JEE Main 2021 — Chemical Bonding And Molecular Structure Question with Solution

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

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

$$\therefore$$ N_b = 10

and N_a = 8