= Solution
Charge balance fixes each <metal> as $\mathrm{M^{II}}$. 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:
$$
\begin{array}{c|c|c|c|c}
&\text{configuration}&n_{\rm unpaired}&S&\mu_{\rm eff}/\mu_B\\\hline
\mathrm{Mn^{II}}&t_{2g}^{3}e_g^{2}&5&5/2&\sqrt{35}=5.92\\
\mathrm{Fe^{II}}&t_{2g}^{4}e_g^{2}&4&2&\sqrt{24}=4.90\\
\mathrm{Cu^{II}}&t_{2g}^{6}e_g^{3}&1&1/2&\sqrt3=1.73
\end{array}
$$
The result follows from $S=n_{\rm unpaired}/2$ and $2\sqrt{S(S+1)}=\sqrt{n_{\rm unpaired}(n_{\rm unpaired}+2)}$. \b[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.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2021/paper-2-crystal-field.png]
{title=Octahedral occupations consistent with the three magnetic moments}
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