Note :

(1) $$\,\,\,\,$$ Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$

(2) $$\,$$ Bond order $$ = {1 \over 2}$$ [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

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

$$\therefore\,\,\,\,$$N_a = 6

N_b = 10

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

Molecular orbital configuration of O$$_2^ + $$ (15 electrons) is

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

$$\therefore\,\,\,\,$$ N_b = 10

N_a = 5

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

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

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

$$\therefore\,\,\,\,$$ N_b = 10

N_a = 4

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

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

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

$$\therefore\,\,\,\,$$ N_b = 10

N_a = 7

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

Molecular orbital configuration of O $$_2^{2 - }$$ (18 electrons) is

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

$$\therefore\,\,\,\,$$ N_b = 10

N_a = 8

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}$$ [ 10 $$-$$ 8] = 1

As Bond length $$ \propto $$ $${1 \over {Bond\,\,order}}$$

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