JEE Main 2025 — Coordination Compounds Question with Solution

(A) $[{\text{Fe(CN)}}_5\text{NO}{]}^{2-}$: This is a heteroleptic complex with iron in the ${\text{Fe}}^{2+}$ oxidation state, resulting in a $3d^6$ electronic configuration. It forms a low spin complex with electron configuration ${\text{t}}_{2g}^6{\text{e}}_g^0$ due to the strong field ligands (SFL).

(B) $[{\text{CoF}}_6{]}^{3-}$: This homoleptic complex contains cobalt in the ${\text{Co}}^{3+}$ oxidation state, leading to a $3d^6$ electron configuration. It is a high spin complex with an ${\text{sp}}^3{\text{d}}^2$ hybridization due to weak field ligands (WFL).

(C) $[{\text{Fe(CN)}}_6{]}^{4-}$: This homoleptic complex has iron in the ${\text{Fe}}^{2+}$ state, giving a $3d^6$ configuration. It forms a low spin complex with hybridization ${\text{d}}^2{\text{sp}}^3$ and electron configuration ${\text{t}}_{2g}^6{\text{e}}_g^0$, due to the strong field nature of the cyanide ligands.

(D) $[{\text{Co(NH}}_3{)}_6{]}^{3+}$: This homoleptic complex contains cobalt in the ${\text{Co}}^{3+}$ state, resulting in a $3d^6$ electron configuration. It forms a low spin complex with hybridization ${\text{d}}^2{\text{sp}}^3$ and configuration ${\text{t}}_{2g}^6{\text{e}}_g^0$ due to the strong field ligands.

(E) $[{\text{Cr(H}}_2\text{O}{)}_6{]}^{2+}$: This homoleptic complex with chromium in the ${\text{Cr}}^{2+}$ state results in a $3d^4$ electronic configuration. It is a high spin complex with hybridization ${\text{d}}^2{\text{sp}}^3$ and configuration ${\text{t}}_{2g}^3{\text{e}}_g^1$ due to weak field ligands.

In summary, the homoleptic complexes that are low spin are:

(C) $[{\text{Fe(CN)}}_6{]}^{4-}$

(D) $[{\text{Co(NH}}_3{)}_6{]}^{3+}$