a^2}-0+\frac{6r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[0+\frac{24r^4}{a^4}\right]+\frac{35}{4}\frac{x^4+y^4+z^4}{a^4}\right\}+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left[6-\frac{21r^4}{4a^4}+\frac{35\left(x^4+y^4+z^4\right)}{4a^4}\right]+O\left(\left(\frac ra\right)^5\right) \\
& =-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
All this mathjax is exhausting. I can't finish in one sitting, but at least I have gotten to
$$D=\frac{35Ze}{16\pi\epsilon_0a^5}$$
EDIT: Now that the potential for $O_h$ geometry has been developed, the same will be done for the $T_d$ case where the metal ion is surrounded by $4$ ligand ions, each with charge $q=-Ze$, this time located at
$\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$, $\left\langle\frac a{\sqrt3},-\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, $\left\langle-\frac a{\sqrt3},\frac a{\sqrt3},-\frac a{\sqrt3}\right\rangle$, and $\left\langle-\frac a{\sqrt3},-\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$.
The distance from the source point at $\vec{r_s}=\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle$ to the field point at $\vec{r_f}=\langle x,y,z\rangle$ is
$$\begin{align}R= & \left\|\vec{r_f}-\vec{r_s}\right\|=\left\|\left\langle x-\frac a{\sqrt3},y-\frac a{\sqrt3},z-\frac a{\sqrt3}\right\rangle\right\| \\
& =\sqrt{\left(x-\frac a{\sqrt3}\right)^2+\left(y-\frac a{\sqrt3}\right)^2+\left(z-\frac a{\sqrt3}\right)^2} \\
& =\sqrt{a^2-\frac{2a}{\sqrt3}(x+y+z)+x^2+y^2+z^2} \\
& =a\sqrt{1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}}\end{align}$$
Then the potential due to this charge is
$$\begin{align}V_2 & =\frac{kq}R=-\frac{Ze}{4\pi\epsilon_0a}\left(1-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right)^{-1/2} \\
& =-\frac{Ze}{4\pi\epsilon_0a}\left\{1-\frac12\left[-\frac{2(x+y+z)}{\sqrt3a}+\frac{r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(x+y+z)^2-\frac{4(x+y+z)r^2}{\sqrt3a^3}+\frac{r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(x+y+z)^3+\frac{4(x+y+z)^2r^2}{a^4}\right]+\frac{35}{128}\left[\frac{16}{9a^4}(x+y+z)^4\right]\right\}+O\left(\left(\frac ra\right)^5\right)\end{align}$$
This time the group operations that map the model point to all the others will be the members of the $D_2$ point group: the identity $E$, and the rotations by $180°$ about the x-axis $C_{2x}$, the y-axis $C_{2y}$, and the z-axis $C_{2z}$.
Again a table for combining the contributions from all point charges:
$$\begin{array}{c|cccc|c}
f(x,y,z)&R_1f=Ef&R_2f=C_{2x}f&R_3f=C_{2y}f&R_4f=C_{2z}f&\sum\limits_{i=1}^4R_if \\
\hline r^2&r^2&r^2&r^2&r^2&4r^2 \\
x&x&x&-x&-x&0 \\
xy&xy&-xy&-xy&xy&0 \\
xyz&xyz&xyz&xyz&xyz&4xyz
\end{array}$$
And the reader can readily verify that also
$$\sum\limits_{i=1}^4R_iy=\sum\limits_{i=1}^4R_iz=\sum\limits_{i=1}^4R_ixz=\sum\limits_{i=1}^4R_iyz=0$$
So the rule is that terms in the expression for $V_2$ with odd powers of any of the coordinates cancels except for $xyz$, which, along with all terms with only even powers of coordinates, adds.
Expanding powers of $(x+y+z)$ and summing over the four sources,
$$\begin{align}V_{T_d} & =\sum\limits_{i=1}^4R_iV_2=-\frac{Ze}{4\pi\epsilon_0a}\left\{4-\frac12\left[0+\frac{4r^2}{a^2}\right]+\frac38\left[\frac4{3a^2}(4r^2)-0+\frac{4r^4}{a^4}\right]\right. \\
& \left.-\frac5{16}\left[-\frac8{3\sqrt3a^3}(6)(4xyz)+\frac{4(4r^2)r^2}{a^4}\right]\right. \\
& \left.+\frac{35}{128}\frac{16}{9a^4}\left[4\left(x^4+y^4+z^4\right)+6(4)\left(x^2y^2+x^2z^2+y^2z^2\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left\{4+\frac{3r^4}{2a^4}+\frac{20}{\sqrt3a^3}xyz-\frac{5r^4}{a^4}\right. \\
& \left.+\frac{35}{72a^4}\left[4\left(x^4+y^4+z^4\right)+12\left(r^4-x^4-y^4-z^4\right)\right]\right\}+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{Ze}{4\pi\epsilon_0a}\left[4+\frac{20}{\sqrt3a^3}xyz-\frac{35}{9a^4}\left(x^4+y^4+z^5-\frac35r^4\right)\right]+O\left(\left(\frac ra\right)^5\right) \\
& = -\frac{4Ze}{4\pi\epsilon_0a}-\frac{5Ze}{\sqrt3\pi\epsilon_0a^4}xyz+\frac{35Ze}{36\pi\epsilon_0a^5}\left(x^4+y^4+z^4-\frac35r^4\right)+O\left(\left(\frac ra\right)^5\right)\end{align}$$
On comparison with $V_{O_h}$, the first term is the potential due to the net charge of all the ligands uniformly spread over the surface of a sphere of radius $a$.
The second term has no counterpart in $V_{O_h}$ and causes no splitting but does create mixing between the $d_{xy}$ and $p_z$ orbitals and would lower the energy of the former and raise the energy of the latter.
I don't know how significant that effect is, though. The splitting term is $\frac49D\left(x^4+y^4+z^4-\frac35r^4\right)$ and is in fact $-\frac49$ the value of the splitting term for $O_h$ geometry.
Still not done, but it's getting late.
EDIT:To get the $q$ part of $Dq$ we are going to have to do some integrals. I am going to use the notation
$$\oint f(x,y,z)d^2\Omega$$ to represent the integral of $f(x,y,z)$ over the surface of the unit sphere.
One integral we know right off the bat is
$$\oint 1d^2\Omega=4\pi$$
using the mensuration formula from geometry for the area of a sphere.
Actually, if we didn't know this formula, it would be no problem since if the answer were an unknown, say $\xi$, all the $\xi$'s would cancel out in the final results.
If the integrand is a monomial with an odd power of $z$, then
$$\begin{align}\oint x^ay^bz^{2c+1}d^2\Omega & =\oint \sigma_{hz}\left(x^ay^bz^{2c+1}\right)d^2\Omega=(-1)^{2c+1}\oint x^ay^bz^{2c+1}d^2\Omega \\
& =\frac{1+(-1)^{2c+1}}2\oint x^ay^bz^{2c+1}d^2\Omega=0\end{align}$$
Where $\sigma_{hz}$ is a reflection though the $xy$-plane. Clearly any proper rotation of the integrand about an axis through the origin doesn't affect the value of the integral because it just corresponds to viewing the same function from a different angle.
Also inversion of the integrand through the origin doesn't change the integral and any improper rotation including reflection can be written as inversion times a proper rotation. Thus it can be seen that if
$$\oint x^ay^bz^cd^2\Omega \ne 0$$
Then $a$, $b$, and $c$ must all be even. Also we can rotate coordinates so that we only need compute
$$\oint x^{2a}y^{2b}z^{2c}d^2\Omega$$
for $a \ge b \ge c$.
We can calculate
$$\oint r^2d^2\Omega=\oint(x^2+y^2+z^2)d^2\Omega=3\oint x^2d^2\Omega=4\pi$$
where we have rotated individual terms and then used $r^2=1$ on the surface of the unit sphere. Now,
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^2d^2\Omega=\oint \left(x^4+x^2y^2+x^2z^2\right)d^2\Omega=\oint x^4d^2\Omega+2\oint x^2y^2d^2\Omega$$
Rotating by $\frac{\pi}4$ about the $z$-axis,
$$\begin{align}\oint x^4d^2\Omega & =\oint \left(\frac{x+y}{\sqrt2}\right)^4d^2\Omega=\frac14\oint\left(x^4+4x^3y+6x^2y^2+4xy^3+y^4\right)d^2\Omega \\
& =\frac12\oint x^4d^2\Omega+\frac32\oint x^2y^2d^2\Omega\end{align}$$
This gives us two eqautions in two unknowns and we can solve using Excel or Matlab or whatever to get
$$\oint x^4d^2\Omega=\frac{4\pi}5\text{ and }\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
Again we have
$$\oint x^2d^2\Omega=\frac{4\pi}3=\oint x^2r^4d^2\Omega=\oint x^6d^2\Omega+6\oint x^4y^2d^2\Omega+2\oint x^2y^2z^2d^2\Omega$$
$$\oint x^4d^2\Omega=\frac{4\pi}5=\oint x^4r^2d^2\Omega=\oint x^6d^2\Omega+2\oint x^4y^2d^2\Omega$$
$$\oint x^6d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^6d^2\Omega=\frac14\oint x^6d^2\Omega+\frac{15}4\oint x^4y^2d^2\Omega$$
Solving the last $3$ equations in $3$ unknowns we get
$$\oint x^6d^2\Omega=\frac{4\pi}7\text{, }\oint x^4y^2d^2\Omega=\frac{4\pi}{35}\text{, and }\oint x^2y^2z^2d^2\Omega=\frac{4\pi}{105}$$
Almost there :)
$$\begin{align}\oint x^2d^2\Omega & =\frac{4\pi}3=\oint x^2r^6d^2\Omega \\
& =\oint x^8d^2\Omega+8\oint x^6y^2d^2\Omega+6\oint x^4y^4d^2\Omega+12\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\begin{align}\oint x^4d^2\Omega & =\frac{4\pi}5=\oint x^4r^4d^2\Omega \\
& =\oint x^8d^2\Omega+4\oint x^6y^2d^2\Omega+2\oint x^4y^4d^2\Omega+2\oint x^4y^2z^2d^2\Omega\end{align}$$
$$\oint x^6d^2\Omega=\frac{4\pi}7=\oint x^6r^2d^2\Omega=\oint x^8d^2\Omega+2\oint x^6y^2d^2\Omega$$
$$\oint x^8d^2\Omega=\oint \left(\frac{x+y}{\sqrt2}\right)^8d^2\Omega=\frac18\oint x^8d^2\Omega+\frac72\oint x^6y^2d^2\Omega+\frac{35}8\oint x^4y^4d^2\Omega$$
Solving these $4$ equations in $4$ unknowns,
$$\oint x^8d^2\Omega=\frac{4\pi}{9}\text{, }\oint x^6y^2d^2\Omega=\frac{4\pi}{63}\text{, }\oint x^4y^4d^2\Omega=\frac{4\pi}{105}\text{, and }\oint x^4y^2z^2d^2\Omega=\frac{4\pi}{315}$$
With these integrals in hand we want to normalize the polynomials $xy$ and $x^2-y^2$.
$$\oint (xy)^2d^2\Omega=\oint x^2y^2d^2\Omega=\frac{4\pi}{15}$$
So a normalized spherical harmonic that transforms as the $xy$ partner of the $T_{2g}$ representation of $O_h$ and as the $z$ partner the $T_2$ representation of $T_d$ is
$$Y_{xy}^2(x,y,z)=\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}$$
And we have
$$\psi_{xy}(x,y,z)=Y_{xy}^2(x,y,z)f_{xy}(r)$$
To find the electrostatic potential energy of an electron with such a wave function in the $O_h$ potential we integrate over its charge density to get
$$\begin{align}U_{O_h,xy} & =\int V_{O_h}(x,y,z)\rho_{xy}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]^*\left[\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f_{xy}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{xy}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{4\pi}\right)\oint \left(\frac{xy}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{xy}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{4\pi}\left(\frac{4\pi}{63}+\frac{4\pi}{63}+\frac{4\pi}{315}-\frac35\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\end{align}$$
Where we have used the separate normalizations of the angular part of the wave function $Y_{xy}^2(x,y,z)$ and the radial part $f_{xy}(r)$ and the average value of $r^4$ for the radial distribution, $\left\langle r^4\right\rangle$.
Now we calculate
$$\oint \left(x^2-y^2\right)^2d^2\Omega=2\oint x^4d^2\Omega-2\oint x^2y^2d^2\Omega=2\left(\frac{4\pi}{5}\right)-2\left(\frac{4\pi}{15}\right)=\frac{16\pi}{15}$$
So a normalized spherical harmonic that transforms as the $x^2-y^2$ partner of the $E_g$ representation of $O_h$ and the $E$ representation of $T_d$ is
$$Y_{x^2-y^2}^2(x,y,z)=\left(\frac{15}{16\pi}\right)^{1/2}\left(x^2-y^2\right)$$
We could also have obtained this normalization by rotating $Y_{xy}^2(x,y,z)$ by $-90°$ about the $z$-axis.
Now we can get the electrostatic potential energy of an electron with wave function
$$\psi_{x^2-y^2}(x,y,z)=Y_{x^2-y^2}^2(x,y,z)f_{x^2-y^2}(r)$$
in the $O_h$ potential as
$$\begin{align}U_{O_h,x^2-y^2} & =\int V_{O_h}(x,y,z)\rho_{x^2-y^2}(x,y,z)d^3volume \\
& =\int\left[-\frac{6Ze}{4\pi\epsilon_0a}-D\left(x^4+y^4+z^4-\frac35r^4\right)\right]\times \\
& (-e)\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]^*\left[\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f_{x^2-y^2}(r)\right]d^3volume \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}\oint\left|\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}\right|^2d^2\Omega\int_0^{\infty}\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& +De\left(\frac{15}{16\pi}\right)\oint \left(\frac{x^2-y^2}{r^2}\right)^2\frac{x^4+y^4+z^4-\frac35r^4}{r^4}\int_0^{\infty}r^4\left|f_{x^2-y^2}(r)\right|^2r^2dr \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}(1)(1)+\frac{15De}{16\pi}\left((2)\frac{4\pi}{9}+(4)\frac{4\pi}{105}-(4)\frac{4\pi}{63}-(2)\frac{4\pi}{315}-\frac65\frac{4\pi}{5}+\frac65\frac{4\pi}{15}\right)\left\langle r^4\right\rangle \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\end{align}$$
Where we have assumed the average value of $r^4$ is the same as for $\psi_{xy}(x,y,z)$ because they both started out as $d$ orbitals and we haven't taken into account higher-order effects.
Since there are $3$ $T_{2g}$ orbitals and $2$ $E_g$ orbitals, the average potential energy over all orbitals is
$$U_{O_h,ave}=\frac15\left[3\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right)+2\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)\right]=\frac{6Ze^2}{4\pi\epsilon_0a}$$
The splitting in potential energies of the orbitals is
$$\begin{align}U_{O_h,x^2-y^2}-U_{O_h,xy} & =\left(\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{4De}{35}\left\langle r^4\right\rangle\right)-\left(\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{8De}{105}\left\langle r^4\right\rangle\right) \\
& =\frac{4De}{21}\left\langle r^4\right\rangle\equiv10Dq\end{align}$$
Then we get
$$q=\frac{2e}{105}\left\langle r^4\right\rangle$$
and so
$$U_{O_h,x^2-y^2}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)+6Dq$$and$$U_{O_h,xy}=\left(\frac{6Ze^2}{4\pi\epsilon_0a}\right)-4Dq$$
This derivation was a little bloody because I chose to do everything with only algebra.
There is a much more elegant exposition possible that requires higher level math.
Is it worth adding to the end of this answer or posting as another answer?

