Which one of the following order is correct for the bond dissociation enthalpy of halogen molecules?

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

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

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

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

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Solution

The correct option is A $C l_{2} > B r_{2} > F_{2} > I_{2}$
Bond dissociation energy of halogen family decreases down the group as the size of atom increases.

The decreasing order for the bond dissociation enthalpy of halogen molecules is $C l_{2} 57 kcal/mol > B r_{2} 45.5 kcal/mol > F_{2} 38 kcal/mol > 35.6 kcal/mol I_{2}$ .
A halogen molecule having larger atoms should have low dissociation energy and vice versa. Fluorine is an exception because of interelectronic repulsion is present in small atom fluorine.

Hence, the correct option is $B$ .

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x

y

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

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