Animated Solution for Chemistry - Salt Analysis: The reagent(s) that can selectively precipiate ${\text{S}}^{2-}$ from a mixture of ${\text{S}}^{2-}$ and ${\text{SO}}_4^{2-}$ in aqueous soltuion is(are) :

Imagine you are a chemical detective handed a beaker containing a mixture of two notorious anions: sulfide (${\text{S}}^{2-}$) and sulfate (${\text{SO}}_4^{2-}$). Your mission is to find a reagent that acts like a highly specific net, catching only the sulfide ions and pulling them out of the solution as a solid precipitate, while leaving the sulfate ions swimming freely. This process is known as selective precipitation.

To solve this, we must rely on our knowledge of solubility rules. Let's interrogate our four suspects (the options) one by one.

Copper(II) Chloride (${\text{CuCl}}_2$)

When we introduce ${\text{Cu}}^{2+}$ ions into the mixture, they encounter both ${\text{S}}^{2-}$ and ${\text{SO}}_4^{2-}$.

Transition metal sulfides are notoriously insoluble. The reaction is swift and decisive:

This forms a striking black precipitate. But what about the sulfate? Most sulfates are soluble, and copper sulfate is no exception. It remains happily dissolved in the water. Therefore, ${\text{CuCl}}_2$ perfectly executes our mission. It selectively precipitates the sulfide.

Barium Chloride (${\text{BaCl}}_2$)

Now let's try Barium. Barium is an alkaline earth metal. Its sulfide, $\text{BaS}$, is actually soluble in water. However, Barium has a famous affinity for sulfate ions:

This forms a dense white precipitate. Barium did exactly the opposite of what we wanted! It caught the sulfate and ignored the sulfide. We can cross this option off our list.

Lead(II) Acetate (${\text{Pb(OOCCH}}_3{)}_2$)

Lead is a heavy hitter. When ${\text{Pb}}^{2+}$ enters the fray, it reacts with both anions:

At first glance, Lead seems to fail our test because it precipitates both. But here is where the magic of physical chemistry comes in. We must look at the Solubility Product Constant ($K_{sp}$).

The $K_{sp}$ of $\text{PbS}$ is approximately $3\times 10^{-28}$, an astronomically small number. In contrast, the $K_{sp}$ of ${\text{PbSO}}_4$ is about $2.5\times 10^{-8}$. Because $\text{PbS}$ is so much less soluble, if we add the lead reagent slowly, the sulfide will precipitate almost entirely before the sulfate even begins to form a solid. Because of this massive difference in solubility, Lead(II) Acetate can practically be used for selective precipitation.

Sodium Nitroprusside (${\text{Na}}_2[{\text{Fe(CN)}}_5\text{NO}]$)

Finally, we test sodium nitroprusside. This is a classic qualitative test for sulfide. When sulfide reacts with the nitroprusside ion, it forms a beautiful purple complex:

However, notice the state of the product. It is a complex ion that remains dissolved in the solution. No precipitate is formed. Since our goal was to precipitate the sulfide, this reagent fails the physical requirement of the task.

Copper(II) chloride is the perfect theoretical answer, cleanly separating the two. Lead(II) acetate is also a correct answer when viewed through the lens of $K_{sp}$ driven selective precipitation. Thus, both (A) and (C) are correct choices for this nuanced JEE Advanced problem.