(c)
The bond order can be calculated as
$\text { B.O }=\frac{1}{2}\left(N_b-N_a\right)$
where, $N_b=$ electrons in bonding orbitals
$N_a=$ electrons in antibonding orbitals.
$(a)$ $CO$ and $O _2^{2-}$
The electronic configuration of $CO$ (14) is
$\sigma 1 s^2\, \sigma^* 1 s^2 \,\sigma 2 s^2 \,\sigma^* 2 s^2 \,\sigma^* 2 p_z^2\, \pi 2 p_x^2\, \pi 2 p^2 y$
$\therefore \quad B . O =\frac{1}{2}(10-4)=\frac{6}{2}=3$
The electronic configuration of $O _2^{2-}(18)$ is
$\sigma 1 s^2\, \sigma^* 1 s^2 \,\sigma 2 s^2\, \sigma^* 2 s^2 \,\sigma 2 p_z^2\, \pi 2 p_x^2\,\pi 2 p y^2 \,\pi^* 2 p_x^2 \,\pi^* 2 p y^2$
$\text { B. } O =\frac{1}{2}(10-8)=1$
$(b)$ $O _2^{-}$ and $CO$
The electronic configuration of $O _2^{-}\,(17)$ is
$\sigma 1 s^2 \,\sigma^* 1 s^2\, \sigma 2 s^2\, \sigma^* 2 s^2\, \sigma 2 p_z^2 \,\pi 2 p_x^2\,\pi 2 p_y^2 \,\pi^* 2 p_x^2 \,\pi^* 2 p_y^1$
$\text { B.O. } =\frac{1}{2}(10-7)=\frac{3}{2}=1.5$
$B.O.$ of $CO$ is $3$
[as calculated in option $(a)$]
$(c)$ B.O of $O _2^{2-}$ is $1$
[as calculated in option $(a)$]
The electronic configuration of $B_2(10)$ is
$\sigma 1 s^2 \,\sigma^* 1 s^2\, \sigma 2 s^2 \,\sigma^* 2 s^2\, \pi 2 p_x^1 \,\pi 2 p_y^1$
$\text { B.O. } =\frac{1}{2}[6-4]=\frac{2}{2}=1$
$(d)$ $B.O.$ of $CO$ is $3$
[as calculated in option $(a)$]
Electronic configuration of $N _2^{+}(13)$ is
$\sigma 1 s^2 \,\sigma^* 1 s^2 \,\sigma 2 s^2\, \sigma^* 2 s^2 \,\pi 2 p_x^2\, \pi 2 p_y^2 \sigma 2 p_z^1$
$\text { B.O. }=\frac{1}{2}[9-4]=\frac{5}{2}=2.5$
Thus, option $(c)$ is correct.