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 question tests knowledge of periodic trends (electronegativity) and oxidation state calculation rules.
Step 2: Detailed Explanation:
Statement I: Electronegativity increases across a period from left to right. Nitrogen, Oxygen, and Fluorine are in the same period. Their values on the Pauling scale are approximately \(N = 3.0, O = 3.5, F = 4.0\). Therefore, the order \(F>O>N\) is correct.
Statement II: In \(OF_2\), fluorine is more electronegative than oxygen (\(4.0\) vs \(3.5\)). Fluorine is always assigned \(-1\). Let oxidation state of O be \(x\): \(x + 2(-1) = 0 \implies x = +2\).
In \(Na_2O\), sodium is a group 1 metal with an oxidation state of \(+1\). Let oxidation state of O be \(y\): \(2(+1) + y = 0 \implies y = -2\).
Thus, Statement II is correct.
Step 3: Final Answer:
Both statements are true.