Classification Of Elements And Periodicity In Properties

Let’s analyze both statements one by one.
(A) OF is paramagnetic while F is diamagnetic: - OF (oxygen fluoride, OF) has an unpaired electron in its molecular orbitals, making it paramagnetic. - F (fluorine molecule), on the other hand, has all of its electrons paired, making it diamagnetic. So, this statement is correct.
(B) OF is more stable towards dissociation into atoms: - OF is more stable than F in terms of dissociation. F has a relatively weak bond compared to OF, making OF less likely to dissociate into atoms. So, this statement is also correct.
Since both statements (A) and (B) are correct, the correct answer is .

Step 1: Understanding the Role of H O as an Oxidizing Agent
Hydrogen peroxide ( ) can act as both an oxidizing and a reducing agent, depending on the reaction conditions.
In acidic and neutral conditions, it tends to act as an oxidizing agent.
In alkaline medium, it can act as either an oxidizing or reducing agent.
Step 2: Completing the Given Reactions
Here, acts as an oxidizing agent by converting to . Here, acts as a reducing agent because it is oxidized to . Here, acts as a reducing agent by reducing iodine ( ) to iodide ( ). Here, oxidizes to , acting as an oxidizing agent.
Step 3: Identifying the Correct Answer
acts as an oxidizing agent in reactions (i) and (iv).

Solution: To solve the matching problem, we will analyze the information given in List I and List II and assign the correct pairings based on our understanding of Chemistry.

Based on this analysis, the correct match from the options is:

• Sodium (Na): Electronegativity ≈ 0.93
• Fluorine (F): Electronegativity ≈ 3.98
• Oxygen (O): Electronegativity ≈ 3.44
• Chlorine (Cl): Electronegativity ≈ 3.16