In the iodination of methane by I2 and light, the propagation step CH4 + I. -> CH3. + HI is

highly endothermic, so the reaction does not proceed to an appreciable extent

highly exothermic, so the reaction is explosive

thermoneutral, so the reaction is as fast as chlorination

impossible because I2 cannot undergo homolysis

highly endothermic, so the reaction does not proceed to an appreciable extent

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JEE Main levelSingle correct optionDifficulty 2 of 3o2074

In the iodination of methane by $I X_{2}$ and light, the propagation step $CH X_{4} + I \cdot CH X_{3} \cdot + HI$ is

Ahighly exothermic, so the reaction is explosive
Bthermoneutral, so the reaction is as fast as chlorination
Cimpossible because $I X_{2}$ cannot undergo homolysis
Dhighly endothermic, so the reaction does not proceed to an appreciable extent

Answer: D

The H-I bond (about 298 kJ/mol) is much weaker than the C-H bond of methane (about 435 kJ/mol), so the step is endothermic by roughly 137 kJ/mol. The chain therefore does not propagate effectively.

I X_{2}

does undergo homolysis readily under light.

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In the iodination of methane by IX2​ and light, the propagation step CHX4​+I⋅​CHX3​⋅+HI is

Answer: D

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In the iodination of methane by $I X_{2}$ and light, the propagation step $CH X_{4} + I \cdot CH X_{3} \cdot + HI$ is