JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Coordination Compounds

For X, $[\mathrm{Pt}(\mathrm{NH}{}_3)(\mathrm{H}{}_2\mathrm{O})\mathrm{BrCl}]$ is square planar with four different ligands; a square-planar $Mabcd$ complex shows three geometrical isomers, so $X=3$. For Y, $[\mathrm{CrCl}{}_2(\mathrm{ox}){}_2{]}^{3-}$ is octahedral with two chloride and two bidentate oxalate ligands: the trans form has a plane of symmetry and is optically inactive, while the cis form is chiral and optically active, so only one isomer is optically inactive and $Y=1$. For Z, $[\mathrm{Co}(\mathrm{NH}{}_3){}_3(\mathrm{NO}{}_2){}_3]$ is an octahedral $M{a}_3{b}_3$ complex that exists as facial and meridional isomers, so $Z=2$. Adding these, $X+Y+Z=3+1+2=6$.