Bond energy of halogens are

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$F_{2} > {Cl}_{2} > {Br}_{2} > I_{2}$

${Cl}_{2} > {Br}_{2} > F_{2} > I_{2}$

${Cl}_{2} > F_{2} > {Br}_{2} > I_{2}$

${Cl}_{2} > F_{2} > I_{2} > {Br}_{2}$

answer is B.

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Detailed Solution

F₂ is expected to have highest bond dissociation enthalpy but it is Cl₂ and the correct decreasing order is:

Cl₂>Br₂>F₂>I₂

Reason is that fluorine atom has very small size due to which there is a high inter electronic repulsion between two fluorine atoms. Therefore, its bond energy reduces.

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