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 Incoming links: Bond-length alternation

Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A4 / c / ii / Solution 2026-10-20

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. Put

β_{S} = β (1 + δ), β_{L} = β (1 - δ), β < 0, 0 < δ < 1.

(1)

Using part (i), the change from the uniform ring is

Δ E_{π} = 4 β δ + 42 β (1 + δ^{2} - 1) = 4 β δ + 22 β δ^{2} + O (δ^{4}) < 0.

(2)

Thus a small alternation lowers the electronic energy, with a leading gain linear in

δ

. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight

π

electrons constitute a

4 n

cyclic count rather than the closed-shell

4 n + 2

case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.

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