# 1.7: Hückel Theory 2 (Eigenvalues)

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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 C_{2} point group. Correlating to D_{2h} point group gives A in C2 → B_{1u} in D_{2h} and B in C_{2} → B_{2g} in D_{2h}:

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 C_{1} and C_{2} carbons of benzene

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

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

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