The sol given below with negatively charged colloidal particles is:

We have to decide in which of the four cases the colloidal particles finally produced carry a net negative charge. The sign of the charge on a colloid is governed by the ion that finally gets adsorbed on the freshly formed precipitate. Whichever ion remains in excess in the dispersion medium is preferentially adsorbed.

First, consider Option A: $$\text{AgNO}_3 \text{ added to KI solution}.$$ Because the KI solution is taken first, $$[ \text{I}^- ]$$ is large. On mixing, the insoluble precipitate $$\text{AgI}$$ is produced by

$$\text{Ag}^+ + \text{I}^- \longrightarrow \text{AgI}\,(\text{solid}).$$

There is still a large excess of $$\text{I}^-$$ ions left in the medium. According to the adsorption theory, these excess $$\text{I}^-$$ ions get adsorbed on the surface of the newly formed $$\text{AgI}$$ particles. So the adsorbed ion is $$\text{I}^-$$, a negative ion, and the charge on each colloidal particle becomes negative.

Now, Option B: $$\text{KI added to AgNO}_3 \text{ solution}.$$ Here $$[ \text{Ag}^+ ]$$ is in excess because the AgNO$$_3$$ solution is taken first. The same precipitate $$\text{AgI}$$ forms, but now the excess ion available is $$\text{Ag}^+$$. These positive $$\text{Ag}^+$$ ions are adsorbed, giving the $$\text{AgI}$$ sol a positive charge, not a negative one.

Option C involves $$\text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O}$$ (hydrated alumina) shaken with pure water. Hydrated alumina has the amphoteric surface $$\text{Al(OH)}_3$$, which preferentially adsorbs $$\text{H}^+$$ ions from water:

$$\text{Al(OH)}_3 + \text{H}^+ \rightarrow \text{Al(OH)}_3 \cdot \text{H}^+.$$

The adsorbed ion is $$\text{H}^+,$$ so the sol is positively charged.

Option D: $$\text{FeCl}_3$$ hydrolyses in hot water, producing a ferric hydroxide sol:

$$\text{Fe}^{3+} + 3\text{H}_2\text{O} \longrightarrow \text{Fe(OH)}_3\,(\text{solid}) + 3\text{H}^+.$$

The medium contains excess $$\text{H}^+$$; these $$\text{H}^+$$ ions adsorb on the $$\text{Fe(OH)}_3$$ particles, again resulting in a positively charged sol.

Summarising, only in Option A do the colloidal particles carry excess adsorbed $$\text{I}^-$$ ions, giving them a negative charge. All the other options produce sols with adsorbed positive ions.

Hence, the correct answer is Option A.