 Solution

Charge balance fixes each metal as ${\mathrm{M}}^{\mathrm{II}}$. In an octahedral crystal field the lower $t_{2g}$ set is $(d_{xy},d_{xz},d_{yz})$ and the upper $e_g$ set is $(d_{z^2},d_{x^2-y^2})$. The measured spin-only magnetic moments select the following occupations:The result follows from $S=n_{\mathrm{unpaired}}/2$ and $2\sqrt{S(S+1)}=\sqrt{n_{\mathrm{unpaired}}(n_{\mathrm{unpaired}}+2)}$. Manganese(II) and iron(II) are high-spin complexes: their octahedral splitting is too small to compensate the additional pairing cost of their low-spin alternatives. Copper(II) is $d^9$, with one unpaired electron regardless of which member of the upper pair is singly occupied. spin-orbit coupling and exchange interactions can produce deviations from the spin-only estimate in real measurements; no such correction is required to explain the supplied values.