Question

Given below are two statements:
Statement I: The correct order of electronegativity of fluorine, oxygen and nitrogen is \(F>O>N\).
Statement II: The oxidation state of oxygen in \(OF_2\) is \(+2\) and in \(Na_2O\) is \(-2\).

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The topic covers Periodic Properties and Redox Reactions.
We need to evaluate the correctness of a statement about electronegativity trends and another about calculating oxidation states.
Step 2: Detailed Explanation:
Analysis of Statement I:
Electronegativity is the tendency of an atom to attract a shared pair of electrons. It generally increases across a period from left to right in the periodic table.
Nitrogen (Group 15), Oxygen (Group 16), and Fluorine (Group 17) are in the second period.
Their electronegativity values on the Pauling scale are approximately F (4.0), O (3.5), and N (3.0).
Therefore, the order \(F>O>N\) is correct.
Analysis of Statement II:
Oxidation states are assigned based on electronegativity differences.
- In \(OF_2\) (oxygen difluoride), Fluorine is the most electronegative element, so it is assigned an oxidation state of \(-1\). Let the oxidation state of Oxygen be \(x\). The sum must be zero: \(x + 2(-1) = 0 \implies x = +2\). This is correct.
- In \(Na_2O\) (sodium oxide), Sodium is an alkali metal (Group 1) and is always assigned an oxidation state of \(+1\) in its compounds. Let the oxidation state of Oxygen be \(y\). The sum must be zero: \(2(+1) + y = 0 \implies y = -2\). This is also correct.
Therefore, Statement II is correct.
Step 3: Final Answer:
Since both statements are true, the correct option is (A).