 Solution

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.