The pair `[Cr(H_(2)O)_(6)]CI_(3)` and `[Cr(H_(2)O)_(4)CI_(2)]CI.H_(2)O` shows ionisation isomerism .

Answer

To determine whether the pair \([Cr(H_2O)_6]Cl_3\) and \([Cr(H_2O)_4Cl_2]Cl \cdot H_2O\) shows ionization isomerism, we need to analyze the compounds and their ionization behavior.

### Step-by-Step Solution:

1. **Identify the Coordination Compounds**:
- The first compound is \([Cr(H_2O)_6]Cl_3\).
- The second compound is \([Cr(H_2O)_4Cl_2]Cl \cdot H_2O\).

2. **Understand Ionization Isomerism**:
- Ionization isomerism occurs when two or more coordination compounds can produce different ions in solution. This typically involves the presence of different anions or cations that can dissociate from the coordination sphere.

3. **Analyze the First Compound**:
- For \([Cr(H_2O)_6]Cl_3\):
- When it ionizes, it dissociates into \(Cr^{3+}\) and \(3Cl^-\) ions.
- The ions produced are \(Cr^{3+}\) and \(Cl^-\).

4. **Analyze the Second Compound**:
- For \([Cr(H_2O)_4Cl_2]Cl \cdot H_2O\):
- When it ionizes, it dissociates into \(Cr^{3+}\), \(2Cl^-\), and \(1Cl^-\) (from the water of crystallization).
- The ions produced are also \(Cr^{3+}\) and \(Cl^-\).

5. **Compare the Ionization Products**:
- In both cases, the ions produced are \(Cr^{3+}\) and \(Cl^-\).
- Since both compounds produce the same ions upon ionization, they do not exhibit ionization isomerism.

6. **Conclusion**:
- Therefore, the statement that the pair \([Cr(H_2O)_6]Cl_3\) and \([Cr(H_2O)_4Cl_2]Cl \cdot H_2O\) shows ionization isomerism is **false**.