The increasing order of stability of the following free radicals is

A free radical is stabilised mainly by two factors:
  • +I and hyperconjugation from attached alkyl groups.
  • Resonance delocalisation over an aromatic ring (benzylic resonance).

Let us analyse each given radical.

Case 1: $$(CH_3)_2\dot{C}H$$ (isopropyl, secondary alkyl)
Only hyperconjugation operates. A secondary radical has fewer hyperconjugative structures than a tertiary one, so it is the least stable among the four listed.

Case 2: $$(CH_3)_3\dot{C}$$ (tert-butyl, tertiary alkyl)
Three $$CH_3$$ groups donate electron density by +I effect and hyperconjugation. Thus it is more stable than the secondary radical but still possesses no resonance.

Case 3: $$(C_6H_5)_2\dot{C}H$$ (diphenylmethyl, benzylic)
The odd electron is adjacent to two phenyl rings. Each ring can delocalise the radical electron over its ortho- and para-positions. Resonance offers much greater stabilisation than hyperconjugation, so this radical is more stable than any purely alkyl radical.

Case 4: $$(C_6H_5)_3\dot{C}$$ (triphenylmethyl, trityl)
Here three phenyl rings participate in resonance, giving the maximum number of canonical forms. Consequently it is the most stabilised radical in the list.

Combining the above observations:

$$\text{secondary alkyl} \; (CH_3)_2\dot{C}H \; \lt \; \text{tertiary alkyl} \; (CH_3)_3\dot{C} \; \lt \; \text{benzylic with 2 rings} \; (C_6H_5)_2\dot{C}H \; \lt \; \text{benzylic with 3 rings} \; (C_6H_5)_3\dot{C}$$

Therefore, the increasing order of stability is
$$(CH_3)_2\dot{C}H \; < \; (CH_3)_3\dot{C} \; < \; (C_6H_5)_2\dot{C}H \; < \; (C_6H_5)_3\dot{C}$$

Option A is correct.