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

The drawn $x,y$ axes pass through opposite sites, rather than along the square edges. In the D4h molecular point group class order $(E,2C_4,C_2,2C_2^{\prime },2C_2^{\prime \prime },i,2S_4,{\sigma }_h,2{\sigma }_v,2{\sigma }_d)$, an s orbital contributes $+1$ whenever its centre is fixed and zero otherwise. ThusFor example, a transverse twofold rotation about the printed $x$ axis fixes sites 1 and 3, while its diagonal counterpart exchanges all four sites. Reflection in the square plane fixes every site. Taking character inner products, with class sizes $(1,2,1,2,2,1,2,1,2,2)$, givesThese have dimensions $1+1+2=4$. The $B_{1g}$ label is tied to the printed diagonal-axis convention; rotating the transverse coordinate labels by $45^{\circ }$ would interchange $B_{1g}$ and $B_{2g}$.

Treat the site atomic orbitals $s_1,\dots ,s_4$ as orthonormal, as in the Hückel method. A normalized set of symmetry-adapted linear combinations isThe first combination is unchanged by every operation; the alternating combination has the transformation signs of $x^2-y^2$ for the printed axes. The last two transform together like $(x,y)$. The squared coefficients sum to one in each case, and their pairwise inner products vanish. Any orthogonal rotation of the two $E_u$ combinations is an equally valid basis for that degenerate subspace.

Use a common Coulomb integral $\alpha$ on each site, a resonance integral $\beta$ on each adjacent pair, zero interaction across a diagonal, and the identity overlap matrix. In site order $(1,2,3,4)$ the Hückel method givesApplying it to the four symmetry-adapted linear combinations givesFor the conventional bonding sign $\beta <0$, the all-in-phase combination is lowest and the alternating combination is highest. The $E_u$ combinations have canceling nearest-neighbour interactions and are nonbonding within this model.

The molecular orbital diagram has one level at $\alpha +2\beta$ labelled $A_{1g}$, two degenerate levels at $\alpha$ labelled $E_u$, and one level at $\alpha -2\beta$ labelled $B_{1g}$, in increasing energy for $\beta <0$. Each spatial molecular orbital admits two opposite-spin electrons. The diagram below also shows how one representative distortion lifts the middle degeneracy; the distorted scale is illustrative, not a fitted molecular spectrum.

Use the trapezium's printed axes, with principal axis $z$ in its plane and $y$ perpendicular to the plane. For either site pair $(1,2)$ or $(3,4)$, the group representation has characters of a representation $(2,0,2,0)$ in the order $(E,C_2(z),\sigma (xz),\sigma (yz))$: identity and the molecular-plane mirror fix both s orbitals, while the other two operations exchange them. The character inner products give ${\Gamma }_{12}={\Gamma }_{34}=A_1\oplus B_1$ in the C2v molecular point group.

Within each pair the in-phase and out-of-phase combinations have $A_1$ and $B_1$ symmetry respectively. One normalized choice isThe sign convention for $d$ is chosen to make the off-diagonal interaction positive in the corresponding two-by-two block. It does not change the physical predictions. These are pair symmetry-adapted linear combinations; the actual molecular orbitals are linear combinations within each of the two equal-symmetry pairs, found by diagonalizing the blocks.

The long edge is $3$–$4$. In the orthonormal pair basis $(a,b)$, its weaker resonance integral $n\beta$ changes the second diagonal element. The two short side edges contribute an inter-pair matrix element $\beta$, givingWith $\lambda =(E-\alpha )/\beta$, its secular equation isThe eigenvalues are thereforeAt $n=1$ they become $\alpha +2\beta$ and $\alpha$, agreeing with the square limit. Symmetry permits mixing of the two $A_1$ pair functions; treating each pair as an isolated bond would miss the off-diagonal term.

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.

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.