JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Coordination Compounds

In ${\mathrm{K}}_3[\mathrm{Co}(\mathrm{C}{\mathrm{O}}_3{)}_3]$ the metal is $\mathrm{C}{\mathrm{o}}^{3+}$ ($d^6$) and carbonate is a weak-field ligand, so the complex is high spin with four unpaired electrons, $s{p}^3{d}^2$ hybridised and octahedral, giving $\mu =\sqrt{4(4+2)}=4.9\mathrm{BM}$; Statement I is correct. For Statement II, $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ ($\mathrm{N}{\mathrm{i}}^{2+}$, strong field) is $ds{p}^2$, square planar, with no unpaired electrons ($0\mathrm{BM}$); $[\mathrm{MnB}{\mathrm{r}}_4{]}^{2-}$ ($\mathrm{M}{\mathrm{n}}^{2+}$, weak field) is $s{p}^3$, tetrahedral, with five unpaired electrons ($5.9\mathrm{BM}$); and $[\mathrm{Co}{\mathrm{F}}_6{]}^{3-}$ ($\mathrm{C}{\mathrm{o}}^{3+}$, weak field) is $s{p}^3{d}^2$, octahedral, with four unpaired electrons ($4.9\mathrm{BM}$), so Statement II is also correct. Both statements are true.