Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

Four elements \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) can form diatomic molecules and monoatomic anions with \(-1\) charge. Consider the following reactions about these. \(2 \mathrm{~B}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{C}^{-}+\mathrm{B}_{2}\) \(\mathrm{B}_{2}+2 \mathrm{D} \longrightarrow 2 \mathrm{~B}^{-}+\mathrm{D}_{2}\) \(2 \mathrm{~A}^{-}+\mathrm{C}_{2}\) no reaction Select correct statement about these. (1) \(\mathrm{A}_{2}\) is strongest oxidizing agent while \(\mathrm{D}\) is strongest reducing agent (2) \(\mathrm{D}_{2}\) is strongest oxidizing agent while \(\mathrm{A}\) is strongest reducing agent (3) \(\mathrm{C}_{2}\) will oxidize \(\mathrm{B}^{-}\)and also \(\mathrm{D}\) - to form \(\mathrm{B}_{2}\) and \(\mathrm{D}_{2}\) (4) \(\mathrm{E}^{\circ} \mathrm{A}_{2} / \mathrm{A}^{-}\)is the lowest (a) 2 and 3 (b) 1 and 3 (c) 2 and 4 (d) 1,2 and 3

Step-by-Step Solution

Answer

(b) 1 and 3

Step 1: Analyze First Reaction

In the reaction \(2 \mathrm{~B}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{C}^{-}+\mathrm{B}_{2}\), \(\mathrm{C}_{2}\) is gaining electrons to become \(\mathrm{C}^{-}\), which makes it an oxidizing agent. Meanwhile, \(\mathrm{B}\) is being oxidized to form \(\mathrm{B}_{2}\), showing \(\mathrm{B}\) acts as a reducing agent.

Step 2: Analyze Second Reaction

The reaction \(\mathrm{B}_{2}+2 \mathrm{D} \longrightarrow 2 \mathrm{~B}^{-}+\mathrm{D}_{2}\) shows \(\mathrm{D}\) being oxidized to \(\mathrm{D}_{2}\), while \(\mathrm{B}\) is reduced to \(\mathrm{B}^{-}\), indicating \(\mathrm{D}\) acts as a reducing agent. Here, \(\mathrm{B}_{2}\) is an oxidizing agent since it is gaining electrons.

Step 3: Analyze Third Reaction

In the reaction \(2 \mathrm{~A}^{-}+\mathrm{C}_{2}\), where no reaction occurs, it implies \(\mathrm{C}_{2}\) is unable to oxidize \(\mathrm{A}^{-}\) back to \(\mathrm{A}_{2}\), suggesting that \(\mathrm{C}_{2}\) is not a stronger oxidizing agent than \(\mathrm{A}_{2}\).

Step 4: Determine Strongest Agents

Since \(\mathrm{C}_{2}\) can convert \(\mathrm{B}\) to \(\mathrm{B}_{2}\) and \(\mathrm{D}\) can convert \(\mathrm{B}_{2}\) to \(\mathrm{B}^{-}\), \(\mathrm{C}_{2}\) is a stronger oxidizing agent than \(\mathrm{A}_{2}\). However, \(\mathrm{A}_{2}\) could be the strongest oxidizing agent due to the inability of \(\mathrm{C}_{2}\) to affect \(\mathrm{A}^{-}\). \(\mathrm{D}\) as a reducing agent transforms \(\mathrm{B}_{2}\) back to \(\mathrm{B}^{-}\).

Step 5: Evaluate Statements

By evaluating the agents, option (3) is false because \(\mathrm{C}_{2}\) cannot oxidize \(\mathrm{B}^{-}\) as in the non-reactive scenario. \(\mathrm{A}_{2}\) could be a stronger oxidizing agent compared to \(\mathrm{C}_{2}\), confirming (1) as potentially correct. Option (2) misconstrues \(\mathrm{A}\) as a reducing agent.\(\mathrm{E}^{\circ} \mathrm{A}_{2} / \mathrm{A}^{-}\) being the lowest is not explicitly verified but is indirect with the reaction information in (4).

