JEE Main 2026 — Coordination Compounds Question with Solution

In Statement I, for $K_3[Co(C{O}_3{)}_3]$, Cobalt is in $+3$ oxidation state ($C{o}^{3+}:3d^6$). Carbonate ($C{O}_3^{2-}$) is a weak field ligand. Thus, no pairing occurs, resulting in 4 unpaired electrons. Hybridisation is $s{p}^3{d}^2$, shape is octahedral, and $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.9$ BM. Statement I is true.
In Statement II:
For $[Ni(CN{)}_4{]}^{2-}$, $N{i}^{2+}$ is $3d^8$. $C{N}^{-}$ is a strong field ligand, causing pairing. Hybridisation is $ds{p}^2$, geometry is square planar, and $\mu =0$ BM.
For $[MnB{r}_4{]}^{2-}$, $M{n}^{2+}$ is $3d^5$. $B{r}^{-}$ is a weak field ligand. Hybridisation is $s{p}^3$, geometry is tetrahedral, and $\mu =\sqrt{5(5+2)}=\sqrt{35}\approx 5.9$ BM.
For $[Co{F}_6{]}^{3-}$, $C{o}^{3+}$ is $3d^6$. $F^{-}$ is a weak field ligand. Hybridisation is $s{p}^3{d}^2$, geometry is octahedral, and $\mu =\sqrt{4(4+2)}=\sqrt{24}\approx 4.9$ BM.
Statement II is true.
Both statements are true.