Increasing order of stability of the resonance structures is :
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Judge each structure by octets first, then by whether each charge sits next to a group that helps it.

1. Rule one. A resonance structure in which every atom has a complete octet is the most stable. An incomplete octet on oxygen, the most electronegative atom here, is the worst case.
2. Rule two. When octets cannot decide, look at where the charges sit. The -NH₂ lone pair stabilises a positive charge next to it. The -CHO group pulls electrons, so it stabilises a negative charge next to it and destabilises a positive one.

   Structure	Octets	Charge placement	Stability
   C	O⁺ has only 6 electrons	positive charge on the most electronegative atom	least stable
   A	one carbon has only 6 electrons	C⁺ next to -CHO, C⁻ next to -NH₂: both wrong way round	low
   B	one carbon has only 6 electrons	C⁻ next to -CHO, C⁺ next to -NH₂: both helped	higher
   D	all atoms complete	positive charge on N as iminium, C⁻ next to -CHO	most stable

3. So. The increasing order of stability is C < A < B < D, which is option (c).

The idea to keep. To rank resonance structures, check complete octets first. Then prefer negative charge on or near electron-withdrawing groups and positive charge on or near lone-pair donors.

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