Question

The correct order of bond dissociation enthalpy of halogens is :

• A. $Cl _{2}> F _{2}> Br _{2}> I _{2}$
• B. $I _{2}> Br _{2}> Cl _{2}> F _{2}$
• C. $Cl _{2}> Br _{2}> F _{2}> I _{2}$
• D. $F _{2}> Cl _{2}> Br _{2}> I _{2}$

Answer 1

The Correct Option is C

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:

\(Cl _{2}> Br _{2}> F _{2}> I _{2}\)

Let's examine why the other options are incorrect:

• \(Cl _{2}> F _{2}> Br _{2}> I _{2}\): Incorrect because F₂ has more repulsion and consequently less bond strength than Br₂, and thus cannot have greater bond enthalpy.
• \(I _{2}> Br _{2}> Cl _{2}> F _{2}\): Incorrect as I₂ cannot have greater enthalpy due to its weak bond strength.
• \(F _{2}> Cl _{2}> Br _{2}> I _{2}\): Incorrect because Cl₂ has higher enthalpy compared to F₂ due to optimal bond length and fewer repulsions.

Thus, the correct answer is indeed \(Cl _{2}> Br _{2}> F _{2}> I _{2}\).