Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 1 / A4

The alternating short and long bonds preserve fourfold, but not eightfold, rotations. Including the molecular plane and inversion symmetry, the planar structure belongs to $D_{4h}$, the D4h molecular point group.

Use the printed axes: the short bonds are $1$–$2$, $3$–$4$, $5$–$6$ and $7$–$8$, and the $x$ and $y$ axes pass through the midpoints of opposite short bonds. Interchanging short and long bonds is not a molecular symmetry operation of this geometry.

For the $p_z$ atomic orbital basis, only orbitals whose centres remain fixed contribute to a character. Each such orbital contributes $+1$ if its $z$ direction is preserved and $-1$ if it is reversed. In the standard D4h molecular point group class order,No nonidentity operation other than ${\sigma }_h$ leaves any carbon site fixed. Reflection in the molecular plane fixes all eight sites but reverses every $p_z$ function.

Molecular character reduction givesThe dimensions sum to $1+1+1+1+2\cdot 2=8$. The two copies of $E_g$ are separate equivalent irreducible representations which may mix with one another; they do not force all four functions to have a common energy.

Write $p_j=p_{zj}$ and use the orthonormal basis of the Hückel method. The four normalized one-dimensional symmetry-adapted linear combinations areFor example, the $xz$ mirror plane exchanges sites $1\leftrightarrow 6$, $2\leftrightarrow 5$, $3\leftrightarrow 4$, $8\leftrightarrow 7$, without changing the sign of an individual $p_z$ orbital. The all-positive combination is unchanged by every vertical mirror but changes sign under ${\sigma }_h$, giving $A_{2u}$. The alternating combination changes sign under both sorts of vertical mirror, as required for $A_{1u}$. The two remaining sign patterns distinguish $B_{1u}$ from $B_{2u}$ in the printed axes. The signs refer to the upper $p_z$ lobes; the lower lobes have opposite quantum phases, and neither colour denotes an electric charge.

For a normalized coefficient vector $\mathbf{c}$, the energy isEvery site has one short-bond and one long-bond neighbour. Equivalently, apply the Hückel method matrix to any coefficient and read off its two neighbouring signs. The results areWith negative resonance integrals and $|{\beta }_S|>|{\beta }_L|$, the $A_{2u}$ and $B_{2u}$ levels are occupied, while $B_{1u}$ and $A_{1u}$ are unoccupied in the closed-shell alternating-bond state.

Use the four supplied opposite-site combinations in the order $({\varphi }_1,{\varphi }_2,{\varphi }_3,{\varphi }_4)$. Their Hückel method matrix isThe negative corner is essential. For example, $⟨{\varphi }_4|H|{\varphi }_1⟩$ includes the long bonds $4$–$5$ and $8$–$1$, each with opposite coefficient signs, giving $-{\beta }_L$.

Because the basis is orthonormal, the secular determinant isIt factors as $[(\alpha -E{)}^2-{\beta }_S^2-{\beta }_L^2{]}^2=0$, giving two doubly degenerate energies, $\alpha \pm \sqrt{{\beta }_S^2+{\beta }_L^2}$.

At ${\beta }_L=0$, the preceding matrix separates into identical $({\varphi }_1,{\varphi }_2)$ and $({\varphi }_3,{\varphi }_4)$ blocks. For ${\beta }_S<0$, a normalized pair at the lower energy $\alpha +{\beta }_S$ isA pair at the upper energy $\alpha -{\beta }_S$ isThe four coefficient vectors are mutually orthonormal. Any orthogonal mixing within either equal-energy pair is an equally valid choice of molecular orbitals. At this limiting parameter value, additional coincidences with the one-dimensional levels occur: the full eight-orbital system is four isolated dimers, with four bonding and four antibonding levels.

Fill four spatial molecular orbitals with the eight $\pi$ electrons. For negative ${\beta }_S,{\beta }_L$ and $|{\beta }_S|>|{\beta }_L|$, the occupied energies are $\alpha +{\beta }_S+{\beta }_L$, $\alpha +{\beta }_S-{\beta }_L$, and the pair at $\alpha -\sqrt{{\beta }_S^2+{\beta }_L^2}$. ThereforeFor equal integrals, ${\beta }_S={\beta }_L=\beta <0$, this becomesAt equal bond lengths, the last two electrons occupy the frontier degeneracy at $E=\alpha$, so their distribution within that degenerate space does not change the bare Hückel method energy.

For ${\beta }_L=0$, four isolated short-bond dimers each contribute $2(\alpha +{\beta }_S)$:These are different coupling limits. Keeping ${\beta }_S=\beta$ while simply removing long-bond coupling does not represent a small distortion at fixed mean bond strength.

The uniform planar eight-electron ring has a partially filled frontier degeneracy. Bond-length alternation splits those levels, permitting the two frontier electrons to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.

A controlled comparison keeps the mean resonance integral fixed. PutUsing part (i), the change from the uniform ring isThus a small alternation lowers the electronic energy, with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.

The eight $\pi$ electrons constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with aromaticity. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of cyclooctatetraene.