Answer:
In the iodination of methane, \[H-I\] is also formed as the product along with iodomethane. Since it is a strong reducing agent, it reduces iodomethane back to methane and makes the reaction reversible. In order to destory \[HI\], oxidising agent like \[HI{{O}_{3}}\] (or\[HN{{O}_{3}}\]) is needed. But HCl and HBr formed in the chlorination and bromination reactions of methane are not in a position to react with the mono substituted product (i.e., \[C{{H}_{3}}-Cl\] and \[C{{H}_{3}}-Br\]) since they are comparatively weak reducing agents. Therefore, no oxidising agent is needed for these reactions.
\[\underset{\text{Methane}}{\mathop{C{{H}_{4}}}}\,+{{I}_{2}}\rightleftharpoons C{{H}_{3}}I+HI\]
\[5HI+HI{{O}_{3}}\xrightarrow[\,]{\,}\,3{{H}_{2}}O+3{{I}_{2}}\]
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