Coordination Compounds Part 4

Concept:Geometrical isomerism in square planar and octahedral coordination complexes.Explanation:For $X$: The complex $[\mathrm{Pt}(\mathrm{NH}_3)(\mathrm{H}_2\mathrm{O})\mathrm{BrCl}]$ is square planar $\mathrm{Pt(II)}$ with four different ligands ($\mathrm{MABCD}$ type).It shows 3 geometrical isomers based on distinct trans pairings.Thus $X = 3$.For $Y$: $[\mathrm{CrCl}_2(\mathrm{ox})_2]^{3-}$ is octahedral with two bidentate oxalate ligands ($\mathrm{AA}$) and two $\mathrm{Cl}$ ligands ($\mathrm{B}_2$), i.e. $[\mathrm{M(AA)}_2\mathrm{B}_2]$ type.It exhibits cis and trans isomers.The cis isomer is optically active (exists as $\Delta$ and $\Lambda$ enantiomers), while the trans isomer is optically inactive.Thus optically inactive isomers = 1 (the trans isomer).So $Y = 1$.For $Z$: $[\mathrm{Co}(\mathrm{NH}_3)_3(\mathrm{NO}_2)_3]$ is octahedral $[\mathrm{MA}_3\mathrm{B}_3]$ type.It shows facial (fac) and meridional (mer) geometrical isomers, giving 2 isomers.Thus $Z = 2$.$X + Y + Z = 3 + 1 + 2 = 6$.Answer:$6$