Answer
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Hint: The bond energy or bond enthalpy can be visualized as the average amount of energy required or produced when a bond in a chemical compound is broken or formed.
Step by step solution:
Bond energy gives an idea of strength of a chemical bond and therefore the stability of the compound.
Hence the more stable the compound, the more energy will be needed to break the bond.
Fluorine$\left( {{F_2}} \right)$, chlorine$\left( {C{l_2}} \right)$, Bromine $\left( {B{r_2}} \right)$ and Iodine $\left( {{I_2}} \right)$ are called halogens compound, they occur as diatomic molecule in which covalent bond is formed after sharing 1 electron with each other.
As it moves down from chlorine to iodine in the halogen group, the atomic size increases and hence bond length also increases.
And as bond length increases, the energy required to the break bonds (dissociation energy) decreases.
Hence we can say among $\left( {C{l_2}} \right)$, $\left( {B{r_2}} \right)$, and $\left( {B{r_2}} \right)$. Chlorine will have the highest bond energy.
Now come to Fluorine:
The bond dissociation energy of $\left( {{F_2}} \right)$ is exceptionally less because Fluorine has small atomic size due to which the electrons are held in a compact volume, consequently there is strong repulsion amongst non-bonded electrons. Hence the bond becomes weak and the dissociation energy is lowest amongst other halogens.
The order of bond dissociation energy is:
$C{l_2} > B{r_2} > {I_2} > {F_2}$
Note: When the bond in a chemical compound is broken, then there is a need for energy hence the bond energy will have a positive value. And when the bond in a chemical compound is formed, then the energy will be released hence the bond energy will have a negative value.
Step by step solution:
Bond energy gives an idea of strength of a chemical bond and therefore the stability of the compound.
Hence the more stable the compound, the more energy will be needed to break the bond.
Fluorine$\left( {{F_2}} \right)$, chlorine$\left( {C{l_2}} \right)$, Bromine $\left( {B{r_2}} \right)$ and Iodine $\left( {{I_2}} \right)$ are called halogens compound, they occur as diatomic molecule in which covalent bond is formed after sharing 1 electron with each other.
As it moves down from chlorine to iodine in the halogen group, the atomic size increases and hence bond length also increases.
And as bond length increases, the energy required to the break bonds (dissociation energy) decreases.
Hence we can say among $\left( {C{l_2}} \right)$, $\left( {B{r_2}} \right)$, and $\left( {B{r_2}} \right)$. Chlorine will have the highest bond energy.
Now come to Fluorine:
The bond dissociation energy of $\left( {{F_2}} \right)$ is exceptionally less because Fluorine has small atomic size due to which the electrons are held in a compact volume, consequently there is strong repulsion amongst non-bonded electrons. Hence the bond becomes weak and the dissociation energy is lowest amongst other halogens.
The order of bond dissociation energy is:
$C{l_2} > B{r_2} > {I_2} > {F_2}$
Note: When the bond in a chemical compound is broken, then there is a need for energy hence the bond energy will have a positive value. And when the bond in a chemical compound is formed, then the energy will be released hence the bond energy will have a negative value.
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