The order of reactivity of different halogens towards halogenation of alkanes is:

F_{2} > C l_{2} > B r_{2} > I_{2}

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B r_{2} > C l_{2} > F_{2} > I_{2}

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I_{2} > B r_{2} > C l_{2} > F_{2}

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C l_{2} > F_{2} > I_{2} > B r_{2}

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Solution

The correct option is A $F_{2} > C l_{2} > B r_{2} > I_{2}$
In halogenation of alkanes, reactivity of halogens decrease with decrease in the electronegativity.
In halogenation of alkanes, the mechansim is radical mechanism. As we go from $F_{2}$ to $I_{2}$ th energy released in the propogating steps decreases which proves the reactivity order.

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