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, 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:

`[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  = `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[pi^"*"(2p_x)]^1[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 ` = 1/2(N_b - N_a)`

= 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^+`  = 1/2 (8 -3)

= 2.5

Thus, the bond order of `O_2^+` is  2.5.

The electronic configuration of `O_2^-` 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^-` = 1/2 (8-5)

= 1.5

Thus, the bond order of `O_2^-` ion is 1.5.