Arrange the halogens in the decreasing order of bond dissociation enthalpy:
(A) \(F_2\)
(B) \(Cl_2\)
(C) \(Br_2\)
(D) \(I_2\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → B > C > A > D
The bond dissociation enthalpy of \( \ce{X2} \) is the energy needed to break one mole of X–X bonds. From chlorine to iodine it falls as the atoms grow larger and the bond grows longer; fluorine does not follow this trend.
(B) \( \ce{Cl2} \) — first (242.6 kJ mol−1). Chlorine has the strongest X–X bond of the four.
(C) \( \ce{Br2} \) — second (192.8 kJ mol−1). The larger Br atoms give a longer, weaker bond than in \( \ce{Cl2} \).
(A) \( \ce{F2} \) — third (158.8 kJ mol−1). The F–F bond is the shortest, yet it is weaker than the Cl–Cl and Br–Br bonds because of the lone pair repulsion between the two small F atoms.
(D) \( \ce{I2} \) — fourth (151.1 kJ mol−1). Iodine has the largest atoms and the longest, weakest bond.
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.
NCERT (Pre-rationalised edition, 2022-23):
Physical Properties – “… One curious anomaly we notice from Table 7.8 is the smaller enthalpy of dissociation of F2 compared to that of Cl2 whereas X–X bond dissociation enthalpies from chlorine onwards show the expected trend: Cl–Cl > Br–Br > I–I. A reason for this anomaly is the relatively large electron-electron repulsion among the lone pairs in F2 molecule where they are much closer to each other than in case of Cl2.”
