Question

Which of the following has the maximum bond energy?
A.${\text{C}}{{\text{l}}_{\text{2}}}$
B.${{\text{F}}_{\text{2}}}$
C.${\text{B}}{{\text{r}}_{\text{2}}}$
D.${{\text{I}}_{\text{2}}}$

Answer 1

The bond energy of any molecules depends on the bond strength of the molecule which in turn depends on the atomic size and the electronic configuration of the atoms. Among the halogens, the fluorine atom has the smallest size while the iodine atom has the largest size.

Complete step by step answer:
In the group 17 or the halogen family of the periodic table, the order of bond energy is as follows:
${\text{C}}{{\text{l}}_{\text{2}}} > {{\text{F}}_2} > {\text{B}}{{\text{r}}_2} > {{\text{I}}_2}$
The bond strength of the molecules is dependent on the size of the molecules as smaller the atom, the more effective will be the bonding and hence better bond strength. But here the case is anomalous because accordingly, fluorine should have the highest bond strength. But, due to the small shape of the fluorine atom and five electrons in the valence shell, there is a shortage of space due to increased nuclear charge. Hence there is a repulsion of lone pairs of the adjacent fluorine atoms.
But in case of chlorine, the presence of the 3d orbital in the valence shell creates more space for the valence electrons and the bond or overlap of electrons occurs smoothly.
Hence, the bond energy is maximum for ${\text{C}}{{\text{l}}_{\text{2}}}$.

So, option A is the correct answer.

Note:
In the case of Bromine and Iodine, there is enough space in the 4d and 5d orbitals to accommodate electrons and avoid steric repulsion but the diffused shape of these orbitals weakens the bond strength and hence, the bond energy is less for them.