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

The radical's magnetic orbital has $\pi$ symmetry. Fe(II)'s partly occupied $e^{\prime \prime }$ set supplies a matching metal $\pi$ magnetic orbital, permitting electron transfer and predominantly antiferromagnetic metal–radical exchange. In Co(II) and Ni(II), that set is filled; their unpaired metal electrons occupy orbitals of different symmetry. Approximate orthogonality suppresses the usual antiferromagnetic kinetic-exchange pathway, favouring ferromagnetic metal–radical coupling through the remaining exchange contributions.

This orbital-symmetry argument assumes the intended idealized binding geometry. It predicts exchange signs, not exact magnitudes; distortion and additional orbital mixing can alter the competition.

Adopt $\mathcal{H}=-2J({\mathbf{S}}_1+{\mathbf{S}}_2)\cdot \mathbf{s}-2J^{\prime }{\mathbf{S}}_1\cdot {\mathbf{S}}_2$, with $S_1=S_2=1$ and ligand spin $s=1/2$. Let ${\mathbf{S}}_{12}={\mathbf{S}}_1+{\mathbf{S}}_2$ and $\mathbf{S}={\mathbf{S}}_{12}+\mathbf{s}$. The angular momentum addition identities giveThe five allowed multiplets areTheir degeneracies $2,2,4,4,6$ sum to 18, matching $3\times 3\times 2$. For dominant ferromagnetic $J$, the ground multiplet has $S_{12}=2$, $S=5/2$. A weak antiferromagnetic $J^{\prime }$ does not overcome the two strong metal–radical bonds.

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.