What is meant by the term bond order ? Calculate the bond order of: $N _{2}, O _{2}, $ $O _{2}^{+}$ and $O _{2}^{-}$

Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.

If $N_{ a }$ is equal to the number of electrons in an anti-bonding orbital, then $N_{ b }$ is equal to the number of electrons in a bonding orbital.

Bond order $=\frac{1}{2}\left(N_{ b }-N_{ a }\right)$

If $N_{ b }\,>\,N_{ a ,}$ then the molecule is said be stable. However, if $N_{ b }\, \leq \,N_{ a }$ then the molecule is considered to be unstable.

Bond order of $N _{2}$ can be calculated from its electronic configuration as:

${[\sigma (1s)]^2}{\left[ {{\sigma ^*}(1s)} \right]^2}{[\sigma (2s)]^2}{\left[ {{\sigma ^*}(2s)} \right]^2}{\left[ {\pi \left( {2{p_x}} \right)} \right]^2}{\left[ {\pi \left( {2{p_y}} \right)} \right]^2}{\left[ {\sigma \left( {2{p_z}} \right)} \right]^2}$

Number of bonding electrons, $N_{b}=10$

Number of anti-bonding electrons, $N_{a}=4$

Bond order of nitrogen molecule $=\frac{1}{2}(10-4)$ $=3$

There are $16$ electrons in a dioxygen molecule, $8$ from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

${[\sigma  - (1s)]^2}{\left[ {{\sigma ^*}(1s)} \right]^2}{[\sigma (2s)]^2}{\left[ {{\sigma ^*}(2s)} \right]^2}{\left[ {\sigma \left( {1{p_z}} \right)} \right]^2}{\left[ {\pi \left( {2{p_x}} \right)} \right]^2}

{\left[ {\pi \left( {2{p_y}} \right)} \right]^2}{\left[ {{\pi ^*}\left( {2{p_x}} \right)} \right]^1}{\left[ {{\pi ^*}\left( {2{p_y}} \right)} \right]^1}$

Since the $1 s$ orbital of each oxygen atom is not involved in boding, the number of bonding electrons $=8=N_{b}$ and the number of anti-bonding electrons $=4=N_{ a }$

Bond order $=\frac{1}{2}\left(N_{ b }-N_{ a }\right)$

$=\frac{1}{2}(8-4)$

$=2$

Hence, the bond order of oxygen molecule is $2$

Similarly, the electronic configuration of $O _{2}^{+}$ can be written as:

$KK{[\sigma (2s)]^2}{\left[ {{\sigma ^*}(2s)} \right]^2}{\left[ {\sigma \left( {2{p_z}} \right)} \right]^2}{\left[ {\pi \left( {2{p_x}} \right)} \right]^2}{\left[ {\pi \left( {2{p_y}} \right)} \right]^2}{\left[ {{\pi ^*}\left( {2{p_x}} \right)} \right]^1}$

$N_{b}=8 N_{a}$

$=3$

Bond order of $O _{2}^{+}=\frac{1}{2}(8-3)$

$=2.5$

Thus, the bond order of $O _{2}^{+}$ is $2.5$

$O _{2}$
The electronic configuration of $O _{2}^{-}$ ion will be:

$KK{[\sigma (2s)]^2}{\left[ {{\sigma ^*}(2s)} \right]^2}{\left[ {\sigma \left( {2{p_z}} \right)} \right]^2}{\left[ {\pi \left( {2{p_x}} \right)} \right]^2}{\left[ {\pi \left( {2{p_y}} \right)} \right]^2}{\left[ {{\pi ^*}\left( {2{p_x}} \right)} \right]^2}

{\left[ {{\pi ^*}\left( {2{p_y}} \right)} \right]^1}$

$N_{b}=8 N_{a}$

$=5$

Bond order of $O _{2}^{-}=\frac{1}{2}(8-5)$

$=1.5$

Thus, the bond order of $O _{2}^{-}$ ion is $1.5$