CBSE Class 11-science Answered

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 =1/2(N_b-N_a)

If N_b > N_a, then the molecule is said be stable. However, if N_b ? N_a, then the molecule is considered to be unstable.

Bond order of N₂ can be calculated from its electronic configuration as:

Number of bonding electrons, N_b = 10

Number of anti-bonding electrons, N_a = 4

Bond order of nitrogen molecule

= 3

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

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=1/2(Nb-Na)

 = 2

Hence, the bond order of oxygen molecule is 2.

Similarly, the electronic configuration of can be written as:

N_b = 8

N_a = 3

Bond order of

= 2.5

Thus, the bond order of is 2.5.

The electronic configuration of ion will be:

N_b = 8

N_a = 5

Bond order of =

= 1.5

Thus, the bond order of ion is 1.5.