= Solution
For two <electrons>, $n+1+D<4$ whenever $0<n<1$: both sides of $D<3-n$ are positive and squaring reduces the inequality to $n<1$. Since $\beta<0$, the trapezium then has a higher total electronic <energy> than the square. \b[The two-electron model favours the square.] For four <electrons>, $D>2$, so $2\beta D<4\beta$: \b[the four-electron model favours the distortion toward a trapezium.] The lower split middle orbital is occupied while the upper one is empty.
The distinction is especially clear near the square. Write $\delta=1-n>0$; then $D=2+\delta^2/4+O(\delta^4)$ and
$$
\begin{aligned}
E_2^{\rm trap}-E_2^{\rm sq}&=\beta[-\delta+\delta^2/4+O(\delta^4)]>0,\\
E_4^{\rm trap}-E_4^{\rm sq}&=\tfrac12\beta\delta^2+O(\delta^4)<0.
\end{aligned}
$$
In particular the four-electron lowering in this specified one-edge model is quadratic near the square; a first-order lowering of the entire occupied sum should not be asserted merely from the middle-level splitting. This <Hückel square-to-trapezium distortion> is an electronic comparison under the stated fixed-$\alpha$, fixed-short-edge-$\beta$ assumptions. Nuclear repulsion, other bonding contributions and the structural cost of distortion are omitted, so an actual equilibrium geometry or optimum $n$ is not determined by this calculation alone.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2b-distortion-energies.png]
{title=Two- and four-electron total-energy changes for the one-edge Hückel distortion model}
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