The complex,[Pt(Py)(NH_3)BrCl],will have how | vedclass

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(A) The complex $[Pt(Py)(NH_3)BrCl]$ is a square planar complex of the type $[M(abcd)]$,where $M = Pt$,$a = Py$,$b = NH_3$,$c = Br$,and $d = Cl$.For a

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$3$

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(A) The complex $[Pt(Py)(NH_3)BrCl]$ is a square planar complex of the type $[M(abcd)]$,where $M = Pt$,$a = Py$,$b = NH_3$,$c = Br$,and $d = Cl$.For a square planar complex of the type $[M(abcd)]$,there are $3$ possible geometrical isomers.These isomers are formed by fixing one ligand (e.g.,$Py$) and varying the positions of the other three ligands $(NH_3, Br, Cl)$ relative to it.Thus,the total number of geometrical isomers is $3$.

The complex,$[Pt(Py)(NH_3)BrCl]$,will have how many geometrical isomers?

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