Enthalpy of dissociation is low for 

Options

A. `F_2`
B. `Cl_2`
C. `Br_2`
D. `I_2`

Answer

To solve the question regarding which halogen has the lowest enthalpy of dissociation, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Enthalpy of Dissociation**:
- The enthalpy of dissociation refers to the energy required to break a bond in a molecule. A lower enthalpy of dissociation indicates that the bond is weaker and requires less energy to break.

2. **Analyzing the Halogen Group**:
- The halogens are found in Group 17 (VIIA) of the periodic table and include fluorine (F), chlorine (Cl), bromine (Br), and iodine (I).

3. **Trend in Bond Dissociation Energy**:
- As we move down the group from fluorine to iodine, the size of the atoms increases. Larger atoms have longer bond lengths, which generally leads to weaker bonds and thus lower bond dissociation energies.

4. **Comparing the Halogens**:
- The order of bond dissociation energies for the halogens is as follows:
- Fluorine (F2) > Chlorine (Cl2) > Bromine (Br2) > Iodine (I2)
- However, fluorine is an exception due to significant repulsion between its lone pairs, resulting in a relatively high bond dissociation energy despite its small size.

5. **Identifying the Lowest Enthalpy of Dissociation**:
- Among the halogens, iodine (I2) has the lowest bond dissociation energy because it is the largest atom in the group, leading to weaker bonds.

6. **Conclusion**:
- Therefore, the enthalpy of dissociation is lowest for iodine (I2).

### Final Answer:
The enthalpy of dissociation is low for iodine (I2).