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

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 - - -



Here N_a = 10

and N_b = 10

(a) Molecular orbital configuration of Ne₂ (20 electrons) is

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

$$\therefore$$N_a = 10

N_b = 10

$$\therefore$$ BO = $$\frac{1}{2}\left[10-10\right]=0$$

(Note : All inert gases has BO = 0 and it does not exist as molecule. Here Ne is also an inert gas.)

(b) Moleculer orbital configuration of $$N_2$$ (14 electrons) 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 = $$\frac{1}{2}\left[10-4\right]=3$$

(d) Molecular orbital configuration of F₂ (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_a = 8

N_b = 10

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

(d) Molecular orbital configuration of O₂ (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_a = 6

N_b = 10

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