Ionization Isomer of Chromium Coordination Complex

To determine the ionization isomer of the given coordination complex:

1. Definition of Ionization Isomerism:
Ionization isomerism arises when the counter-ion in a complex salt acts as a ligand, displacing a ligand from the coordination sphere, which then becomes the counter-ion. Consequently, the two isomers yield different ions in aqueous solution despite having the same overall empirical formula.

2. Analysis of the Given Complex:
The given coordination compound is:
$[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{NO}_2)]\text{Cl}$

Upon dissociation in an aqueous solution, it produces a chloride ion ($\text{Cl}^-$) as the counter-ion:
$[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{NO}_2)]\text{Cl} \xrightarrow{\text{H}_2\text{O}} [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{NO}_2)]^+ + \text{Cl}^-$

3. Formation of the Ionization Isomer:
Exchanging the ion outside the coordination sphere ($\text{Cl}^-$) with the nitrite ligand ($\text{NO}_2^-$) inside the coordination sphere gives:
$[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2](\text{NO}_2)$

Upon dissociation in water, this isomer yields a nitrite ion ($\text{NO}_2^-$):
$[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2](\text{NO}_2) \xrightarrow{\text{H}_2\text{O}} [\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]^+ + \text{NO}_2^-$

Since both complexes have the same molecular formula but produce different ions in solution, $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2](\text{NO}_2)$ is the ionization isomer.

(Note on other options:)

• $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}(\text{ONO})]\text{Cl}$ represents a linkage isomer.
• $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2(\text{NO}_2)]\cdot\text{H}_2\text{O}$ represents a hydrate/solvate isomer.

Therefore, the correct option is B.