# MOs of Benzene using group theory

I am trying to analyze the $$\sigma$$ orbitals of benzene molecule using group theory, doing the same thing that is done to the $$\pi$$ orbitals. The symmetry group of benzene is the $$D_{6h}$$; considering the $$sp^{2}$$ orbitals of the benzene:

and applying the symmetry operations to these orbitals, I can find the caracters of the representation with which they are transformed, and I can decompose this representation in the irreducible representation of the group. Doing this, I found

$$\Gamma = 2 A_{1g} \oplus 2E_{2g} \oplus B_{1u} \oplus B_{2u} \oplus 2E_{1u}$$

Now, I can construct the projection operators and with them I get the symmetrized orbitals for each representation. The molecular orbitals will be the linear combinations of these orbitals. My question is: In which function should I apply the projection operators? In the $$sp^{2}$$ function?

• There's no reason to apply hybrids to this problem. You should pick an appropriate set of coordinates and use the atomic orbitals instead.
– Zhe
Commented Jun 7, 2019 at 19:48
• Then again, whatever happens next can be told in terms of $sp^2$ orbitals all right, as well as it can be told in any other terms. Also, like I said, it is not much fun. Commented Jun 7, 2019 at 20:07
• Zhe, can you be more specific? So, can I choose any orbital (let's say, the $p_{x}$ orbital of Carbon) and apply the projectors to it? Ivan Neretin yes, I agree, but it's only an educational problem. Commented Jun 7, 2019 at 20:29

Your decomposition is for the hybrid orbitals that point from carbon to carbon, omitting the ones that point out towards the hydrogens, so you really want to project starting from one of the hybrid orbitals. First off we correct your decomposition: $$\Gamma=A_{1g}\oplus A_{2g}\oplus B_{1u}\oplus B_{2u}\oplus2 E_{1u}\oplus 2E_{2g}$$ We only need half the operations of $$D_{6h}$$ because all the hybrid orbitals transform the same under $$\sigma_h$$, so we write out the matrices for the group operations of the subgroup $$D_6$$ and a set of matrix representations for each irreducible representation and the result of each group operation acting on $$\phi_1$$, first half: $$\begin{array}{c|cccccc}R&E&C_6&C_6^2&C_6^3&C_6^4&C_6^5\\\hline A_{1g}&1&1&1&1&1&1\\ A_{2g}&1&1&1&1&1&1\\ B_{1u}&1&-1&1&-1&1&-1\\ B_{2u}&1&-1&1&-1&1&-1\\ E_{1u}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix} &\begin{bmatrix}-1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\ E_{2g}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix} &\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\ R\phi_1&\phi_1&\phi_3&\phi_5&\phi_7&\phi_9&\phi_{11}\end{array}$$ Ummm... yeah, I assume the $$x$$-axis points right in your illustration and the $$y$$-axis points up. Also I number the hybrid orbitals $$1$$ to $$12$$ counterclockwise starting with the one just above the $$x$$-axis in the first quadrant. I will label the dihedral $$C_2$$ rotation axes $$a$$ to $$f$$ with $$a$$ parallel to the $$x$$-axis counterclockwise in $$30°$$ increments. So now we can produce the second half of our table: $$\begin{array}{c|cccccc}R&C_{2a}&C_{2b}&C_{2c}&C_{2d}&C_{2e}&C_{2f}\\\hline A_{1g}&1&1&1&1&1&1\\ A_{2g}&-1&-1&-1&-1&-1&-1\\ B_{1u}&1&-1&1&-1&1&-1\\ B_{2u}&-1&1&-1&1&-1&1\\ E_{1u}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix} &\begin{bmatrix}-1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\ E_{2g}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix} &\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\ R\phi_1&\phi_{12}&\phi_2&\phi_4&\phi_6&\phi_8&\phi_{10}\end{array}$$ Then it remains to apply projection operators: $$\Gamma_j^{(\mu,i)}=\sum_{R\in G}D_{ij}^{(\mu)}\left(R^{-1}\right)R\phi_1$$ (Note the $$R^{-1}$$ in the above formula) $$\Gamma_{A_{1g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}+\phi_{12}+\phi_2+\phi_4+\phi_6+\phi_8+\phi_{10}$$ $$\Gamma_{A_{2g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}-\phi_{12}-\phi_2-\phi_4-\phi_6-\phi_8-\phi_{10}$$ $$\Gamma_{B_{1u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}+\phi_{12}-\phi_2+\phi_4-\phi_6+\phi_8-\phi_{10}$$ $$\Gamma_{B_{2u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}-\phi_{12}+\phi_2-\phi_4+\phi_6-\phi_8+\phi_{10}$$ $$\Gamma_{E_{1u},x}^{(1)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}+\phi_{12}+\frac12\phi_2-\frac12\phi_4-\phi_6-\frac12\phi_8+\frac12\phi_{10}$$ $$\Gamma_{E_{1u},y}^{(1)}=\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$ $$\Gamma_{E_{1u},x}^{(2)}=-\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$ $$\Gamma_{E_{1u},y}^{(2)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}-\phi_{12}-\frac12\phi_2+\frac12\phi_4+\phi_6+\frac12\phi_8-\frac12\phi_{10}$$ $$\Gamma_{E_{2g},x^2-y^2}^{(1)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}+\phi_{12}-\frac12\phi_2-\frac12\phi_4+\phi_6-\frac12\phi_8-\frac12\phi_{10}$$ $$\Gamma_{E_{1u},2xy}^{(1)}=\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$ $$\Gamma_{E_{1u},x^2-y^2}^{(2)}=-\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$ $$\Gamma_{E_{2g},2xy}^{(2)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}-\phi_{12}+\frac12\phi_2+\frac12\phi_4-\phi_6+\frac12\phi_8+\frac12\phi_{10}$$
• Thank you so much, you helped me a lot! The last functions we wrote are the symmetrized functions, right? Can I write them in function of the orbitals $s$, $p_{x}$ and $p_{y}$ of the Carbons? I would use the relations for the $sp^{2}$ orbitals: $$\phi_{1} = (1/\sqrt{3}) s + (2/\sqrt{6}) p_{x},$$ $$\phi_{2} = (1/\sqrt{3}) s - (1/\sqrt{6}) p_{x} + (1/\sqrt{2}) p_{y},$$ $$\phi_{3} = (1/\sqrt{3}) s - (1/\sqrt{6}) p_{x} - (1/\sqrt{2}) p_{y}.$$ Commented Jun 8, 2019 at 14:24
• My $$\phi_1=\frac1{\sqrt3}s_1-\frac2{\sqrt6}p_{1y}$$ $$\phi_2=\frac1{\sqrt3}s_1-\frac1{\sqrt2}p_{1x}+\frac1{\sqrt6}p_{1y}$$ $$\phi_3=\frac1{\sqrt3}s_2+\frac1{\sqrt2}p_{2x}-\frac1{\sqrt6}p_{2y}$$ I think you would be better off writing in terms of the $p$ orbitals parallel and perpendicular to the ring, $p_{i\parallel}$ and $p_{i,\perp}$ along with the $s_i$ orbitals, at least until you get the SALCs just to keep the algebra as simple as possible. The SALCs for the $p_{i\perp}$, $s_i$, and for the hydrogen $s_j$ orbital sets will be the same, so that makes your life easier. Commented Jun 9, 2019 at 2:26