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

Options

A. `Cl_(2) gt Br_(2) gt F_(2) ge I_(2)`
B. `Br_(2) gt I_(2) gt F_(2) gt Cl_(2)`
C. `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`
D. `I_(2) gt Br_(2) gt Cl_(2) gt F_(2)`

Explanation

To determine the correct order for the bond dissociation enthalpy of halogen molecules, we need to analyze the halogens in Group 17 of the periodic table, which includes Fluorine (F2), Chlorine (Cl2), Bromine (Br2), and Iodine (I2).

### Step-by-Step Solution:

1. **Understanding Bond Dissociation Enthalpy**:
- Bond dissociation enthalpy is the energy required to break a bond between two atoms in a molecule. Higher bond dissociation enthalpy means a stronger bond.

2. **Trend in the Group**:
- As we move down the group from Fluorine to Iodine, the atomic size increases. This increase in size leads to longer bond lengths and weaker bonds due to reduced overlap of atomic orbitals.

3. **Analyzing Each Halogen**:
- **Fluorine (F2)**: Despite being the smallest atom, F2 has a low bond dissociation enthalpy due to significant lone pair-lone pair repulsion (there are three lone pairs on each F atom). This repulsion weakens the bond.
- **Chlorine (Cl2)**: Chlorine has a higher bond dissociation enthalpy than Fluorine because the lone pair-lone pair repulsion is less significant compared to Fluorine.
- **Bromine (Br2)**: Bromine has an even lower bond dissociation enthalpy than Chlorine due to its larger atomic size and longer bond length.
- **Iodine (I2)**: Iodine has the lowest bond dissociation enthalpy among the halogens due to the largest atomic size and weakest bond.

4. **Conclusion**:
- The general trend for bond dissociation enthalpy in halogens is:
**F2 > Cl2 > Br2 > I2**.

### Final Answer:
The correct order for the bond dissociation enthalpy of halogen molecules is:
**F2 > Cl2 > Br2 > I2**.