A complex of iron and cyanide ions is `100%`ionised ar `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)`

Explanation

To solve the problem, we need to determine the complex of iron and cyanide ions based on the given information about the boiling point elevation. Let's break down the solution step by step.

### Step 1: Understand the Formula for Boiling Point Elevation
The formula for boiling point elevation (\( \Delta T_b \)) is given by:
\[ \Delta T_b = K_b \cdot m \cdot i \]
where:
- \( \Delta T_b \) = elevation in boiling point
- \( K_b \) = ebullioscopic constant
- \( m \) = molality of the solution
- \( i \) = Van't Hoff factor (number of particles the solute dissociates into)

### Step 2: Identify Given Values
From the problem, we have:
- \( \Delta T_b = 2.08 \, \text{°C} \)
- \( K_b = 0.52 \, \text{°C kg/mol} \)
- \( m = 1 \, \text{mol/kg} \)

### Step 3: Rearrange the Formula to Find \( i \)
We can rearrange the formula to solve for \( i \):
\[ i = \frac{\Delta T_b}{K_b \cdot m} \]

### Step 4: Substitute the Values
Now, we can substitute the values into the equation:
\[ i = \frac{2.08}{0.52 \cdot 1} \]

### Step 5: Calculate \( i \)
Calculating the above expression:
\[ i = \frac{2.08}{0.52} = 4 \]

### Step 6: Interpret the Result
The Van't Hoff factor \( i = 4 \) indicates that the complex dissociates into 4 ions in solution.

### Step 7: Identify the Complex
Now we need to find a complex of iron that, when dissociated, gives a total of 4 ions.

1. **Option 1: \( K_3[Fe(CN)_6] \)**
- Dissociates into \( 3K^+ \) and \( [Fe(CN)_6]^{3-} \)
- Total ions = \( 3 + 1 = 4 \)

2. **Other Options (if provided)**
- We would check other complexes to see if they also yield 4 ions. However, since we have already found one that works, we can conclude.

### Conclusion
The complex that corresponds to the given boiling point elevation and dissociates into 4 ions is \( K_3[Fe(CN)_6] \).

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