Compound `PdCl_(4). 6H_(2)O` is a hydrated complex, 1 molal aqueous solution of it has freezing point 269.28 K. Assuming 100% ionization of complex, calculate the molecular formula of the complex (`K_(f)` for water = 1.86 K `mol^(-1)`)

Options

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

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Calculate the change in freezing point (ΔTf)
The freezing point of pure water is 273 K. The freezing point of the solution is given as 269.28 K.

\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 273 \, \text{K} - 269.28 \, \text{K} = 3.72 \, \text{K} \]

### Step 2: Use the freezing point depression formula
The formula for freezing point depression is:

\[ \Delta T_f = i \cdot K_f \cdot m \]

Where:
- \( \Delta T_f \) = change in freezing point
- \( i \) = van 't Hoff factor (number of particles the solute breaks into)
- \( K_f \) = freezing point depression constant for water = 1.86 K kg/mol
- \( m \) = molality of the solution = 1 mol/kg

Substituting the known values:

\[ 3.72 = i \cdot 1.86 \cdot 1 \]

### Step 3: Solve for the van 't Hoff factor (i)
Rearranging the equation to solve for \( i \):

\[ i = \frac{3.72}{1.86} \approx 2 \]

### Step 4: Determine the number of particles (n)
The van 't Hoff factor \( i \) is related to the number of particles \( n \) produced from the dissociation of the solute. For the complex \( PdCl_4 \), we can assume it dissociates into \( PdCl_3 \) and \( Cl^- \) ions.

Using the relation:

\[ i = \alpha \cdot (n - 1) + 1 \]

Where \( \alpha \) is the degree of dissociation. Assuming 100% ionization, \( \alpha = 1 \):

\[ 2 = 1 \cdot (n - 1) + 1 \]

This simplifies to:

\[ n - 1 = 1 \implies n = 2 \]

### Step 5: Identify the molecular formula of the complex
From the previous steps, we can deduce that the complex \( PdCl_4 \) dissociates to form \( PdCl_3 \) and one chloride ion. Since we also have 6 water molecules in the hydrated complex, we can write the molecular formula as:

\[ PdCl_3 \cdot 6H_2O \]

### Final Answer
The molecular formula of the complex is \( PdCl_3 \cdot 6H_2O \).

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