Question

The no. of geometrical isomer(s) possible for $[CoC{l}_{2}B{r}_2{]}^{2-}$ (tetrahedral) is/are :

Answer 1

The correct answer is 0. $[CoC{l}_{2}B{r}_2{]}^{2-}$ is tetrahedral complex ion, it has no geometrical isomers.

In tetrahedral complex, the relative position of the unidentate ligand attached to the central metal atom are the same with respect to each other and lacks plane of symmetry.