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.

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