Question

[Co(NH\(_3\))\(_{4}\)(NO\(_2\))\(_{2}\)]Cl exhibits.
(A) Ionisation isomerism
(B) Linkage isomerism
(C) Geometrical isomerism
(D) Coordination isomerism
(E) Solvate isomerism
Choose the correct answer from the options given below:

• A. (A) and (B) only
• B. (B) and (C) only
• C. (A), (B) and (C) only
• D. (B), (D) and (E) only

Hint:

To check for isomerism in a coordination compound, follow a checklist: 1. Is there an ambidentate ligand (like NO\(_2\), SCN)? \(\rightarrow\) Linkage. 2. Is there a counter-ion that can also act as a ligand? \(\rightarrow\) Ionisation. 3. Does the geometry (e.g., MA\(_4\)B\(_2\), MA\(_3\)B\(_3\), M(AA)\(_2\)B\(_2\)) allow for different spatial arrangements? \(\rightarrow\) Geometrical.

Answer 1

The Correct Option is C

Step 1: Core Idea:
Isomers share the same chemical formula but differ in atom arrangement. We'll analyze \([Co(NH_3)_4(NO_2)_2]Cl\) for isomerism types.
Step 2: Detailed Breakdown:
(A) Ionisation Isomerism: This involves a ligand inside the coordination sphere swapping places with a counter-ion outside. In this complex, NO\(_2^-\) and Cl\(^-\) can switch, forming \([Co(NH_3)_4(NO_2)Cl]NO_2\). Thus, it's possible.
(B) Linkage Isomerism: This arises from ambidentate ligands, which can bind through different atoms. The nitro ligand (NO\(_2^-\)) is ambidentate. It can bond via nitrogen (-NO\(_2\), nitro) or oxygen (-ONO, nitrito). Therefore, an isomer like \([Co(NH_3)_4(ONO)_2]Cl\) can exist. Linkage isomerism is possible.
(C) Geometrical Isomerism: A type of stereoisomerism. The complex has an octahedral geometry, MA\(_4\)B\(_2\) (M=Co, A=NH\(_3\), B=NO\(_2\)). The two 'B' ligands can be adjacent (90\(^\circ\)), forming the cis isomer, or opposite (180\(^\circ\)), forming the trans isomer. Therefore, geometrical isomerism is possible.
(D) Coordination Isomerism: This occurs when both cation and anion are complex ions, exchanging ligands. Here, the anion is Cl\(^-\), so this type is not possible.
(E) Solvate Isomerism: A form of ionisation isomerism involving solvent molecules. No solvent is present, so it's not applicable.
Step 3: Conclusion:
The complex exhibits ionisation (A), linkage (B), and geometrical (C) isomerism. The correct option is (3).