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 formed when 1 mole of CrCl₃·6H₂O is treated with excess AgNO₃, we can follow these steps:

### Step 1: Understand the Reaction
When CrCl₃·6H₂O is treated with excess AgNO₃, it leads to the formation of AgCl. The reaction can be summarized as:
\[ \text{CrCl}_3 \cdot 6\text{H}_2\text{O} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{other products} \]

### Step 2: Identify the Chloride Ions
From the question, we know that 3 moles of AgCl are formed. Since each mole of AgCl requires one mole of Cl⁻ ions, this means that 3 moles of Cl⁻ ions are involved in the reaction.

### Step 3: Determine the Chloride in the Complex
The original compound, CrCl₃·6H₂O, contains 3 chloride ions (Cl⁻). When treated with AgNO₃, all 3 chloride ions are released as AgCl. This indicates that all the chloride ions in CrCl₃·6H₂O are available for reaction with AgNO₃.

### Step 4: Write the Formula of the Complex
Since all 3 chloride ions are released and the chromium ion (Cr³⁺) remains coordinated with the water molecules, the formula of the complex can be represented as:
\[ \text{[Cr(H}_2\text{O)}_6]^{3+} \text{Cl}_3 \]

### Final Formula
Thus, the formula of the complex is:
\[ \text{[Cr(H}_2\text{O)}_6]^{3+} \text{Cl}_3 \]