= Solution
The radical's magnetic orbital has $\pi$ symmetry. Fe(II)'s partly occupied $e''$ set supplies a matching metal $\pi$ magnetic orbital, permitting electron transfer and \b[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 \b[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.
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