Among the given compounds I-III, the correct order of bond dissociation energy of C–H bond marked with * is :

The correct answer is II > I > III.

The bond dissociation energy (BDE) of a C–H bond is the energy required for its homolytic cleavage to form a carbon radical and a hydrogen radical:
$R-H\to R^{•}+H^{•}$

The strength of a C–H bond and its corresponding bond dissociation energy depend primarily on the hybridization of the carbon atom holding the hydrogen atom:

1. Hybridization and s-Character:
As the s-character of the hybridized carbon orbital increases, the electrons in the C–H bond are held closer to the carbon nucleus, resulting in a shorter, stronger bond that is harder to break.

• sp Carbon (50% s-character): Forms the strongest C–H bond with the highest BDE.
• sp² Carbon (33% s-character): Forms a bond of intermediate strength and BDE.
• sp³ Carbon (25% s-character): Forms the weakest C–H bond with the lowest BDE.

Let us analyze the given compounds based on the images:

• Compound I (Benzene ring): The marked hydrogen is attached to a benzene ring carbon, which is sp² hybridized. Phenyl radicals are relatively unstable compared to alkyl radicals, resulting in a high BDE of approximately 111 kcal/mol.
• Compound II (Phenylacetylene): The marked hydrogen is attached to a terminal alkyne carbon, which is sp hybridized. Because of the high s-character (50%), this C–H bond is extremely strong, with a very high BDE of approximately 131 kcal/mol.
• Compound III (Cyclopropene): The marked hydrogen is attached to the methylene carbon of cyclopropene, which is sp³ hybridized. Although there is ring strain, the C–H bond involving an sp³ orbital is inherently much weaker than sp² and sp hybridized C–H bonds, resulting in the lowest BDE (around 85–95 kcal/mol).

Comparing the bond dissociation energies, we get the order:
$\mathrm{sp}\left(\mathrm{II}\right)>{\mathrm{sp}}^2\left(I\right)>{\mathrm{sp}}^3\left(\mathrm{III}\right)$

Therefore, the correct order of bond dissociation energy of the marked C–H bond is II > I > III.