JEE Main 2022 — Chemical Bonding And Molecular Structure Question with Solution

Note :

(1) Bond order $$=\frac{1}{2}$$ [N_b $$-$$ N_a]

N_b = No of electrons in bonding molecular orbital

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

(2) upto 14 electrons, molecular orbital configuration is



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

(3) 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.

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

(A) 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

N_a = 8

$$\therefore$$ BO = $$\frac{1}{2}$$ [ 10 $$-$$ 8] = 1

(B)   $$C_2^{2-}$$ has 14 electrons.

Moleculer orbital configuration of $$C_2^{2-}$$ is

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

$$\therefore$$N_a = 4

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-4\right]=3$$

(C)   $$N_2^{2-}$$ has 16 electrons.

Moleculer orbital configuration of $$N_2^{2-}$$ is

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

$$\therefore$$N_a = 6

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-6\right]=2$$

The correct order of bond orders of $${C_2}^{2-}$$, $${N_2}^{2-}$$ and $${O_2}^{2-}$$

$${O_2}^{2-}$$ < $${N_2}^{2-}$$ < $${C_2}^{2-}$$