Key Concepts

Oxidizing Agent

In an oxidation-reduction reaction, an oxidizing agent facilitates the oxidation of another substance by accepting electrons from it. The oxidizing agent is reduced as it gains these electrons. In the context of the given reactions,

• For the reaction \(2 \mathrm{~B}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{C}^{-}+\mathrm{B}_{2}\), \(\mathrm{C}_{2}\) accepts electrons to become \(\mathrm{C}^{-}\), thereby acting as the oxidizing agent.
• In \(\mathrm{B}_{2}+2 \mathrm{D} \longrightarrow 2 \mathrm{~B}^{-}+\mathrm{D}_{2}\), \(\mathrm{B}_{2}\) gains electrons to become \(\mathrm{B}^{-}\), showing that it is the oxidizing agent here.

Understanding the role of oxidizing agents is crucial to analyzing any redox reaction, as it helps identify which compound gains electrons and thus is reduced during the process.

Reducing Agent

A reducing agent in a redox reaction donates electrons to another substance, causing the reduction of that substance. The reducing agent itself gets oxidized. Let's look at the given reactions:

• In \(2 \mathrm{~B}+\mathrm{C}_{2} \longrightarrow 2 \mathrm{C}^{-}+\mathrm{B}_{2}\), \(\mathrm{B}\) is oxidized as it forms \(\mathrm{B}_{2}\), so \(\mathrm{B}\) is the reducing agent.
• Similarly, in the reaction \(\mathrm{B}_{2}+2 \mathrm{D} \longrightarrow 2 \mathrm{~B}^{-}+\mathrm{D}_{2}\), \(\mathrm{D}\) loses electrons to form \(\mathrm{D}_{2}\), making \(\mathrm{D}\) a reducing agent.

Identifying the reducing agents helps us predict which elements or molecules are capable of donating electrons in a reaction, and thereby undergoing oxidation.

Reaction Analysis

Analyzing reactions involves determining which substances are oxidized and which are reduced, as well as identifying the oxidizing and reducing agents. In the given problem:

• First, \(\mathrm{C}_{2}\) oxidizes \(\mathrm{B}\) to \(\mathrm{B}_{2}\), while \(\mathrm{B}\) reduces \(\mathrm{C}_{2}\) to \(\mathrm{C}^{-}\).
• In the second reaction, \(\mathrm{B}_{2}\) is reduced to \(\mathrm{B}^{-}\) by\(\mathrm{D}\), and \(\mathrm{D}\) is oxidized to \(\mathrm{D}_{2}\).
• The third case shows that \(\mathrm{C}_{2}\) cannot oxidize \(\mathrm{A}^{-}\) back to \(\mathrm{A}_{2}\), indicating it is not a strong enough oxidizing agent compared to \(\mathrm{A}_{2}\).

This detailed step-by-step analysis of reactions helps in understanding the flow of electrons during the process, aiding in determining relative strengths of oxidizing and reducing agents.

Diatomic Molecules

Diatomic molecules consist of two atoms, which may be the same or different. In the context of the problem, elements \(\mathrm{A}\), \(\mathrm{B}\), \(\mathrm{C}\), and \(\mathrm{D}\) can all form diatomic molecules (\(\mathrm{A}_{2}\), \(\mathrm{B}_{2}\), \(\mathrm{C}_{2}\), \(\mathrm{D}_{2}\)). Diatomic molecules are often involved in oxidation-reduction reactions as they can be either oxidizing or reducing agents, as we've seen:

• \(\mathrm{C}_{2}\) acts as an oxidizing agent by converting \(\mathrm{B}\) into \(\mathrm{B}_{2}\).
• \(\mathrm{D}_{2}\) results from the oxidation of \(\mathrm{D}\).

Diatomic molecules play an essential role in redox chemistry, and understanding their formation and reactive properties allows us to better understand the chemical behavior of these elements in various reactions.