The bond dissociation energy is highest for

Options

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

Answer

To determine which halogen has the highest bond dissociation energy, we will analyze the bond dissociation energies of the halogens: fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2).

### Step-by-Step Solution:

1. **Understanding Bond Dissociation Energy**:
Bond dissociation energy is the energy required to break a bond between two atoms in a molecule. Generally, shorter bonds are stronger and require more energy to break.

2. **Analyzing the Halogens**:
The halogens in question are:
- Fluorine (F2)
- Chlorine (Cl2)
- Bromine (Br2)
- Iodine (I2)

3. **Trends in Bond Length and Size**:
As we move down the group from fluorine to iodine, the atomic size increases due to the addition of electron shells. This increase in size leads to longer bond lengths.

4. **Effect of Bond Length on Bond Dissociation Energy**:
- Longer bonds are generally weaker, meaning they have lower bond dissociation energies.
- Therefore, we can expect that as we go from F2 to I2, the bond dissociation energy will decrease.

5. **Special Case of Fluorine**:
- Although fluorine is the smallest halogen, it has significant inter-electronic repulsion due to its small size and high electronegativity. This results in the F-F bond being longer than expected, making it weaker than the Cl-Cl bond.
- Thus, the bond dissociation energy of F2 is actually lower than that of Cl2.

6. **Comparison of Bond Dissociation Energies**:
- F2 < Cl2 < Br2 < I2
- Therefore, Cl2 has a higher bond dissociation energy than F2, and both Br2 and I2 have even lower bond dissociation energies.

7. **Conclusion**:
The bond dissociation energy is highest for Cl2 among the halogens.

### Final Answer:
The bond dissociation energy is highest for **Cl2**.