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(D) $1$. $[PtCl_2Br_2]^{2-}$ is a square planar complex of type $[M(a)_2(b)_2]$. It has $2$ geometrical isomers (cis and trans).$2$. $[CoCl_2Br_2]^{2-

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$[Co(NH_3)_2(Gly)_2]^+$

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(D) $1$. $[PtCl_2Br_2]^{2-}$ is a square planar complex of type $[M(a)_2(b)_2]$. It has $2$ geometrical isomers (cis and trans).$2$. $[CoCl_2Br_2]^{2-}$ is a tetrahedral complex of type $[M(a)_2(b)_2]$. Tetrahedral complexes do not show geometrical isomerism.$3$. $[Pt(NH_3)(Gly)Br]$ is a square planar complex of type $[M(abcd)]$. It has $3$ geometrical isomers.$4$. $[Co(NH_3)_2(Gly)_2]^+$ is an octahedral complex of type $[M(a)_2(b)_2]$ where $b$ is a bidentate ligand. It has $5$ geometrical isomers (cis-cis-cis,cis-cis-trans,cis-trans-cis,trans-cis-cis,and trans-trans-trans).Therefore,the complex with the highest number of geometrical isomers is $[Co(NH_3)_2(Gly)_2]^+$,which has $5$ isomers.

Which of the following complex compounds has the highest number of geometrical isomers?

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