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PHY.K02UF Molecular and Solid State Physics

## Molecular orbitals of benzene

Benzene (C6H6) consists of 6 carbon atoms in a ring. A hydrogen atom is attached to each carbon atom. The carbon-carbon bond length is 1.40 Å and the carbon-hydrogen bond length is 1.10 Å.

The molecular orbital Hamiltonian is,

$$H_{\text{mo}}= - \frac{\hbar^2}{2m_e}\nabla^2 -\sum\limits_{a=1}^A \frac{Z_ae^2}{4\pi\epsilon_0 |\vec{r}-\vec{r}_a|} .$$

Here $a$ sums over all of the atoms in the molecule. For bezene there is one Coulomb term for each hydrogen atom with $Z=1$ and one Coulomb term for each carbon atom with $Z=3.25$. The effective nuclear charge for carbon is reduced because of screening.

Carbon has 6 electrons. Two electrons occupy the 1s orbital. Three electrons participate in sp² bonds with the neighboring carbon atoms or with a hydrogen atom. The sixth electron occupies the 2pz orbital which is half filled. The valence orbitals are the 6 carbon 2pz orbitals. Using the LCAO method, we guess that a good solution to the molecular orbital Hamiltonian can be found in terms of a linear combination of the valence orbitals.

$$$\psi_{\text{mo}}= c_1\phi^C_{2p_z}(\vec{r}-\vec{r}_1)+c_2\phi^C_{2p_z}(\vec{r}-\vec{r}_2)+c_3\phi^C_{2p_z}(\vec{r}-\vec{r}_3)+c_4\phi^C_{2p_z}(\vec{r}-\vec{r}_4)+c_5\phi^C_{2p_z}(\vec{r}-\vec{r}_5)+c_6\phi^C_{2p_z}(\vec{r}-\vec{r}_6).$$$

Here $c_i$ are constants that need to be determined and $\phi^C_{2p_z}(\vec{r})$ is the 2pz atomic orbital with $Z=3.25$. This wavefunction is inserted into the time-independent Schrödinger equation,

$$$H_{\text{mo}}\psi_{\text{mo}}=E\psi_{\text{mo}}.$$$

Multipling the Schrödinger equation from the left by each of the atomic orbitals results in a set of $N$ algebraic equations called the Roothaan equations.

$$\begin{matrix} \langle \phi^C_{2pz}(\vec{r}-\vec{r}_1) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_1)|\psi_{\text{mo}} \rangle \\ \langle \phi^C_{2pz}(\vec{r}-\vec{r}_2) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_2)|\psi_{\text{mo}} \rangle \\ \langle \phi^C_{2pz}(\vec{r}-\vec{r}_3) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_3)|\psi_{\text{mo}} \rangle \\ \langle \phi^C_{2pz}(\vec{r}-\vec{r}_4) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_4)|\psi_{\text{mo}} \rangle \\ \langle \phi^C_{2pz}(\vec{r}-\vec{r}_5) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_5)|\psi_{\text{mo}} \rangle \\ \langle \phi^C_{2pz}(\vec{r}-\vec{r}_6) | H_{\text{mo}} |\psi_{\text{mo}}\rangle =E\langle\phi^C_{2pz}(\vec{r}-\vec{r}_6)|\psi_{\text{mo}} \rangle \end{matrix}$$

The Roothaan equations can be written in matrix form,

$$$\left[ \begin{matrix} H_{11} & H_{12} & H_{13} & H_{14} & H_{15} & H_{16} \\ H_{21} & H_{22} & H_{23} & H_{24} & H_{25} & H_{26} \\ H_{31} & H_{32} & H_{33} & H_{34} & H_{35} & H_{36} \\ H_{41} & H_{42} & H_{43} & H_{44} & H_{45} & H_{46} \\ H_{51} & H_{52} & H_{53} & H_{54} & H_{55} & H_{56} \\ H_{61} & H_{62} & H_{63} & H_{64} & H_{65} & H_{66} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ c_6 \end{matrix} \right] = E\left[ \begin{matrix} S_{11} & S_{12} & S_{13} & S_{14} & S_{15} & S_{16} \\ S_{21} & S_{22} & S_{23} & S_{24} & S_{25} & S_{26} \\ S_{31} & S_{32} & S_{33} & S_{34} & S_{35} & S_{36} \\ S_{41} & S_{42} & S_{43} & S_{44} & S_{45} & S_{46} \\ S_{51} & S_{52} & S_{53} & S_{54} & S_{55} & S_{56} \\ S_{61} & S_{62} & S_{63} & S_{64} & S_{65} & S_{66} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ c_6 \end{matrix} \right].$$$

Here the elements of the Hamiltonian matrix and the overlap matrix are,

$$$H_{ij}= \langle \phi^C_{2p_z}(\vec{r}-\vec{r}_i) |H_{\text{mo}}|\phi^C_{2pz}(\vec{r}-\vec{r}_j)\rangle \hspace{1cm}\text{and}\hspace{1cm} S_{ij}= \langle \phi^C_{2p_z}(\vec{r}-\vec{r}_i)|\phi^C_{2pz}(\vec{r}-\vec{r}_j)\rangle .$$$