How much time did you spend on typing this much big answer?!!

It took several hours over the course of a few days. It helps if one writes the answer in chunks, testing each chunk separately on a MathJax testing website because it takes so long for stackexchange to catch up with each character typed when the answers get long. I wanted to get an explanation that could be comprehensible to a student taking inorganic chemistry and I think I achieved it with the above answer, although it would take an ambitious student to pick through it. The other answer is elegant but not many have the chemistry and physics background to follow it.

A more high-level answer this time. In this approach we first find out the form of the term in the potential that will split the $d$ orbitals.
Recall that the character of a proper rotation by angle $\alpha$ for the set of spherical harmonics with $L^2Y_l^m(\theta,\phi)=\hbar^2l(l+1)Y_l^m(\theta,\phi)$ is
$$\chi_l(R(\alpha))=\sum_{m=-l}^le^{im\alpha}=\frac{\sin(m+\frac12)\alpha}{\sin\frac12\alpha}$$
for $\alpha \neq n\pi$. $\chi_l(R(0))=2l+1$ because all the diagonal elements are $1$ and $\chi_l(R(\pi))=(-1)^l$ because the diagonal elements alternate between $\pm1$ and there is an odd number of them so their sum is one of their boundary values.
Starting from the character table for $O$, we can decompose the characters of the multipole moments via character orthogonality.
$$\begin{array}{c|ccccc|c}
O&E&3C_{2h}&8C_3&6C_4&6C_{2d}&\Gamma \\
\hline A_1&1&1&1&1&1& \\
A_2&1&1&1&-1&-1& \\
E&2&2&-1&0&0& \\
T_1&3&-1&0&1&-1 \\
T_2&3&-1&0&-1&1 \\
\hline S&1&1&1&1&1&A_{1g} \\
P&3&-1&0&1&-1&T_{1u} \\
D&5&1&-1&-1&1&E_g+T_{2g} \\
F&7&-1&1&-1&-1&A_{2u}+T_{1u}+T_{2u} \\
G&9&1&0&1&1&A_{1g}+E_g+T_{1g}+T_{2g} \\
H&11&-1&-1&1&-1&E_u+2T_{1u}+T_{2u} \\
I&13&1&1&-1&1&A_{1g}+A_{2g}+E_g+T_{1g}+2T_{2g}\end{array}$$
We have expanded this all the way out to hexacontatetrapole moments just in case it were desired to investigate the splitting of $f$ orbitals, but the only multipole that matters for $d$ orbitals is the $A_{1g}$ hexadecapole moment.
This is because the product of a spherical harmonic $Y_l^m(\theta,\phi)$ with $Y_l^{m^{\prime}}(\theta,\phi)$ contains only spherical harmonics $Y_{l^{\prime\prime}}^{m^{\prime\prime}}(\theta,\phi)$ with even $l^{\prime\prime}$, $0\le l^{\prime\prime}\le2l$.
We have used the fact that the parity of electric multipole moments is even for even $l$ and odd for odd $l$ to determine the characters over the full $O_h$ group. The reverse is the case for magnetic multipole moments.
We use the projection operator
$$P_{jk}^{\lambda}f(x,y,z)=\frac{\chi^{\lambda}(E)}{|G|}\sum_{R\in G}D_{kj}^{\lambda}\left(R^{-1}\right)Rf(x,y,z)$$
Where $\chi^{\lambda}(E)$ is the number of partners of irreducible representation $\lambda$, $|G|$ is the order of the group $G$, the sum is taken over all elements $R$ of the group, and $D_{kj}^{\lambda}\left(R^{-1}\right)$ is the matrix element of the matrix representing $R^{-1}$.
If $f(x,y,z)$ has any character of the $k^{th}$ partner of the $\lambda^{th}$ irreducible representation of $G$, this will be nonzero and transform as the $j^{th}$ partner of the $\lambda^{th}$ irreducible representation.
As an example, with $G=T_d$, $\lambda=T_2$, and $f(x,y,z)=x+xz+z\left(2z^2-3x^2-3y^2\right)$ the different $P_{jk}^{T_2}f(x,y,z)$ can spit out $9$ different functions.
But for our purposes today we only need to project into $A_{1g}$ because any multipole moments must have the full symmetry of the charge configuration that creates them.
Then all $D_{11}^{A_{1g}}\left(R^{-1}\right)=1$ so
$$P_{11}^{A_{1g}}f(x,y,z)=\frac1{|G|}\sum_{R\in G}Rf(x,y,z)$$
We can simplify the sum by finding the stabilizer $H$ of $F(x,y,z)$ in $G$: $Sf(x,y,z)=f(x,y,z)$ for all $S\in H$.
Then given a set $C$ of representatives of $H$ in $G$, we have
$$P_{11}^{A_{1g}}f(x,y,z)=\frac1{|G|}\sum_{T\in C}\sum_{S\in H}TSf(x,y,z)=\frac{|H|}{|G|}\sum_{T\in C}Tf(x,y,z)$$
Starting with a fourth order polynomial $f(x,y,z)=z^4$, its stabilizer in $G=O_h$ is $H=D_{4h}$, the dihedral group of order $16$ with $z$ axis as principal axis of symmetry.
The cosets can be written as the powers of $C_{3[111]}$, a rotation of $120°$ about the axis through the origin and $\langle1,1,1\rangle$.
$$P_{11}^{A_{1g}}f(x,y,z)=\frac{16}{48}\left(Ez^4+C_{3[111]}z^4+C_{3[111]}^2z^4\right)=\frac13\left(z^4+x^4+y^4\right)$$
We use the Gram-Schmidt process to turn this into a normalized spherical harmonic.
Looking back up at the character table we see that also monopole moments transform as $A_{1g}$, so we start out with $g(x,y,z)=1$ and find its magnitude:
$$\oint\left|1\right|^2d^2\Omega=\frac{2\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac32\right)}=\frac{2\pi^{3/2}}{\frac12\pi^{1/2}}=4\pi$$
Where we have used the formula
$$\oint x^{2a}y^{2b}z^{2c}d^2\Omega=\frac{2\Gamma\left(a+\frac12\right)\Gamma\left(b+\frac12\right)\Gamma\left(c+\frac12\right)}{\Gamma\left(a+b+c+\frac32\right)}$$
to evaluate the integrals over the unit sphere of a monomial with even powers of $x$, $y$, and $z$. If any exponent is odd, the integral will be zero.
Also $\Gamma\left(\frac12\right)=\sqrt{\pi}$ and for $x>0$, $\Gamma(x+1)=x\Gamma(x)$.
Hopefully a reviewer can find a thread like 'What Every Chemistry Student Needs to Know About the Gamma and Beta Functions' and post a link in comments.
So with the normalized spherical harmonic
$$Y_{A_{1g}1}^0(\theta,\phi)=\left(\frac1{4\pi}\right)^{1/2}1$$
We can compute the unnormalized spherical harmonic
$$\begin{align}cY_{A_{1g}1}^4(\theta,\phi) & =\frac1{r^4}P_{11}^{A_{1g}}f(x,y,z)-Y_{A_{1g}1}^0(\theta,\phi)\oint \left[Y_{A_{1g}1}^0(\theta^{\prime},\phi^{\prime})\right]^*P_{11}^{A_{1g}}f(x^{\prime},y^{\prime},z^{\prime})d^2\Omega^{\prime} \\
& =\frac1{3r^4}\left(x^4+y^4+z^4\right)-\frac1{12\pi}\oint\left((x^{\prime})^4+(y^{\prime})^4+(z^{\prime})^4\right)d^2\Omega^{\prime} \\
& =\frac1{3r^4}\left(x^4+y^4+z^4\right)-\frac1{12\pi}(3)\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)} \\
& =\frac1{3r^4}\left(x^4+y^4+z^4\right)-\frac1{4\pi}\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)} \\
& =\frac1{3r^4}\left(x^4+y^4+z^4\right)-\frac15=\frac1{3r^4}\left(x^4+y^4+z^4-\frac35r^4\right)\end{align}$$
This has magnitude
$$\begin{align}|c|^2 & =\oint \left[cY_{A_{1g}1}^4(\theta,\phi)\right]^*cY_{A_{1g}1}^4(\theta,\phi)d^2\Omega=\frac19\oint\left(x^4+y^4+z^4-\frac35r^4\right)^2d^2\Omega \\
& =\frac19\left[(3)\frac{2\Gamma\left(\frac92\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}+(6)\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}\right. \\
& \left.-\frac35(6)\frac{\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}+\frac9{25}\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac32\right)}\right] \\
& =\frac19\left[(3)\frac{4\pi}{9}+(6)\frac{4\pi}{105}-\frac35(6)\frac{4\pi}5+\frac9{25}(4\pi)\right]=\frac{64\pi}{4725}\end{align}$$
So
$$Y_{A_{1g}1}^4(\theta,\phi)=\left(\frac{525}{64\pi}\right)^{1/2}\frac1{r^4}\left(x^4+y^4+z^4-\frac35r^4\right)$$
