Past exam of the chemistry B course of the University of Cambridge / 2024 / Paper 2 / B6 / e / Solution

The dimer has two unpaired electrons and a triplet ground state, $S=1$. Its approximately threefold-symmetric Fe${}_2$(CO)${}_3$ core supplies a pair of degenerate frontier molecular orbitals. The last two electrons occupy separate members with parallel spins, favored by exchange, rather than pairing in one. A useful qualitative picture is a sigma bond plus two perpendicular, singly occupied pi half-bonds: together these support a formal Fe–Fe double bond while retaining paramagnetism. The spin-only magnetic moment for $S=1$ is $\sqrt{8}{\mu }_B\simeq 2.83{\mu }_B$; real moments can include orbital and spin–orbit effects.

The formal 18-electron count does not require every molecular orbital to be doubly occupied. It is therefore compatible with an open-shell dimer. The primary structural/magnetic study establishes this unusual triplet state; an isolated-atom high-spin $d^7$ diagram would not describe the coupled metal–ligand system.^[ref]