`PtCl_(4). 6H_(2)O` can exist as hydrated complex 1 molal aq. Solution has depression in freezing point of `3.72^(@)` . Assume `100%` ionisation and `K_(f)(H_(2)O) = 1.86^(@) mol^(-1)` kg , then complex is :

Options

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

Answer

To solve the problem, we need to determine the number of ions produced when the hydrated complex PtCl4·6H2O dissociates in solution. We will use the depression in freezing point to find the van 't Hoff factor (i) and then identify the correct hydrated complex.

### Step-by-Step Solution:

1. **Identify the given data:**
- Depression in freezing point (ΔTf) = 3.72 °C
- Freezing point depression constant (Kf) for water = 1.86 °C kg/mol
- The solution is 1 molal.

2. **Use the formula for depression in freezing point:**
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- ΔTf = depression in freezing point
- i = van 't Hoff factor (number of particles the solute dissociates into)
- Kf = freezing point depression constant
- m = molality of the solution

3. **Substitute the known values into the formula:**
\[ 3.72 = i \cdot 1.86 \cdot 1 \]

4. **Solve for i:**
\[ i = \frac{3.72}{1.86} = 2 \]

5. **Interpret the value of i:**
The van 't Hoff factor (i) of 2 indicates that the complex dissociates into 2 particles in solution.

6. **Analyze the dissociation of the complex:**
The hydrated complex PtCl4·6H2O can dissociate as follows:
\[ \text{PtCl}_4^{2-} + 2\text{Cl}^- \rightarrow \text{Pt}^{2+} + 4\text{Cl}^- \]
In this case, if we consider the dissociation:
- 1 PtCl4 ion produces 1 Pt2+ ion and 4 Cl- ions, giving a total of 5 ions (i = 5), which does not match our value of i.
- Alternatively, if we consider the complex dissociating to produce 1 PtCl2 ion and 2 Cl- ions, we get:
\[ \text{PtCl}_2^{2-} + 2\text{Cl}^- \rightarrow \text{Pt}^{2+} + 2\text{Cl}^- \]
This gives us a total of 3 ions (i = 3), which is also not correct.

7. **Find the correct dissociation:**
The only way to achieve i = 2 is if the complex dissociates into:
\[ \text{PtCl}_2 + 2\text{Cl}^- \rightarrow \text{Pt}^{2+} + 2\text{Cl}^- \]
Here, we have 1 PtCl2 ion and 2 Cl- ions, leading to a total of 3 ions (i = 3), which is still incorrect.

8. **Conclusion:**
Based on the calculations, the only feasible dissociation that gives i = 2 is:
\[ \text{PtCl}_4^{2-} \rightarrow \text{Pt}^{2+} + 2\text{Cl}^- \]
Therefore, the complex that exists is **PtCl2·6H2O**.