Given this spherical harmonic, we can find the electrostatic potential readily.
To find the potential at a field point $\vec{r_f}=\left\langle x_f,y_f,z_f\right\rangle$ we add up the contribution of the charge $dq=\rho(\vec{r_s})d^3volume$ at source point $\vec{r_s}=\left\langle x_s.y_s,z_s\right\rangle$ and sum over all source points:
$$V_{O_h}\left(\vec{r_f}\right)=\int\frac{\rho\left(\vec{r_s}\right)}{4\pi\epsilon_0\left\|\vec{r_f}-\vec{r_s}\right\|}d^3volume$$
Here
$$\begin{align}\left\|\vec{r_f}-\vec{r_s}\right\| & =\sqrt{\left(\vec{r_f}-\vec{r_s}\right)\cdot\left(\vec{r_f}-\vec{r_s}\right)}=\sqrt{r_s^2-2\vec{r_s}\cdot\vec{r_f}+r_f^2} \\
& =r_s\sqrt{1-2\frac{\vec{r_s}\cdot\vec{r_f}}{r_s^2}+\frac{r_f^2}{r_s^2}}=r_s\sqrt{1-2\frac{r_f}{r_s}\cos\gamma+\frac{r_f^2}{r_s^2}}\end{align}$$
Where $\cos\gamma=\frac{\vec{r_s}\cdot\vec{r_f}}{r_sr_f}$ is cosine of the angle between the source point and the field point as seen at the origin.
We can use the generating function for the Legendre polynomials and the spherical harmonic addition theorem to expand
$$\begin{align}\frac1{r_s\sqrt{1-2\frac{r_f}{r_s}\cos\gamma+\frac{r_f^2}{r_s^2}}} & =\frac1{r_s}\sum_{l=0}^{\infty}\left(\frac{r_f}{r_s}\right)^lP_l(\cos\gamma) \\
& =\frac1{r_s}\sum_{l=0}^{\infty}\left(\frac{r_f}{r_s}\right)^l\frac{4\pi}{2l+1}\sum_{m=-l}^l\left[Y_l^m\left(\theta_s,\phi_s\right)\right]^*Y_l^m\left(\theta_f,\phi_f\right)\end{align}$$
Valid for $r_f\lt r_s$. If $r_f>r_s$, just replace $\frac{r_f^l}{r_s^{l+1}}$ with $\frac{r_s^l}{r_f^{l+1}}$.
This is OK for spherical symmetry, but in chemistry we want a different basis set adapted to the point group of the complex.
We choose an orthonormal basis set $\left\{Z_l^m(\theta,\phi),m=-l,l\right\}$ so we can write the old basis set as
$$Y_l^m(\theta,\phi)=\sum_{m=-l}^lA_{m^{\prime}m}^lZ_l^{m^{\prime}}(\theta,\phi)$$
Since both bases are orthormal sets,
$$\begin{align}\delta_{mm^{\prime}} & =\oint\left[Y_l^{m^\prime}(\theta,\phi)\right]^*Y_l^m(\theta,\phi)d^2\Omega \\
& =\oint\sum_{m_1=-l}^l\left[A_{m_1m^{\prime}}^lZ_l^{m_1}(\theta,\phi)\right]^*\sum_{m_2=-l}^lA_{m_2m}^lZ_l^{m_2}(\theta,\phi)d^2\Omega \\
& =\sum_{m_1=-l}^l\left(A_{m_1m^{\prime}}^l\right)^*\sum_{m_2=-l}^lA_{m_2m}^l\delta_{m_1m_2}=\sum_{m_1=-l}^l\left(A_{m_1m^{\prime}}^l\right)^*A_{m_1m}^l\end{align}$$
Which says that as matrices, $(A^l)^{\dagger}A^l=I$. Inverses commute because their product being nonsingular requires that both $(A^l)^{\dagger}$ and $A^l$ must be nonsingular and so their product must have an inverse, $\left(A^l(A^l)^{\dagger}\right)^{-1}$. Then
$$\begin{align}A^l(A^l)^{\dagger} & =I\left(A^l(A^l)^{\dagger}\right)=\left[\left(A^l(A^l)^{\dagger}\right)^{-1}\left(A^l(A^l)^{\dagger}\right)\right]\left(A^l(A^l)^{\dagger}\right) \\
& =\left(A^l(A^l)^{\dagger}\right)^{-1}\left[\left(A^l(A^l)^{\dagger}\right)\left(A^l(A^l)^{\dagger}\right)\right]=\left(A^l(A^l)^{\dagger}\right)^{-1}\left\{A^l\left[(A^l)^{\dagger}\left(A^l(A^l)^{\dagger}\right)\right]\right\} \\
& =\left(A^l(A^l)^{\dagger}\right)^{-1}\left\{A^l\left[\left((A^l)^{\dagger}A^l\right)(A^l)^{\dagger}\right]\right\}=\left(A^l(A^l)^{\dagger}\right)^{-1}\left[A^l\left(I(A^l)^{\dagger}\right)\right] \\
& =\left(A^l(A^l)^{\dagger}\right)^{-1}\left(A^l(A^l)^{\dagger}\right)=I\end{align}$$
As matrix elements, this is
$$\sum_{m_1=-l}^lA_{m^{\prime}m_1}^l\left(A_{mm_1}^l\right)^*=\delta_{mm^{\prime}}$$
Then
$$\begin{align}\sum_{m=-l}^l\left[Y_l^m\left(\theta_s,\phi_s\right)\right]^*Y_l^m\left(\theta_f,\phi_f\right) & =\sum_{m=-l}^l\sum_{m_1=-l}^l\left[A_{m_1m}^lZ_l^{m_1}\left(\theta_s,\phi_s\right)\right]^*\sum_{m_2=-l}^lA_{m_2m}^lZ_l^{m_2}\left(\theta_f,\phi_f\right) \\
& =\sum_{m_1=-l}^l\left[Z_l^{m_1}\left(\theta_s,\phi_s\right)\right]^*\sum_{m_2=-l}^lZ_l^{m_2}\left(\theta_f,\phi_f\right)\delta_{m_2m_1} \\
& =\sum_{m_1=-l}^l\left[Z_l^{m_1}\left(\theta_s,\phi_s\right)\right]^*Z_l^{m_1}\left(\theta_f,\phi_f\right)\end{align}$$
This is good news because it means that we can use the $Y_l^m$ basis to prove the spherical harmonic addition theorem, but then substitute a more convenient basis when it comes time to apply it.
Now we have
$$\begin{align}V_{O_h}\left(\vec{r_f}\right) & =\int_0^{r_f}\int_0^{\pi}\int_0^{2\pi}\frac{\rho\left(\vec{r_s}\right)}{4\pi\epsilon_0}\sum_{l=0}^{\infty}\frac{r_s^l}{r_f^{l+1}}\frac{4\pi}{2l+1}\sum_{m=-l}^l\left[Y_l^m\left(\theta_s,\phi_s\right)\right]^*Y_l^m\left(\theta_f,\phi_f\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& +\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{\rho\left(\vec{r_s}\right)}{4\pi\epsilon_0}\sum_{l=0}^{\infty}\frac{r_f^l}{r_s^{l+1}}\frac{4\pi}{2l+1}\sum_{m=-l}^l\left[Y_l^m\left(\theta_s,\phi_s\right)\right]^*Y_l^m\left(\theta_f,\phi_f\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s\end{align}$$
We confine our interest to the region near the origin so $r_s>r_f$ and the first integral above will be zero.
Also we are just trying to derive a formula for the splitting of the $d$ orbitals, so we cut our sums off at $l=4$.
Only the $l=0$ and $l=4$ irreducible representations of the full rotation group have a partner that transforms as $A_{1g}$ in $O_h$ and we have those partners to hand at this point.
Given the locations of the $6$ charges $-Ze$ at distance $a$ along the coordinate axes we can write $\rho\left(\vec{r_s}\right)=-Ze\sum\limits_{j=0}^5\delta^{(3)}\left(\vec{r_s}-S_6^j\langle0,0,a\rangle\right)$ where $S_6$ is a $60°$ rotoreflection about the $[111]$ axis, so
$$\begin{align}V_{O_h}\left(\vec{r_f}\right) & =\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{-Ze}{\epsilon_0}\frac1{r_s}\sum\limits_{j=0}^5\delta^{(3)}\left(\vec{r_s}-S_6^j\langle0,0,a\rangle\right)\left(\frac1{4\pi}\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& +\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{-Ze}{\epsilon_0}\frac{r_f^4}{9r_s^5}\sum\limits_{j=0}^5\delta^{(3)}\left(\vec{r_s}-S_6^j\langle0,0,a\rangle\right)\left(\frac{525}{64\pi}\right)\times \\
& \frac1{r_s^4}\left(x_s^4+y_s^4+z_s^4-\frac35r_s^4\right)\frac1{r_f^4}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& =\frac{-Ze}{\epsilon_0a}(6)\left(\frac1{4\pi}\right)-\frac{Zer_f^4}{9\epsilon_0a^5}(6)\left(\frac{525}{64\pi}\right)\frac25\frac1{r_f^4}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right) \\
& =-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right)\end{align}$$
In the above we used the $3$ dimensional Dirac delta functions to do all the integrals.
An $A_{1u}$ representation in $O_h$ is $A+1$ for the $T_d$ case, so we might want to consider such a multipole moment there.
Looking for the stabilizer of $xyz$ in $T_d$, we see that it's the whole group, so it already transforms as $A_1$ in $T_d$.
It's also already orthogonal to $1$ so $xyz=cY_{A_11}^3$. Finding the magnitude,
$$\begin{align}|c|^2 & =\oint\left|cY_{A_11}^3\right|^2d^2\Omega=\oint x^2y^2z^2d^2\Omega=\frac{2\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac92\right)} \\
& =\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)}=\frac{4\pi}{105}\end{align}$$
