The total number of isomers for square planar | vedclass

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Question

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

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) For a square planar complex of the type $[M(abcd)]$,where $a, b, c,$ and $d$ are four different ligands,the number of geometrical isomers is $3$.These isomers correspond to the arrangements where one ligand is fixed,and the other three are permuted relative to it.Let $M = Pd$,$a = H_2O$,$b = Br$,$c = NH_3$,and $d = CN$.Fixing $H_2O$ at one position,the possible arrangements for the other three ligands $(Br, NH_3, CN)$ are:$1$. $Br$ trans to $NH_3$ (with $CN$ trans to $H_2O$)$2$. $Br$ trans to $CN$ (with $NH_3$ trans to $H_2O$)$3$. $Br$ trans to $H_2O$ (with $NH_3$ trans to $CN$)Thus,there are $3$ geometrical isomers.

The total number of isomers for square planar complex $[Pd(H_2O)(Br)(NH_3)(CN)]$ is/are

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