JEE Main 2025 Jan 22 Shift 2, Chemistry Q17: Colour, Magnetism And Stability Of Co-ordination Compounds

A homoleptic complex has only one type of ligand. A complex is low-spin if the ligand field splitting ${\Delta }_o$ is large enough to force pairing (typical for strong-field ligands on 3d metal ions, especially in higher oxidation states).

(A) $[Fe(CN{)}_{5}NO{]}^{2-}$: contains two different ligands ($C{N}^{-}$ and $NO$), hence heteroleptic. Rule it out for the homoleptic criterion.

(B) $[Co{F}_6{]}^{3-}$: $C{o}^{3+}$ ($d^6$) with $F^{-}$ (weak field) $\Rightarrow$ high spin.

(C) $[Fe(CN{)}_6{]}^{4-}$: $F{e}^{2+}$ ($d^6$) with $C{N}^{-}$ (strong field) $\Rightarrow$ homoleptic, low spin.

(D) $[Co(N{H}_3{)}_6{]}^{3+}$: $C{o}^{3+}$ ($d^6$) with $N{H}_3$. For $C{o}^{3+}$, even $N{H}_3$ produces a sufficiently large ${\Delta }_o$ to give pairing $\Rightarrow$ homoleptic, low spin.

(E) $[Cr(H_{2}O{)}_6{]}^{2+}$: $C{r}^{2+}$ ($d^4$) with $H_{2}O$ (weak field) $\Rightarrow$ high spin.

$\therefore$ The homoleptic low-spin complexes are (C) and (D).