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

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

We know, 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

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

Configuration of $$H{e}_2^{+}$$ (3 electrons) is = $${\sigma }_{1s^2}{\sigma }_{1s^1}^{\ast }$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (2 $$-$$1) = 0.5

Configuration of $$H{e}_2^{-}$$ (5 electrons) is = $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^1}$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (3 $$-$$2) = 0.5

Configuration of $$B{e}_2$$ (4 electrons) is = $${\sigma }_{1s^2}$$ $${\sigma }_{1s^2}^{\ast }$$

$$\therefore$$ Bond order = $$\frac{1}{2}$$ (2 $$-$$ 2) = 0