Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 10 / b / i / Solution

Use the stated local axes: $z$ is radial along O–Ru, $x$ is tangential within the Ru–O plane, and $y$ is perpendicular to that plane. The oxygen's three $s{p}^2$ hybrids make the in-plane sigma framework. Its remaining $p_y$ orbital is perpendicular to the plane and can overlap with one $d_{yz}$ orbital on each Ru.

Choose the local orbital phases so their O-facing lobes agree. Of the three metal orbitals, onlyhas the appropriate combination to interact with oxygen $p_y$. The other two independent combinations have coefficients summing to zero and are nonbonding with respect to the central oxygen. Mixing $D$ with $p_y$ gives one lower pi-bonding orbital and one upper pi-antibonding orbital. Each Ru also retains its $d_{xy}$ and $d_{xz}$ orbitals as nonbonding with respect to this central-oxo pi interaction. Thus the complete low-energy diagram contains one bonding level, eight metal-nonbonding levels, and one antibonding level. The eight consist of two $d_{yz}$ combinations, three $d_{xy}$ and three $d_{xz}$ orbitals. They need not be exactly degenerate once other ligands are considered.

For the neutral formula as printed, oxide contributes $-2$ and six acetates contribute $-6$. The Ru oxidation numbers therefore sum to $+8$, giving sixteen metal d electrons overall, formally two Ru(III) and one Ru(II), possibly delocalized. Including the oxygen pi pair gives eighteen electrons, filling the bonding orbital and all eight nonbonding orbitals, leaving pi* empty. If the often encountered singly charged cation were intended instead, there would be seventeen electrons and one singly occupied nonbonding level. This charge distinction changes occupancy, not the orbital connectivity.