Note :

(1) $$\,\,\,\,$$ Bond strength $$ \propto $$ Bond order

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

(3) $$\,$$ 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 - - -



Molecular orbital configuration of NO (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

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

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

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

N_a = 4

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

Similarly Molecular orbital configuration of NO²⁺ (13 electrons) is

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

$$\therefore\,\,\,\,$$ N_b = 9

N_a = 4

$$\therefore\,\,\,\,$$ BO = $${1 \over 2}\left[ {9 - 4} \right]$$ = 2.5

Molecular orbital configuration of NO^- (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_b = 10

N_a = 6

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

As Bond strength $$ \propto $$ Bond order

$$ \therefore $$ NO^– will have minimum bond strength.