Problem 118 - A Complete Resource Book in Chemistry for JEE Main

Question

In the following sequence of reactions, identify the end product (D). \(\mathrm{Na}_{2} \mathrm{CO}_{3} \stackrel{\mathrm{SO}_{2}}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Na}_{2} \mathrm{CO}_{3}}{\longrightarrow}(\mathrm{B})\) Elemental \(\mathrm{S} \longrightarrow{\Delta}{\longrightarrow}(\mathrm{C}) \longrightarrow{\mathrm{I}_{2}}{\longrightarrow}\) (D) (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) (c) \(\mathrm{Na}_{2} \mathrm{~S}\) (d) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The end product D is \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \) (option b).

Step 1: Identify compound A from Na2CO3 and SO2

Reacting sodium carbonate (\( \mathrm{Na}_2 \mathrm{CO}_3 \)) with sulfur dioxide (\( \mathrm{SO}_2 \)) in aqueous solution forms sodium sulfite (\( \mathrm{Na}_2 \mathrm{SO}_3 \)). Therefore, compound A is \( \mathrm{Na}_2 \mathrm{SO}_3 \).

Step 2: Identify compound B from Na2SO3 and Na2CO3

When sodium sulfite (\( \mathrm{Na}_2 \mathrm{SO}_3 \)) is treated with additional \( \mathrm{Na}_2 \mathrm{CO}_3 \), no additional reaction occurs given this set of reagents, so \( \mathrm{Na}_2 \mathrm{SO}_3 \) remains as compound B.

Step 3: Convert compound B to compound C with elemental S

Sodium sulfite (\( \mathrm{Na}_2 \mathrm{SO}_3 \)) reacts with elemental sulfur (\( \mathrm{S} \)) and under heating forms sodium thiosulfate (\( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \)). Therefore, compound C is \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \).

Step 4: Convert compound C to D with I2

Sodium thiosulfate (\( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \)) reacts with iodine (\( \mathrm{I}_2 \)) to form sodium tetrathionate (\( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \)). Therefore, compound D is \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \).

Key Concepts

Sodium Compounds

Sodium is an abundant element in the Earth’s crust, often appearing in compounds rather than in its pure form. Among the most important sodium compounds is sodium carbonate, or \( \text{Na}_2\text{CO}_3 \), commonly known as washing soda. It's widely used in various chemical reactions due to its basicity and solubility in water. This makes it an ideal substrate in chemical processes.

In the chemical process discussed, sodium carbonate reacts with sulfur dioxide \( (\text{SO}_2) \) to form sodium sulfite \( (\text{Na}_2\text{SO}_3) \). This transformation is indicative of sodium compounds’ ability to act as a base, neutralizing acids like sulfur dioxide in aqueous environments, showcasing the versatile nature of sodium salts. Sodium compounds such as these play vital roles in numerous industrial applications ranging from glass-making to paper production.

Sulfur Chemistry

Sulfur is a versatile non-metallic element integral to various chemical reactions, often forming complex anions. In the described sequence, sulfur chemistry is at the core of the transformations.

Initially, sulfur dioxide \( (\text{SO}_2) \) reacts with sodium carbonate, leading to the formation of sodium sulfite \( (\text{Na}_2\text{SO}_3) \). Subsequently, elemental sulfur \( (\text{S}) \) modifies the sulfite ion further when combined under heat to yield sodium thiosulfate \( (\text{Na}_2\text{S}_2\text{O}_3) \). This conversion illustrates how sulfur can bridge oxygen atoms through its variable oxidation states, creating complex and stable structures essential in both natural and synthetic processes.

Step-by-Step Chemical Conversion

Chemical reactions often occur in steps, allowing scientists to precisely control and predict outcomes. This sequence demonstrates chemical conversions in a controlled manner, beginning with \( \text{Na}_2\text{CO}_3 \) and \( \text{SO}_2 \) to form \( \text{Na}_2\text{SO}_3 \). This first step sets the stage for subsequent transformations.

The second step remains unchanged since no further reaction occurs immediately. Following that, elemental sulfur adds to \( \text{Na}_2\text{SO}_3 \) under heat, converting it to \( \text{Na}_2\text{S}_2\text{O}_3 \), demonstrating the crucial role of temperature and elemental reactivity.

Finally, iodine reacts with \( \text{Na}_2\text{S}_2\text{O}_3 \) to produce \( \text{Na}_2\text{S}_4\text{O}_6 \), a more complex thiosulfate. This sequential approach reflects real-world laboratory processes, where each step must be understood for effective manipulation and application.

Inorganic Chemistry

Inorganic chemistry studies compounds primarily without carbon, focusing on metals and nonmetals like sodium and sulfur. The sequence provided delves into this field by featuring cleansing and oxidation-reduction reactions.

Sodium compounds act effectively as bases and oxidizing agents, as seen with \( \text{Na}_2\text{CO}_3 \). Sulfur compounds, in turn, display dynamic chemistry by transitioning from sulfite to thiosulfate and eventually tetrathionate forms. These transformations are essential illustrations of inorganic chemistry's diversity, showcasing reactions that can produce a range of products with distinct properties.

This chemical transformation sequence highlights essential principles of inorganic chemistry, such as acid-base interactions, redox reactions, and compound formation, making it an indispensable area of study in understanding materials around us.