JEE Main 2026 — Apr 5, Shift 1 — Chemistry solution

JEE Main2026Apr 5, Shift 1Chemistry

Statement I : The correct order of electronegativity of fluorine, oxygen, nitrogen is F > O > N.

Statement II : The oxidation state of oxygen in OF $_{2}$ is +2 and in Na $_{2}$ O is $- 2$ .

In the light of the above statements, choose the correct answer from the options given below :

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

Solution

Pauling electronegativities: F (4.0) > O (3.5) > N (3.0), so Statement I is correct.

In OF $_{2}$ , fluorine being the most electronegative element forces oxygen to a $+ 2$ state. In Na $_{2}$ O, the more electropositive sodium leaves oxygen in the usual $- 2$ state. Statement II is also correct.

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