# 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 researchgate.net/post/… implies that Introduction to Ligand Field Theory By B.N.Figgis: Chap-2 discusses this is detail – MaxW Sep 20 '16 at 16:27
• 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. – porphyrin Sep 21 '16 at 8:36

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 neel.cnrs.fr/UserFiles/file/MCBT/Equipe25/Elsa/… – DavePhD Nov 9 '16 at 14:37
• 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. – gannex Nov 12 '16 at 8:29
• 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 Nov 19 '17 at 0:01
• @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. – orthocresol Nov 19 '17 at 0:46

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 $$U_{T_dx^2-y^2}=\frac{4Ze^2}{4\pi\epsilon_0a}+6\left(-\frac49\right)Dq=\frac{4Ze^2}{4\pi\epsilon_0a}-\frac83Dq$$

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?!! – Apoorv Potnis Jan 30 '18 at 15:10
• 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. – user5713492 Jan 30 '18 at 19:21