Recall the trend in bond dissociation enthalpy of halogen molecules

Bond dissociation enthalpy generally decreases down the group because atomic size increases, so the bond length increases and the bond becomes weaker.

So, one may expect: $F_2 > Cl_2 > Br_2 > I_2$ but fluorine is an exception.

Why is $F_2$ an exception?

In $F_2$, each fluorine atom is very small and has three lone pairs. Because of the very short $F-F$ bond length, there is strong lone pair-lone pair repulsion between the two fluorine atoms.

This repulsion weakens the $F-F$ bond significantly.

Hence, even though fluorine is small, the bond dissociation enthalpy of $F_2$ is less than that of $Cl_2$ and $Br_2$.

Actual order

The known bond dissociation enthalpy order is: $Cl_2 > Br_2 > F_2 > I_2$

Check each option

• A: $F_2 > Cl_2 > Br_2 > I_2$
  Incorrect, because $F_2$ does not have the highest bond dissociation enthalpy.

• B: $Cl_2 > Br_2 > F_2 > I_2$
  Correct.

• C: $I_2 > Br_2 > Cl_2 > F_2$
  Incorrect, because $I_2$ has the weakest bond among these.

• D: $Cl_2 > F_2 > Br_2 > I_2$
  Incorrect, because $Br_2$ has greater bond dissociation enthalpy than $F_2$.

Final answer

Therefore, the correct order is: $Cl_2 > Br_2 > F_2 > I_2$ So the correct option is B.