When silver nitrate solution is added to potassium iodide solution then the sol produced is:

When silver nitrate solution is added to potassium iodide solution, the double displacement reaction $$\text{AgNO}_3 + \text{KI} \to \text{AgI}\downarrow + \text{KNO}_3$$ takes place, producing a precipitate of silver iodide ($$\text{AgI}$$).

Since silver nitrate is added to potassium iodide, the iodide ion ($$\text{I}^-$$) is in excess. In colloid chemistry, the precipitated particles preferentially adsorb the ion that is in excess and is common to the lattice. Here, $$\text{I}^-$$ is a lattice ion of $$\text{AgI}$$ and is present in excess, so the colloidal particles of $$\text{AgI}$$ adsorb $$\text{I}^-$$ ions on their surface. This gives the sol a primary layer of $$\text{I}^-$$ ions.

Therefore, the sol produced is represented as $$\text{AgI} / \text{I}^-$$, meaning silver iodide particles with negatively charged iodide ions adsorbed on their surface.

The answer is option (1).