In general, the Roothaan equations must be solved numerically. However, benzene is a speical case where some progress can be made analytically. Elements of the Hamiltonian matrix and the overlap matrix corresponding to orbitals that are not on the same atom or on neighboring atoms are negligibly small and are set to zero. Because of the symmetry of the molecule all of the diagonal elements are the same and the elements corresponding to orbitals on neighboring sites are the same. The only integrals that need to be calculated are $S_{11}$, $S_{12}$, $H_{11}$, and $H_{12}$. The Roothaan equations take the form,

$$$\left[ \begin{matrix} H_{11} & H_{12} & 0 & 0 & 0 & H_{12} \\ H_{12} & H_{11} & H_{12} & 0 & 0 & 0 \\ 0 & H_{12} & H_{11} & H_{12} & 0 & 0 \\ 0 & 0 & H_{12} & H_{11} & H_{12} & 0 \\ 0 & 0 & 0 & H_{12} & H_{11} & H_{12} \\ H_{12} & 0 & 0 & 0 & H_{12} & H_{11} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ c_6 \end{matrix} \right] = E\left[ \begin{matrix} 1 & S_{12} & 0 & 0 & 0 & S_{12} \\ S_{12} & 1 & S_{12} & 0 & 0 & 0 \\ 0 & S_{12} & 1 & S_{12} & 0 & 0 \\ 0 & 0 & S_{12} & 1 & S_{12} & 0 \\ 0 & 0 & 0 & S_{12} & 1 & S_{12} \\ S_{12} & 0 & 0 & 0 & S_{12} & 1 \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \\ c_6 \end{matrix} \right].$$$

The matrix on each side of this equation can be written in terms of the identity matrix $\textbf{I}$, the translation operator $\textbf{T}$, and the inverse of the translation operator $\textbf{T}^{-1}$.

$$$H=H_{11}\textbf{I}+H_{12}\textbf{T}+H_{12}\textbf{T}^{-1}\hspace{1cm}S=\textbf{I}+S_{12}\textbf{T}+S_{12}\textbf{T}^{-1}$$$

The eigenvectors of these matrices are also eigenvectors of the translation operator,

$$$\left[ \begin {matrix} e^{i\pi/3} \\ e^{i2\pi/3} \\ -1 \\ e^{i4\pi/3} \\ e^{i5\pi/3} \\ 1 \end{matrix} \right],\, \left[ \begin {matrix} e^{i2\pi/3} \\ e^{i4\pi/3} \\ 1 \\ e^{i8\pi/3} \\ e^{i10\pi/3} \\ 1 \end{matrix} \right],\, \left[ \begin {matrix} -1 \\ 1 \\ -1 \\ 1 \\ -1 \\ 1 \end{matrix} \right],\, \left[ \begin {matrix} e^{i4\pi/3} \\ e^{i8\pi/3} \\ 1 \\ e^{i16\pi/3} \\ e^{i20\pi/3} \\ 1 \end{matrix} \right],\, \left[ \begin {matrix} e^{i5\pi/3} \\ e^{i10\pi/3} \\ -1 \\ e^{i20\pi/3} \\ e^{i25\pi/3} \\ 1 \end{matrix} \right],\, \left[ \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{matrix} \right].$$$

The eigen energies are,

$$$E_{\text{mo},j}=\frac{H_{11}+2H_{12}\cos\left(\frac{\pi j}{3}\right)}{1+2S_{12}\cos\left(\frac{\pi j}{3}\right)}\hspace{2cm}j=1,2,\cdots,6.$$$

The molecular orbitals are,

$$$\psi_{\text{mo},j}=\frac{1}{\sqrt{6}}\sum\limits_{n=1}^6 \exp\left(\frac{i\pi nj}{3}\right)\phi^C_{2pz}(\vec{r}-\vec{r}_n)\hspace{2cm}j=1,2,\cdots,6.$$$

There are 6 valence electrons and we have calculated 6 molecular orbitals. In the ground state, the 6 electrons will occupy the 3 molecular orbitals with the lowest energies. Because $H_{12}<0$, the occupied orbitals are $\psi_{\text{mo},6}$, $\psi_{\text{mo},1}$ and $\psi_{\text{mo},5}$. $\psi_{\text{mo},6}$ has the lowest energy and $\psi_{\text{mo},1}$ and $\psi_{\text{mo},5}$ have the same energy.

The results of a more detailed numerical calculation of the energy levels of benzene can be found here.

A description of a Hückel model for benzene can be found in Molecular Physics & Elementary of Quantum Chemistry by H. Haken and H. C. Wolf (The German version, Molekülphysik und Quantenchemie, is available as an ebook from the TU Graz library).