Arrange the halogens in the increasing order of bond dissociation enthalpy:

(A) \(F_2\)

(B) \(Cl_2\)

(C) \(Br_2\)

(D) \(I_2\)

Choose the correct answer from the options given below:

B > C > A > D

The correct answer is option 4. B > C > A > D.

Bond dissociation enthalpy is the energy required to break a bond between two atoms in a molecule. For diatomic halogen molecules (e.g., \( F_2 \), \( Cl_2 \), \( Br_2 \), \( I_2 \)), BDE refers to the energy needed to break the bond between the two halogen atoms in each molecule.

General Trend in Halogens

As we move down the group of halogens from fluorine to iodine, the bond dissociation enthalpy generally decreases. This trend can be explained by considering several factors:

Atomic Size and Bond Length:

Atomic Size: As you move down the group from fluorine to iodine, the size of the atoms increases. Fluorine atoms are much smaller than iodine atoms.

Bond Length: The bond length increases as the size of the atoms increases. For example, the bond length in \( I_2 \) is much longer than in \( F_2 \) because iodine atoms are larger.

Bond Strength:

Weaker Bonds: A longer bond is generally weaker because the bonding electrons are shared over a larger distance, leading to less effective bonding.

As the bond length increases from \( F_2 \) to \( I_2 \), the bond strength decreases, resulting in lower BDE. Therefore, it requires less energy to break the bond in larger halogen molecules.

Exception: Fluorine

Fluorine is an exception to the general trend due to specific reasons:

Electron-Electron Repulsion:

Small Size and Lone Pairs: In \( F_2 \), the fluorine atoms are very small, and the lone pairs of electrons on each fluorine atom are very close together.

Repulsion: The close proximity of these lone pairs results in significant electron-electron repulsion. This repulsion weakens the \( F-F \) bond compared to what would be expected based on the general trend.

Weakened Bond:

Bond Strength in \( F_2 \): Despite fluorine’s small size generally favoring stronger bonds, the lone pair repulsion in \( F_2 \) weakens the bond significantly.

Comparison to Chlorine: In contrast, chlorine atoms are larger, so the lone pairs are further apart, resulting in less repulsion and a stronger \( Cl-Cl \) bond compared to \( F-F \).

Let us compare the BDE of each diatomic halogen molecule:

A. \( F_2 \) (Fluorine):

The \( F-F \) bond is weakened by strong electron-electron repulsion between the lone pairs on the small fluorine atoms. This results in a lower bond dissociation enthalpy compared to chlorine.

B. \( Cl_2 \) (Chlorine):

The \( Cl-Cl \) bond is stronger than the \( F-F \) bond due to less lone pair repulsion. The bond dissociation enthalpy of \( Cl_2 \) is higher than that of \( F_2 \).

C. \( Br_2 \) (Bromine):

The \( Br-Br \) bond has even less bond dissociation enthalpy than \( Cl_2 \) due to the larger size of bromine atoms compared to chlorine.

D. \( I_2 \) (Iodine):

The \( I-I \) bond is the weakest among the halogens due to the large size of iodine atoms, which results in a very long bond length and thus lower bond strength.

Conclusion:

Therefore, the correct order of bond dissociation enthalpy from highest to lowest is:

\(\text{Cl}_2 > \text{Br}_2 > \text{F}_2 > \text{I}_2\)

This order reflects the balance between increasing bond length and decreasing bond strength as we move down the group, with the exception of fluorine’s lower BDE due to lone pair repulsion.

Thus, the correct answer is: 4. B > C > A > D.