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The bond dissociation energy of $F_2$ is very low due to:
A) Low density.
B) Repulsions between non bonding electrons.
C) Its Low atomic number.
D) Attractions between non bonding electrons.

Last updated date: 13th Jun 2024
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Hint: Bond dissociation energy is the energy required in an endothermic process to break down a chemical bond to form two molecules or atomic structure with either alone or shared pair of electrons. It means large dissociation energy is required to break a stable chemical bond. The bond dissociation energy increases with the increase in the difference of electronegativities of the bonded atoms.

Complete answer:
Fluorine atoms being the smallest molecule has the small radii and theoretically should have the greatest attraction of their nuclei to the shared pair of electrons making it a covalent bond very hard to break. Such a strong nuclear attraction to the valence electrons ( due to high effective nuclear charge ) causes less shielding of valence electrons and leads to interelectronic repulsions between the small F - F molecules. These molecules repel each other because of the valence electrons being closer to each other which makes F - F the bond weak and very low dissociation energy is required.

Hence, the correct option is (B), Repulsions between non bonding electrons.

Note: Bond dissociation is also known as bond enthalpy which calculates standard enthalpy change when a bond is cleaved between reactants and products of the reaction at 0 K (absolute zero). Since fluorine as a molecule exhibits very high electronegative property despite its low size requires less bond dissociation energy to break the bond.