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

(A) I2 > Br2 > Cl2 > F2

(B) Cl2 > Br2 > F2 > I2

(C) Br2 > I2 > F2 > Cl2

(D) F2 > Cl2 > Br2 > I2

To determine the correct order of bond dissociation enthalpy for halogen molecules, we need to consider the strength of the bonds formed between the atoms in these diatomic molecules: I2, Br2, Cl2, and F2. The bond dissociation enthalpy is essentially the energy required to break a bond between two atoms in a molecule, and it can be influenced by several factors, including atomic size and electronegativity.

Understanding Bond Strength in Halogens

Halogens are located in Group 17 of the periodic table, and as you move up the group from iodine (I) to fluorine (F), the atomic size decreases. This decrease in size generally leads to stronger bonds due to the increased overlap of atomic orbitals. However, there are exceptions due to other factors, such as lone pair repulsion.

Analyzing Each Molecule

• Iodine (I2): Iodine has the largest atomic radius, leading to a relatively weak bond. The bond dissociation enthalpy is lower compared to the other halogens.
• Bromine (Br2): Bromine is smaller than iodine, resulting in a stronger bond than I2, but still not as strong as Cl2 or F2.
• Chlorine (Cl2): Chlorine has a smaller atomic radius than bromine, which allows for better orbital overlap, leading to a stronger bond than both I2 and Br2.
• Fluorine (F2): Although fluorine is the smallest halogen, the bond in F2 is weaker than Cl2 due to significant lone pair repulsion between the two fluorine atoms, which destabilizes the bond.

Comparing Bond Dissociation Enthalpies

Based on the above analysis, we can summarize the bond dissociation enthalpies in the following order:

• Cl2 has a higher bond dissociation enthalpy than Br2.
• Br2 has a higher bond dissociation enthalpy than I2.
• F2, despite being the smallest, has a lower bond dissociation enthalpy than Cl2 due to lone pair repulsion.

Final Order of Bond Dissociation Enthalpy

Putting this all together, the correct order from highest to lowest bond dissociation enthalpy is:

• Cl2 > Br2 > F2 > I2

Thus, the correct answer to your question is (B) Cl2 > Br2 > F2 > I2. This order reflects the balance between atomic size, bond strength, and the effects of lone pair repulsion in these halogen molecules.