When mercuric iodide is added to the aqueous solution of potassium iodide, the _____:

Freezing point is raised

The correct answer is option 1. Freezing point is raised.

When mercuric iodide or $HgI_2$ (solute) reacts with aqueous potassium iodide or $KI$ (solvent), there appears the odourless yellow crystals of the complex $K_2[HgI_4]$ or mercuric potassium iodide. Its IUPAC name is potassium tetraiodomercurate (II).

When mercuric iodide is added to an aqueous solution of potassium iodide, we have seen the reaction. Let us relate this to the freezing point of the potassium iodide solution:

When potassium iodide ionizes in water as $KI \rightarrow K^+ + I^-$, and when it reacts with $HgI_2$, the number of moles of potassium iodide used is 2. So, the number of ionizable particles is 4.

The reaction is

$2KI + HgI_2 \rightarrow K_2[HgI_4]$

But after the reaction, the number of ionizable particles is 3. The particles ionized are $2K^+ + [HgI_4]^-$.

So, after the reaction, the number of ionizable particles or ions is decreased. Hence, the colligative properties will change, and thus the freezing point of the solution will also change. The number of particles has decreased. The freezing point $\propto \frac{1}{\text{decrement of particles}}$. So, the freezing point of the solution will increase.

When mercuric iodide is added to an aqueous solution of $KI$, the freezing point is raised.