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Chemistry Question 58 – JEE-MAIN 2026

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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 OF2 is +2 and in Na2O is -2. In the light of the above statements, choose the correct answer from the options given below

Recall the general trend of electronegativity across a period and down a group in the periodic table.

Step 1: Evaluate Statement I: Electronegativity Order✦ Active

Electronegativity generally increases across a period from left to right and decreases down a group. Fluorine (F), Oxygen (O), and Nitrogen (N) are all in the second period. Fluorine is the most electronegative element. The order of electronegativity in the second period is F > O > N. Therefore, Statement I is true.

Step 2: Evaluate Statement II: Oxidation States of Oxygen○ Expand

For OF2: Fluorine is more electronegative than oxygen, so its oxidation state is −1. Let the oxidation state of oxygen be x. The sum of oxidation states must be zero: x+2(−1)=0⟹x=+2. So, oxygen in OF2 has an oxidation state of +2.

For Na2O: Sodium (Na) is an alkali metal, so its oxidation state is +1. Let the oxidation state of oxygen be y. The sum of oxidation states must be zero: 2(+1)+y=0⟹y=−2. So, oxygen in Na2O has an oxidation state of −2.

Both parts of Statement II are correct. Therefore, Statement II is true.

💡 Teacher's Secret Hint

Remember that fluorine always takes a -1 oxidation state in compounds, overriding oxygen's usual -2.

Step 3: Conclusion○ Expand

Since both Statement I and Statement II are true, the correct option is 'Both Statement I and Statement II are true'.

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