Question

Explain`why
${N_2}$ has greater bond energy than $N_2^ +$ but ${O_2}$ has lower bond dissociation energy then $O_2^ +$.

Answer 1

Hint: The bond dissociation energy is related with the bond order of the molecule. When the bond order is high, more energy is required to break the bond between the atom of the molecule.

Complete step by step answer:
The bond dissociation energy is defined as the energy required to break the chemical bond between the two atoms in the molecule.
Bond order is defined as the number of bonds present between the atoms of the molecule. The bond order indicates the stability of the bond in the molecule.
The bond order of any molecule is calculated by the formula shown below.
$B.O = \dfrac{{{N_A} - {N_B}}}{2}$
Where,
B.O is the bond order
${N_B}$ is the number of electrons in a bonding molecular orbital.
${N_A}$ is the number of electrons in a non-bonding molecular orbital.
The bond dissociation energy is related to the bond order of the molecule. When the bond order is high, then high energy is required to break the bond so the bond dissociation energy will be high.
According to molecular orbital theory, the number of electrons in an antibonding molecular orbital of ${N_2}$ is 10 and the number of electrons in bonding orbital is 4.
The bond order is calculated as shown below.
$\Rightarrow B.O = \dfrac{{10 - 4}}{2}$
$\Rightarrow B.O = \dfrac{6}{2}$
$\Rightarrow B.O = 3$
In $N_2^ +$, the number of electrons in antibonding molecular orbital is 9 and the number of electrons in bonding orbital is 4.
The bond order is calculated as shown below.
$\Rightarrow B.O = \dfrac{{9 - 4}}{2}$
$\Rightarrow B.O = \dfrac{5}{2}$
$\Rightarrow B.O = 2.5$
As, the bond order of ${N_2}$ is more than $N_2^ +$ therefore then bond dissociation energy of ${N_2}$ is more than $N_2^ +$.
The number of electrons in the antibonding molecular orbital of ${O_2}$ is 10 and the number of electrons in bonding orbital is 6.
The bond order is calculated as shown below.
$\Rightarrow B.O = \dfrac{{10 - 6}}{2}$
$\Rightarrow B.O = \dfrac{4}{2}$
$\Rightarrow B.O = 2$
In $O_2^ +$, the number of electrons in antibonding molecular orbital is 10 and the number of electrons in bonding orbital is 5.
The bond order is calculated as shown below.
$\Rightarrow B.O = \dfrac{{10 - 5}}{2}$
$\Rightarrow B.O = \dfrac{5}{2}$
$\Rightarrow B.O = 2.5$
As, the bond order of $O_2^ +$ is more than ${O_2}$ therefore then bond dissociation energy of $O_2^ +$ is more than ${O_2}$.

Note: The plus charge on the molecule shows that one electron is lost from the bonding molecular orbital which results in the change in the bond order from the normal bond order.