Problem 34 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The reagent(s) that can selectively precipitate \(\mathrm{S}^{2-}\) from a mixture of \(\mathrm{S}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\) in aqueous solution is (are) (a) \(\mathrm{CuCl}_{2}\) (b) \(\mathrm{BaCl}_{2}\) (c) \(\mathrm{Pb}\left(\mathrm{OOCCH}_{3}\right)_{2}\) (d) \(\mathrm{Na}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}\right]\)

Step-by-Step Solution

Answer

The reagent is (d) \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \).

Step 1: Identifying Selective Precipitation Concept

We need to determine which reagent can selectively precipitate sulfide ions \( \mathrm{S}^{2-} \) while leaving sulfate ions \( \mathrm{SO}_4^{2-} \) in solution. This involves finding a reagent that forms an insoluble compound with \( \mathrm{S}^{2-} \) and not with \( \mathrm{SO}_4^{2-} \).

Step 2: Analyzing Each Option

(a) \( \mathrm{CuCl}_2 \): Copper(II) chloride reacts with sulfide ions to form copper(II) sulfide \( \mathrm{CuS} \), a very insoluble compound, precipitating \( \mathrm{S}^{2-} \). However, it doesn't react significantly with \( \mathrm{SO}_4^{2-} \).(b) \( \mathrm{BaCl}_2 \): Barium chloride reacts with both sulfide ions and sulfate ions. While it forms barium sulfide \( \mathrm{BaS} \), it also forms barium sulfate \( \mathrm{BaSO}_4 \), thus not being selective.(c) \( \mathrm{Pb(OOCCH}_3)_2 \): Lead(II) acetate can react to form lead(II) sulfide \( \mathrm{PbS} \) which precipitates \( \mathrm{S}^{2-} \). However, it also can form \( \mathrm{PbSO}_4 \), and thus is not entirely selective.(d) \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \): Also known as sodium nitroprusside, this reagent reacts specifically with sulfide ions to form a black precipitate, not affecting \( \mathrm{SO}_4^{2-} \).

Step 3: Evaluating the Selectivity of Reagents

Reagents that react to form a precipitate with \( \mathrm{S}^{2-} \) more favorably than with \( \mathrm{SO}_4^{2-} \) are selective for sulfide. Based on the analysis, sodium nitroprusside \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \) only reacts with sulfide to form a precipitate, making it selective in aqueous solutions.

Step 4: Conclusion

The reagent that selectively precipitates \( \mathrm{S}^{2-} \) from the mixture is sodium nitroprusside \( \mathrm{Na}_2[\mathrm{Fe(CN)}_5 \mathrm{NO}] \). It does not form a precipitate with \( \mathrm{SO}_4^{2-} \), hence it's the correct choice for selective precipitation.

Key Concepts

Sulfide Ion Precipitation

In chemistry, selective precipitation is a key method of separating ions in a solution based on their different solubilities. When dealing with sulfide ions (\( \mathrm{S}^{2-} \)), certain reagents can cause them to form insoluble compounds and precipitate out of the solution. Let's examine the problem at hand: choosing a reagent that will selectively precipitate \( \mathrm{S}^{2-} \) from a mixture containing both sulfide and sulfate ions (\( \mathrm{SO}_{4}^{2-} \)).

The selective precipitation process hinges on finding a reagent that:

• Forms an insoluble compound with sulfide ions.
• Does not react, or reacts very minimally, with sulfate ions.

Among common reagents, \( \mathrm{CuCl}_{2} \) (Copper(II) chloride) stands out as a candidate because it forms copper(II) sulfide (\( \mathrm{CuS} \)), which is highly insoluble. However, copper does not form a similar insoluble product with sulfates. This property allows sulfides to be separated effectively in both qualitative and quantitative analysis.

Sodium Nitroprusside Reaction

Sodium nitroprusside, with the chemical formula \( \mathrm{Na}_{2} \left[ \mathrm{Fe(CN)}_{5} \mathrm{NO} \right] \), is known for its selective reaction with sulfide ions. This reagent specifically interacts with \( \mathrm{S}^{2-} \) ions and not with \( \mathrm{SO}_{4}^{2-} \) ions. In the reaction with sulfide ions, sodium nitroprusside forms a black-colored precipitate. This distinctive color change makes it a particularly useful qualitative test for the presence of sulfide ions.The reaction is primarily employed because:

• The resulting precipitate's color is starkly different from the original solution, facilitating easy visualization.
• Sodium nitroprusside does not interfere with other common anions like sulfates, making it extraordinarily selective.

This selectivity allows chemists to confidently detect and precipitate sulfide ions without the concern of cross-reactions with other ions in the solution.

Solubility of Sulfide and Sulfate

Understanding the solubility rules is critical when predicting the outcome of reactions. Sulfide ions (\( \mathrm{S}^{2-} \)) and sulfate ions (\( \mathrm{SO}_{4}^{2-} \)) exhibit different solubility behaviors, which can be leveraged in laboratory settings to achieve selective precipitation.

In general:

• Sulfides tend to form highly insoluble compounds with metal cations. Examples include copper(II) sulfide (\( \mathrm{CuS} \)) and lead sulfide (\( \mathrm{PbS} \)).
• Sulfates are usually more soluble, though there are exceptions such as barium sulfate (\( \mathrm{BaSO}_{4} \)), which is very insoluble.

This differential solubility is pivotal in methods that separate ions because it allows chemists to add specific reagents knowing which ions will stay dissolved and which will precipitate. In solutions where precise detection or measurement is required, this solubility contrast underpins effective separation techniques, thus playing a crucial role in fields like analytical chemistry.