Calculate the spin-only magnetic moment of ${\text{[Fe(H}}_2{\text{O)}}_6{]}^{2+}$.

Solution

1. The oxidation state of Fe in ${\text{[Fe(H}}_2{\text{O)}}_6{]}^{2+}$ is +2, which corresponds to a $3d^6$ configuration.
2. Water is a weak field ligand, so the electrons remain unpaired in the $t_{2g}$ and $e_g$ orbitals.
3. The configuration is $t_{2g}^4{e}_g^2$, resulting in $n=4$ unpaired electrons.
4. The spin-only magnetic moment is $$\begin{array}{rl}\mu & =\sqrt{n(n+2)}\\ & =\sqrt{4(4+2)}\\ & =\sqrt{24}\approx 4.90\text{ BM}\end{array}$$.
5. Hence the answer is (C).