The correct order of bond dissociation enthalpy of halogens is :

Solution

Chemistry ma’am · 3 steps

Bonds weaken down the group as atoms grow, but fluorine is the exception: tiny and crowded with lone pairs.

1. General trend. Going down group 17, the atoms get bigger and the bond gets longer. The orbital overlap gets weaker, so the bond dissociation enthalpy falls from Cl₂ to Br₂ to I₂.
2. Fluorine breaks it. The F atom is very small, so the F–F bond is short. The lone pairs on the two F atoms sit very close and repel each other strongly, and this weakens the bond. So F₂ falls far below Cl₂, and even below Br₂.

   Halogen	Bond dissociation enthalpy (kJ mol⁻¹)	Rank
   Cl₂	242.6	1st
   Br₂	192.8	2nd
   F₂	158.8	3rd
   I₂	151.1	4th

3. So. The order is Cl₂ > Br₂ > F₂ > I₂, which is option (b).

The idea to keep. For X₂ bond enthalpy, Cl₂ is the strongest and I₂ the weakest. F₂ is anomalously weak because its lone pairs repel across a very short bond.

Answer from the NTA final answer key. Solution by PrepWiser.