JEE Main 2025 Jan 29 Shift 1, Chemistry Q8: Colour, Magnetism And Stability Of Co-ordination Compounds

Use VBT: count $d$-electrons, decide spin (based on ligand strength), then assign geometry/hybridisation.

(A) $[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$: ${\mathrm{Mn}}^{2+}$ is $d^5$; ${\mathrm{Br}}^{-}$ is a weak-field ligand and the complex is tetrahedral $\Rightarrow s{p}^3$, 5 unpaired electrons, paramagnetic. Matches (IV).

(B) $[\mathrm{Fe}{\mathrm{F}}_6{]}^{3-}$: ${\mathrm{Fe}}^{3+}$ is $d^5$; ${\mathrm{F}}^{-}$ is weak field so it is high-spin octahedral $\Rightarrow s{p}^3{d}^2$, 5 unpaired, paramagnetic. Matches (II).

(C) $[\mathrm{Co}({\mathrm{C}}_2{\mathrm{O}}_4{)}_3{]}^{3-}$: ${\mathrm{Co}}^{3+}$ is $d^6$; with ${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$ acting as a chelating ligand the complex is low-spin octahedral $\Rightarrow d^{2}s{p}^3$, no unpaired electrons, diamagnetic. Matches (I).

(D) $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: ${\mathrm{Ni}}^0$ is $3d^{8}4s^2$; CO is a strong field $\pi$-acceptor that pairs the $3d$ electrons and promotes the $4s$ pair, leaving an empty $4s$ and three $4p$ orbitals $\Rightarrow s{p}^3$ tetrahedral, all paired, diamagnetic. Matches (III).

$\therefore$ (A)-(IV), (B)-(II), (C)-(I), (D)-(III).