Calculate CFSE of light pink compound formed, when KMnO4 is reduced by acidified H2S. — Coordination Compounds Chemistry Question

💡 Solution & Explanation

Step 1: Diamagnetic complexes contain zero unpaired electrons (all d-electrons are paired). Analyse each option using CFT. Step 2: (I) K4[Fe(CN)6] contains Fe^2+ (3d^6) with strong-field CN^-, which pairs all electrons (t_2g^6 eg^0, n = 0). (II) K3[Cr(CN)6] contains Cr^3+ (3d^3), which always has 3 unpaired electrons (paramagnetic). (III) K3[Co(CN)6] has Co^3+ (3d^6) with CN^-, forming a low-spin paired complex (t_2g^6 eg^0, n = 0). Step 3: (IV) K2[Ni(CN)4] features Ni^2+ (3d^8) with CN^-, forming a square planar dsp^2 hybridized complex with all 8 electrons paired (n = 0). Therefore, I, III, and IV are diamagnetic, matching option (b).