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

Bond order of CO = 3

Bond order of NO⁺ = 3

Difference = 0 = $$\frac{x}{2}$$

$$\Rightarrow$$ x = 0

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)    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

(B)   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