The correct oredr of bond dissociation enthalpy of halogens is :

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

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

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Solution

The correct option is C $C l_{2} > B r_{2} > F_{2} > I_{2}$
Bond energy is inversely dependent on bond length. The order of bond length is $B r_{2} > C l_{2} > F_{2} . .$

Hence, the order of bond energy should be $B r_{2} < C l_{2} < F_{2} . .$ But, bond dissociation enthalpy of $F_{2}$ is less than $C l_{2}$ because of small size of $F_{2}$ and large inter-electronic repulsions.
Thus, order of bond energy is $C l_{2}$ > $F_{2}$ and $B r_{2}$ < $C l_{2}$ .
So, option (d) is correct.

Compound Enthalpy of dissociation
(in kJ/mol)

F-F 158.8

Cl-Cl 242.6

Br-Br 192.8

I-I 151.1

Suggest Corrections

Compound	Enthalpy of dissociation (in kJ/mol)
F-F	158.8
Cl-Cl	242.6
Br-Br	192.8
I-I	151.1