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

Moleculer orbital configuration of CN^– (14 electrons) 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 = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Moleculer orbital configuration of CN (13 electrons) is

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

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

N_a = 4

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

Moleculer orbital configuration of CN⁺ (12 electrons) is

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

$\therefore\,\,\,\,$ N_b = 8

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right]$ = 2

Hence, CN^– has highest bond order.