Question

The freezing point depression of $0.001 \,M$ of $A_{x} B_{y}\left[F e(C N)_{6}\right]$ is $5.58 \times 10^{-3} \,K$. If the oxidation state of $Fe$ is $+2$ and $K_{f}=1.86 \,K\,kg\, mol ^{-1}$, then the total number of possibilities for different types of $A$ and $B$ cations are

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

The oxidation state of $Fe$ is $+2$ in the complex. On ionisation, reaction occur as follows :

Total number of moles of equilibrium
$=1-\alpha+4 \alpha+\alpha=1+4 \alpha$
$\therefore i=\frac{\text { Total number of moles at equili. }}{\text { Initial moles }}=\frac{1+4 \alpha}{1}$
$=1+4 \times 0.5=1+2=3$
Thus, the total number of possibilities for different types of $A$ and $B$ cations are $3$ .