Presence of $S^{2-}$ ion gives violet colour with which of the following?

Question

Accepted Answer

Presence of $S^{2-}$ ion gives violet color with $Na_{2}[Fe(CN)_{5}NO]$, which is Option 4. Here's why this is the correct answer:

1. **Explanation for Option 4 ($Na_{2}[Fe(CN)_{5}NO]$):**
   - The compound $Na_{2}[Fe(CN)_{5}NO]$ is known as sodium nitroprusside. When $S^{2-}$ ions are present, they react with sodium nitroprusside to form a complex that exhibits a violet color. This reaction is specific and is used as a qualitative test for the presence of sulfide ions ($S^{2-}$). The formation of this violet-colored complex is due to the interaction between the sulfide ions and the nitroprusside ion, which leads to a change in the electronic structure and results in the observed color change.

2. **Explanation for why other options are incorrect:**

- **Option 1 ($K_{2}Cr_{2}O_{7}$):**
     - Potassium dichromate ($K_{2}Cr_{2}O_{7}$) is an oxidizing agent and does not specifically react with $S^{2-}$ ions to produce a violet color. Instead, it may oxidize sulfide ions to sulfur or sulfate, but this reaction does not result in a violet color. Therefore, it is not the correct choice for detecting $S^{2-}$ ions through a color change.

- **Option 2 ($KMnO_{4}$):**
     - Potassium permanganate ($KMnO_{4}$) is another strong oxidizing agent. It can oxidize $S^{2-}$ ions, but the reaction typically results in the formation of manganese dioxide (MnO2) and sulfate ions, which do not produce a violet color. The color change associated with $KMnO_{4}$ is usually due to its own reduction, not the formation of a violet complex with $S^{2-}$ ions.

- **Option 3 ($Pb(CH_{3}COO)_{2}$):**
     - Lead acetate ($Pb(CH_{3}COO)_{2}$) reacts with $S^{2-}$ ions to form lead sulfide (PbS), which is a black precipitate. This reaction is used to detect sulfide ions, but it does not produce a violet color. The formation of a black precipitate is distinct from the violet color observed with sodium nitroprusside.

In conclusion, Option 4 ($Na_{2}[Fe(CN)_{5}NO]$) is the correct answer because it specifically reacts with $S^{2-}$ ions to produce a violet color, which is a characteristic and well-known test for the presence of sulfide ions. The other options do not produce this specific color change with $S^{2-}$ ions.