Form four complexes with four different ligands which are ,

CHEMISTRY · Single correct · CBSE · Chemistry · Coordination Compounds · Coordination Compounds

Problem

Cr3+\text{Cr}^{3+} form four complexes with four different ligands which are [CrCl6]3−[\text{CrCl}_6]^{3-}, [Cr(H2O)6]3+[\text{Cr}(\text{H}_2\text{O})_6]^{3+}, [Cr(NH3)6]3+[\text{Cr}(\text{NH}_3)_6]^{3+} and [Cr(CN)6]3−[\text{Cr}(\text{CN})_6]^{3-}. The order of CFSE (Δ0\Delta_0) in these complexes is in the order: (1) [CrCl6]3−=[Cr(H2O)6]3+=[Cr(NH3)6]3+=[Cr(CN)6]3−[\text{CrCl}_6]^{3-} = [\text{Cr}(\text{H}_2\text{O})_6]^{3+} = [\text{Cr}(\text{NH}_3)_6]^{3+} = [\text{Cr}(\text{CN})_6]^{3-} (2) [CrCl6]3−<[Cr(H2O)6]3+<[Cr(NH3)6]3+<[Cr(CN)6]3−[\text{CrCl}_6]^{3-} < [\text{Cr}(\text{H}_2\text{O})_6]^{3+} < [\text{Cr}(\text{NH}_3)_6]^{3+} < [\text{Cr}(\text{CN})_6]^{3-} (3) [CrCl6]3−>[Cr(H2O)6]3+>[Cr(NH3)6]3+>[Cr(CN)6]3−[\text{CrCl}_6]^{3-} > [\text{Cr}(\text{H}_2\text{O})_6]^{3+} > [\text{Cr}(\text{NH}_3)_6]^{3+} > [\text{Cr}(\text{CN})_6]^{3-} (4) [CrCl6]3−<[Cr(H2O)6]3+=[Cr(NH3)6]3+<[Cr(CN)6]3−[\text{CrCl}_6]^{3-} < [\text{Cr}(\text{H}_2\text{O})_6]^{3+} = [\text{Cr}(\text{NH}_3)_6]^{3+} < [\text{Cr}(\text{CN})_6]^{3-}

Related theory

Browse the problem library