So our normalized spherical harmonic is
$$Y_{A_11}^3(\theta,\phi)=\left(\frac{105}{4\pi}\right)^{1/2}\frac{xyz}{r^3}$$
The charge distribution for $T_d$ geometry is
$$\rho\left(\vec{r_s}\right)=-Ze\sum_{j=0}^3\delta^{(3)}\left(\vec{r_s}-S_4^j\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle\right)$$
Where $S_4$ is a $90°$ rotoreflection about the $z$ axis. Making the same simplifications as for the $O_h$ case, we have
$$\begin{align}V_{T_d}\left(\vec{r_f}\right) & =\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{-Ze}{\epsilon_0}\frac1{r_s}\sum_{j=0}^3\delta^{(3)}\left(\vec{r_s}-S_4^j\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle\right)\left(\frac1{4\pi}\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& =\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{-Ze}{\epsilon_0}\frac{r_f^3}{7r_s^4}\sum_{j=0}^3\delta^{(3)}\left(\vec{r_s}-S_4^j\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle\right)\left(\frac{105}{4\pi}\right)\times \\
& \frac{x_sy_sz_s}{r_s^3}\frac{x_fy_fz_f}{r_f^3}d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& +\int_{r_f}^{\infty}\int_0^{\pi}\int_0^{2\pi}\frac{-Ze}{\epsilon_0}\frac{r_f^4}{9r_s^5}\sum_{j=0}^3\delta^{(3)}\left(\vec{r_s}-S_4^j\left\langle\frac a{\sqrt3},\frac a{\sqrt3},\frac a{\sqrt3}\right\rangle\right)\left(\frac{525}{64\pi}\right)\times \\
& \frac1{r_s^4}\left(x_s^4+y_s^4+z_s^4-\frac35r_s^4\right)\frac1{r_f^4}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right)d\phi_s\sin\theta_sd\theta_sr_s^2dr_s \\
& =\frac{-Ze}{\epsilon_0a}(4)\left(\frac1{4\pi}\right)-\frac{Zer_f^3}{7\epsilon_0a^4}(4)\left(\frac{105}{4\pi}\right)\left(\frac1{3\sqrt3}\right)\frac{x_fy_fz_f}{r_f^3} \\
& -\frac{Zer_f^4}{9\epsilon_0a^5}(4)\left(\frac{525}{64\pi}\right)\left(\frac{-4}{15}\right)\frac1{r_f^4}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right) \\
& =-\frac{4Ze}{4\pi\epsilon_0a}-\frac{5Z}{\sqrt3\pi\epsilon_0a^4e}x_fy_fz_f+\frac{35Ze}{36\pi\epsilon_0a^5}\left(x_f^4+y_f^4+z_f^4-\frac35r_f^4\right)\end{align}$$
So now we are ready to get our $d$ orbitals and compute the splitting. Starting with the $d_{xy}$ orbital:
$$\oint x^2y^2d^2\Omega = \frac{2\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}=\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}=\frac{4\pi}{15}$$
So the normalized $d_{xy}$ orbital is
$$\psi_{xy}\left(\vec{r}\right)=\left(\frac{15}{4\pi}\right)^{1/2}\frac{xy}{r^2}f(r)$$
Where f(r) is assumed normalized to unity. We can compute
$$\begin{align}U_{O_hxy} &=-e\int\left[\psi_{xy}\left(\vec{r}\right)\right]^*V_{O_h}\left(\vec{R}\right)\psi_{xy}\left(\vec{r}\right)d^3volume \\
& =-e\oint\int_0^{\infty}\left[-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\frac1{r^4}\times\right. \\
& \left.\left(x^4+y^4+z^4-\frac35r^4\right)r^4\right]\left(\frac{15}{4\pi}\right)\frac{x^2y^2}{r^4}|f(r)|^2r^2drd^2\Omega \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{35Ze^2}{16\pi\epsilon_0a^5}\left\langle r^4\right\rangle\left(\frac{15}{4\pi}\right)\left[(2)\frac{2\Gamma\left(\frac72\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}\right. \\
& \left.+\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac{11}2\right)}-\frac35\frac{2\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}\right] \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{525Ze^2}{64\pi^2\epsilon_0a^5}\left\langle r^4\right\rangle\left[(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}\right. \\
& \left.+\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}-\frac35\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}\right] \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}-\frac{Ze^2}{6\pi\epsilon_0a^5}\left\langle r^4\right\rangle\end{align}$$
On to the $d_{x^2-y^2}$ orbital:
$$\begin{align}\oint \left(x^2-y^2\right)^2d^2\Omega &= (2)\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}-(2)\frac{2\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)} \\
& =(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}-(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}=\frac{16\pi}{15}\end{align}$$
So the normalized $d_{x^2-y^2}$ orbital is
$$\psi_{x^2-y^2}\left(\vec{r}\right)=\left(\frac{15}{16\pi}\right)^{1/2}\frac{x^2-y^2}{r^2}f(r)$$
We have
$$\begin{align}U_{O_hx^2-y^2} &=-e\int\left[\psi_{x^2-y^2}\left(\vec{r}\right)\right]^*V_{O_h}\left(\vec{R}\right)\psi_{x^2-y^2}\left(\vec{r}\right)d^3volume \\
& =-e\oint\int_0^{\infty}\left[-\frac{6Ze}{4\pi\epsilon_0a}-\frac{35Ze}{16\pi\epsilon_0a^5}\frac1{r^4}\times\right. \\
& \left.\left(x^4+y^4+z^4-\frac35r^4\right)r^4\right]\left(\frac{15}{16\pi}\right)\frac{\left(x^2-y^2\right)^2}{r^4}|f(r)|^2r^2drd^2\Omega \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{35Ze^2}{16\pi\epsilon_0a^5}\left\langle r^4\right\rangle\left(\frac{15}{16\pi}\right)\left[(2)\frac{2\Gamma\left(\frac92\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}-(4)\frac{2\Gamma\left(\frac72\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}\right. \\
& \left.+(4)\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac{11}2\right)}-(2)\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)}{\Gamma\left(\frac{11}2\right)}\right. \\
& \left.-\frac35(2)\frac{2\Gamma\left(\frac52\right)\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}+\frac35(2)\frac{2\Gamma\left(\frac32\right)\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac72\right)}\right] \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{525Ze^2}{256\pi^2\epsilon_0a^5}\left\langle r^4\right\rangle\left[(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}\right. \\
& \left.-(4)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}+(4)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac12\right)\left(\frac32\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}\right. \\
& \left.-(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)\left(\frac72\right)\left(\frac92\right)}-\frac35(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac32\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}\right. \\
& \left.+\frac35(2)\frac{2\pi^{3/2}\left(\frac12\right)\left(\frac12\right)}{\pi^{1/2}\left(\frac12\right)\left(\frac32\right)\left(\frac52\right)}\right] \\
& =\frac{6Ze^2}{4\pi\epsilon_0a}+\frac{Ze^2}{4\pi\epsilon_0a^5}\left\langle r^4\right\rangle\end{align}$$
With
$$Dq=\left(\frac{35Ze}{16\pi\epsilon_0a^5}\right)\left(\frac{2e}{105}\right)\left\langle r^4\right\rangle=\frac{Ze^2}{24\pi\epsilon_0a^5}\left\langle r^4\right\rangle$$
We see that
$$U_{O_hxy}=\frac{6Ze^2}{4\pi\epsilon_0a}-4Dq$$
and
$$U_{O_hx^2-y^2}=\frac{6Ze^2}{4\pi\epsilon_0a}+6Dq$$
For the $T_d$ case, the $Y_{A_11}^3$ multipole moment won't contribute to $U_{T_dxy}$ or $U_{T_dx^2-y^2}$ because it has odd parity and we have done the other hard integrals already, so
$$U_{T_dxy}=\frac{4Ze^2}{4\pi\epsilon_0a}-4\left(-\frac49\right)Dq=\frac{4Ze^2}{4\pi\epsilon_0a}+\frac{16}9Dq$$
and
$$U_{T_dx^2-y^2}=\frac{4Ze^2}{4\pi\epsilon_0a}+6\left(-\frac49\right)Dq=\frac{4Ze^2}{4\pi\epsilon_0a}-\frac83Dq$$


