Discussion
\(\psi_{1}, \psi_{2}, \psi_{3}\), and \(\psi_{4}\) are four Hückel molecular orbitals of benzene with orbital energies \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\), respectively.
\[ \begin{aligned} \psi_{1} &= \tfrac{1}{2}(\phi_{B} + \phi_{C} - \phi_{E} - \phi_{F}) \\ \psi_{2} &= 6^{-1/2}(\phi_{A} - \phi_{B} + \phi_{C} \\ &\qquad - \phi_{D} + \phi_{E} - \phi_{F}) \\ \psi_{3} &= 6^{-1/2}(\phi_{A} + \phi_{B} + \phi_{C} \\ &\qquad + \phi_{D} + \phi_{E} + \phi_{F}) \\ \psi_{4} &= 12^{-1/2}(2\phi_{A} + \phi_{B} - \phi_{C} \\ &\qquad - 2\phi_{D} - \phi_{E} + \phi_{F}) \end{aligned} \]
The correct order of the orbital energies is
(The six carbon atoms of benzene are denoted by A to F and \(\phi_{J}\) is the \(2p_{z}\) orbital of Jth carbon of benzene.)
Answer: C
The order is \(E_{3} \lt E_{1} = E_{4} \lt E_{2}\).
In Hückel theory the energy of a benzene MO is \(\alpha + 2\beta\cos(2\pi k/6)\), with \(\beta \lt 0\), so fewer nodes means lower energy.
- \(\psi_{3}\) has all coefficients equal, no nodes: \(\alpha + 2\beta\), the lowest.
- \(\psi_{1}\) (pattern 0, 1, 1, 0, -1, -1) and \(\psi_{4}\) (2, 1, -1, -2, -1, 1) each change sign twice around the ring: they are the degenerate pair at \(\alpha + \beta\).
- \(\psi_{2}\) alternates in sign at every atom: \(\alpha - 2\beta\), the highest.
So \(E_{1} = E_{4}\) lie between \(E_{3}\) and \(E_{2}\). The other orders either break the degeneracy of \(\psi_{1}\) and \(\psi_{4}\) or put the fully alternating orbital too low.
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