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NEET CHEMISTRYMedium

Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?

A

Cl2>Br2>F2>I2\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2

B

Br2>I2>F2>Cl2\text{Br}_2 > \text{I}_2 > \text{F}_2 > \text{Cl}_2

C

F2>Cl2>Br2>I2\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2

D

I2>Br2>Cl2>F2\text{I}_2 > \text{Br}_2 > \text{Cl}_2 > \text{F}_2

Step-by-Step Solution

Generally, as we move down the halogen group, the atomic size increases, which leads to longer and weaker bonds, so bond dissociation enthalpy should decrease . However, fluorine (F2\text{F}_2) is anomalous. Because of the extremely small size of the fluorine atom, the lone pairs of electrons on the two adjacent bonded atoms experience strong interelectronic repulsions , making the F-F\text{F-F} bond weaker than expected. As a result, the bond dissociation enthalpy of F2\text{F}_2 is lower than that of both Cl2\text{Cl}_2 and Br2\text{Br}_2. According to the standard values , the bond dissociation enthalpies are roughly Cl2\text{Cl}_2 (243\approx 243 kJ/mol) > Br2\text{Br}_2 (192\approx 192 kJ/mol) > F2\text{F}_2 (155\approx 155 kJ/mol) > I2\text{I}_2 (151\approx 151 kJ/mol). Therefore, the correct order is Cl2>Br2>F2>I2\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2.

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