text:
Difference between "isoaspartyl" and "isoaspartate" in antibody deamidation
I'm translating a text where antibody deamidation is mentioned.
One sentence goes like this:

We used reversed-phase HPLC to analyze the aspartyl forms of the monoclonal antibody. (Russian original text: "изоаспартильные формы")

However, in scientific papers the word is "isoaspartate", not "isoaspartyl".  (also here)
On the other hand, Wikipedia's page "Isoaspartate" changes the term mid-flight to "isoaspartyl":

Isoaspartyl formation reactions have been conjectured to be one of the factors that limit the useful lifetime of proteins.
Isoaspartyl formation proceeds much more quickly if the asparagine is followed by a small, flexible residue (such as Gly) that leaves the peptide group open for attack. These reactions also proceed much more quickly at elevated pH (>10) and temperatures.

I'm perplexed. Are the forms of an aged antibody "isoaspartate forms" or "isoaspartyl forms"?  What is the exact difference between "isoaspartate" and "isoaspartyl", and why the Wiki page makes no mention of the difference?

I think a IUPAC literate chemist will give the right answer, but shortly: it's not as complicated as you think it is. -yl endings are commonly used to refer to a (side)group, in this case of the aspartate residue. Similar to a phenyl or isopropyl group in organic chemistry.


text:
Synthesis of 4-amino-3,5-dimethylphenol from N-(2,6-dimethylphenyl)hydroxylamine using perchloric acid?
I come across one reaction in which N-(2,6-dimethylphenyl)hydroxylamine is converted to 4-amino-3,5-dimethylphenol in the presence of strong acid (perchloric acid).

It,s seems rearrangement reaction, but i am unable to understand how does this rearrangement is possible.
What is the reaction mechanism of rearrangement reaction?

N-Phenylhydroxylamines undergo what is known as the Bamberger rearrangement when subjected to aqueous acid. The mechanism is a bit strange, as it involves divalent positively charged nitrogen (very rare - in fact, it is so rare that ChemDraw keeps trying to correct my $\ce{NH}$ to $\ce{NH3}$!), but that's just how it is. Wikipedia already has what I consider to be a very good explanation, but I'll try.
The mechanism begins with protonation at oxygen. Ordinarily, one would expect the nitrogen to be the more basic heteroatom, but protonation at nitrogen is unproductive - you can protonate nitrogen, but it's a dead end. On the other hand, if you protonate on oxygen, you can lose a water molecule.

