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Arrange the following molecules in increasing order of energy: ${N_2}, {O_2}, C{l_2}, {F_2}$.

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Last updated date: 13th Jun 2024
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Answer
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Hint: Bond dissociation energy is the energy required to break the chemical bond present between atoms of a molecule. Bond dissociation energy is directly proportional to the number of bonds. The more the number of the bonds the more energy will be required to break the bonds.

Step-by-Step Solution-
We have to arrange the following molecules in increasing order of energy-
${N_2}, {O_2}, C{l_2}, {F_2}$
In${N_2}$, nitrogen has atomic number $7$ and electronic configuration-$1{s^2},2{s^2}2{p^3}$ so to complete its octet it forms triple bond with the other nitrogen which is written as-$N \equiv N$.
In${O_2}$, oxygen has atomic number $8$ and electronic configuration- $1{s^2},2{s^2}2{p^4}$ so to complete its octet oxygen forms double bond with the other oxygen atom which is written as-$O = O$
In $C{l_2}$, chlorine has atomic number- $17$ and electronic configuration- $\left[ {{\text{Ne}}} \right]3{s^2}3{p^5}$ so it forms single bond with other chlorine atom which is written as-$Cl - Cl$ .
In ${F_2}$, fluorine has atomic number-$9$ and electronic configuration $1{s^2},2{s^2}2{p^5}$ so it also forms single bond with other fluorine atom which is written as $F - F$.
Now we know that Bond dissociation energy is directly proportional to the number of bonds.
According to this ${N_2}$ has the highest energy as it has a triple bond so it will require more energy to break the bond. The next highest is ${O_2}$ as it has a double bond.
Now both $C{l_2}$ and ${F_2}$ have a single bond. Here $C{l_2}$ will have more bond dissociation energy than ${F_2}$ because ${F_2}$ has smaller size as compared with $C{l_2}$ so there will be more repulsion between electron and nuclei of the atom as compared to chlorine molecule. Due to this, the bond formed between fluorine and fluorine is weak and requires only little energy to break the bond.
Answer- Hence the increasing order of the energy is given as- ${F_2} < C{l_2} < {O_2} < {N_2}$

Note: Here the general rule is that-
As the radius becomes bigger, the distance between electrons of the outermost shell and nucleus also increases. And we know that the outermost shell electrons form bonds. So covalent bonds become weak due to which less energy is required to break the bond.
But ${F_2}$ has small size due to which the nuclei and electrons are very close. This causes repulsion between the electrons due to which the covalent bond weakens.