Question

The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is (are):
A. $[Cr{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}$ and $[Cr{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl$
B. ${{[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]}^{+}}$ and ${{[Pt{{(N{{H}_{3}})}_{2}}({{H}_{2}}O)Cl]}^{+}}$
C. ${{[CoB{{r}_{2}}C{{l}_{2}}]}^{2-}}$ and ${{[PtB{{r}_{2}}C{{l}_{2}}]}^{2-}}$
D. $[Pt{{(N{{H}_{3}})}_{3}}(N{{O}_{3}})]Cl$ and $[Pt{{(N{{H}_{3}})}_{3}}Cl]Br$

Answer 1

To find out the correct option or options of the given question, we should refer to different types of isomerism. Remember, the compounds given should be isomer not among themselves but they should show the same type of isomerism. So, look out for which compounds are having the same type of isomerism.

Complete step by step answer:
Before proceeding to find the appropriate option or options, let us look into what is isomerism and what are the types of isomerism.
Two or more compounds that have the same chemical formula but a different arrangement of atoms is known as isomers. Due to this difference in the arrangement of atoms, coordination compounds predominantly exhibit two types of isomerism namely, stereo-isomerism and structural isomerism. It has different types like optical, geometrical, linkage, coordination, ionisation, solvate and ligand isomerism.
So, let us look on each option and state what type of isomerism they have,

In first option, $[Cr{{(N{{H}_{3}})}_{5}}Cl]C{{l}_{2}}$ complex does not have any replaceable ions or ambidentate ligands. And it has only one chlorine within the complex, so no optical isomerism is seen. So, it will not show any structural isomerism and stereoisomerism. While in $[Cr{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]Cl$, because of presence of two chlorine within the complex, it will show geometrical isomerism. So, we can infer that both the complexes do not have the same isomerism.

In the second option, both the complexes will not show any structural isomerism, as it does not have any ambidentate ligand or any replaceable ions. But they will show geometrical isomerism as ${{[Co{{(N{{H}_{3}})}_{4}}C{{l}_{2}}]}^{+}}$ is octahedral and ${{[Pt{{(N{{H}_{3}})}_{2}}({{H}_{2}}O)Cl]}^{+}}$ is square planar (Octahedral and square planar will show geometrical isomerism). So, both the complexes will have the same type of isomerism.

In third option, ${{[CoB{{r}_{2}}C{{l}_{2}}]}^{2-}}$ is tetrahedral and it will not show any structural isomerism and stereoisomerism. While ${{[PtB{{r}_{2}}C{{l}_{2}}]}^{2-}}$ is tetrahedral and it will show geometrical isomerism but no structural isomerism. Thus, both complexes are not the same.

Finally, in the fourth option in $[Pt{{(N{{H}_{3}})}_{3}}(N{{O}_{3}})]Cl$ complex, $N{{O}_{3}}$ and $Cl$ are replaceable. So, it will show structural isomerism (i.e. ionization isomerism) while in $[Pt{{(N{{H}_{3}})}_{3}}Cl]Br$ complex, chlorine and bromine are replaceable. So, it will also show ionization isomerism, which is a type of structural isomerism. So, both the complexes have the same type of isomerism.
So, the correct answer is “Option B and D”.

Note: Try to remember different types of isomerism. Isomers forming non-superimposable mirror images are referred to as optical isomerism. Square planar and octahedral structures can only form geometrical isomerism. Complexes having ligand group can show linkage isomerism while when there is interchange of ligands between cationic and anionic entities of different metal in a coordination compound, it is referred to as coordination isomerism. When a counter ion in a complex salt which acts as a potential ligand replaces the ligand, it is said to be the ionisation isomerism.