= Solution
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 $J'/k_B=-8$ K. In the maximally aligned <ground state> the two metal–radical products are $1\times1/2$ and the metal–metal product is $1\times1$. Therefore
$$
\boxed{F_C=-\frac{2(800)(1/2)-8}{2(800)(1/2)+8}=-\frac{792}{808}=-0.9802.}
$$
Here $F_C=-1$ denotes complete satisfaction of all exchange interactions; the small deviation $1+F_C=0.0198$ quantifies the weak frustrated contribution in this normalization. \b[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.
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