$\,$ 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

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 = 14

$\therefore\,\,\,\,$ 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 $O_2^ -$ (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 $_2^{2 + }$ (14 electrons) is

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

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

N_a = 4

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

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