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

Put $D=\sqrt{n^2-2n+5}=\sqrt{(n-1{)}^2+4}$. For $0<n<1$, $D>n+1$, since $D^2-(n+1{)}^2=4(1-n)>0$. Remembering $\beta <0$, the two lowest trapezium molecular orbital energies areThe remaining levels lie at $2\alpha -{ϵ}_2$ and $2\alpha -{ϵ}_1$. Filling the lowest spatial orbitals with opposite-spin pairs givesFor 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.