alleryThe spin-only magnetic moment of $\mathrm{Fe}^{3+}$ (ground state) is approximately:
A$1.73\,\text{BM}$
B$2.83\,\text{BM}$
C$4.90\,\text{BM}$
D$5.92\,\text{BM}$
Answer & Solution
Correct answer: D. $5.92\,\text{BM}$
$\mathrm{Fe}^{3+}: [\mathrm{Ar}]3d^5$, so $n = 5$ unpaired electrons (high-spin, weak-field assumption).
$\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\,\text{BM}$.
The other options correspond to $n = 1, 2$ and $4$ unpaired electrons.
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