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(B) For a square planar complex of the type $[M(abcd)]$,where $a, b, c, d$ are four different ligands,the number of geometrical isomers is $3$.In this

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$3$

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(B) For a square planar complex of the type $[M(abcd)]$,where $a, b, c, d$ are four different ligands,the number of geometrical isomers is $3$.In this case,the complex is $[Pt(NH_3)(Br)(Cl)(Py)]^{2+}$.We can fix one ligand (e.g.,$NH_3$) at one position and then arrange the other three ligands $(Br, Cl, Py)$ in the remaining three positions.$1$. $NH_3$ trans to $Br$ (with $Cl$ and $Py$ trans to each other).$2$. $NH_3$ trans to $Cl$ (with $Br$ and $Py$ trans to each other).$3$. $NH_3$ trans to $Py$ (with $Br$ and $Cl$ trans to each other).Thus,there are $3$ possible geometrical isomers.

How many geometrical isomers are possible for the square planar complex $[Pt(NH_3)(Br)(Cl)(Py)]^{2+}$?

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