JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Coordination Compounds

A homoleptic complex has only one kind of ligand, and whether it is low spin depends on whether that ligand sits high enough in the spectrochemical series to force pairing.

(A) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}\right]}^{2-}$ has two different ligands ($\mathrm{C}{\mathrm{N}}^{-}$ and $\mathrm{NO}$) — it is heteroleptic, so it is excluded regardless of spin state.

(B) ${\left[\mathrm{Co}{\mathrm{F}}_6\right]}^{3-}$ is homoleptic. $\mathrm{C}{\mathrm{o}}^{3+}$ is $3d^6$, and ${\mathrm{F}}^{-}$ is a weak-field ligand, giving the high-spin $t_{2g}^4{e}_g^2$ configuration.

(C) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4-}$ is homoleptic. $\mathrm{F}{\mathrm{e}}^{2+}$ is $3d^6$, and $\mathrm{C}{\mathrm{N}}^{-}$ is a strong-field ligand, forcing all six electrons into $t_{2g}^6{e}_g^0$ — low spin.

(D) ${\left[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_6\right]}^{3+}$ is homoleptic. $\mathrm{C}{\mathrm{o}}^{3+}$ is $3d^6$, and $\mathrm{N}{\mathrm{H}}_3$ is a strong enough field ligand for a $+3$ metal to give the low-spin $t_{2g}^6{e}_g^0$ configuration.

(E) ${\left[\mathrm{Cr}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{2+}$ is homoleptic. $\mathrm{C}{\mathrm{r}}^{2+}$ is $3d^4$, and ${\mathrm{H}}_2\mathrm{O}$ is a weak-field ligand, giving high-spin $t_{2g}^3{e}_g^1$.

Only (C) and (D) are both homoleptic and low spin.