Question

Among $\mathrm{A}.-\mathrm{D}.$, the complexes that can display geometrical isomerism are:

$\mathrm{A}.{\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_3\mathrm{Cl}\right]}^{+}$

$\mathrm{B}.\left[\mathrm{Pt}\right({\mathrm{NH}}_3\left){\mathrm{Cl}}_5\right]$

$\mathrm{C}.\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_2\mathrm{Cl}\right({\mathrm{NO}}_2\left)\right]$

$\mathrm{D}.[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4\mathrm{ClBr}{]}^{2+}$

• A. $\mathrm{A}.\mathrm{and}\mathrm{B}.$
• B. $\mathrm{D}.\mathrm{and}\mathrm{A}.$
• C. $\mathrm{C}.\mathrm{and}\mathrm{D}.$
• D. $\mathrm{B}.\mathrm{and}\mathrm{C}.$

Answer 1

The correct option is C

$\mathrm{C}.\mathrm{and}\mathrm{D}.$

Explanation for correct option:

(C) $\mathbf{C}\mathbf{.}\mathbf{}\mathbf{and}\mathbf{}\mathbf{D}\mathbf{.}$

1. Geometrical isomerism is a type of isomerism in which two compounds have same molecular formula but differ in relative arrangement of atoms.
2. The complexes of type ${\mathrm{Ma}}_4\mathrm{bc}$ and ${\mathrm{Ma}}_2\mathrm{bc}$ or ${\mathrm{Ma}}_2{\mathrm{b}}_2$ can show geometrical isomerism, provided that ${\mathrm{Ma}}_2\mathrm{bc}$ or ${\mathrm{Ma}}_2{\mathrm{b}}_2$ are square planar.
3. The complexes in option A are this type so they will show geometrical isomerism, this can be illustrated from the following diagrams:

Explanation for incorrect options:

(A) $\mathbf{A}\mathbf{.}\mathbf{}\mathbf{and}\mathbf{}\mathbf{B}\mathbf{.}$

The compounds $A.andB.$ do not show geometrical isomerism as these compounds do not have ${\mathrm{Ma}}_4\mathrm{bc}$ or ${\mathrm{Ma}}_2\mathrm{bc}$ structures.

(B) $\mathit{D}\mathbf{.}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{A}\mathbf{.}$

Compound $\mathrm{A}.$ do not show geometrical isomerism as it do not show${\mathrm{Ma}}_4\mathrm{bc}$ or ${\mathrm{Ma}}_2\mathrm{bc}$ structure.

(D) $\mathbf{B}\mathbf{.}\mathbf{}\mathbf{and}\mathbf{}\mathbf{C}\mathbf{.}$

Compound $\mathrm{B}.$ do not show geometrical isomerism as it do not have ${\mathrm{Ma}}_4\mathrm{bc}$ or ${\mathrm{Ma}}_2\mathrm{bc}$ structures.

Hence, option A is correct. Compounds $\mathbf{C}\mathbf{.}\mathbf{}\mathbf{and}\mathbf{}\mathbf{D}\mathbf{.}$ shows geometrical isomers.