When 1 mol `CrCl_(3).6H_(2)O` is treated with excess of `AgNO_(3)`, 3 mol of `AgCl` are obtained. The formula of the coplex is

Options

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

Explanation

To determine the formula of the complex when 1 mol of CrCl₃·6H₂O is treated with excess AgNO₃, resulting in the formation of 3 mol of AgCl, we can follow these steps:

### Step 1: Understand the dissociation of CrCl₃·6H₂O
When CrCl₃·6H₂O is dissolved in water, it dissociates into its constituent ions. The dissociation can be represented as:
\[ \text{CrCl}_3 \cdot 6\text{H}_2\text{O} \rightarrow \text{Cr}^{3+} + 3\text{Cl}^- + 6\text{H}_2\text{O} \]

### Step 2: Analyze the formation of AgCl
When excess AgNO₃ is added to the solution, the chloride ions (Cl⁻) react with Ag⁺ ions to form AgCl precipitate. The reaction can be represented as:
\[ \text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl} \]

### Step 3: Determine the number of Cl⁻ ions involved
From the dissociation of CrCl₃·6H₂O, we see that there are 3 Cl⁻ ions produced. Since 3 moles of AgCl are formed, this indicates that all 3 Cl⁻ ions are available to react with Ag⁺ ions.

### Step 4: Identify the structure of the complex
Since all 3 Cl⁻ ions are outside the coordination sphere and can react with AgNO₃, we can conclude that the complex has 3 chloride ions as counter ions. The remaining part of the complex consists of Cr and 6 water molecules. Thus, the complex can be represented as:
\[ \text{[Cr(H}_2\text{O)}_6]^{3+} + 3\text{Cl}^- \]

### Step 5: Write the final formula of the complex
Combining the coordination complex and the counter ions, the formula of the complex is:
\[ \text{CrCl}_3 \cdot 6\text{H}_2\text{O} \]

### Conclusion
The formula of the complex is \(\text{[Cr(H}_2\text{O)}_6]^{3+} \text{Cl}^-_3\).

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