Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Aqueous solution of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) on reaction with \(\mathrm{Cl}_{2}\) gives \(-\)(a) \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) (b) \(\mathrm{NaHSO}_{4}\) (c) \(\mathrm{NaCl}\) (d) \(\mathrm{NaOH}\)

Step-by-Step Solution

Answer

The products are \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), \( \mathrm{NaHSO}_4 \), and \( \mathrm{NaCl} \).

Step 1: Determine the Balanced Chemical Equation

When chlorine gas \( \mathrm{Cl}_2 \) reacts with sodium thiosulfate \( \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 \), it forms sodium tetrathionate \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \) as well as sodium chloride \( \mathrm{NaCl} \) and sodium bisulfate \( \mathrm{NaHSO}_4 \). The balanced reaction is:\[ \mathrm{Cl}_2 + 2 \mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 + 2 \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 + 2 \mathrm{NaCl} + 2 \mathrm{NaHSO}_4 \]

Step 2: Identify the Products

From the balanced chemical equation, we can identify the products of the reaction: \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \), \( \mathrm{NaCl} \), and \( \mathrm{NaHSO}_4 \).

Step 3: Match the Products with Given Options

The products that match the given options in the problem are \( \mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6 \) (option a), \( \mathrm{NaHSO}_4 \) (option b), and \( \mathrm{NaCl} \) (option c). Option (d), \( \mathrm{NaOH} \), is not a product of this reaction.

Key Concepts

Chemical Reactions

Chemical reactions are processes where one or more substances, known as reactants, are transformed into different substances, called products. These reactions involve the breaking of bonds in reactants and the formation of new bonds in products. Chemical reactions are represented as chemical equations, which show the reactants on the left and the products on the right. In the reaction between chlorine and sodium thiosulfate, several types of substances interact chemically to yield new products.

Different types of reactions include combination, decomposition, single displacement, double displacement, and redox reactions. Understanding these types helps in predicting the products of a reaction. For instance, when chlorine gas interacts with the aqueous solution of sodium thiosulfate, it is an example of a redox reaction, where electron transfer occurs.

Balancing Equations

In chemical reactions, balancing the equation is crucial because it represents the conservation of mass, indicating that the number of atoms for each element is the same in the reactants and the products. Balancing ensures that the reaction abides by the law of conservation of mass.

To balance a chemical equation, you adjust the coefficients—the numbers placed before molecules in the equation—until you have an equal number of each type of atom on both sides. For example, in the equation given for the reaction between chlorine and sodium thiosulfate:

• Each element has equal atoms on both sides: Cl, Na, S, O, and H are balanced based on their presence in the reactants and products.
• The reaction produces multiple products, which must reflect the same elemental count as found in initial substances.

This involves careful examination of each element, ensuring the same number is present as reactants and in the products.

Chlorine Reactions

Chlorine is a highly reactive halogen and participates in numerous chemical reactions. As a strong oxidizing agent, it readily reacts with many substances, often leading to electron transfer reactions, also called redox reactions.

When chlorine gas (\(\mathrm{Cl}_2\)) is introduced into an aqueous solution of sodium thiosulfate (\(\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3\)), it oxidizes sodium thiosulfate to form sodium tetrathionate (\(\mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6\)), sodium chloride (\(\mathrm{NaCl}\)), and sodium bisulfate (\(\mathrm{NaHSO}_4\)). This reaction showcases the powerful oxidative nature of chlorine, where it plays a crucial role in converting thiosulfate ions to tetrathionate ions. Understanding how chlorine interacts with other compounds provides insight into its broad applications, from disinfectants to analytical chemistry.

Sodium Thiosulfate Reactions

Sodium thiosulfate is known for its reaction with oxidizing agents, displaying its versatility in various chemical contexts. Reacting with chlorine, sodium thiosulfate serves as a reducing agent, donating electrons during the formation of its products.

This specific reaction involves the conversion of sodium thiosulfate (\(\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3\)) into sodium tetrathionate (\(\mathrm{Na}_2 \mathrm{S}_4 \mathrm{O}_6\)), alongside producing sodium chloride (\(\mathrm{NaCl}\)) and sodium bisulfate (\(\mathrm{NaHSO}_4\)).

• The reaction showcases its application in analytical chemistry, often used to neutralize chlorine in pools.
• It highlights the significance of sodium thiosulfate as a multi-purpose chemical in various industries, from photography to gold extraction.

The ability of sodium thiosulfate to act effectively with diverse substances makes it a valuable compound in both theoretical and practical chemistry settings.