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

Question

The compound that cannot act both as oxidising and reducing agent is: [Main Jan. 09, 2020 (I)] (a) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (b) \(\mathrm{HNO}_{2}\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Step-by-Step Solution

Answer

\( \mathrm{H}_{3} \mathrm{PO}_{4} \) cannot act as both an oxidizing and reducing agent.

Step 1: Identify Oxidation States

Determine the oxidation state of the central atom in each compound. \( \mathrm{H}_{3} \mathrm{PO}_{4} \) contains phosphorus in the +5 oxidation state, \( \mathrm{HNO}_{2} \) contains nitrogen in the +3 oxidation state, \( \mathrm{H}_{2} \mathrm{SO}_{3} \) contains sulfur in the +4 oxidation state, and \( \mathrm{H}_{2} \mathrm{O}_{2} \) contains oxygen in the -1 oxidation state.

Step 2: Determine Possibility of Acting as Oxidizing and Reducing Agent

For a substance to act as both an oxidizing and reducing agent, it must have the ability to either increase or decrease its oxidation state. \( \mathrm{H}_{3} \mathrm{PO}_{4} \) has the central atom phosphorus in its maximum oxidation state (+5), so it can only act as an oxidizing agent and cannot be further oxidized.

Step 3: Analyze Other Compounds

\( \mathrm{HNO}_{2} \) can be oxidized to \( \mathrm{HNO}_{3} \) (nitrogen to +5) or reduced to \( \mathrm{N}_{2} \) or \( \mathrm{NH}_{3} \) (nitrogen to 0 or -3). \( \mathrm{H}_{2} \mathrm{SO}_{3} \) can be oxidized to \( \mathrm{H}_{2} \mathrm{SO}_{4} \) (sulfur to +6) or reduced to \( \mathrm{H}_{2} \mathrm{S} \) (sulfur to -2). \( \mathrm{H}_{2} \mathrm{O}_{2} \) can be oxidized to \( \mathrm{O}_{2} \) (oxygen to 0) or reduced to \( \mathrm{H}_{2} \) (oxygen to -2).

Step 4: Conclude with the Compound That Cannot Act Both Ways

Based on the analysis, \( \mathrm{H}_{3} \mathrm{PO}_{4} \) is the compound that cannot act as both an oxidizing and a reducing agent because the phosphorus atom is already in its highest oxidation state.

Key Concepts

Oxidation States

Understanding oxidation states is key when studying redox reactions. An oxidation state is a number that represents the total number of electrons that an atom either gains or loses to form a chemical bond. It helps you determine how elements transition through chemical reactions.
For example, consider the compound \(\mathrm{H}_{3} \mathrm{PO}_{4}\).Here, phosphorus (P) is in the +5 oxidation state, which means it has lost five electrons.Conversely, in \(\mathrm{HNO}_{2}\), nitrogen has a +3 oxidation state.This implies nitrogen can either gain or lose electrons, being more flexible in reactions.
Knowing the oxidation state helps predict if a substance can act as an oxidizing or reducing agent.

Oxidizing Agents

An oxidizing agent is a substance that gains electrons in a chemical reaction, causing another substance to be oxidized. This means the oxidizing agent itself is reduced. In redox chemistry, substances with high oxidation states often serve as oxidizing agents.

• Take \(\mathrm{H}_{3} \mathrm{PO}_{4} \) (phosphoric acid), for example. Phosphorus is in a +5 oxidation state, the highest for this element. Therefore, it can easily gain electrons from other substances, making it a strong oxidizing agent.
• In contrast, a compound like \(\mathrm{HNO}_{2}\) has nitrogen in a +3 state, which allows it to act as both oxidizing and reducing agent, depending on the reaction conditions.

Understanding how oxidizing agents function helps clarify why phosphorus in \(\mathrm{H}_{3} \mathrm{PO}_{4}\) cannot act as both an oxidizing and reducing agent.

Reducing Agents

Reducing agents are substances that donate electrons in chemical reactions, causing other substances to be reduced. When a substance acts as a reducing agent, it itself is oxidized because it loses electrons.

• For example, \(\mathrm{H}_{2} \mathrm{SO}_{3}\) can function as a reducing agent. With sulfur in a +4 oxidation state, it can lose electrons to form \(\mathrm{H}_{2} \mathrm{SO}_{4}\) with sulfur in a +6 state, demonstrating its ability to donate electrons.
• Similarly, \(\mathrm{H}_{2} \mathrm{O}_{2}\) can serve as both an oxidizing and reducing agent due to its ability to be easily converted to \(\mathrm{O}_{2}\) or \(\mathrm{H}_{2}\) by donating or accepting electrons.

Comprehending how reducing agents work is essential to understanding complex redox reactions and predicting possible reaction pathways.