Calculate the bond order of $N_2, O_2, O_2^+$ and $O_2^-$.

For $N_2$: Total electrons $= 14$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. Bond Order $(BO) = \frac{1}{2}(10 - 4) = 3$.
For $O_2$: Total electrons $= 16$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2$.
For $O_2^+$: Total electrons $= 15$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$.
For $O_2^-$: Total electrons $= 17$. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$.