Question

The number of geometrical isomers for octahedral ${\left[CoC{l}_4{\left(N{H}_3\right)}_2\right]}^{-}$, square planar ${\left[AuB{r}_{2}C{l}_2\right]}^{-}$ and
$\left[PtC{l}_2\left(en\right)\right]$ are :

• A. 2, 2, 2
• B. 2, 2, no isomerism
• C. 3, 2, 2
• D. 2, 3, no isomerism

Answer 1

The correct option is B 2, 2, no isomerism

The octahedral complex ${\left[CoC{l}_4{\left(N{H}_3\right)}_2\right]}^{-}$ shows two geometrical isomers. In the cis isomer, the two ammonia ligands are adjacent to each other and in the trans isomer, they are opposite to each other.
The square planar complex ${\left[AuB{r}_{2}C{l}_2\right]}^{-}$ shows two geometrical isomers. In the cis isomer, the two chloride ligands are adjacent to each other and in the trans isomer, they are opposite to each other.
The geometrical isomers are not possible for the complex $\left[PtC{l}_2\left(en\right)\right]$. Both the chloride ligands are adjacent to each other.