Question

Which of the following do not show geometrical isomerism ? (Assume all ligands are unidentate )

• A. Square planar $\left[MX{L}_3\right]$
• B. Square planar $\left[M{X}_2{L}_2\right]$
• C. Octahedral $\left[M{X}_2{L}_4\right]$
• D. Octahedral $\left[M{X}_3{L}_3\right]$

Answer 1

The correct option is A Square planar $\left[MX{L}_3\right]$

In square planar complexes, the type $\left[MX{L}_3\right]$ does not have any isomers as there is no pair of ligand (X) that can be arranged adjacent to or opposite to each other to form cis or trans respectively. All possible combinations of square planar $\left[MX{L}_3\right]$ result in the same spatial arrangement and hence cannot exhibit geometrical isomerism.
Hence, option A is the correct answer.