Using the spin-only formula, compute the sum of the squares of the magnetic moments (in BM^2) of the following four species: [Fe(CN)6]^3-, [CoF6]^3-…

The spin-only moment is $\mu=\sqrt{n(n+2)}$ BM, so $\mu^2=n(n+2)$. Fe$^{3+}$ is $d^5$ and with strong-field $\mathrm{CN^-}$ gives low-spin $t_{2g}^5$ with $n=1$, so $\mu^2=3$. Co$^{3+}$ is $d^6$ and with weak-field $\mathrm{F^-}$ gives high-spin $t_{2g}^4e_g^2$ with $n=4$, so $\mu^2=24$. In $\mathrm{Ni(CO)_4}$ nickel is Ni(0); CO is a strong-field ligand, so the ten valence electrons pair up as $3d^{10}$ in a tetrahedral $sp^3$ complex, with no unpaired electrons, hence $\mu^2=0$. Mn$^{2+}$ is $d^5$ and tetrahedral with weak-field $\mathrm{Cl^-}$, so it is high spin with $n=5$ and $\mu^2=35$. The sum is $3+24+0+35=62$.