Observe the given compounds I, II, and III. Identify the correct...

The bond dissociation energy (BDE) of a $\mathrm{C-H}$ bond is inversely proportional to the stability of the carbon free radical formed after its homolytic cleavage.

Let us analyze the radicals formed from each compound:

Compound I: Homolytic cleavage of the marked $\mathrm{C-H}$ bond generates an allylic radical ($\mathrm{CH_2=CH-CH_2^\bullet}$). This radical is highly stabilized by resonance.

Compound II: Cleavage of the marked $\mathrm{C-H}$ bond generates an $\mathrm{sp}$ hybridized carbon radical ($\mathrm{CH_3-C\equiv C^\bullet}$). The unpaired electron resides in an $\mathrm{sp}$ orbital with 50% s-character, making it highly unstable as the electron is held tightly by the nucleus. Thus, this bond is the strongest.

Compound III: Cleavage of the marked $\mathrm{C-H}$ bond generates a secondary alkyl radical ($\mathrm{(CH_3)_2CH^\bullet}$). This radical is stabilized by hyperconjugation (6 $\alpha$-hydrogens) but is less stable than the resonance-stabilized allylic radical.

The order of stability of the resulting free radicals is: I $>$ III $>$ II.

Since a more stable radical corresponds to a lower bond dissociation energy, the decreasing order of BDE is the reverse of the radical stability order.

Therefore, the decreasing order of BDE is II $>$ III $>$ I.

Answer: II $>$ III $>$ I