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Which of the following has least bond dissociation energy?
(a)- ${{F}_{2}}$
(b)- $C{{l}_{2}}$
(c)- $B{{r}_{2}}$
(d)- ${{I}_{2}}$

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Last updated date: 21st Jul 2024
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Answer
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Hint: All the molecules given in the options are diatomic molecules so, when they are separated into radicals then there some energy must be applied. Therefore, this energy is known as bond dissociation energy. As the size of the atom increases, the bond strength decreases and the energy to break the bond also decreases.

Complete answer:
Some of the molecules have strong bonds which require energy to break bonds. The bond dissociation energy is the energy that is used to break the bond between the molecules. All the molecules given in the options are diatomic molecules so, when they are separated into radicals then some energy must be applied because the bonds between the molecules are strong.
The molecules in the options are fluorine, chlorine, bromine, and iodine. We know that these are the elements of group 17, i.e., the halogen family. These are highly electronegative atoms, so they require high energy to break the bond between them.
As the size of the atoms increases, the bond between the atoms weakens. The size of the atom increases down the group in the halogen family. So, the size of the iodine atom is the largest so the bond between ${{I}_{2}}$. Therefore, the energy required to break the bond will be the least.

Hence, the correct answer is an option (d).

Note:
Fluorine's bond dissociation energy is lower than that of chlorine and bromine due to the fluorine's higher interelectronic repulsions than those of bromine and chlorine.