Statement I : The correct order of electronegativity of fluorine, oxygen nitrogen is \(\mathrm{F}>\mathrm{O}>\mathrm{N}\). Statement II : The oxidation state of oxygen in \(\mathrm{OF}_{2}\) is +2 and in \(\mathrm{Na}_{2} \mathrm{O}\) 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

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Classification of Elements and Periodicity

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

Answer: (A)

Across the second period the effective nuclear charge increases while the radius stays small, so electronegativity rises as

N < O < F

, giving

F > O > N

and making Statement I correct. In

O F_{2}

fluorine is the more electronegative atom and takes

- 1

, forcing oxygen to

+ 2

; in

N a_{2} O

each sodium is

+ 1

, leaving oxygen at

- 2

, so Statement II is correct. Both statements are correct.

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option