Problem 72 - A Complete Resource Book in Chemistry for JEE Main

Question

Fluorine oxidizes \(\mathrm{HSO}_{4}^{-}\)to \(\ldots \ldots\) (a) \(\mathrm{SO}_{2}\) (b) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}\)

Step-by-Step Solution

Answer

Fluorine oxidizes \( \mathrm{HSO}_{4}^{-} \) to \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \).

Step 1: Identify the Reaction

Fluorine is a strong oxidizing agent, capable of oxidizing other species. In this exercise, we're examining which substance is formed by the oxidation of \( \mathrm{HSO}_{4}^{-} \) by fluorine.

Step 2: Determine Possible Products

Consider the possible products of the oxidation of \( \mathrm{HSO}_{4}^{-} \). The oxidizing action of fluorine can lead to the formation of higher oxidation state compounds such as \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \), which is known to be a peroxodisulfate, a common oxidation product of sulfur in high oxidation states.

Step 3: Analyze Oxidation State Changes

Check the oxidation state of sulfur in the reactant and products. In \( \mathrm{HSO}_{4}^{-} \), sulfur has an oxidation state of +6. The possible products are: \( \mathrm{SO}_{2} \) (oxidation state +4), \( \mathrm{S}_{4} \mathrm{O}_{6}^{2-} \) (average oxidation state +2.5), \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \) (average oxidation state +2), and \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) (oxidation state +7). Since the sulfur oxidation state should increase, \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) is the only logical product.

Step 4: Confirm the Correct Answer

Given the oxidation states and typical reaction pathways, it is confirmed that the product formed is \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) as this involves an increase in the oxidation state of sulfur, consistent with an oxidation process.

Key Concepts

Fluorine as Oxidizing Agent

Fluorine is the most electronegative element on the periodic table. This means it has a strong tendency to attract electrons from other substances. Acting as a powerful oxidizing agent, fluorine can strip electrons from other elements, causing them to become oxidized.

When fluorine is involved in a chemical reaction, it readily takes electrons from other atoms, changing the chemical structure and oxidation state of the involved elements. As an oxidizing agent, fluorine facilitates the oxidation process, typically leading to the formation of compounds with higher oxidation states. This capability makes fluorine a valuable reactant in certain chemical processes where oxidation is necessary.

• Fluorine's high electronegativity makes it unique among halogens.
• It has a substantial impact on oxidation reactions by altering oxidation states.
• It's a key component in chemical reactions needing strong oxidation.

Oxidation State Changes

In a chemical reaction, the oxidation state of an element can change as it gains or loses electrons. This shift in the oxidation state is a crucial aspect of redox (reduction-oxidation) reactions. For instance, in the exercise given, sulfur changes its oxidation state as it undergoes oxidation by fluorine.

Originally, sulfur in the hydrogen sulfate ion ( \( \mathrm{HSO}_{4}^{-} \) ) has an oxidation state of +6. During the oxidation process with fluorine, sulfur is converted into \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \), a compound where the oxidation state of sulfur is raised to +7. This elevation in oxidation state confirms that sulfur has lost electrons, which is the core mechanism of an oxidation reaction.

• Oxidation involves an increase in the oxidation state.
• Reduction is the opposite, where the state decreases as electrons are gained.
• Sulfur compounds often exhibit multiple stable oxidation states.

Sulfur Compounds

Sulfur is a versatile element that forms a wide variety of compounds, often distinguished by their varying oxidation states. In chemical reactions, sulfur can exist in several forms, each with unique properties and uses.

The exercise highlights sulfur's ability to form diverse compounds by showing the transition from \( \mathrm{HSO}_{4}^{-} \) to \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \). Sulfur's oxidation states can range from -2, as in sulfides, to +6, as in sulfates. In these reactions, sulfur is unique for its ability to adjust its oxidation state significantly, forming compounds like sulfites, thiosulfates, and peroxodisulfates.

• Sulfur's flexible oxidation state facilitates its role in varied chemical processes.
• Common sulfur compounds include \( \mathrm{SO}_{2} \), \( \mathrm{H}_{2} \mathrm{SO}_{4} \), and \( \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \).
• Peroxodisulfates \( \mathrm{S}_{2} \mathrm{O}_{8}^{2-} \) are often used in industrial processes.