JEE Main · 2025 · 28 Jan 2025 · Shift 1 · Chemical Bonding and Molecular Structure

$Cl{F}_3$ is $s{p}^{3}d$ hybridised with a trigonal bipyramidal electron-pair arrangement; to minimise lone pair–bond pair repulsion, both lone pairs sit in equatorial positions, giving the familiar bent (T-shaped) molecule, so $n=2$. Checking unpaired electrons: $V^{3+}$ is $[Ar]3d^2$ (2 unpaired), $T{i}^{3+}$ is $[Ar]3d^1$ (1 unpaired), $C{u}^{2+}$ is $[Ar]3d^9$ (1 unpaired), $N{i}^{2+}$ is $[Ar]3d^8$ (2 unpaired), and $T{i}^{2+}$ is $[Ar]3d^2$ (2 unpaired). The ions with exactly 2 unpaired electrons are $V^{3+}$, $N{i}^{2+}$ and $T{i}^{2+}$ — A, D and E.