Chemistry

Problem

\text{Cr}^{3+}

form four complexes with four different ligands which are

[\text{CrCl}_6]^{3-}

[\text{Cr}(\text{H}_2\text{O})_6]^{3+}

[\text{Cr}(\text{NH}_3)_6]^{3+}

and

[\text{Cr}(\text{CN})_6]^{3-}

. The order of CFSE (

\Delta_0

) in these complexes is in the order: a.

[\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-}

[\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-}

[\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-}

[\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-}