JEE Main 2025 Apr 2 Shift 2, Chemistry Q8: Bonding Of Co-ordination Compounds

In $[\mathrm{MnC}{\mathrm{l}}_6{]}^{3-}$ the metal is in the +3 oxidation state, giving ${\mathrm{Mn}}^{3+}:[\mathrm{Ar}]3d^4$.

Chloride is a weak-field ligand, so pairing does not occur. The four d electrons stay unpaired in the 3d set, and the empty $4s,4p,4d$ orbitals participate in the hybrid scheme:

Hybridisation $=s{p}^3{d}^2$ (outer-orbital, high-spin complex).

Number of unpaired electrons $=4\Rightarrow$ paramagnetic.

$\therefore$ $s{p}^3{d}^2$, paramagnetic with four unpaired electrons.