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

$\therefore \,B.O. = \frac{{10 - 4}}{2} = 3$

$M.O.$ electronic configuration of $O_2^-$ is 

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

$\therefore \,B.O. = \frac{{10 - 7}}{2} = 1.5$

$M.O.$ electronic configuration of $CN^+$

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

$\therefore \,B.O. = \frac{{9 - 4}}{2} = 2.5$

$M.O.$ electronic configuration of $NO^+$ is

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

$\therefore \,B.O. = \frac{{10 - 4}}{2} = 3$

$\therefore $ $CN^-$ and $NO^+$ have bond order equal to $3$