JEE Main 2020 — Chemical Bonding And Molecular Structure Question with Solution

Note :

(1) Bond strength $$\propto$$ Bond order

(2) Bond length $$\propto$$ $$\frac{1}{Bondorder}$$

(3) 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

(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 - - -



Molecular orbital configuration of NO (15 electrons) is

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

$$\therefore$$ N_b = 10

N_a = 5

$$\therefore$$ BO = $$\frac{1}{2}\left[10-5\right]$$ = 2.5

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

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

$$\therefore$$ N_b = 10

N_a = 4

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

Similarly Molecular orbital configuration of NO²⁺ (13 electrons) is

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

$$\therefore$$ N_b = 9

N_a = 4

$$\therefore$$ BO = $$\frac{1}{2}\left[9-4\right]$$ = 2.5

Molecular orbital configuration of NO^- (16 electrons) is

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

$$\therefore$$ N_b = 10

N_a = 6

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

As Bond strength $$\propto$$ Bond order

$$\therefore$$ NO^– will have minimum bond strength.