The correct order of increasing bond dissociation enthalpy among the following is:

Question

Accepted Answer

The correct order of increasing bond dissociation enthalpy among the given diatomic molecules is:

1. $I_{2}<F_{2}<Br_{2}<Cl_{2}$

### Explanation:

1. **Bond Dissociation Enthalpy (BDE):** This is the energy required to break a bond in a molecule to form two separate atoms. The strength of the bond is directly related to the bond dissociation enthalpy; stronger bonds have higher dissociation enthalpies.

2. **Iodine ($I_{2}$):** Iodine has the weakest bond among the halogens due to its large atomic size, which results in a longer bond length and weaker bond strength. Therefore, it has the lowest bond dissociation enthalpy.

3. **Fluorine ($F_{2}$):** Although fluorine is a small atom and typically forms strong bonds, the $F-F$ bond is relatively weak compared to $Br_{2}$ and $Cl_{2}$ due to the high electron-electron repulsion between the lone pairs on the fluorine atoms. This repulsion weakens the bond, resulting in a lower bond dissociation enthalpy than $Br_{2}$ and $Cl_{2}$.

4. **Bromine ($Br_{2}$):** Bromine has a stronger bond than iodine and fluorine due to its moderate atomic size, which allows for a balance between bond length and bond strength. Thus, its bond dissociation enthalpy is higher than that of $I_{2}$ and $F_{2}$.

5. **Chlorine ($Cl_{2}$):** Chlorine has the strongest bond among the halogens listed here. Its smaller atomic size compared to bromine and iodine, combined with less electron-electron repulsion compared to fluorine, results in a strong bond with the highest bond dissociation enthalpy.

### Why Other Options Are Incorrect:

- **Option 2 ($I_{2}<Br_{2}<Cl_{2}<F_{2}$):** This option incorrectly places $F_{2}$ with the highest bond dissociation enthalpy. As explained, the $F-F$ bond is weakened by electron repulsion, making it weaker than $Br_{2}$ and $Cl_{2}$.

- **Option 3 ($I_{2}<Br_{2}<F_{2}<Cl_{2}$):** This option incorrectly suggests that $F_{2}$ has a higher bond dissociation enthalpy than $Br_{2}$. However, due to the reasons mentioned above, $F_{2}$ has a lower bond dissociation enthalpy than $Br_{2}$.

- **Option 4 ($I_{2}<Cl_{2}<Br_{2}<F_{2}$):** This option incorrectly places $Cl_{2}$ before $Br_{2}$ and $F_{2}$ as having a lower bond dissociation enthalpy. In reality, $Cl_{2}$ has the highest bond dissociation enthalpy among the given molecules.

Thus, Option 1 is the correct order based on the bond dissociation enthalpies of the halogen molecules.