Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Reaction of an inorganic sulphite \(X\) with dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) generates compound Y. Reaction of \(\mathrm{Y}\) with NaOH gives \(\mathrm{X}\). Further, the reaction of \(X\) with \(Y\) and water affords compound \(Z . Y\) and \(Z\), respectively, are: [Main Sep. 06, 2020 (II)] (a) \(\mathrm{SO}_{2}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) (b) \(\mathrm{SO}_{3}\) and \(\mathrm{NaHSO}_{3}\) (c) \(\mathrm{SO}_{2}\) and \(\mathrm{NaHSO}_{3}\) (d) \(\mathrm{S}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{2}\)

Step-by-Step Solution

Answer

Option (c): \(\mathrm{SO}_2\) and \(\mathrm{NaHSO}_3\).

Step 1: Analyze Reaction of X with H2SO4

The question states that an inorganic sulfite \(X\) reacts with dilute \(\mathrm{H}_2\mathrm{SO}_4\) to produce compound \(Y\). Considering the chemical properties of sulfites, \(X\) is likely \(\mathrm{Na}_2\mathrm{SO}_3\). The reaction is as follows: \(\mathrm{Na}_2\mathrm{SO}_3 + \mathrm{H}_2\mathrm{SO}_4 \rightarrow 2\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O} + \mathrm{Na}_2\mathrm{SO}_4\). \(Y\) is \(\mathrm{SO}_2\).

Step 2: Analyzing Reaction of Y with NaOH

Now, \(Y\) (which is \(\mathrm{SO}_2\)) reacts with \(\mathrm{NaOH}\) to yield \(X\) again. The reaction is: \(\mathrm{SO}_2 + 2\mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_3 + \mathrm{H}_2\mathrm{O}\). This confirms that \(X\) is indeed \(\mathrm{Na}_2\mathrm{SO}_3\).

Step 3: Analyzing Reaction of X with Y and Water to Form Z

\(X = \mathrm{Na}_2\mathrm{SO}_3\) and \(Y = \mathrm{SO}_2\). When these react in the presence of water, the reaction provides compound \(Z\). This reaction is: \(\mathrm{Na}_2\mathrm{SO}_3 + \mathrm{SO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{NaHSO}_3\), which indicates that \(Z\) is \(\mathrm{NaHSO}_3\).

Step 4: Identifying Compounds Y and Z

From the reactions, it's clear \(Y = \mathrm{SO}_2\) and \(Z = \mathrm{NaHSO}_3\). Therefore, based on the options given, we have: \(Y\) is \(\mathrm{SO}_2\), and \(Z\) is \(\mathrm{NaHSO}_3\).

Key Concepts

Sulfite Reaction

Inorganic sulfites, like sodium sulfite (Na₂SO₃), often engage in chemical reactions with acids. Such reactions are prominent in chemistry due to their ability to release gases. Sulfite ions (SO₃²⁻) are known for their reductive properties.
This is why, when sodium sulfite reacts with dilute sulfuric acid (H₂SO₄), sulfur dioxide (SO₂) gas and water are produced. The equation governing this process is:

• Na₂SO₃ + H₂SO₄ → 2SO₂ + 2H₂O + Na₂SO₄

This transformation illustrates the well-known sulfite reaction where a sulfite compound decomposes to release sulfur dioxide gas upon exposure to a strong acid.
The release of SO₂ is a characteristic feature in reactions involving sulfites, often resulting in the restoration of the original sulfite via subsequent reactions.

Sulfur Dioxide

Sulfur dioxide (SO₂) is a simple, yet pivotal molecule in inorganic chemistry. This gas is colorless, with a distinct, penetrating odor recognizable by many as the smell of burnt matches.
It plays a major role in chemical reactions, particularly when it acts as a reducing agent. In the context of our exercise, the sulfur dioxide generated from sulfite reactions can further engage in reactions of its own, such as with sodium hydroxide (NaOH).
When sulfur dioxide is bubbled through sodium hydroxide solution, sodium sulfite can be regenerated:

• SO₂ + 2NaOH → Na₂SO₃ + H₂O

This reaction is crucial because it illustrates how SO₂ can revert to its original sulfite form. Consequently, this cycle of reactions highlights the reversible nature of these chemical processes in inorganic chemistry.

Sodium Bisulfite

Sodium bisulfite, commonly referred to as NaHSO₃, is a compound with significant applications in both industrial and laboratory settings. It acts as a preservative, antioxidant, and reducing agent in various scenarios.
In our exercise, sodium bisulfite emerges from the interaction of sodium sulfite, sulfur dioxide, and water. The chemical equation representing this reaction is:

• Na₂SO₃ + SO₂ + H₂O → 2NaHSO₃

This reaction exemplifies how sodium bisulfite forms in environments rich with sulfur dioxide and moisture.
Sodium bisulfite's ability to form through the reaction of sulfite and SO₂ in water illustrates the compound's adaptive nature, influenced by the surrounding chemical environment. This feature makes sodium bisulfite a valuable component in many biochemical and industrial processes.