The affinity for hydrogen decreases in the group from fluorine to iodine. Which of the hydrogen halides should have lowest bond dissociation enthalpy?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → HI
Down group 17 the halogen atom becomes larger, its orbital overlaps less with the small 1s orbital of hydrogen, and the H–X bond becomes longer and weaker. This is why the affinity for hydrogen decreases from fluorine to iodine. Bond dissociation enthalpies (kJ mol−1):
H–F 574 > H–Cl 432 > H–Br 363 > H–I 295
So HI has the lowest bond dissociation enthalpy. For the same reason HI is the strongest acid of the four and HF the weakest.
Why the other options are incorrect
Option (1) HCl — Incorrect. 432 kJ mol−1, stronger than H–Br and H–I.
Option (2) HBr — Incorrect. 363 kJ mol−1, still stronger than H–I.
Option (4) HF — Incorrect. 574 kJ mol−1, the strongest H–X bond, because F is the smallest halogen.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.