JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Coordination Compounds

In triamminetrinitrocobalt(III), $[\mathrm{Co}(\mathrm{N}{\mathrm{H}}_3{)}_3(\mathrm{N}{\mathrm{O}}_2{)}_3]$, the octahedron carries three ligands of each kind. The three like ligands can occupy one triangular face or lie along a meridian, which gives the facial and meridional forms, so $X=2$.
In trioxalatochromate(III), $[\mathrm{Cr}({\mathrm{C}}_2{\mathrm{O}}_4{)}_3{]}^{3-}$, all three ligands are identical symmetrical bidentate ligands and each must span a pair of adjacent positions. Every such arrangement can be brought onto any other by rotating the octahedron, so no facial, meridional, cis or trans distinction survives and $Y=0$. The two forms this ion does possess are non-superimposable mirror images, which is optical isomerism and not geometrical.
Hence $X+Y=2$.