d and f Block Elements: Speed-Solving Techniques (6)

$\mu = \sqrt{n(n + 2)}\,\text{BM}$

where $n$ is the number of unpaired electrons.

$Fe^{2+}$: Atomic configuration of Fe is $[Ar]\,3d^6\,4s^2$. Removing 2 electrons gives $Fe^{2+}: [Ar]\,3d^6$. The $3d^6$ configuration has $n = 4$ unpaired electrons (Hund’s rule: 4 unpaired in 5 d-orbitals before pairing begins, then one electron pairs).

$Mn^{2+}$: Mn is $[Ar]\,3d^5\,4s^2$. Removing 2 electrons gives $Mn^{2+}: [Ar]\,3d^5$. Half-filled, so $n = 5$.

$Cu^{2+}$: Cu is $[Ar]\,3d^{10}\,4s^1$. Removing 2 electrons gives $Cu^{2+}: [Ar]\,3d^9$. So $n = 1$.

$\mu(Fe^{2+}) = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90\,\text{BM}$

$\mu(Mn^{2+}) = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92\,\text{BM}$

$\mu(Cu^{2+}) = \sqrt{1 \times 3} = \sqrt{3} \approx 1.73\,\text{BM}$

$Mn^{2+}$ has the largest $\mu$ (5.92 BM) — it is the most paramagnetic. This is no coincidence — half-filled d⁵ is the most stable AND most paramagnetic configuration.

Final: $Fe^{2+}$ = 4.90 BM, $Mn^{2+}$ = 5.92 BM, $Cu^{2+}$ = 1.73 BM. $Mn^{2+}$ most paramagnetic.