Problem 24 - Chemistry in 30 days

Question

The reaction of sodium thiosulphate with \(\mathbf{I}_{2}\) gives (a) sodium sulphide (b) sodium sulphite (c) sodium sulphate (d) sodium tetrathionate

Step-by-Step Solution

Answer

(d) sodium tetrathionate

Step 1: Understanding the Reaction

The chemical reaction of sodium thiosulphate with iodine is given by the balanced equation: \[ 2 ext{Na}_2 ext{S}_2 ext{O}_3 + ext{I}_2 ightarrow 2 ext{NaI} + ext{Na}_2 ext{S}_4 ext{O}_6 \] Sodium thiosulphate (\( ext{Na}_2 ext{S}_2 ext{O}_3\)) reacts with iodine (\( ext{I}_2\)) to produce sodium iodide (\( ext{NaI}\)) and sodium tetrathionate (\( ext{Na}_2 ext{S}_4 ext{O}_6\)).

Step 2: Identifying the Main Product

From the balanced reaction equation, it's clear that one of the products formed is sodium tetrathionate (\( ext{Na}_2 ext{S}_4 ext{O}_6\)). This is the compound listed in the options.

Step 3: Choosing the Correct Option

Out of the given options: - Sodium sulphide - Sodium sulphite - Sodium sulphate - Sodium tetrathionate The correct product of sodium thiosulphate with iodine is sodium tetrathionate, which matches option (d).

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is like a recipe for a chemical reaction. It shows us exactly how much of each reactant is needed and how much of each product will be formed as a result. This balance is essential because it obeys the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction.

In the reaction between sodium thiosulphate (\(\text{Na}_2\text{S}_2\text{O}_3\)) and iodine (\(\text{I}_2\)), the balanced equation is: \[ 2 \text{Na}_2 \text{S}_2 \text{O}_3 + \text{I}_2 \rightarrow 2 \text{NaI} + \text{Na}_2 \text{S}_4 \text{O}_6 \]

• There are 4 sodium (Na) atoms, 4 sulfur (S) atoms, 6 oxygen (O) atoms, and 2 iodine (I) atoms on both sides of the equation.
• This balancing means that all the atoms from the reactants are accounted for in the products.
• It ensures that the same number of each type of atom is present before and after the reaction, preserving mass.

Balancing equations helps us to predict the outcome of reactions and determines how much of each product is formed.

Iodine Reaction

Iodine is a fascinating element that participates in reactions crucial for various chemical processes. In this reaction, iodine (\(\text{I}_2\)) plays the role of an oxidizing agent. This means that it helps to bring about the oxidation reaction, which involves the loss of electrons from another substance—in this case, sodium thiosulphate (\(\text{Na}_2\text{S}_2\text{O}_3\)).

Here’s how iodine contributes to the reaction:

• When iodine reacts with sodium thiosulphate, it triggers the conversion of sodium thiosulphate into sodium tetrathionate and sodium iodide.
• The iodine becomes reduced, changing from its diatomic state (\(\text{I}_2\)) to iodide ions (2\(\text{I}^–\)).

This reduction of iodine makes it possible for the sodium thiosulphate to transform into another complex compound, illustrating how iodine can facilitate significant chemical changes. In the reaction, iodine's exact amount is crucial to ensure complete conversion of sodium thiosulphate.

Sodium Tetrathionate Formation

The formation of sodium tetrathionate (\(\text{Na}_2\text{S}_4\text{O}_6\)) is the key outcome in this chemical reaction with iodine. This compound is a thiosulfate derivative and signifies the addition and expansion of the sulfur chain.

Understanding sodium tetrathionate:

• It consists of a complex structure comprising multiple sulfur and oxygen atoms.
• As a tetrathionate, it extends beyond the original thiosulphate form by linking two additional sulfur atoms.
• Its formation demonstrates the ingenuity of sulfur chemistry and the ability of sulfur atoms to form extended chain-like structures.

To summarize, sodium tetrathionate's creation exemplifies the intriguing chemical transformations achievable when substances like sodium thiosulphate react with oxidizing agents like iodine. This compound is typically the compound of interest in reactions similar to this, as it represents a substantial change in chemical identity.