 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2b/section-b/36/c/i/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2B / Section B / 36 / c / i / Solution by

Codex  0 2026-10-20

Put

D = n^{2} - 2 n + 5 = (n - 1)^{2} + 4

. For

0 < n < 1

,

D > n + 1

, since

D^{2} - (n + 1)^{2} = 4 (1 - n) > 0

. Remembering

β < 0

, the two lowest trapezium molecular orbital energies are

ϵ_{1} = α + \frac{β}{2} (n + 1 + D) (A_{1}), ϵ_{2} = α + \frac{β}{2} (D - n - 1) (B_{1}) .

The remaining levels lie at

2 α - ϵ_{2}

and

2 α - ϵ_{1}

. Filling the lowest spatial orbitals with opposite-spin pairs gives

2 electrons 4 electrons square 2 α + 4 β 4 α + 4 β trapezium 2 α + β (n + 1 + D) 4 α + 2 β D

For the square four-electron case, two electrons occupy the lowest orbital and two occupy the degenerate

E_{u}

set. Their summed one-electron energy is fixed here, but this Hückel method calculation alone does not decide the spin or correlation within that open shell.

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