Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 1 / A5

The planar para-quinone framework has three mutually perpendicular twofold axes, inversion through its centre, and the associated mirror planes. With $x$ horizontal, $y$ along the two C=O bonds and $z$ perpendicular to the molecule, it belongs to the D2h point group, $\boxed{D_{2h}}$.

Write $p_i$ for the orthonormal $2p_z$ atomic orbital on numbered site $i$, with every positive lobe pointing along $+z$. A normalized set of symmetry-adapted linear combinations isEach has unit squared coefficient sum, and every distinct pair has zero inner product. The orbital representation therefore reduces to $\boxed{3B_{1u}+3B_{3g}+B_{2g}+A_u}$. For example, the fully symmetric site sum still has $B_{1u}$ orbital symmetry because a $p_z$ orbital itself changes sign under operations that reverse $z$. Ignoring that sign would assign the wrong representations.

Set $E=\alpha +\lambda \beta$, with the usual bonding convention $\beta <0$. In the Hückel method, the dimensionless matrix $M=(H-\alpha I)/\beta$ has diagonal entries one on oxygen sites 1 and 8, zero on carbon, and unit entries between bonded neighbours. In the bases $(O_{+},C_{+},R_{+})$ and $(O_{-},C_{-},R_{-})$ above, its blocks areThe one-dimensional $B_{2g}$ and $A_u$ blocks have $\lambda =1$ and $-1$, respectively. The characteristic polynomial of the $B_{1u}$ block factors asso its three molecular orbital energies areFor the $B_{3g}$ block the matrix trace is zero. Thus its three $\lambda$ values sum to zero, and the pair supplied in the paper gives the remaining one as $-(1.86-2.12)\simeq 0.26$. Directly, its secular equation is ${\lambda }^3-4\lambda +1=0$, whose roots are approximately $1.860806$, $0.254102$ and $-2.114908$. The supplied pair is approximate, so $0.26$ from its trace and $0.2541$ from the exact block should not be mistaken for an additional physical splitting.

All eight energies, ordered from low to high for $\beta <0$, areThe eight $\pi$ electrons fill the four lowest molecular orbitals. The equality at $\lambda =1$ is an accidental degeneracy of this parameterized model, not a multidimensional representation of $D_{2h}$.

The extra electron occupies the lowest unoccupied molecular orbital of the neutral molecule, the $B_{3g}$ orbital at the middle root ${\lambda }_L=0.25410169\dots$ of ${\lambda }^3-4\lambda +1=0$. WriteThe first and third rows of the $B_{3g}$ block giveChoosing the overall phase so that site 1 is positive gives $(a,b,c)\simeq (0.664623,-0.495741,-0.559033)$. In the original site order,The squared coefficients sum to one. Within this fixed-orbital Hückel orbital charge density approximation, the added electron contributes site populationsThus the oxygen sites $\boxed{1\text{ and }8}$ receive the largest added negative charge, each receiving an increment of approximately $-0.221e$. Including the four doubly occupied neutral orbitals gives an anion $\pi$ population of about $1.621$ on each oxygen, $0.870$ on sites 2 and 7, and $1.005$ on each remaining carbon. Thus oxygen also has the largest total $\pi$ population in this model. A coefficient's sign is an orbital phase, not the sign of its electric charge. Predicting an actual charge density would additionally require orbital relaxation and a population-analysis convention; the stated result is the requested orbital model.