Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / Section B / 20 / c / iii

A frustrated triangle has two ferromagnetic metal–radical bonds and an antiferromagnetic metal–metal bond. With the Hamiltonian convention above, take $J/k_B=+800$ K and $J^{\prime }/k_B=-8$ K. In the maximally aligned ground state the two metal–radical products are $1\times 1/2$ and the metal–metal product is $1\times 1$. ThereforeHere $F_C=-1$ denotes complete satisfaction of all exchange interactions; the small deviation $1+F_C=0.0198$ quantifies the weak frustrated contribution in this normalization. The numerical answer assumes antiferromagnetic metal–metal exchange. The printed magnitudes alone do not establish that sign. If the metal–metal exchange were instead ferromagnetic, all three bonds would be satisfied and $F_C=-1$. The ground-state scalar products also follow directly from $S_{12}=2$, $S=5/2$, so a separate classical approximation is unnecessary for this fully polarized multiplet.