Deriving the irreps of the d-orbitals under C3 and S6 operations in octahedral symmetry

Question

Contextual background: I am trying to reproduce the symmetries in the graph on this page (MO diagram for $\ce{ML6}$ complex). I have reproduced the metal s and p orbital representations as well as the linear combination of representations for the $\ce{L6}$ fragment. All that remain are the metal d orbitals.

So I know the five d-orbitals collectively reduce to the $\mathrm{e_g}$ + $\mathrm{t_{2g}}$ representation. I put my five d-orbitals at the origin and went through the operations for the reducible representation. Everything else adds up, except I can't find that $-1$ in the $C_3$ and $S_6$ operations.

Character table

$$\small\begin{array}{c|cccccccccc|cc}\hline O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline \mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2+z^2 \\ \mathrm{A_{2g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & 1 & 1 & -1 & & \\ \mathrm{E_g} & 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\ \mathrm{T_{1g}} & 3 & 0 & -1 & 1 & -1 & 3 & 1 & 0 & -1 & -1 & (R_x,R_y,R_z) & \\ \mathrm{T_{2g}} & 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 & & (xy,xz,yz) \\ \mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\ \mathrm{A_{2u}} & 1 & 1 & -1 & -1 & 1 & -1 & 1 & -1 & -1 & 1 & & \\ \mathrm{E_u} & 2 & -1 & 0 & 0 & 2 & -2 & 0 & 1 & -2 & 0 & & \\ \mathrm{T_{1u}} & 3 & 0 & -1 & 1 & -1 & -3 & -1 & 0 & 1 & 1 & (x,y,z) & \\ \mathrm{T_{2u}} & 3 & 0 & 1 & -1 & -1 & -3 & 1 & 0 & 1 & -1 & & \\ \hline \end{array}$$

Work

$$\begin{array}{c|cccccccccc} \hline O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & 3C_2 & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline \mathrm{E_g} & 2 & -1 & 0 & 0 & 2 & 2 & 0 & -1 & 2 & 0 \\ \mathrm{T_{2g}} & 3 & 0 & 1 & -1 & -1 & 3 & -1 & 0 & -1 & 1 \\ \hline \Gamma_{\text{d-orbitals}} & 5 & ? & 1 & -1 & 1 & 5 & -1 & ? & 1 & 1 \\ \hline \end{array}$$

As can be seen, everything matches up but the $C_3$ and $S_6$ operations. I tried using Molecule Viewer and rotating the d-orbitals, but I still don't see the $-1$. They all look like zero to me.

I would expect the $\mathrm d_{x^2-y^2}$ orbital to correspond to $E_g$ from the character table, but I only see it rotate into a "$\mathrm d_{y^2-z^2}$" orbital, with the orbitals along the y and z axes instead of x and y axes. The $\mathrm d_{z^2}$ orbital rotates into a "$\mathrm d_{x^2}$" orbital, lying along the x axis instead of the z axis.

Anyone know which d-orbital rotates into $-1$ for a $C_3$ operation? I assume a $-1$ for $S_6$ is the same orbital, but if not then does anyone which that one is as well?

Thank you

Answer 1

The $C_3$ operation rotates $x \rightarrow y$, $y \rightarrow z$ and $z \rightarrow x$. Hence, $$\phi_1=2z^2-(x^2+y^2)$$ rotates into $$-(1/2)\phi_1 +(3/2)\phi_2$$ and $$\phi_2=x^2-y^2$$
rotates into $$-(1/2)\phi_1 -(1/2)\phi_2$$ so, in the basis $( \phi_1, \phi_2 )$ the matrix that represents $C_3$ is

$$\left [\begin{matrix} -1/2 & -1/2 \\ 3/2 & -1/2 \end{matrix}\right]$$

Denoting this 2 by 2 matrix by M what we have is $C_3(\phi_1 \phi_2) = (\phi_1 \phi_2)M$. The trace of this matrix is $-1$.

This is what you find in the character table.

Since $C_3^3=I$ (identity) its eigenvalues are the cubic roots of unity: $1, \lambda$ and $ \lambda^*$ where $\lambda = \exp(2\pi i/3)=-1/2+i\sqrt 3/2$, and $\lambda^*$ is the complex conjugate of $\lambda $.

