JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Coordination Compounds

In $[Co(N{H}_3{)}_6{]}^{3+}$, $C{o}^{3+}$ ($3d^6$) meets the strong-field ligand $N{H}_3$, which forces all six d-electrons to pair into the lower $t_{2g}$ set ($t_{2g}^6{e}_g^0$); this uses the inner $(n-1)d$ orbitals for $d^{2}s{p}^3$ hybridisation, giving a low-spin, spin-paired inner-orbital complex. In $[Co{F}_6{]}^{3-}$, the weak-field $F^{-}$ ligand cannot force pairing, so the electrons spread out as $t_{2g}^4{e}_g^2$ with maximum unpaired spins, using the outer $nd$ orbitals for $s{p}^3{d}^2$ hybridisation — a high-spin, spin-free outer-orbital complex. $[Co(N{H}_3{)}_6{]}^{3+}$ is therefore the spin-paired complex and $[Co{F}_6{]}^{3-}$ the spin-free complex.