In a colloidal state the particle size of the dispersed phase range between \(10^3\) to \(10^6\) pm and colloidal sol is heterogenous in nature. Based on the nature of interaction between the dispersed phase and dispersion medium colloidal sol are classified as lyophilic and lyophobic. Stability of sol is due to presence of charge on the sol particles and the neutralization of the same is known as coagulation or precipitation.

Based on this answer the following:

When \(KI\) solution is added to \(AgNO_3\) solution such that \(AgNO_3\) amount exceeds \(KI\) amount.

The precipitated \(AgI\) absorbs \(Ag^+\) ions from \(KI\) solution and positively charged colloidal sol result.

The correct answer is option 2. The precipitated \(AgI\) absorbs \(Ag^+\) ions from \(KI\) solution and positively charged colloidal sol result.

When \(KI\) solution is added to \(AgNO_3\) solution in excess (meaning \(AgNO_3\) has more moles than \(KI\)), a precipitation reaction occurs. Silver \((Ag^+)\) ions from \(AgNO_3\) react with iodide \((I^-)\) ions from \(KI\) to form insoluble silver iodide \((AgI)\).

However, since \(AgNO_3\) is in excess, there will be some leftover \(Ag^+\) ions in the solution.

These leftover \(Ag^+\) ions are attracted to the surface of the already formed \(AgI\) precipitate due to electrostatic interactions. Silver iodide \((AgI)\) has a slightly positive lattice and can adsorb some additional \(Ag^+\) ions, creating a positively charged colloidal sol.

Let us break down why the other options are incorrect:

1. Negatively charged colloidal sol: This is the opposite of what happens. The precipitate adsorbs positive \(Ag^+\) ions, not negative \(I^-\) ions.

3. Precipitated KI: The precipitate is \(AgI\), not \(KI\). \(KI\) remains dissolved in the solution until the \(Ag^+\) ions are used up.

4. No colloidal sol formation: A colloidal sol does form in this scenario due to the adsorption of \(Ag^+\) ions by the \(AgI\) precipitate.