It is easy to find eigenvectors with the eigenvalue 1 [e.g., an s orbital or x+y+z].

You can find a linear combination of $\phi_1$ and $\phi_2$ that will be an eigenvector of $C_3$ with the eigenvalue $\lambda$ (transform into itself multiplied by $\lambda$) and another linear combination that will have eigenvalue $\lambda^*$. Now, the matrix representing $C_3$ will be diagonal with $ \lambda$ and $\lambda^*$ along the diagonal. The trace remains $-1$. Something similar should work for $S_6$.

Answer 2

From reading what you wrote I think you don't entirely understand how degenerate irreps work. Firstly, it's incorrect to say "$\mathrm{d}_{x^2-y^2}$ corresponds to $\mathrm{E_g}$". The $\mathrm{d}_{x^2-y^2}$ orbital by itself does not transform as anything. It only makes sense to say that the two orbitals transform together as $\mathrm{E_g}$. But anyway, let's go from the start.

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For non-degenerate irreps, the behaviour of an orbital is that the symmetry operation transforms it into some multiple of itself:

$$\hat{R}\psi = \chi^R\psi \tag{1}$$

where $\chi^R$ is the character under the symmetry operation $\hat{R}$. For example if $\psi$ is an s-orbital, then all the symmetry operations leave it untouched: $\hat{R}\psi = \psi$. Therefore $\chi^R = 1$ for all symmetry operations, and $\psi$ transforms as the totally symmetric irrep ($\mathrm{A_{1g}}$ in this case).

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Taking this logic and applying it to degenerate irreps is very dangerous. When two orbitals $(\psi_1, \psi_2)$ transform together as a degenerate irrep, what happens is that

$$\begin{align} \hat{R}\pmatrix{\psi_1 & \psi_2} &= \pmatrix{\psi_1 & \psi_2}\pmatrix{\Gamma_{11} & \Gamma_{12} \\ \Gamma_{21} & \Gamma_{22}} \\[8pt] &= \pmatrix{\Gamma_{11}\psi_1 + \Gamma_{21}\psi_2 & \Gamma_{12}\psi_1 + \Gamma_{22}\psi_2} \tag{2} \end{align}$$

where $\Gamma = \pmatrix{\Gamma_{11} & \Gamma_{12} \\ \Gamma_{21} & \Gamma_{22}}$ is some matrix that you have to find out by investigating the effect of $\hat{R}$ on the two functions.

$\Gamma$ is called a "matrix representation" of the symmetry operation. The character under this symmetry operation (which is what is listed in the character table) is the trace of this matrix $\Gamma$, i.e.

$$\chi^R = \Gamma_{11} + \Gamma_{22}. \tag{3}$$

(Notice that in the one-dimensional case the matrix $\Gamma$ simply becomes a number and the character is simply that number itself, which regenerates equation $(1)$.)

Now let's say $\psi_1 = \mathrm{d}_{x^2-y^2}$, $\psi_2 = \mathrm{d}_{z^2}$, and $\hat{R} = \hat{C}_3$. The issue is that you seem to expect that the rotation $C_3$ to have a nicely well-defined effect on $\psi_1$:

$$\hat{R}\psi_1 = k_1\psi_1 \tag{4}$$

But for a degenerate irrep this is simply not true. From equation $(2)$ one can see that

$$\hat{R}\psi_1 = \Gamma_{11}\psi_1 + \Gamma_{21}\psi_2 \tag{5}$$

Furthermore there is no guarantee that any of these two numbers $\Gamma_{11}$ or $\Gamma_{21}$ are related to the number in the character table. The only requirement is that $\Gamma_{11} + \Gamma_{22}$ (the latter of which we haven't even started talking about!) is equal to the character under the $C_3$ operation, i.e. $-1$. So now you see why asking "which d-orbital rotates into $−1$ for a $C_3$ operation?" doesn't make much sense. Neither of those two d-orbitals do.

Hopefully with this description you are in a better position to understand the other answer, which is essentially telling you that $\Gamma_{11} = -1/2$ and $\Gamma_{22} = -1/2$.