Question

Calculate the bond order of $\text{ O}_{\text{2}}^{\text{+}}\text{ }$.

Answer 1

Hint: First draw a molecular orbital diagram (MOT) where the atomic orbitals combine to form molecular orbitals. The total electrons associated with the molecules are filled in the MOT diagram.
To solve this question, we need to write the molecular orbital configuration. To find out the bond order from the molecular orbital configuration is:
$\text{Bond order =}\dfrac{\text{1}}{\text{2}}\left[ \text{Bonding electrons -antibonding electrons} \right]$

Complete answer:
The atomic orbitals of the oxygen and one electron less oxygen atom$\text{ }{{\text{O}}^{\text{+}}}\text{ }$ combine to form the molecular orbitals of $\text{ O}_{\text{2}}^{\text{+}}\text{ }$the molecule. the electronic configuration of oxygen and $\text{ }{{\text{O}}^{\text{+}}}\text{ }$can be written as follows,
Electronic configuration of O :
${{\text{ }}_{8}}\text{O = 1}{{\text{s}}^{\text{2}}}\text{ , 2}{{\text{s}}^{\text{2}}}\text{ , 2p}_{x}^{2}=\text{2p}_{y}^{1}=\text{2p}_{z}^{1}\text{ }$
The oxygen loses one electron and forms$\text{ }{{\text{O}}^{\text{+}}}\text{ }$. Thus, Electronic configuration of $\text{ }{{\text{O}}^{\text{+}}}\text{ }$is,
${{\text{ }}_{7}}{{\text{O}}^{\text{+}}}\text{ = 1}{{\text{s}}^{\text{2}}}\text{ , 2}{{\text{s}}^{\text{2}}}\text{ , 2p}_{x}^{1}=\text{2p}_{y}^{1}=\text{2p}_{z}^{1}\text{ }$
Thus, 8 electrons from the O atom and 7 electrons from the $\text{ }{{\text{O}}^{\text{+}}}\text{ }$are needed to fill in the molecular orbitals. Therefore, the MOT diagram $\text{O}_{\text{2}}^{\text{+}}$ contains the 15 electrons (8 electrons of O + 7 electrons of $\text{ }{{\text{O}}^{\text{+}}}\text{ }$).
The MOT diagram for the $\text{O}_{\text{2}}^{\text{+}}$ molecule is as shown below,

Therefore, the molecular orbital configuration of $\text{O}_{\text{2}}^{\text{+}}$ is as follows:$\text{O}_{\text{2}}^{\text{+}}$
$\text{ }\!\!\sigma\!\!\text{ 1}{{\text{s}}^{\text{2}}}\text{,}{{\text{ }\!\!\sigma\!\!\text{ }}^{\text{*}}}\text{1}{{\text{s}}^{\text{2}}}\text{, }\!\!\sigma\!\!\text{ 2}{{\text{s}}^{\text{2}}}\text{, }{{\text{ }\!\!\sigma\!\!\text{ }}^{\text{*}}}\text{2}{{\text{s}}^{\text{2}}}\text{, }\!\!\sigma\!\!\text{ 2}{{\text{p}}^{\text{2}}}_{z}\text{, 2p}_{\text{x}}^{\text{2}}\text{ }\!\!\pi\!\!\text{ = 2p}_{\text{y}}^{\text{2}}\text{ }\!\!\pi\!\!\text{ , 2p}_{\text{x}}^{\text{2}}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{*}}}\text{=2p}_{\text{y}}^{0}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{*}}}$
There are 10 bonding (including molecular orbitals formed by the $\text{1s}$ orbitals) and 5 electrons in the antibonding molecular orbitals. The number of bonding and antibonding electrons can be determined by simply calculating the electrons in the orbital.
We know that the bond order of a molecule depends on the total number of electrons in bonding and antibonding MO. The bonding electrons stabilize the molecule while antibonding electrons contribute towards the destabilization of the molecule. Therefore bond order is written as,
$\text{Bond order =}\dfrac{\text{1}}{\text{2}}\left[ \text{Bonding-antibonding} \right]\text{ }$
Substitute the number of electrons. We have,
$\text{Bond order }=\dfrac{1}{2}\text{ }\left[ 10-5 \right]=2.5$
Therefore, the bond order of $\text{O}_{\text{2}}^{\text{+}}$ is $2.5$.

Note:
Note that, 2.5. Bond order may look confusing. The two oxygen atoms have two complete bonds and one partial bond thus it is 2.5.

Bond order is indirectly proportional to the length of the bond. The higher the bond order, the shorter and stronger will be the bond. The addition of each electron in the antibonding molecular orbital will decrease the bond order.