`PtCl_(4).6H_(2)O` can exist as a hydrated complex. `1 m` aqueous solution has the depression in freezing point of `3.72^(@)`. Assume `100%` ionization and `K_(f)(H_(2)O)=1.86^(@)mol^(-1) kg`, then the complex is

Options

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

Explanation

To solve the problem, we need to determine the nature of the hydrated complex of PtCl₄·6H₂O based on the given depression in freezing point of a 1 m aqueous solution.

### Step-by-Step Solution:

1. **Understand the Depression in Freezing Point**:
The depression in freezing point (\( \Delta T_f \)) is given as 3.72°C. The formula to calculate the depression in freezing point is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_f \) = cryoscopic constant of the solvent (for water, \( K_f = 1.86 \, \text{°C kg/mol} \))
- \( m \) = molality of the solution (given as 1 mol/kg)

2. **Substitute the Known Values**:
Substitute the known values into the equation:
\[ 3.72 = i \cdot 1.86 \cdot 1 \]

3. **Solve for \( i \)**:
Rearranging the equation to solve for \( i \):
\[ i = \frac{3.72}{1.86} = 2 \]

4. **Interpret the Value of \( i \)**:
The value of \( i = 2 \) indicates that the hydrated complex dissociates into 2 particles in solution. This means that the complex can release 2 ions when it ionizes.

5. **Determine the Composition of the Complex**:
The complex is \( \text{PtCl}_4 \cdot 6\text{H}_2\text{O} \). Given that it releases 2 ions, we can deduce that:
- The platinum ion \( \text{Pt}^{2+} \) contributes 1 ion.
- The complex must also release 1 chloride ion \( \text{Cl}^- \).

Therefore, the dissociation can be represented as:
\[ \text{PtCl}_4 \cdot 6\text{H}_2\text{O} \rightarrow \text{Pt}^{2+} + 2\text{Cl}^- + 6\text{H}_2\text{O} \]
However, since we only need 2 ions, we can conclude that one of the chlorides remains in the complex.

6. **Final Conclusion**:
The hydrated complex \( \text{PtCl}_4 \cdot 6\text{H}_2\text{O} \) exists as \( \text{[PtCl}_3\text{(H}_2\text{O)}_6]^{+} + \text{Cl}^- \).

### Final Answer:
The complex is \( \text{[PtCl}_3\text{(H}_2\text{O)}_6]^{+} + \text{Cl}^- \).