Question

Which one of the following halogens has the highest bond energy?

(A) ${{F}_{2}}$

(B) $C{{l}_{2}}$

(C) $B{{r}_{2}}$

(D) ${{I}_{2}}$

Answer 1

Hint: In halogens, the bond energy is closely associated with the bond dissociation energy. As the amount of energy needed to break the bond will define the strength of the bond between the atoms.

Complete step by step solution:

The halogens formed by the group 17 elements are present as diatoms, with the outermost shell in the halogen having half-filled p-orbital which undergoes overlapping with adjacent atoms to form the covalent bond. This attraction faced by the bonding pair from both the nuclei holds the molecule together. Thus, the strength of the bond depends on the distance between the bond pair and the nuclei.

Then, with the increase in the size of the atoms down the halogen group, the distance of the nuclei from the bond pair increases leading to the decrease in the effective attraction and thus the bond weakens.

So, the bond dissociation enthalpy also decreases with the decrease of the bond strength down the group. That is, the bond formation energy is directly proportional to the bond dissociation energy.

Therefore, the order for the dissociation enthalpy is as follows:

\[Cl-Cl\,\,\,>Br-Br\,\,>F-F\,\,>I-I\]

With the fluorine diatom having the less bond strength is due to its small size, having only s and p-orbitals. It has three lone pairs of electrons in the outermost shell which face a significant amount of repulsion from the adjacent atom. Thus, this repulsion counters the attraction force and hence, weakens the bond.

Therefore, the halogen with the highest bond energy to be option (B)- $C{{l}_{2}}$.

Note: Except, for fluorine, the presence of the d-orbitals in the other higher halogen atoms. This involves them in the  formation of multiple bonding which leads to their high bond energy.