The correct order of stability of following carbocations is :

Solution

Chemistry ma’am · 4 steps

Check for aromaticity first; after that, the more rings sharing the positive charge, the more stable the ion.

1. Aromatic cation. C is the tropylium ion. Its ring is planar, cyclic and fully conjugated, with 6 π electrons (4n + 2 with n = 1). So it is aromatic, and the positive charge is spread equally over all 7 carbons.
2. Benzylic resonance. In A and B the charge spreads into each phenyl ring at the ortho and para positions. Each ring adds 3 resonance structures, so A (Ph₃C⁺) has 10 and B (Ph₂CH⁺) has 7. More delocalisation means more stability, so A is more stable than B.

   Cation	What stabilises it	Rank
   C, tropylium	aromatic, 6 π electrons over 7 carbons	1
   A, Ph₃C⁺	resonance into 3 rings, 10 structures	2
   B, Ph₂CH⁺	resonance into 2 rings, 7 structures	3
   D, 2-butyl	only hyperconjugation (5 α-H) and +I	4

3. Plain alkyl. D is a simple secondary carbocation with no resonance at all. Hyperconjugation and the +I effect are much weaker than resonance, so D is the least stable.
4. So. The order is C > A > B > D, which is option (d).

The idea to keep. For carbocation stability, aromaticity beats everything. Next comes resonance, and more rings means more delocalisation. Hyperconjugation and inductive effects only decide between plain alkyl ions.

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