Question

Which of the following compounds show optical isomerism?

• A. $[\mathrm{Cu}{({\mathrm{NH}}_3)}_4{]}^{2+}$
• B. ${\left[{\mathrm{ZnCl}}_4\right]}^{2-}$
• C. $[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3{]}^{3-}$
• D. $[\mathrm{Co}{\left(\mathrm{CN}\right)}_6{]}^{3-}$

Answer 1

The correct option is C

$[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3{]}^{3-}$

The explanation for the correct option

(C) $\mathbf{[}\mathbf{Cr}{\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{}{\mathbf{]}}^{\mathbf{3}\mathbf{-}}$

• For a compound to show optical isomerism, its mirror image should be non-superimposable on it.
• ${\mathrm{C}}_2{{\mathrm{O}}_4}^{2-}$ (oxalate) ion is a bidentate ligand. So its mirror image is not superimposable.

• Thus option C is correct.

The explanation for incorrect options

(A) $[\mathrm{Cu}{({\mathrm{NH}}_3)}_4{]}^{2+}$

• Here all the ligands are the same and monodentate.
• Thus mirror images will be superimposable on each other.

• Thus option A is incorrect.

(B) ${\left[{\mathrm{ZnCl}}_4\right]}^{2-}$

• All ligands are the same here.
• ${\left[{\mathrm{ZnCl}}_4\right]}^{2-}$ is tetrahedral.
• Since the compound is tetrahedral and the ligands attached are the same, this compound also will not show optical isomerism.
• Hence option B is incorrect

(D) $[\mathrm{Co}{\left(\mathrm{CN}\right)}_6{]}^{3-}$

• Here also, all the ligands are the same
• Therefore the mirror image will be superimposable.

• Therefore $[\mathrm{Co}{\left(\mathrm{CN}\right)}_6{]}^{3-}$ will not show optical isomerism.
• Hence option D is incorrect.

Thus the correct option is (C) $\mathbf{[}\mathbf{Cr}{\mathbf{\left(}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{}{\mathbf{]}}^{\mathbf{3}\mathbf{-}}$