Question

The freezing point of an aqueous solution of $KCN$ containing $0.1892$ mol kg${}^{-1}$ was $-{0.704}^{o}C$. On adding $0.045$ mole of $Hg(CN{)}_2$, the freezing point of the solution was $-{0.620}^{o}C$. If whole of $Hg(CN{)}_2$ is used in complex formation according to the equation,

$Hg(CN{)}_2+mKCN\to K_m[Hg\left(CN{)}_{m+2}\right]$

Assume $\left[Hg\right(CN{)}_{m+2}{]}^{m-}$ is not ionised and the complex molecule is 100% ionised. $K_f\left(H_{2}O\right)$ is $1.86$ kg mol${}^{-1}$K.
The formula of the complex is :

• A. $K_2\left[Hg\right(CN{)}_4]$
• B. $K_3\left[Hg\right(CN{)}_6]$
• C. $K_2\left[Hg\right(CN{)}_3]$
• D. none of these

Answer 1

The correct option is A $K_2\left[Hg\right(CN{)}_4]$

The van't Hoff factor,

$i=\frac{\Delta T_f}{K_{f}m}$

$i=\frac{0.704}{1.86\times 0.1892}=2$

$i=\frac{0.620}{1.86\times 0.1892}=1.76$

The van't Hoff factor $i$ decreases from 2 to 1.76. The complex formation can be represented as:

$Hg(CN{)}_2+mKCN\to K_m[HG\left(CN{)}_{m+2}\right]$

$Hg(CN{)}_2+mC{N}^{-}\to [HG(CN{)}_{m+2}{]}^{m-}$

$m=2$

Hence, the formula of the complex is $K_2\left[Hg\right(CN{)}_4]$.