The correct order of stability for the following alkoxides is :

Solution

Chemistry ma’am · 3 steps

Follow where the negative charge can go: the further it spreads, the more stable the alkoxide.

1. What stabilises. An alkoxide is more stable when its negative charge is spread over more atoms. Resonance (−M) spreads the charge much better than the inductive effect (−I) alone.

   Alkoxide	Effect on O⁻	Where the charge can go
   (C) ⁻O-CH=CH-NO₂	Resonance through C=C into NO₂ (−M), plus −I	Onto the nitro oxygens, which are very electronegative
   (B) CH₂=C(NO₂)-O⁻	Resonance with C=C, plus −I of NO₂ on the same carbon	Only onto the terminal CH₂ carbon
   (A) CH₃-CH(NO₂)-O⁻	−I of NO₂ only	Stays on oxygen, no resonance

2. Why C beats B. In (C) the O⁻, the C=C and the NO₂ form one conjugated chain, so the charge reaches the nitro oxygens. In (B) the NO₂ sits on the same carbon as O⁻, so it is cross-conjugated: resonance pushes the charge only onto CH₂, a carbon, and the NO₂ helps only by −I.
3. So. (B) still has some resonance and an sp² carbon next to O⁻, so it beats (A), which has only −I. The order is (C) > (B) > (A), option (a).

The idea to keep. For anion stability, check resonance first and induction second. A group stabilises by −M only if it lies in conjugation with the charge, not just close to it.

Answer from the NTA final answer key. Solution by PrepWiser.