1. We need a reagent that selectively precipitates $S^{2-}$ from a mixture of $S^{2-}$ and $SO_4^{2-}$.

2. So the reagent must form an insoluble compound with $S^{2-}$, but should not precipitate sulfate under these conditions.

3. Check each option:

Option A: $CuCl_2$

$Cu^{2+}$ reacts with sulfide ion to form highly insoluble copper sulfide: $Cu^{2+} + S^{2-} \rightarrow CuS \downarrow$

Copper sulfate is soluble in water, so $SO_4^{2-}$ is not precipitated by $Cu^{2+}$.

Hence, $CuCl_2$ selectively precipitates $S^{2-}$.

✅ A is correct.

Option B: $BaCl_2$

$Ba^{2+}$ gives an insoluble sulfate: $Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4 \downarrow$

Barium sulfide is soluble, so it does not selectively precipitate sulfide.

Thus $BaCl_2$ precipitates sulfate, not sulfide.

❌ B is incorrect.

Option C: $Pb(OOCCH_3)_2$

$Pb^{2+}$ forms insoluble lead sulfide: $Pb^{2+} + S^{2-} \rightarrow PbS \downarrow$

Although lead sulfate is also sparingly soluble, in qualitative analysis lead acetate is commonly used to detect/precipitate sulfide as black $PbS$. The intended selective precipitation here is of sulfide.

✅ C is correct.

Option D: $Na_2[Fe(CN)_5NO]$

This is sodium nitroprusside. It gives a violet coloration with sulfide ion: $S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow \text{violet complex}$

It is a test reagent, not a precipitating reagent.

❌ D is incorrect.

4. Therefore, the reagents that precipitate $S^{2-}$ are: $\boxed{A,\ C}$

5. Comparison with stored correct answer:

• Derived answer: $A, C$
• Stored correct answer: $A, C$

They match.