Question

Which of the following arrangements of bond dissociation enthalpy of halogen is correct?
A $F - F > Cl - Cl > Br - Br$
B. $F - F < Cl - Cl < Br - Br$
C.$F - F > Cl - Cl < Br - Br$
D.$F - F < Cl - Cl > Br - Br$

Answer 1

Hint: The bond dissociation energy is defined as the amount of energy required to break the bond between the atoms so that they split into the individual atom. The bond dissociation energy depends on the atomic size of the element as well as the bond length.

Complete step by step answer:
The halogen molecule is a diatomic molecule where the atom is joined with the other atom by a covalent bond. The covalent bond is formed when the overlapping of p-orbital takes place.
As the atomic size increases from the chlorine atom to the iodine atom, the bond length also increases. With an increase in the bond length from chlorine atom to iodine atom, there is a decrease in the Bond dissociation energy.
But the bond dissociation energy in fluorine molecules is less than the chlorine molecule because of the small size of the fluorine atom. In a fluorine atom, the 2p orbital is very compact and very near to the nucleus. Because of the small size, the electrons are localized at compact volume, and this results in strong repulsion between the non-bonding electrons. Thus, the bond becomes weak even though the bond length is short. Therefore, the bond dissociation energy of fluorine is less than chlorine and another halogen atom.
Thus, the order of bond dissociation energy is $F - F < Cl - Cl > Br - Br$.

Thus,the correct option is D.

Note: In chlorine molecule, bromine molecule and iodine molecule multiple bonding can be seen due to the presence of d-orbital. As d-orbital is absent in fluorine, multiple bonding cannot be seen. This is why the bond dissociation energy is high for chlorine and other halogen molecules.