The bond dissociation enthalpy refers to the energy required to break the bond between two atoms in a molecule. In the case of diatomic halogens (X2 molecules), this enthalpy helps determine the strength of the bond between two halogen atoms. The order of bond dissociation enthalpy for halogens is determined by a combination of atomic size, bond length, and bond strength.
Let's analyze the bond dissociation enthalpies of the halogens:
- Fluorine (F2): Despite being a small atom with high electronegativity, its bond dissociation enthalpy is relatively lower because of the electron-electron repulsions in its small molecule.
- Chlorine (Cl2): Exhibits the highest bond dissociation enthalpy among halogens due to optimal bond length and strong bond strength with minimal repulsion.
- Bromine (Br2): Has a moderate bond dissociation enthalpy due to increased atomic size compared to Cl2, resulting in a weaker bond than chlorine but stronger than iodine.
- Iodine (I2): Has the lowest bond dissociation enthalpy among the common halogens due to the large atomic size, resulting in a longer and weaker bond.
From the above analysis, the correct order of bond dissociation enthalpy of halogens is:
Let's examine why the other options are incorrect:
- : Incorrect because F2 has more repulsion and consequently less bond strength than Br2, and thus cannot have greater bond enthalpy.
- : Incorrect as I2 cannot have greater enthalpy due to its weak bond strength.
- : Incorrect because Cl2 has higher enthalpy compared to F2 due to optimal bond length and fewer repulsions.
Thus, the correct answer is indeed .