The correct order of bond dissociation enthalpy of halogen is:

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

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

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

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

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Solution

The correct option is C $C l_{2} > B r_{2} > F_{2} > I_{2}$
Bond dissociation energy of halogen family decreases down the group as the size of atom increases. But $F_{2}$ has $F — F$ , $F_{2}$ involves repulsion of non-bonding electrons and more over its size is small and hence due to high repulsion its bond dissociation energy is very low.

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C - X

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

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