Question

Identify the ionization isomer of \([\text{Cr}(\text{H}_2\text{O})_4\,\text{Cl}(\text{NO}_2)]\,\text{Cl}\) from the following.

• A. \([\text{Cr}(\text{H}_2\text{O})_4\,(\text{O}_2\text{N})]\,\text{Cl}_2\)
• B. \([\text{Cr}(\text{H}_2\text{O})_4\,\text{Cl}_2]\,(\text{NO}_2)\)
• C. \([\text{Cr}(\text{H}_2\text{O})_4\,\text{Cl}(\text{ONO})]\,\text{Cl}\)
• D. \([\text{Cr}(\text{H}_2\text{O})_4\,\text{Cl}_2(\text{NO}_2)]\,\text{H}_2\text{O}\)

Hint:

Ionization isomers swap an ion between the coordination sphere and the counter ion.

Answer 1

The Correct Option is B

Step 1: Understand ionization isomerism
Ionization isomers have the same formula but give different ions in solution, because a ligand and the counter ion are exchanged.

Step 2: Original complex
\([Cr(H_2O)_4Cl(NO_2)]Cl\) has \(Cl^-\) and \(NO_2^-\) in the sphere and a free \(Cl^-\) outside.

Step 3: Exchange
Swap the outside \(Cl^-\) with the inside \(NO_2^-\). The result is \([Cr(H_2O)_4Cl_2]NO_2\), option (B).

Step 4: Why others are wrong
(A) has a different formula, with a different count of chlorine. (C) is a linkage isomer, because \(ONO\) is the nitrito form of the ligand. (D) has a different number of water molecules and a different formula.

Final Answer:
Option B is the ionization isomer. \[ \boxed{\text{(B)}\ [Cr(H_2O)_4Cl_2]NO_2} \]