Species	B.O.
O$$_2^{2 - }$$	1
CO	3
NO$$^ \oplus $$	3

Note :

(1) $$\,$$ Bond order $$ = {1 \over 2}$$ [N_b $$-$$ N_a]

N_b = No of electrons in bending 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) In O atom 8 electrons present, so in O₂, 8 $$ \times $$ 2 = 16 electrons present.

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

$$ \therefore $$ 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

(B)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

$$\therefore$$ N_b = 10

N_a = 4

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

(C)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

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

$$\therefore$$ N_b = 10

N_a = 4

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