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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

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  • $\begingroup$ Please don't use MathJax in titles. And if you can, please use the native image uploader, that helps to prevent issues with dead image links. $\endgroup$ – orthocresol Dec 18 '16 at 16:22
  • $\begingroup$ Thank you for the corrections. I found both yours and Jacob's answers helpful for different reasons. Is it only possible to award a check to one response? $\endgroup$ – Blaise Dec 19 '16 at 16:21
  • $\begingroup$ Yes, you can only accept one answer; just accept whichever one you like, don't worry about it. $\endgroup$ – orthocresol Dec 19 '16 at 16:31
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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 $S_6$.

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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.


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).


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 $\Gamma_{22} = -1/2$.

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  • $\begingroup$ This makes a lot of sense. In basic chemistry they gloss over this theory, and we (or at least in my case) are taught a shortcut: performing the operation for all relevant orbitals, then add up which ones transform into themselves. I repeated this for t2g and noticed that "transforming into themselves" corresponds to the trace. Hence it works most of the time. Aiming deliberately for the trace component, it seems like "eyeballing" it may still be possible for eg. Incidentally, is $d_{z^2}$ always shorthand for $d_{2z^2-x^2-y^2}$? I always treated it as $z^2$, which fails here for the matrix. $\endgroup$ – Blaise Dec 19 '16 at 16:16
  • $\begingroup$ Yes, it is indeed shorthand. $\endgroup$ – orthocresol Dec 19 '16 at 16:29

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