When only a little quantity of ${\mathrm{HgCl}}_2\left(\mathrm{s}\right)$ is added to excess KI (aq) to obtain a clear soltion, which of the following is true for this solution? (no volume change on mixing) 

Due to this addition a complex ${\left[{\mathrm{HgI}}_4\right]}^{2-}$ will get formed as follows 

${\mathrm{Hg}}^{2+}+4{\mathrm{I}}^{-}\Rightarrow {\left[{\mathrm{HgI}}_4\right]}^{2-}$

or ${\mathrm{Hg}}^{2+}+2{\mathrm{Cl}}^{-}+4{\mathrm{I}}^{-}\rightleftharpoons {\left[{\mathrm{HgI}}_4\right]}^{2-}+2{\mathrm{Cl}}^{-}$

on adding we can see that $4{\mathrm{I}}^{-}$ ions are consumed but resultant number of particles are three. 

Hence boiling point gets decreased. 

freezing point gets increased.