When one mole `CrCl_3 .6H_2O` is treated with excess of `AgNo_3`, 3 mole of AgCl is obtained. The formula of complex is

Options

A. `[Cr(H_2O)_5Cl]Cl_2.H_2O`
B. `[Cr(H_2O)_3Cl_3]3H_2O`
C. `[Cr(H_2O)_4Cl_2]Cl.2H_2O`
D. `[Cr(H_2O)_6]Cl_3`

Answer

To determine the formula of the complex formed when one 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 produces `AgCl`. The reaction indicates that chlorine ions are being released, which will react with silver ions to form silver chloride (AgCl).

### Step 2: Analyze the Moles of AgCl Produced
According to the question, 3 moles of `AgCl` are produced from 1 mole of `CrCl3.6H2O`. This implies that 3 moles of chloride ions (Cl⁻) are released into the solution.

### Step 3: Identify the Chloride Ions in the Complex
In the complex `CrCl3.6H2O`, there are 3 chloride ions. Since we are producing 3 moles of `AgCl`, it indicates that all 3 chloride ions are ionizable. This means that they are not part of the coordination sphere of the complex but are instead outside of it.

### Step 4: Determine the Coordination Number
The coordination number of chromium in this complex is determined by the number of water molecules coordinated to it. Since there are 6 water molecules, we can conclude that these are the ligands attached to the chromium ion.

### Step 5: Write the Formula of the Complex
Based on the above analysis, we can write the formula of the complex as:
\[ \text{[Cr(H}_2\text{O)}_6] \text{Cl}_3 \]
This indicates that there are 6 water molecules coordinated to chromium and 3 chloride ions outside the coordination sphere.

### Final Answer
The formula of the complex is:
\[ \text{[Cr(H}_2\text{O)}_6] \text{Cl}_3 \]

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