The correct order of bond dissociation enthalpy of halogen is :
Question:

The correct order of bond dissociation enthalpy of halogen is :

1. $\mathrm{F}_{2}>\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}$

2. $\mathrm{Cl}_{2}>\mathrm{F}_{2}>\mathrm{Br}_{2}>\mathrm{I}_{2}$

3. $\mathrm{Cl}_{2}>\mathrm{Br}_{2}>\mathrm{F}_{2}>\mathrm{I}_{2}$

4. $\mathrm{I}_{2}>\mathrm{Br}_{2}>\mathrm{Cl}_{2}>\mathrm{F}_{2}$

Correct Option: , 3

Solution:

Fact based $\mathrm{F}_{2}$ has $\mathrm{F}-\mathrm{F}, \mathrm{F}_{2}$ involves repulsion of non-bonding electrons $\backslash$ more over its size is small $\&$ hence due to high repulsion its bond dissociation energy in very low.