How many geometrical isomers are shown by the | vedclass

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(B) The given complex is of the type $[Mabcd]$,where $M = Pt^{2+}$,and $a, b, c, d$ are four different monodentate ligands $(NH_3, NH_2OH, H_2O, Py)$.

Vedclass · Accepted Answer

$3$

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(B) The given complex is of the type $[Mabcd]$,where $M = Pt^{2+}$,and $a, b, c, d$ are four different monodentate 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 are formed by keeping one ligand fixed and changing the positions of the other three ligands relative to the fixed one.If we fix $a$,then $b, c, d$ can be arranged in $3$ ways: $1$. $b$ is trans to $a$ ($c$ and $d$ are trans to each other).$2$. $c$ is trans to $a$ ($b$ and $d$ are trans to each other).$3$. $d$ is trans to $a$ ($b$ and $c$ are trans to each other).Thus,there are $3$ geometrical isomers.

How many geometrical isomers are shown by the complex $[Pt(NH_3)(NH_2OH)(H_2O)(Py)]^+$?

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