Statement I: The correct order of electronegativity of fluorine, oxygen and nitrogen is:

F > O > N

Statement II: The oxidation state of oxygen in OF₂ is +2 and in Na₂O is -2.

Choose the correct answer.

• (1) Both Statement I and Statement II are correct.
• (2) Both Statement I and Statement II are incorrect.
• (3) Statement I is correct but Statement II is incorrect.
• (4) Statement I is incorrect but Statement II is correct.

This question examines two important concepts of chemistry: electronegativity and oxidation number. Let us analyze both statements individually.

Electronegativity is the tendency of an atom to attract shared electrons towards itself in a chemical bond. According to the Pauling electronegativity scale:

• Fluorine = 3.98
• Oxygen = 3.44
• Nitrogen = 3.04

Therefore, the correct order is:

F > O > N

Hence Statement I is correct.

Now consider Statement II.

In OF₂, fluorine is the most electronegative element and always has an oxidation state of -1. Let the oxidation state of oxygen be x.

x + 2(-1) = 0

x = +2

Therefore, oxygen has an oxidation state of +2 in OF₂.

For Na₂O:

Sodium belongs to Group 1 and has an oxidation state of +1. Let oxygen have oxidation state x.

2(+1) + x = 0

x = -2

Thus oxygen has an oxidation state of -2 in sodium oxide.

Both statements are scientifically correct. Questions of this type are frequently asked in NEET and JEE examinations because they test conceptual understanding of periodic properties and oxidation number rules.

Correct Option: (1) Both Statement I and Statement II are correct.