Discussion
The possible values of \(E\) are
Answer: A
The energies are \(\alpha + \sqrt{2}\beta,\ \alpha,\ \alpha - \sqrt{2}\beta\).
Dividing each row of the determinant by \(\beta\) and writing \(x = (\alpha - E)/\beta\):
\[ \begin{aligned} \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} &= x(x^{2}-1) - x \\ &= x^{3} - 2x = 0 \\ x &= 0,\ \pm\sqrt{2} \end{aligned} \]
So \(E = \alpha - x\beta\) gives \(\alpha + \sqrt{2}\beta\) (bonding), \(\alpha\) (non-bonding) and \(\alpha - \sqrt{2}\beta\) (antibonding). The radical's odd electron sits in the non-bonding level.
\(\alpha \pm \beta\) are the ethylene levels, \(\alpha \pm 2\beta\) are the extreme levels of benzene, and \(2\sqrt{2}\) doubles the root of \(x^{2} = 2\).
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