JEE Main 2023 — Chemical Bonding And Molecular Structure Question with Solution

Species	B.O.
O$${}_2^{2-}$$	1
CO	3
NO$${}^{\oplus }$$	3

Note :

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

N_b = No of electrons in bending molecular orbital

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

(4) upto 14 electrons, molecular orbital configuration is



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

(5) After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

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

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

$$\therefore$$ 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)    CO has 14 electrons.

Moleculer orbital configuration of CO 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_b = 10

N_a = 4

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

(C)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

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

$$\therefore$$ N_b = 10

N_a = 4

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