Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

In the following reaction sequence in aqueous solution, the species \(X\), \(Y\) and \(Z\), respectively, are [Adv. 2016] \(\stackrel{\text { with time }}{\longrightarrow} \begin{gathered}Z \\ \text { black precipitate }\end{gathered}\)(a) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3}, \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{Ag}_{2} \mathrm{~S}\) (b) \(\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{3}\right]^{5-}, \mathrm{Ag}_{2} \mathrm{SO}_{3}, \mathrm{Ag}_{2} \mathrm{~S}\) (c) \(\left[\mathrm{Ag}\left(\mathrm{SO}_{3}\right)_{2}\right]^{3-}, \mathrm{Ag}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}, \mathrm{Ag}\) (d) \(\left[\mathrm{Ag}\left(\mathrm{SO}_{3}\right)_{3}\right]^{3}, \mathrm{Ag}_{2} \mathrm{SO}_{4}, \mathrm{Ag}\)

Step-by-Step Solution

Answer

The correct answer is option (a).

Step 1: Understand the Reaction Products

The problem provides a reaction sequence leading to a black precipitate labeled as \( Z \). Typically in inorganic chemistry, a black precipitate is indicative of silver sulfide, \( \text{Ag}_2\text{S} \). Thus, the product \( Z \) is \( \text{Ag}_2\text{S} \).

Step 2: Analyze Possible Starting Materials

We need to identify the species from which \( \text{Ag}_2\text{S} \) can be derived. Among the options, species containing silver (Ag) and thiosulfate \( \text{S}_2\text{O}_3 \) (or sulfite \( \text{SO}_3 \)) could reasonably lead to a silver sulfide precipitate.

Step 3: Eliminate Improbable Options

Options (a) and (b) suggest a complex ion, \( \left[\text{Ag}(\text{S}_2\text{O}_3)_{2}\right]^{3-} \) or \( \left[\text{Ag}(\text{S}_2\text{O}_3)_{3}\right]^{5-} \), which can decompose into silver sulfide via the release of \( \text{Ag}^+ \) ions. Options (c) and (d) suggest complexes that do not typically decompose to \( \text{Ag}_2\text{S} \), as they involve sulfite rather than thiosulfate.

Step 4: Confirm Correct Sequence

Since \( \text{Ag}_2\text{S} \) (black precipitate) is the product \( Z \), choice (a) makes logical sense when considering the decomposition of thiosulfate complexes. Specifically, \( \left[\text{Ag}(\text{S}_2\text{O}_3)_{2}\right]^{3-} \) can convert to \( \text{Ag}_2\text{S}_2\text{O}_3 \), releasing \( \text{Ag}^+ \), which then produces \( \text{Ag}_2\text{S} \).

Step 5: Select the Correct Option

From the analysis, the correct sequence is (a) \([\text{Ag}(\text{S}_2\text{O}_3)_2]^3, \text{Ag}_2\text{S}_2\text{O}_3, \text{Ag}_2\text{S}\), where \( Z \) is \( \text{Ag}_2\text{S} \).

Key Concepts

Reaction Mechanism

A reaction mechanism is an ordered sequence of steps that describes the detailed progress of a chemical reaction. It's almost like a movie showing each transition and transformation of molecules and atoms throughout the reaction process. In the case of the exercise given, the reaction mechanism involves the gradual decomposition and transformation of complex ions in solution, resulting in specific chemical changes.

Understanding reaction mechanisms allows chemists to predict what products might form and under what conditions. In complex reactions, such as those involving silver and thiosulfate, intermediate species play a crucial role in how the end product is formed. Recognizing these intermediates can help better understand reaction kinetics and predict subsequent reactions.

For students learning about complex ion decomposition and silver compounds, this approach offers a step-by-step understanding that aids in grasping more challenging concepts like those in inorganic chemistry. It’s like piecing together a puzzle; once you know how each piece fits, the whole picture becomes clearer.

Silver Compounds

Silver compounds hold a special place in chemistry due to their unique reactive properties and colorful precipitates. Silver sulfide, known chemically as \( \text{Ag}_2\text{S} \), is a common product in reactions involving silver ions and sulfide donors. Known for its characteristic black color, silver sulfide forms when silver ions \( \text{Ag}^+ \) meet sulfide ions \( \text{S}^{2-} \) or decomposing thiosulfate complexes.

Silver compounds are frequently encountered in photographic processes, electronic devices, and antimicrobial applications. In reactions, silver ions often collaborate with other ions to form complex ions or precipitates. For example, in the presence of thiosulfate, silver can form complex ions like \( \left[\text{Ag}(\text{S}_2\text{O}_3)_2\right]^{3-} \).

Learning about silver compounds involves not just recognizing their chemical formulas but understanding the conditions under which they form, stabilize, or break down in reactions. This knowledge is valuable in experimental chemistry, where silver's unique reactiveness is harnessed.

Complex Ion Decomposition

Complex ion decomposition is all about breaking down an intricate ion arrangement into simpler ions or molecules. Often, this process involves an intermediate that temporarily changes form or loses a part of its structure. In aqueous solutions, like those in the exercise, the stability of complex ions is influenced by factors like concentration, pH levels, and temperature.

In the specific case of silver thiosulfate, the complex ion \( \left[\text{Ag}(\text{S}_2\text{O}_3)_2\right]^{3-} \) decomposes to release silver ions \( \text{Ag}^+ \). These silver ions can then interact with other reactive species to form new compounds, such as silver sulfide, the black precipitate.

Understanding complex ion decomposition assists students in predicting chemical behavior and potential reaction pathways. It unlocks the secrets behind how complex ions can transform under various conditions, offering insight into both natural and industrial chemical processes.

Thiosulfate Chemistry

Thiosulfate ions \( \text{S}_2\text{O}_3^{2-} \) play a significant role in redox reactions and complex ion chemistry. Known for their ability to form stable complexes with silver ions, thiosulfates are landmarks in many chemical applications, including photography and gold extraction.

In the context of the exercise, thiosulfate acts as a ligand in forming silver complexes. These silver thiosulfate complexes can temporarily bind silver ions, preventing them from directly forming precipitates with sulfide. Over time, these complexes can decompose, which is crucial in the formation of black silver sulfide \( \text{Ag}_2\text{S} \).

Thiosulfate chemistry exemplifies how ligands can stabilize metal ions and delay their precipitation, even in complex systems. For chemistry students, mastering these concepts is invaluable for understanding broader topics like kinetics and equilibria in redox reactions.