Question

Which of the following complex can show geometrical isomerism?

• A. $\left(\left[Pt \left(gly\right) \left(Cl\right)_{2}\right]\right)^{-}$
• B. $\left[Pt \left(en\right) ClBr\right]$
• C. $\left(\left[Pt \left(gly\right) ClBr\right]\right)^{-}$
• D. $\left[Pt \left(en\right) \left(Cl\right)_{2}\right]$

Answer 1

Answer: (C)

Geometrical Isomers is also known as "Cis-Trans" isomerism. It is observed in molecules where rotation is restricted such as in alkenes where cis and Trans isomers can be formed along the double bond. Coordination complexes also witness Cis and Trans isomers.
We see here that $Pt$ is attached to bidentate ligands' glycine (gly) and ethylenediamine (en). We must note that in bidentate ligands the adjacent donor atoms are always $90^{^\circ }$ to each other, they would never be $180^{^\circ }$ to each other since it would break the ring.
$Pt$ forms square planar complex and with reference to structure, Option $2\text{and}4$ can be discarded as no cis or Trans isomer can be formed as we have same adjacent donor atoms in "en" ligands. Even though if $ClandBr$ change their positions they would always be $90^{^\circ }$ to each other and their would be no change in the configuration.
Next we have option $1$ , even though gly has different donor atoms, but it has $2Cl$ ligands attached to $Pt$ which limits the formation of cis and Trans isomers.
Last option $3$ is correct as all the $4$ atoms attached to $Pt$ are different as it is of the form $\left[\right.M\left(\right.abcd\left.\right)\left]\right.$ . Changing the positions of ligands would form different Cis- Trans Isomers.