JEE Main 2024 · 9 April · morning · Single correct · +4 / −1
Relative stability of the contributing structures is :
- (a) (II) > (I) > (III)
- (b) (III) > (II) > (I)
- (c) (I) > (III) > (II)
- (d) (I) > (II) > (III)
Answer and solution
Answer (d) (I) > (II) > (III)
Solution
Chemistry ma’am · 3 steps
Neutral beats charge-separated, and a minus charge belongs on oxygen, never a plus.
- Rules to apply. A neutral structure with every octet complete is the most stable. Among charge-separated structures, the better one has the negative charge on the more electronegative atom.
Structure Charges Octets Verdict (I) none all complete most stable (II) - on O, + on CH₂ terminal C has only 6 middle: charge separated, but the minus sits on electronegative O (III) + on O, - on CH₂ O has only 6 least stable: plus on O and minus on C are both wrong - Compare II and III. Both separate charges and both leave one atom with 6 electrons. (II) puts the negative charge on oxygen, which holds it well. (III) puts the positive charge on oxygen and leaves oxygen electron-deficient, which is very unfavourable.
- So. The order is (I) > (II) > (III), option (d).
The idea to keep. Rank resonance structures in this order of checks: no charge separation first, then complete octets, then negative charge on the more electronegative atom.
Answer from the NTA final answer key. Solution by PrepWiser.