The number of geometric isomers that can exis | vedclass

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(D) The complex $[Pt(Cl)(py)(NH_3)(NH_2OH)]^+$ is of the type $[M(abcd)]$,where $M=Pt$,$a=Cl^-$,$b=py$,$c=NH_3$,and $d=NH_2OH$.For a square planar com

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$3$

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(D) The complex $[Pt(Cl)(py)(NH_3)(NH_2OH)]^+$ is of the type $[M(abcd)]$,where $M=Pt$,$a=Cl^-$,$b=py$,$c=NH_3$,and $d=NH_2OH$.For a square planar complex of the type $[M(abcd)]$,the number of geometric isomers is given by $3$.These isomers are formed by fixing one ligand (e.g.,$Cl^-$) at one position and arranging the other three ligands ($py$,$NH_3$,$NH_2OH$) in the remaining three positions relative to the fixed ligand.Thus,there are $3$ possible geometric isomers.

The number of geometric isomers that can exist for the square planar complex $[Pt(Cl)(py)(NH_3)(NH_2OH)]^+$ is ($py=$ pyridine):

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