JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Coordination Compounds

Working out hybridisation and magnetism ligand by ligand:

$[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$: ${\mathrm{Mn}}^{2+}$ is $d^5$; ${\mathrm{Br}}^{-}$ is a weak-field ligand, and the tetrahedral geometry never pairs electrons regardless of field strength, so it remains high spin with $s{p}^3$ hybridisation and 5 unpaired electrons — paramagnetic.

$[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$: ${\mathrm{Fe}}^{3+}$ is $d^5$; ${\mathrm{F}}^{-}$ is a weak-field ligand, giving the high-spin octahedral configuration $t_{2g}^3{e}_g^2$, which requires outer d orbitals — $s{p}^3{d}^2$ hybridisation, paramagnetic.

$[\mathrm{Co}({\mathrm{C}}_2{\mathrm{O}}_4{)}_3{]}^{3-}$: ${\mathrm{Co}}^{3+}$ is $d^6$; the chelating oxalate is a strong-field ligand, forcing the low-spin configuration $t_{2g}^6{e}_g^0$ with every electron paired — inner-orbital $d^{2}s{p}^3$ hybridisation, diamagnetic.

$[\mathrm{Ni}(\mathrm{CO}{)}_4]$: the CO ligands hold nickel at the zero oxidation state, so ${\mathrm{Ni}}^0$ is $d^{10}$; with the d subshell completely filled, the complex hybridises using the 4s and 4p orbitals — $s{p}^3$, and a filled $d^{10}$ shell has no unpaired electrons, so it is diamagnetic.

Matching: (A)–$s{p}^3$ & paramagnetic, (B)–$s{p}^3{d}^2$ & paramagnetic, (C)–$d^{2}s{p}^3$ & diamagnetic, (D)–$s{p}^3$ & diamagnetic.