$K_{2}Hg{I}_4$ is $40\%$ ionised in aqueous solution. The value of its van't Hoff factor (i) is:
(JEE-Main 2019)

The correct option is D $1.8$
$K_{2}Hg{I}_4$ is $40\%$ ionised.

$\therefore \alpha =\frac{40}{100}=0.4$

$K_2\left[Hg{I}_4\right]\to 2K^{+}+[Hg{I}_4{]}^{2-}$

$N=\frac{2+1}{1}=3$

$i=1+(N-1)\alpha$

$i=1+(3-1)\left(0.4\right)$

$i=1+2\times 0.4$

$i=1.8$