A six coordination complex of formula `CrCl_(3)*6H_(2)O` has green colour. A 0.1 M solution of the complex when treated with excess of `AgNO_(3)` gave 28.7g of white precipitate. The formula of the complex would be:

Options

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

Answer

To determine the formula of the complex `CrCl3·6H2O` based on the information provided, we can follow these steps:

### Step 1: Calculate the moles of AgCl precipitate formed.
Given that the mass of the white precipitate (AgCl) formed is 28.7 g, we can calculate the number of moles of AgCl using its molar mass.

**Molar mass of AgCl:**
- Ag = 107.87 g/mol
- Cl = 35.45 g/mol
- Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol

**Calculating moles of AgCl:**
\[ \text{Moles of AgCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{28.7 \, \text{g}}{143.32 \, \text{g/mol}} \approx 0.200 \, \text{moles} \]

### Step 2: Relate moles of AgCl to the chloride ions in the complex.
Since each mole of AgCl corresponds to one mole of Cl⁻ ions, the formation of 0.200 moles of AgCl indicates that there are 0.200 moles of Cl⁻ ions released from the complex.

### Step 3: Determine the number of ionizable chloride ions.
The complex is a coordination compound with the formula `CrCl3·6H2O`. The total number of chloride ions in this formula is 3. However, we need to determine how many of these are ionizable.

Given that the solution is 0.1 M and produces 0.200 moles of AgCl, we can conclude that the complex must contain 2 ionizable chloride ions (Cl⁻) because the remaining chloride ion is likely involved in the coordination sphere.

### Step 4: Identify the coordination number and remaining water molecules.
The coordination number of the complex is 6. Since we have established that 2 chloride ions are ionizable, the remaining coordination sites must be occupied by water molecules. Therefore, we have:

- 2 Cl⁻ ions (ionizable)
- 4 H₂O molecules (coordinated)

This gives us a coordination sphere of 6, which is consistent with the coordination number.

### Step 5: Write the final formula for the complex.
Based on the analysis:
- The complex has 2 Cl⁻ ions outside the coordination sphere and 4 H₂O molecules coordinated to the chromium ion.

Thus, the formula of the complex can be written as:
\[ \text{Cr(H}_2\text{O})_4\text{Cl}_2\cdot 2\text{Cl}\cdot 6\text{H}_2\text{O} \]

### Conclusion:
The correct formula of the complex is:
\[ \text{[Cr(H}_2\text{O})_4\text{Cl}_2] \cdot 2\text{Cl} \cdot 6\text{H}_2\text{O} \]