Question

The number of geometric isomers that can exist for square planar ${\left[Pt(Cl)(py)\left(N{H}_3\right)\left(N{H}_{2}OH\right)\right]}^{+}$is________ (py = pyridine)

Answer 1

Answer: 3

For square planar complexes of the type $[Mabcd{]}^{n\pm }$ where all the four ligands are different, three geometrical isomers are possible.
So, for the given square planar complex ${\left[Pt(Cl)(py)\left(N{H}_3\right)\left(N{H}_{2}OH\right)\right]}^{+}$, following three geometrical isomers are possible:

where $a=Cl,b=py,c=N{H}_3$ and $d=N{H}_{2}OH$.