JEE Main 2009 — Chemical Bonding And Molecular Structure Question with Solution

Note :

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

(2) 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

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

Then in $$O_2^{+}$$ no of electrons = 15

in $$O_2^{-}$$ no of electrons = 17

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

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

Molecular orbital configuration of O$${}_2^{+}$$ (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

Molecular orbital configuration of O$${}_2^{2+}$$ (14 electrons) is

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^0}^{\ast }={\pi }_{2p_y^o}^{\ast }$$

$$\therefore$$ N_b = 10

N_a = 4

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

Molecular orbital configuration of $$O_2^{-}$$ (17 electrons) is

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

$$\therefore$$ N_b = 10

N_a = 7

$$\therefore$$ BO = $$\frac{1}{2}\left[10-7\right]$$ = 1.5

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

As Bond length $$\propto$$ $$\frac{1}{Bondorder}$$

So $$O_2^{2+}$$ has the shortest bond length.