Animated Solution for Chemistry - Surface Chemistry: The sol given below with negatively charged colloidal particles is

The behavior of colloids often feels like magic, but it is governed by elegant rules of physical chemistry. In this problem, we are tasked with identifying which chemical reaction produces a negatively charged colloidal sol.

To solve this, we must dive into the concept of preferential adsorption, a phenomenon that dictates the electrical destiny of a colloidal particle.

When two solutions are mixed to form a precipitate, the resulting solid particles don't just sit there; they interact with their environment. The surface of the precipitate has a strong affinity for its own constituent ions.

If one of the reacting solutions is present in excess, the precipitate will preferentially adsorb the common ion from that excess solution onto its surface. This adsorbed layer of ions gives the entire colloidal particle its net electrical charge.

Let's quickly eliminate the obvious wrong answers. When ${\text{FeCl}}_3$ is added to hot water, it undergoes hydrolysis to form hydrated ferric oxide (${\text{Fe}}_2{\text{O}}_3\cdot x{\text{H}}_2\text{O}$). Metal oxides and hydroxides like this, as well as ${\text{Al}}_2{\text{O}}_3\cdot x{\text{H}}_2\text{O}$, naturally form positively charged sols. Therefore, options (a) and (d) are out.

The Positive Sol

Now, let's look at the subtle difference between options (b) and (c).

In option (b), the phrasing is "$\text{KI}$ added to ${\text{AgNO}}_3$ solution". This implies that we have a beaker full of ${\text{AgNO}}_3$ (it is in excess), and we are dropping a small amount of $\text{KI}$ into it.

The reaction forms a precipitate of silver iodide ($\text{AgI}$):

Because ${\text{AgNO}}_3$ is in excess, the solution is swimming with extra ${\text{Ag}}^{+}$ and ${\text{NO}}_3^{-}$ ions. The $\text{AgI}$ precipitate looks for a common ion to adsorb. It finds the ${\text{Ag}}^{+}$ ions and strongly adsorbs them onto its surface.

This creates a positively charged sol, represented as $\text{AgI}/{\text{Ag}}^{+}$.

The Negative Sol

In option (c), the scenario is flipped: "${\text{AgNO}}_3$ added to $\text{KI}$ solution".

This time, the beaker is full of $\text{KI}$, meaning potassium iodide is in excess. The precipitate formed is still $\text{AgI}$. However, the environment is completely different. The solution is now rich in ${\text{K}}^{+}$ and ${\text{I}}^{-}$ ions.

The $\text{AgI}$ precipitate again looks for a common ion. This time, it finds the ${\text{I}}^{-}$ ions in excess. It adsorbs these iodide anions onto its surface, acquiring a net negative charge.

This creates a negatively charged sol, represented as $\text{AgI}/{\text{I}}^{-}$.

By simply changing the order of mixing—and thereby changing which reactant is in excess—we completely reversed the electrical charge of the resulting colloid. Since the question specifically asked for the negatively charged sol, the correct choice is the one where the anion is in excess.

Therefore, the correct option is (c).