Question

What is meant by the term bond order? Calculate the bond order of: $N_2$, $O_2$, $O_2^+$ and $O_2^-$ .

Answer 1

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, thenNb is equal to the number of electrons in a bonding orbital.
Bond order = $\frac{1}{2}(N_b-N_a)$
If $Nb > Na$ , then the molecule is said be stable. However, if $Nb \leq Na$, then the molecule is considered to be unstable.
Bond order of $N_2$ can be calculated from its electronic configuration as:

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

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:

$[\sigma-(1s)]^2[\sigma^*(1s)]^2[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(1p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[π*(2px)]2[\pi^*(2p_y)]^1$

Since the $1s$ 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 =$\frac{1}{2}(N_b-N_a)$
=$\frac{1}{2}(8-4)$
=$2$

Hence, the bond order of oxygen molecule is $2$.
Similarly, the electronic configuration of $O_2^+$ can be written as:
$KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^1$

$N_b$ = $8$
$N_a$ = $3$
Bond order of $O_2^+=\frac{1}{2}(8-3)$
= $2.5$
Thus, the bond order of $O_2^+$ is $2.5$.

The electronic configuration of ion will be:

$KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^2[\pi^*(2p_y)]^1$
$N_b$ = $8$
$N_a$ = $5$
Bond order of $O_2^-$=$\frac{1}{2}(8-5)$
= $1.5$
Thus, the bond order of $O_2^-$ ion is $1.5$.