`1 xx 10^(-3)` m solution of `Pt(NH_(3))_(4)Cl_(4)` in `H_(2)O` shows depression in freezing point of `0.0054^(@)C`. The formula of the compound will be [Given `K_(f) (H_(2)O) = 1.86^(@)C m^(-1)`]

Options

A. `[Pt(NH_3)_4]Cl_4`
B. `[PtNH_3)_3Cl]Cl_3`
C. `[Pt(NH_3)_4 Cl_2]Cl_2`
D. `Pt(NH_3)Cl_3]Cl`

Answer

To solve the problem, we need to find the formula of the compound `Pt(NH₃)₄Cl₄` based on the given information about the depression in freezing point and the van't Hoff factor.

### Step-by-Step Solution:

1. **Understand the Formula for Depression in Freezing Point**:
The depression in freezing point (\( \Delta T_f \)) is given by the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]
where:
- \( \Delta T_f \) = depression in freezing point
- \( K_f \) = freezing point depression constant
- \( m \) = molality of the solution
- \( i \) = van't Hoff factor (number of particles the solute dissociates into)

2. **Substitute the Given Values**:
We know:
- \( \Delta T_f = 0.0054 \, ^\circ C \)
- \( K_f = 1.86 \, ^\circ C \, m^{-1} \)
- \( m = 1 \times 10^{-3} \, m \)

Plugging these values into the formula:
\[ 0.0054 = 1.86 \cdot (1 \times 10^{-3}) \cdot i \]

3. **Solve for the van't Hoff Factor (i)**:
Rearranging the equation to solve for \( i \):
\[ i = \frac{0.0054}{1.86 \times 1 \times 10^{-3}} \]
Calculating the right side:
\[ i = \frac{0.0054}{0.00186} \approx 2.903 \approx 3 \]

4. **Interpret the van't Hoff Factor**:
The van't Hoff factor \( i \) indicates the number of particles the solute dissociates into. Since \( i \approx 3 \), this means the compound dissociates into 3 ions in solution.

5. **Determine the Possible Formulas**:
The compound \( Pt(NH₃)₄Cl₄ \) can be analyzed:
- The coordination complex \( Pt(NH₃)₄ \) does not dissociate.
- The 4 chloride ions \( Cl^- \) are the counter ions that dissociate.

Therefore, the dissociation can be represented as:
\[ Pt(NH₃)₄Cl₄ \rightarrow Pt(NH₃)₄ + 4Cl^- \]
This results in 5 ions, which does not match our \( i \).

6. **Check Other Possible Formulas**:
We need to find a compound that gives exactly 3 ions upon dissociation.

- For \( Pt(NH₃)₄Cl_2 \):
\[ Pt(NH₃)₄Cl_2 \rightarrow Pt(NH₃)₄ + 2Cl^- \]
This gives 3 ions (1 + 2).

- For \( Pt(NH₃)Cl_3 \):
\[ Pt(NH₃)Cl_3 \rightarrow Pt(NH₃) + 3Cl^- \]
This gives 4 ions (1 + 3).

- For \( Pt(NH₃)Cl \):
\[ Pt(NH₃)Cl \rightarrow Pt(NH₃) + Cl^- \]
This gives 2 ions (1 + 1).

7. **Conclusion**:
The only formula that gives a van't Hoff factor of 3 (i.e., dissociates into 3 ions) is \( Pt(NH₃)₄Cl_2 \).

### Final Answer:
The formula of the compound is \( Pt(NH₃)₄Cl_2 \).