Question -
Answer -
Bond order is defined asone half of the difference between the number of electrons present in thebonding and anti-bonding orbitals of a molecule.
If Na is equal to thenumber of electrons in an anti-bonding orbital, then Nb is equal to thenumber of electrons in a bonding orbital.
Bondorder = 
If Nb > Na, then the molecule issaid be stable. However, if Nb ≤ Na, then the molecule isconsidered to be unstable.
Bond order of N2 can be calculatedfrom its electronic configuration as:
Number of bonding electrons, Nb = 10
Number of anti-bonding electrons, Na = 4
Bondorder of nitrogen molecule 
= 3
There are 16 electrons ina dioxygen molecule, 8 from each oxygen atom. The electronic configuration ofoxygen molecule can be written as:
Since the 1s orbital of eachoxygen atom is not involved in boding, the number of bonding electrons = 8= Nb and the number of anti-bonding electrons = 4= Na.
Bondorder 
= 2
Hence, the bond order ofoxygen molecule is 2.
Similarly,the electronic configuration of 
can be written as:
Nb = 8
Na = 3
Bondorder of 
= 2.5
Thus,the bond order of is 2.5.Theelectronic configuration of 
ion will be:Nb = 8
Na = 5
Bondorder of =
= 1.5
Thus,the bond order of ionis 1.5.