JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Coordination Compounds

The spin-only magnetic moment is $\mu =\sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons.

${\mathrm{Ti}}^{3+}$ is $[\mathrm{Ar}]3d^1$: one unpaired electron gives $\mu =\sqrt{3}=1.73$ B.M.

${\mathrm{V}}^{2+}$ is $[\mathrm{Ar}]3d^3$: three unpaired electrons give $\mu =\sqrt{15}=3.87$ B.M.

${\mathrm{Ni}}^{2+}$ is $[\mathrm{Ar}]3d^8$: filling the five d-orbitals with 8 electrons leaves two unpaired, giving $\mu =\sqrt{8}=2.84$ B.M.

${\mathrm{Sc}}^{3+}$ is $[\mathrm{Ar}]3d^0$: no unpaired electrons, so $\mu =0.00$ B.M.

The correct pairing is ${\mathrm{Ti}}^{3+}$–1.73, ${\mathrm{V}}^{2+}$–3.87, ${\mathrm{Ni}}^{2+}$–2.84, ${\mathrm{Sc}}^{3+}$–0.00.