# 1.7: Hückel Theory 2 (Eigenvalues)

The energies (eigenvalues) may be determined by using the Hückel approximation.

$E \left( \psi_{B_{2g}} \right) = \dfrac{1}{6}(6)( \alpha - 2\beta ) = \alpha - 2\beta$

The energies of the remaining LCAO’s are:

$E \left( \psi_{E_{1g}}^a \right) = \left( \psi_{E_{1g}}^b \right) = \alpha + \beta$

$E \left( \psi_{E_{2u}}^a \right) = \left( \psi_{E_{2u}}^b \right) = \alpha - \beta$

Note the energies of the E orbitals are degenerate. Constructing the energy level diagram, we set α = 0 and β as the energy parameter (a negative quantity, so an MO whose energy is positive in units of β has an absolute energy that is negative),

The energy of benzene based on the Hückel approximation is

$E_{total} = 2(2\beta) + 4(\beta) = 8\beta$

What is the delocalization energy (i.e. π resonance energy)?

To determine this, we consider cyclohexatriene, which is a six-membered cyclic ring with 3 localized π bonds; in other terms, cyclohexatriene is the product of three condensed ethylene molecules. For ethylene,

Following the procedures outlined above, we find,

\begin{aligned}
&\mathrm{E}\left(\psi_{1}\right)=\left\langle\frac{1}{\sqrt{2}}\left(\phi_{1}+\phi_{2}\right)|\mathrm{H}| \frac{1}{\sqrt{2}}\left(\phi_{1}+\phi_{2}\right)\right\rangle=\frac{1}{2}(2 \alpha+2 \beta)=\beta \\
&\mathrm{E}\left(\psi_{2}\right)=\left\langle\frac{1}{\sqrt{2}}\left(\phi_{1}-\phi_{2}\right)|\mathrm{H}| \frac{1}{\sqrt{2}}\left(\phi_{1}-\phi_{2}\right)\right\rangle=\frac{1}{2}(2 \alpha-2 \beta)=-\beta
\end{aligned}

The above was determined in the C2 point group. Correlating to D2h point group gives A in C2 → B1u in D2h and B in C2 → B2g in D2h:

The Hückel energy of ethylene is,

$E_{total} = 2(\beta) = 2\beta$

Therefore, the energy of cyclohexatriene is 3(2β) = 6β. The resonance energy is therefore,

The bond order is given by,

Consider the B.O. between the C1 and C2 carbons of benzene

$[ \psi_{1}(A_{2u})] = 2( \dfrac{1}{ \sqrt{6}} )( \dfrac{1}{ \sqrt{6}}) = \dfrac{1}{3}$

$[ \psi_{3}(E_{1g}^a)] = 2( \dfrac{1}{ \sqrt{12}} )( \dfrac{1}{ \sqrt{12}}) = \dfrac{1}{3}$

$[ \psi_{4}(E_{1g}^b)] = \dfrac{1}{2}(0)( \dfrac{1}{2} ) = \dfrac{0}{ \dfrac{2}{3} }$