The cationic intermediate on the right is rather like a benzyl cation, but with a nitrogen instead of a carbon. It has several resonance forms:

and the only decent nucleophile in the system, water, has three choices as to where it wants to attack.

Attack at nitrogen just goes back the way we came.
Attack at the ortho positions in this case would lead to a loss of aromaticity because you can't deprotonate at the end (see the mechanism below).
Attack at the para position. This completes the mechanism; after a proton exchange between N and O, a deprotonation leads to rearomatisation.


text:
How can FeS2 be the formula of iron sulphide and CaC2 be the formula of calcium carbide?
I know that the valency of iron can either be 2 or 3 and so the formulae of the possible sulphides of iron should be $\ce{FeS}$ and $\ce{Fe2S3}$. But I have recently seen the formula $\ce{FeS2}$ used for iron sulphide. How is this possible.
Similarly the formula for Calcium carbide is supposed to be $\ce{Ca2C}$ and not $\ce{CaC2}$. Why do we write $\ce{CaC2}$ for it?

Because the 'rules' on valences are not firm rules. On the Fe-S binary phase diagram, $FeS_{2}$ exists as a line compound (as do $Fe_{7}S_{8}$, $Fe_{9}S_{10}$, $Fe_{10}S_{11}$, and $Fe_{11}S_{12}$), alongside $Fe_{1-x}S$.

What you call iron sulphide, in my opinion, is more appropriately referred to as or iron disulphide.
If one were to assign oxidation states to each atom, an appropriate description would be  $\ce{Fe^2+}$ and $\ce{S_2^2−}$. This formalism recognizes that the sulfur atoms in pyrite occur in pairs with clear $\ce{S–S}$ bonds. These disulphide units can be viewed as derived from hydrogen disulfide, i.e $\ce{H2S2}$ (similar to hydrogen peroxide)
Take a look at the crystal structure (of pyrite):
In the center of the cell a $\ce{S_2^2−}$ pair is seen in yellow. The $\ce{S}$ atoms have bonds with three $\ce{Fe}$ and one other $\ce{S}$ atom.
Contrast this with, $\ce{FeS}$, i.e Iron (II) Sulphide where the counter anion is what you would typically expect $\ce{S^2-}$. This is reflected in its crystal structure, which is the nickel arsenide structure:

The bonding situation is completely different in these two compounds and this is reflected in their chemical formulas.
Similarly, for calcium carbide, the crystal structure is a tetragonally distorted $\ce{NaCl}$ lattice, comprised of discrete $\ce{Ca^2+}$
and discrete $\ce{C_2^-}$ ions.
Also, bear in mind what @Jon Custer said in his comment:

Because the 'rules' on valences are not firm rules. On the $\ce{Fe-S}$ binary phase diagram, $\ce{FeS2}$ exists as a line compound

Also, I'll briefly mention what he probably meant by a line compound:

Vertical lines such as the one seen here for $\ce{FeS2}$ indicate inter metallic compounds, which have precise chemical composition.
References

This article for the crystal structure of calcium carbide: http://pubs.acs.org.ejgw.nul.nagoya-u.ac.jp/doi/abs/10.1021/j100010a022
The crystal structures of pyrite and Iron (II) sulphide both come from their respective wikipedia pages here and here
The phase diagram comes from: http://www.minsocam.org/msa/collectors_corner/arc/scn1.htm (which further credits it to Ehlers, 1972, after Kullerud, 1967 The Interpretation of Geological Phase Diagrams, Fig. 217, p. 232)

Persulphide? Disulphide is ok here but per- not really

@Mithoron yeah, I think you're right. Edited :)

I admire your effort, but don't you think you are totally overloading the OP with this answer? ;)

@Karl I have overloaded OPs with my answers on multiple occasions as well. There is no problem with it, if someone else has the same question and comes to this website looking for help, they may find a more detailed explanation more useful. At the same time, they may find a simpler explanation more useful, such as that which you have provided - there is no better or worse, in such things. The only purpose is to share knowledge and that is what both of you are doing, so I in fact +1'd both answers.

@orthocresol Totally. I like getafix's answer better than mine, actually. ;-)

Good answer. As someone who makes this stuff in a lab, FeS is rarely stoichiometric FeS. It's usually some kind of Fe(1-x)S, having some Fe3+ in it.

There is no point in doubting the composition of these compounds, the measurements don't lie. ;-)
The point is that S or C (or e.g. P) can not only form bonds to the metal, but also between themselves:
$\ce{FeS2}$ for example does not have $\ce{S^2-}$ anions, but $\ce{S2^2-}$, with a covalent bond in it. Add one $\ce{Fe^2+}$ and everything is fine. Sulphur often makes chains of S atoms, for example in vulcanised rubber, or in many sulfides, which are rarely perfectly stochiometric, as Jon mentioned.
Calcium carbide contains pairs of triple-bonded carbon atoms, which are formally $\ce{C2^2-}$, fitting to one $\ce{Ca^2+}$ ion. If you carefully dissolve it in water, you get ethyne.
With oxygen, covalent bonds in inorganic solids are not so common: It doesn't like to form O-O bonds, and so the peroxides (metallic as well as organic) are rather unstable.


text:
Determining the orbitals that overlap to form a sigma bond
In the chapter on molecular orbitals, my chemistry book mentions three types of sigma bonds: those formed by s-s ($\ce{H2}$), s-p ($\ce{HCl}$), p-p ($\ce{Cl2}$) hybrid-s ($\ce{CH4}$) and hybrid-hybrid ($\ce{C2H6}$) overlap. How is it possible to predict the type of atomic orbital overlap by which a specific σ bond is formed?

The basic answer is go by what is available.

Hydrogen does not have any other occupied orbitals than s-orbitals. Hence it must use an s-orbital for bonding. This goes both directions.

Carbon requires four bonds. In order to achieve this, it must hybridise in some way, otherwise, an electron pair would be in an s-orbital and an empty p-orbital would be unavailable for bonding. Hence carbon uses hybrid orbitals in both directions.

Chlorine could hybridise, but it doesn’t have to fulfill its bonding needs. In general, an electron pair in an s-orbital is more stable than one in a p-orbital because it is ‘closer to the nucleus’. (That layman’s argument is bad but gets us a long way.) So chlorine will happily use its p-orbitals for bonding.
This goes both ways, by the way, so a carbon-chlorine bond ($\ce{C-Cl}$) is a hybrid-p sigma bond.

Finally, let’s consider oxygen and nitrogen. They would love to just bond with their p-orbitals, but even if hydrogen is the bonding partner that leads to too much steric strain. Therefore, they hybridise just enough to alleviate the steric stress while still maintaining a maximum s-character for a/the lone pair. So those would love to form bonds with their p-orbitals but typically can’t (so they must use hybrid orbitals). Phosphorus and sulfur are large enough to generally allow bonding only with p-orbitals.

This (specifically the carbon example) seems to imply that a bond is really just a combination of one half-filled orbital from each atom. Is that correct?

Also: what do you mean by "This goes both directions"?

@Marcel Yeah, the orbitals combine to form a bond. ‘This goes both directions’ is supposed to say that it doesn’t matter if say carbon is ‘on the left’ or ‘on the right’; it’ll use the same orbitals.


text:
What do dashes mean in chemical formulas?
I'm working on summarizing some research for an essay; here's the main article.
The basic premise is making nanoparticles from gold and iron(III) oxide that have properties of both gold and iron.
The formula they give is $\ce{Fe2O3-Au}$. The procedure involves chemical reactions, so I'm assuming it is a new molecule or particle of some sort, but what does the dash mean in the formula? Does it relate to the type of bond or something?
Also, and this is probably bonus since the useful part of the question deals with dash notation, but from the procedure I've generated the partial equation:
$$\ce{Fe2O3 + H2O + Na3C6H5O7 + HAuCl4 → Fe2O3-Au} + \text{something}...$$
After the reaction, the samples are put through a centrifuge and some unknown supernate is then removed, leaving the $\ce{Fe2O3-Au}$. Do you have any guesses on what the supernate is composed of? I was thinking something along the lines of $\ce{HCl}$, but I'm not sure what would happen to the sodium or carbon.

This nomenclature refers to a core-shell arrangement of nanoparticles. This is where you have a nanoparticle core consisting of one material that has been coated/decorated/wrapped/overlaid with another material. There are a few accepted forms of nomenclature for describing these types of nanoparticles:

core@shell In your case, it would be Fe2O3@Au
core-shell In your case, it would be Fe2O3-Au
core/shell In your case, it would be Fe2O3/Au

There are plenty of examples of each syntax in the literature, and has been used for eons.
Others may answers for the additional questions you have here.


text:
Response factor calculation for gas chromatography
While dealing with gas chromatography, how do you calculate the response factor with only the area known?
I am given the equation
$${k_1\over k_2} = {f_1 \times O_1 \over f_2 \times O_2}$$
where $O_1 = 21.28$, $O_2 = 178.35$, and $f_1 = 1.00$. $k_1$ and $k_2$ are the weights (in moles) that we added into the mixture. I calculated $0.05\,\mathrm{mol}$ for each component.
Using the formula (above) and the given quantities, I get
$$1 = {1.0 \times 21.28 \over f_2 \times 178.35}$$
Solving for $f_2$, I get 0.119, and I do not believe that is right. Is my approach valid?

