 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. The two-electron model favours the square. For four electrons, $D>2$, so $2\beta D<4\beta$: 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 $D=2+{\delta }^2/4+O({\delta }^4)$ andIn 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.