JEE Main 2025 Apr 8 Shift 2, Chemistry Q19: Bonding Of Co-ordination Compounds

$[Ni(CO{)}_4]$: nickel is in the zero oxidation state, $[Ar]3d^{8}4s^2$; CO is a strong field ligand that causes the 4s electrons to pair within the d-orbitals, freeing the 4s and 4p orbitals for sp${}^3$ hybridisation, giving a tetrahedral, diamagnetic (0 BM) complex: A-III.

$[Ni(CN{)}_4{]}^{2-}$: Ni${}^{2+}$ is $[Ar]3d^8$; CN${}^{-}$ is a strong field ligand that pairs up the two unpaired d-electrons, freeing one d-orbital for dsp${}^2$ hybridisation, giving a square planar, diamagnetic (0 BM) complex: B-II.

$[NiC{l}_4{]}^{2-}$: Ni${}^{2+}$ is $[Ar]3d^8$; Cl${}^{-}$ is a weak field ligand, so no pairing occurs, giving sp${}^3$ hybridisation with two unpaired electrons and $\mu =\sqrt{8}\approx 2.83$ BM, tetrahedral: C-I.

$[MnB{r}_4{]}^{2-}$: Mn${}^{2+}$ is $[Ar]3d^5$; with a weak field ligand this high-spin ion retains all five unpaired electrons in sp${}^3$ (tetrahedral) hybridisation, giving $\mu =\sqrt{35}\approx 5.92$ BM: D-IV.

Combining these gives A-III, B-II, C-I, D-IV, matching option (3).