Alkanes can be iodinated in the presence of:

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I_{2}

and P

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H I O_{3}

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P I_{3}

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Solution

The correct option is C $H I O_{3}$
This is because iodination is reversible and thus needs a reducing agent to drive the reaction to forward direction according to Le Chatelier's principle.

The reaction is as follow:
$C H_{4} + I_{2} \leftrightarrow C H_{3} I + H I$

The HI formed in a product makes the reaction reversible.Thus, $H I O_{3}$ is added which reduce HI to $I_{2}$ and stop the reversible reaction.

$H I O_{3} + 5 H I \to 3 H_{2} O + 3 I_{2}$

Option C is correct.

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