The correct stability order of the following resonance structures of CH₃-CH=CH-CHO is
CH₃-C(⊖)H-CH=C(|(:O:⊕))-H ↔ CH₃-C(⊕)H-CH=C(|(:O:⊖))-H ↔ CH₃-CH=CH-C(∥(:O:))-H

Solution

Chemistry ma’am · 3 steps

Check each structure for full octets, no charge separation, and the negative charge on oxygen.

1. Rules to apply. A resonance structure is more stable when every atom has a complete octet and there is no charge separation. If charges must appear, the negative charge should sit on the more electronegative atom, which here is oxygen.

   Structure	Octets	Charges	Verdict
   III	All complete	None, neutral	most stable
   II	Carbon with ⊕ has only 6	Separated, but ⊖ on electronegative O	intermediate
   I	O with ⊕ and one bond has only 6	Separated, and ⊕ on electronegative O	least stable

2. Why II beats I. Both I and II have separated charges. In II, oxygen holds the negative charge, which it can hold well, and the CH₃ group helps the positive carbon through hyperconjugation. In I, the charges are the wrong way round: oxygen is positive and carbon is negative.
3. So. The stability order is III > II > I, which is option (c).

The idea to keep. Rank resonance structures in this order: neutral with full octets first, then charge-separated with the negative charge on the more electronegative atom, and last, charges on the wrong atoms.

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