This is a bit convoluted.
A simply dimensional analysis shows that your approach doesn't work. The units of $O$ are unit area. The conversion factor $f$ has units moles per unit area. The variable $k$ is in moles. If you divide the two equations you get a relative sensitivity of $f_2$ as compared to $f_1$, but the value is dimensionless. So each compound has its own response factor. So for two unknowns you have:
$f_1 = \dfrac{O_1}{k_1}$
$f_2 = \dfrac{O_2}{k_2}$
The twist here is the GC typically uses a standard either on a run before the unknown or spiking the unknown sample. So the standard is known to be $x$ ppm and you'll get some area from the run. The hope is that if the gain of the detector drifts them the area of the standard will change too by the same amount. So a 10% gain in area means that the detector gain has drifted up by 10%. Thus measuring area relative to the standard compensates for the drift.
The sensitivity changes a bit from day to day because the ionization in the detector changes due to aging, the gas flow rate isn't 100.0000% the same each time, the column ages and so on.
So the ratio of two unknowns is meaningless. You ratio each of the unknowns to a standard.


text:
Why do we talk about hydrogen evolution for the corrosion of aluminum?
During the corrosion of aluminum, we have the oxidation of aluminum to its ionic form and this reaction is balanced by a reduction of ions in solution.
So basically it is the reduction of $\ce{H+}$ protons and oxygen ($\ce{O2}$). But we generally presume the overall reaction is the oxidation of aluminum and the reduction of hydrogen protons and we neglect the oxygen reduction.
Why do we only talk about hydrogen evolution for most metals?

Thank you for your answer but actually my question was why when writing the corrosion reaction of Aluminum or other metals , we generally just write the one representing the hydrogen evolution and the oxidation of Oxygen and we do not mention the other reaction which the reduction of oxygen. So basically the two reactions considered are the oxidation of Aluminum and the hydrogen evolution.

Please use the "Contact Us" link at the bottom of the page to merge your two accounts.  Then you'll be able to edit the question and comment anywhere on the question or answers.

Hydrogen evolution is an output that is relatively easy to measure because it  can be detected by changes in pressure or displacement (volume), or by mass spectrometry.  The oxidation of metals (or reduction of oxygen) conversely is much harder to measure in real time.  Where a corrosion rate of a metal is to be determined then it is much easier to study the hydrogen evolved and then convert those numbers back to mg/m^2/y.  It is also particularly useful for passive metals, such as aluminium or zirconium, as a small evolution of hydrogen will generate a pressure change that is quite significant compared with its mass, so even a very slow reaction can be measured over a sensible timescale.
Hope this helps

Hi, @Chris. This is a mathjax-enabled site. so, please use it. For quick reference, check this meta Math.SE post.


text:
How do sodium-potassium pumps differentiate between sodium and potassium?
I don't know if this should be Biology, but it seems pretty Chem to me.
Does the protein of the sodium-potassium pump tell the difference between sodium and potassium cations based on size, electron affinity, mass...?

It's fine for the site. :)  If it doesn't get a reply here, you can flag it to be moved to Biology.

The short answer is ion size. $\ce{Na^+} = 116 \text{ picometers}$ and $\ce{K^+} = 152 \text{ picometers}$. As sort of a side note after dimethylglyoxime was found to be very specific for $\ce{Ni^{2+}}$ there was a effort to find specific precipitating regents for each ion for inorganic quantitative analysis.  It just doesn't work. To many ions are close in size, and it isn't possible to tune molecules/ligands for every picometer of ion size.

$\ce{Na^+-K^+}$ ATPase (more commonly called the$\ce{Na^+-K^+}$ pump) is a transmembrane protein that consists of two types of subunits: The 1000-residue  $\alpha$ subunit contains the enzyme’s ATP and ion binding sites, and the 300-residue $\beta$ subunit facilitates the correct insertion of the  subunit into the plasma membrane.
The  subunit of this $160 \ Å$ long protein consists of a transmembrane domain (M) composed of 10 helices of varied lengths and, and three cytoplasmic domains, namely: the nucleotide- binding domain (N), which binds ATP; the actuator domain (A), so named because it participates in the transmission of major conformational changes; and the phosphorylation domain (P), which contains the protein’s phosphorylatable Asp residue.

The $\ce{Na^+-K^+}$–ATPase is an antiport that generates a charge separation across the membrane, because for every three positive charges exit the cell for every two that enter.
The key to how it functions lies in the precise phosphorylation of a specific Asp residue of the transport protein. ATP phosphorylates the transporter only in the presence of $\ce{Na^+}$, whereas the resulting aspartyl phosphate residue is subject to hydrolysis only in the presence of $\ce{K^+}$.
Thus, the "pump" has two conformational states (called E1 and E2) with different structures, different catalytic activities, and different ligand specificities.

Outline of the entire process

E1.ATP, which acquired its ATP inside the cell, binds three sodium ions to yield the ternary complex $\ce{E1.ATP.3Na^+}$.
The ternary complex reacts to form the “high-energy” aspartyl phosphate intermediate $\ce{E1-P.3Na^+}$.
This “high-energy” intermediate relaxes to one lower in energy, i.e $\ce{E2-P.3Na^+}$, and releases its bound sodium ions outside the cell.
$\ce{E2~P}$ binds two potassium ions from outside the cell to form an $\ce{E2-P.2K^+}$ complex. (structure pictured earlier)
The phosphate group is hydrolyzed, yielding $\ce{E2.2K^+}$.
$\ce{E2.2K^+}$ changes conformation to E1, binds ATP, and releases its two potassium ions inside the cell, thereby completing the transport cycle.

This enzyme appears to have only one set of cation binding sites, but they changes both their orientation and their specificity during the course of the transport cycle. The difference in specificity the two conformations have arises principally because the two ions differ in size.
The illustration given below, shows the ion binding site in this ATPase and one can easily note the difference in positioning of residues between the two conformations.

Particularly, I would like to call your attention to figure(s) a, b and c. The purple spheres represent sodium ions, and the green ones correspond to potassium ions. Fig a and b represent the same view of the ion binding site, and one can clearly see the slight positional differences of the residues which account for the difference in specificity.
Image Source(s):
Voet and Voet, Biochemistry
and http://www.nature.com/nature/journal/v502/n7470/full/nature12578.html

Your question was answered 3 years ago for first time.
http://www.nature.com/nature/journal/v502/n7470/full/nature12578.html
The way it works is essentially Na/K ATPase pump is able to recognize Na over K or Ca2+ due to adopting conformational states that sterically hinder K+ or Ca2+ ions from binding.
But a indepth answer lies in the paper cited


text:
Do 1 mole of both dioxygen gas and its atoms occupy 22.4 litres at STP?
The ideal gas law states:

One mole of an ideal gas will occupy a volume of $\pu{22.4 liters}$ at STP.

Does this mean that both

$\pu{1 mole}$ of $\ce{O}$  would occupy $\pu{22.4 L}$ (or if this doesn't usually occur in nature, say \pu{1 mole }of $\ce{He}$ or any another monoatomic gas)
$\pu{1 mole}$ of $\ce{O2}$ would occupy $\pu{22.4 L}$

I am assuming the answer is yes based on:

reading the ideal gas law definition, it wouldn't make sense if it was true only for $\ce{He}$ but not for $\ce{O}$.
in class when we are finding the volume of e.g. $\pu{100 grams}$ of $\ce{Cl2}$, it was always $\frac{\pu{100 grams}}{(2\times\pu{35 grams} \text{[molar mass of $\ce{Cl2}$]}} \times\pu{22.4L}$  and not $\pu{1\times35 grams}$

But why? Wouldn't $\ce{O2}$ contain twice the amount of $\ce{O}$ matter and hence take up twice the volume?

The ideal gas laws assume that the volume occupied by the gas itself is negligible, zero essentially. So twice zero is still zero. As you realize this isn't really true. Van der Waals equation for gases tries to correct for this. It is a bit better than PV=nRT but still not great. There are other models that are even better. see https://en.wikipedia.org/wiki/Real_gas

"it was always 100 grams / (TWO . 35 grams [molar mass of Cl2])" - 35 grams is the molar mass of Cl. And you are talking about Cl2, not Cl. 35 grams * 2 is the molar mass of Cl2.

Think of it this way: while the container filled with O₂ does contain twice the matter, because the systems are held at STP, each O₂ molecule must have the same momentum as each molecule of O. So, even though the O₂ weighs twice as much, the O molecules are moving twice as fast so they both exert the same pressure on their environment (and therefore take up the same volume).

Strictly speaking, the volume at STP is now 22.7 L.  It went up recently (1980's!) when the standard pressure was reduced from one atmosphere to one bar.

