Which of the following complexes produces three moles of silver chloride when its mole is treated with excess of silver nitrate ?

Options

A. `[Cr(H_(2)O)_(3)Cl_(3)]`
B. `[Cr(H_(2)O)_(4)Cl_(2)]Cl`
C. `[Cr(H_(2)O)_(5)Cl_(2)]Cl_(2)`
D. `[Cr(H_(2)O)_(6)]Cl_(3)`

Explanation

To solve the question of which complex produces three moles of silver chloride (AgCl) when treated with excess silver nitrate (AgNO3), we will analyze each complex option step by step.

### Step 1: Understanding the Coordination Complexes
We need to identify the coordination complexes and their dissociation into ions when treated with silver nitrate. The key is to find the number of chloride ions (Cl⁻) that can react with silver ions (Ag⁺) to form AgCl.

### Step 2: Analyzing Each Complex
1. **Complex 1**: Assume this complex does not contain any ionizable ions (e.g., Cl⁻). Therefore, it will not produce any AgCl when treated with AgNO3.
- **Result**: 0 moles of AgCl.

2. **Complex 2**: This complex has one Cl⁻ ion. When it dissociates, it produces one Cl⁻ ion that can react with one Ag⁺ ion from AgNO3.
- **Result**: 1 mole of AgCl.

3. **Complex 3**: This complex contains two Cl⁻ ions. Upon dissociation, it produces two Cl⁻ ions which can react with two Ag⁺ ions.
- **Result**: 2 moles of AgCl.

4. **Complex 4**: This complex has three Cl⁻ ions. When it dissociates, it produces three Cl⁻ ions that can react with three Ag⁺ ions from AgNO3.
- **Result**: 3 moles of AgCl.

### Step 3: Conclusion
From the analysis, we find that **Complex 4** produces three moles of silver chloride when treated with excess silver nitrate.

### Final Answer
The correct option is **Complex 4**.

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