How many geometrical isomers will [Pt(NH_3)(N | vedclass

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(B) The complex $[Pt(NH_3)(NH_2OH)(H_2O)(py)]^+$ is a square planar complex of the type $[Mabcd]$,where $M = Pt$ and $a, b, c, d$ are four different l

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$3$

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(B) The complex $[Pt(NH_3)(NH_2OH)(H_2O)(py)]^+$ is a square planar complex of the type $[Mabcd]$,where $M = Pt$ and $a, b, c, d$ are four different ligands $(NH_3, NH_2OH, H_2O, py)$.For a square planar complex of the type $[Mabcd]$,the number of geometrical isomers is $3$.These isomers correspond to the three possible pairs of ligands that can be trans to each other: $1$. $a$ trans to $b$ ($c$ trans to $d$)$2$. $a$ trans to $c$ ($b$ trans to $d$)$3$. $a$ trans to $d$ ($b$ trans to $c$)Thus,there are $3$ geometrical isomers.

How many geometrical isomers will $[Pt(NH_3)(NH_2OH)(H_2O)(py)]^+$ form?

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