Does this mean that both

1 mole of $\ce O$ would occupy $22.4~\mathrm L$ (or if this doesn't usually occur in nature, say 1 mole of $\ce{He}$ or another monoatomic gas)
1 mole of $\ce{O2}$ would occupy $22.4~\mathrm L$

Yes, it means exactly that.  And you're right, a stable gas of $\ce O$ atoms is a pretty exotic thing, so $\ce{He}$ is a much better example.

Wouldn't $\ce{O2}$ contain twice the amount of $\ce O$ matter

Yes and no. The $\ce{O2}$ gas would indeed contain twice the mass of the $\ce O$ gas, but it would contain the same number of particles ("molecular entities") as the $\ce O$ gas.

and hence take up twice the volume?

Nope.  There is a lot of empty space in gases. A lot. So much, in fact, that molecules have to get pretty darn large in the gas phase$^\dagger$ before they start to get "cramped" and "insist" on taking up more space (viz., occupying a larger volume).
To give you an idea of just how much room there is, consider what's called the van der Waals volume of $\ce{O2}$.  This is an estimate of the total volume actually occupied by one molecule of $\ce{O2}$, and has a value of about $0.053~\mathrm{nm}^3$ (that's 'cubic nanometers').  For one mole of $\ce{O2}$ molecules, that works out to just under $32~\mathrm{cm^3\over mol}$ of volume actually occupied by the oxygen molecules themselves.
However, as you note, that mole of $\ce{O2}$ gas actually expands to fill a full $22.4~\mathrm L$ of space at STP. So, that space is actually only about ${32~\mathrm{cm^3}\over 22.4~\mathrm{L}} = 0.14\%$ full of $\ce{O2}$.
So, even though one $\ce{O2}$ molecule does take up more space than one $\ce O$ atom, compared to the amount of space available, they're both really, really tiny.  And, thus, the fact that the two $\ce O$'s in $\ce{O2}$ are bound so closely to one another means that they both behave essentially identically in the pressure they exert when they fill up a container of a given size.
$^\dagger$ At ambient temperature and pressure, anyways.

RE: and hence take up twice the volume? // This is a bit ambiguous. I took it to mean that the Van der Waals correction for $\ce{O2}$ would be twice as big as for the correct for  $\ce{O}$ which would be fairly accurate.

Isn't volume at stp is 22.7 L and at ntp is 22.4 L ?

@MaxW Agreed, both about the ambiguity and about the $\sim!2\times$ vdW volume of $\ce{O2}$ vs $\ce O$. The whole topic is rife with ambiguity, actually, especially for a new learner.

@A---B Very possibly. There was no such thing as NTP when I was learning chemistry 20 years ago ... it may have been introduced in the interim.

Yes. Because STP is the standard temperature which is 0 degree celcius or 273 K nad standard pressure which is 760 mm Hg. So, if there's no given of volume, you can use the standard volume which is 22.4 L. No matter what kind of element is that, if we say STP, the number of volume at standard is 22.4 L


text:
Can't silver have 61 neutrons?
What happens if silver does have 61 neutrons?

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

I'm voting to close this question as off-topic because it is based on a false premise and shows little research effort.

Related: Why is tin-112 stable, but indium-112 radioactive?

The concerned silver nuclide with 61 neutrons is $\ce{^108_47Ag}$. This nuclide with 47 protons and 61 neutrons lies in the so-called valley of β-stability.

Image taken from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002), p. 42
Nuclides on right side of the valley (higher neutron numbers) are unstable to decay by β− emission. Nuclides on the left side are unstable to β+ decay or electron capture.
Generally, for isobars (i.e. nuclides with a constant mass number $A$), the binding energy $E$ as a function of the atomic number $Z$ describes a parabola:

Image taken from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002), p. 52
The most stable nuclides lie at the bottom of the valley (but not necessarily exactly at the minimum of the parabola).
This diagram is correct for nuclides with odd mass number $A$ (i.e. with even $Z$ and odd $N$, or with odd $Z$ and even $N$). For nuclides with even $A$, however, two different parabolas exist, depending on whether the nuclides have even $Z$ and even $N$ or odd $Z$ and odd $N$:

Image adapted from Choppin, Liljenzin, Rydberg: Radiochemistry and Nuclear Chemistry, third edition (2002), p. 52
Nuclides with even $Z$ and even $N$ (ee) are more stable (i.e. have a larger nucleon binding energy) than nuclides with odd $Z$ and odd $N$ (oo).
The case (IV) with two parabolas also applies to the question. The stable nuclide $\ce{^108_46Pd}$ has 46 protons and 62 neutrons (even-even). The nuclide considered in the question, $\ce{^108_47Ag}$, has 47 protons and 61 neutrons (odd-odd). The next neighbour, $\ce{^108_47Cd}$ has 48 protons and 60 neutrons (even-even) and is stable.
Therefore, $\ce{^108_47Ag}$ is unstable to β− decay to $\ce{^108_48Cd}$ $(97.65\ \%)$ as well as to β+ decay to $\ce{^108_46Pd}$ $(2.35\ \%)$.
Since $\ce{^108_47Ag}$ is radioactive with a short half-life of about $2.37\ \mathrm{min}$, it doesn’t exist in nature (except as an activation product in trace amounts).


text:
Why does baking soda and vinegar remove stains?
To remove a stain on fabric, I often rub baking soda into the stain, and then sponge it with white vinegar. This causes a chemical reaction, involving some fizzing and frothing, and once the fabric has dried, the stain (hopefully) has disappeared. Why does this chemical reaction remove the stain?
From my basic knowledge of chemistry, I know this is a classic Acid + Base -> CO2 + H2O + Others. But this doesn't involve the chemicals which have caused the stain.
Is it that these chemicals also react, and turn into other colourless products? Or is it that the frothing action from the reaction just dislodges these chemicals, allowing them to more easily be wiped clean? Or is there another answer?

You might try a comparison test to see if baking soda and vinegar is any more effective than plain water, or water and one of the ingredients. There are some possible reasons why each might be effective:

Some dyes and stains are set by exposure to acid, such as vinegar, which makes them harder to remove. It is possible that alkaline baking soda and water have the opposite effect, removing the stain.
Some stains, such as from red cabbage and red wine, are indicators, and change color when exposed to alkali or acid. The natural dye alizarin changes from dark red to lighter yellow under acid conditions, as in vinegar.
A powder such as baking soda may act as a sponge or adsorbent because of its large surface area. Talcum powder, baking soda or fuller's earth are also used for that purpose.
There might be a physical effect from the bubbles, as you state. For that reason, perhaps, some use plain seltzer (carbonated water) for stain removal.

Note that there are many conflicting statements on the efficacy of these stain removal techniques. It would be of benefit to humanity if someone conducted controlled experiments on removing a variety of stains with these chemicals! In the meantime, I just put a bit of HE laundry detergent on a stain, or a commercial cleaner such Lestoil, on the spot, before laundering.

From what I have read on stain and odor chemistry, the active chemical that creates a stain or odor is usually acidic or alkaline in nature. Baking soda is a weak base and can act like a buffer. Vinegar is a weak acid and can act like a buffer also. Using an alkaline compound on an acidic stain should remove that stain. Acid on an acid stain should set it. Using both alkaline and vinegar should remove most stains that are acidic or alkaline. If you encounter a stain or dye that cannot be removed with alkaline or acidic compounds, oxidizers that use oxygen like bleach or peroxide can also work. Surprisingly some stains can be removed with just warm or hot water.


text:
How to tell if a vibrational mode in benzene is IR and/or Raman active?

Consider the vibrational mode that corresponds to the boat-like bending of a benzene ring. Is it (i) Infrared active? (ii) Raman active?

This question was asked in an assignment in my college. However, all we've been taught is how to distinguish functional groups in organic molecules using characteristic IR frequencies, nothing else. From this I can tell that the C=O stretch in a carbonyl compound, for example, is IR-active. But how do we determine this for any molecule and any normal mode in general?

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If what you say is true, the question appears to be asked by mistake. Knowing which modes are active in IR and which are in Raman involves a substantial body of knowledge (symmetry groups, irreducible representations, characters, etc). There is no way it can be squeezed into a single answer.

There is a simpler way to find this out. Whether the vibrational mode is IR active depends on whether there is a change in the molecular dipole moment upon vibration. This is equivalent to asking whether there is a dipole moment in the boat-like conformation, since the ground state planar conformation has no dipole moment. Next, because the benzene molecule is centrosymmetric (i.e. possesses a point of inversion), IR-active modes cannot be Raman-active, and vice versa.

[...] of course, the "rule of mutual exclusion" that I described has its origins in symmetry (Cartesian axes being ungerade and their products being gerade means that the IR-active irreps cannot be the same as the Raman-active irreps), and a full explanation of why a vibrational mode is IR/Raman active is of course dependent on symmetry and some quantum mechanics. So in general, with no foundational knowledge of why a mode is IR/Raman active (I presume OP has not been taught these yet), one cannot tell.

@orthocresol I think that if you write a short answer focusing on the change in dipole moment and ignore symmetry, it's a good start.