A complex iron and cyanide ions is `100%` ionised at 1 m (molal) . If its elevation in b.p is 2.08 . Then the complex is `(K_(b)=0.52^(@) mol^(-1) kg)`:

Options

A. `K_(3)[Fe(CN)_(6)]`
B. `Fe(CN)_(2)`
C. `K_(4)[Fe(CN)_(6)]`
D. `Fe(CN)_(4)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
We are given:
- Boiling point elevation (ΔT_b) = 2.08 °C
- Molality (m) = 1 m (molal)
- Boiling point elevation constant (K_b) = 0.52 °C kg/mol

### Step 2: Use the formula for boiling point elevation
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \cdot m \cdot i \]
Where:
- ΔT_b = boiling point elevation
- K_b = ebullioscopic constant
- m = molality of the solution
- i = van 't Hoff factor (number of particles the solute dissociates into)

### Step 3: Rearrange the formula to find the van 't Hoff factor (i)
Rearranging the formula to solve for i:
\[ i = \frac{\Delta T_b}{K_b \cdot m} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ i = \frac{2.08}{0.52 \cdot 1} \]

### Step 5: Calculate the value of i
Calculating the value:
\[ i = \frac{2.08}{0.52} = 4 \]

### Step 6: Interpret the value of i
The van 't Hoff factor (i) is equal to 4, which means that the complex dissociates into 4 ions in solution.

### Step 7: Identify the complex
Now we need to find a complex that dissociates into 4 ions. The options provided are:
1. K₃[Fe(CN)₆]
2. Fe₂(CN)₆
3. Fe(CN)₆
4. K₄[Fe(CN)₆]

Let's analyze the first option:
- K₃[Fe(CN)₆] dissociates as follows:
\[ K₃[Fe(CN)₆] \rightarrow 3K^+ + [Fe(CN)₆]^{3-} \]
This results in a total of 4 ions (3 K⁺ ions + 1 [Fe(CN)₆]³⁻ ion).

### Conclusion
The complex that corresponds to the given data is **K₃[Fe(CN)₆]**.