Question

Which of the following will give a pair of enantiomorphs?
A.$\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{l_6}} \right]$
B.$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_{_2}}} \right]N{O_2}$
C.$\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]\left[ {Co{{\left( {CN} \right)}_6}} \right]$
D.$\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]{\text{ }}\left( {en = N{H_2}C{H_2}C{H_2}N{H_2}} \right)$

Answer 1

Hint:Enantiomorphs are optical isomers which are commonly known to us as enantiomers. Enantiomers or enantiomorphs are coordination compounds that have the same molecular formula but differ in their abilities to rotate directions of plane polarized light.

Complete step by step solution:
For a compound to show optical isomerism, it is essential that it does not have a plane of symmetry in its structure. Thus, square planar structures are very rarely optically active since they have an axis of symmetry.
Considering option A, that is $\left[ {Pt{{\left( {N{H_3}} \right)}_4}} \right]\left[ {PtC{l_6}} \right]$
The given coordination complex has two different coordination spheres and thus exhibits coordination isomerism. But, since all the ligands are the same, that is, either ammine or chloro group, thus, the complex will not exhibit optical isomerism.
Considering option B, that is $\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_{_2}}} \right]N{O_2}$
The given coordination compound has different ligands present and the central metal ion has a coordination number of 6. The shape of the complex is octahedral. But there exists a plane of symmetry in the molecule and thus, it does not form a pair of enantiomorphs.
Considering option C, that is $\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]\left[ {Co{{\left( {CN} \right)}_6}} \right]$.
The given coordination complex has two different coordination spheres and thus exhibits coordination isomerism. But, since all the ligands are the same, that is, either ammine or cyano group, thus, the complex will not exhibit optical isomerism.
Considering option D, that is $\left[ {Co{{\left( {en} \right)}_2}C{l_2}} \right]$
The given coordination compound has different ligands present and the central metal ion has a coordination number of 6. The shape of the complex is octahedral. There are no planes of symmetry present in the cis configuration of the molecule. It can have a non-super imposable mirror image and thus, will give a pair of enantiomorphs.

Hence, the correct answer is D.

Note:
Enantiomers are a pair of molecules which are non-super imposable mirror images of each other. They possess the property of chirality. Optical isomers of a compound have identical physical and chemical properties. Those isomers which rotate the plane of polarized light to the left side are called as laevo rotatory and those isomers which rotate the plane of polarized light to right direction are known as dextro rotatory.