Consider the following compound (X)
H-C(I)≡C-CH₂(II)-CH(III)-CH₃(IV) (X)
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C-H bond are :

Solution

Chemistry ma’am · 4 steps

Check for resonance first, then hyperconjugation, and look at which orbital holds the lone electron.

1. Two stabilising effects. A carbon radical is stabilised by resonance (delocalising the odd electron) and by hyperconjugation with neighbouring C–H bonds. Resonance is the stronger effect.
2. Carbon II. Removing H from C-II gives a secondary propargylic radical, next to the triple bond. The odd electron delocalises into the π bond: HC≡C-ĊH-R ↔ HĊ=C=CH-R. It also has hyperconjugation, so it is the most stable.
3. Carbon I. Removing H from C-I leaves the odd electron in an sp orbital of the alkyne carbon. This orbital lies in the plane of the molecule, at right angles to the π bonds, so there is no resonance and no hyperconjugation. High s character also holds the electron close to the nucleus, which destabilises the radical. That makes it the least stable radical, even less stable than a primary one.

   Carbon	Radical formed	Stabilisation	Stability
   II	secondary, propargylic	resonance with C≡C + hyperconjugation	most stable
   III	tertiary alkyl	hyperconjugation only (8 α-H)	second
   IV	primary alkyl	hyperconjugation only (1 α-H)	third
   I	alkynyl (sp carbon)	none	least stable

4. So. The order is II > III > IV > I. The most stable radical is at II and the least stable is at I, which is option (d).

The idea to keep. Resonance-stabilised radicals (allylic, propargylic, benzylic) beat even tertiary ones. Radicals on sp or sp² carbons (vinyl, alkynyl, aryl) are the least stable of all.

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