N_b = No of electrons in bonding molecular orbital

N_a $$=$$ No of electrons in anti bonding molecular orbital

(1) $$\,\,\,\,$$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $$=$$ 4 and N_b = 10

(2) $$\,\,\,\,$$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $$ \times $$ 2 = 16 electrons present.

Then in $$O_2^ + $$ no of electrons = 15

in $$O_2^ - $$ no of electrons = 17

in $$O_2^{2 - }$$ no of electrons = 18

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$

$$\therefore\,\,\,\,$$ N_b = 10

and N_a = 8