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

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

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

Question

Accepted Answer

The correct order of bond dissociation enthalpy among the given diatomic molecules is: $Cl_{2}>Br_{2}>F_{2}>I_{2}$, which corresponds to Option 3. Let's delve into the reasoning behind this order and why the other options are incorrect.

### Explanation for Option 3:

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 influenced by factors such as atomic size and electron-electron repulsion.

2. **Chlorine ($Cl_{2}$):** Chlorine has a relatively high bond dissociation enthalpy due to its optimal balance between atomic size and bond strength. The bond length in $Cl_{2}$ is shorter than in $Br_{2}$ and $I_{2}$, leading to a stronger bond.

3. **Bromine ($Br_{2}$):** Bromine has a lower bond dissociation enthalpy than chlorine because of its larger atomic size, which results in a longer bond length and weaker bond strength compared to $Cl_{2}$.

4. **Fluorine ($F_{2}$):** Although fluorine is smaller than chlorine, the bond dissociation enthalpy is lower than that of $Cl_{2}$ and $Br_{2}$ due to significant electron-electron repulsion in the small $F_{2}$ molecule. This repulsion weakens the bond.

5. **Iodine ($I_{2}$):** Iodine has the lowest bond dissociation enthalpy among the halogens listed due to its large atomic size, which results in a long bond length and weak bond strength.

### Why Other Options are Incorrect:

- **Option 1 ($F_{2}>Cl_{2}>Br_{2}>I_{2}$):** This option incorrectly places $F_{2}$ with the highest bond dissociation enthalpy. As explained, $F_{2}$ has significant electron-electron repulsion, making its bond weaker than $Cl_{2}$ and $Br_{2}$.

- **Option 2 ($Cl_{2}>F_{2}>Br_{2}>I_{2}$):** While this option correctly places $Cl_{2}$ above $F_{2}$ and $I_{2}$, it incorrectly places $F_{2}$ above $Br_{2}$. The electron-electron repulsion in $F_{2}$ makes its bond weaker than that of $Br_{2}$.

- **Option 4 ($Cl_{2}>Br_{2}>I_{2}>F_{2}$):** This option incorrectly places $F_{2}$ with the lowest bond dissociation enthalpy. Although $F_{2}$ is weaker than $Cl_{2}$ and $Br_{2}$, it is stronger than $I_{2}$ due to iodine's larger atomic size and longer bond length.

In conclusion, Option 3 is the correct order of bond dissociation enthalpy for the given diatomic molecules, based on the balance of atomic size, bond length, and electron repulsion factors.