To determine the bond order of $O_2^-, O_2,$ and $O_2^+,$ we will use molecular orbital theory.$\\$The molecular orbital configuration for $O_2$ is:$\\ \sigma_{1s}^2 \sigma_{1s}^2 \sigma_{2s}^2 \sigma_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1 \pi_{2p_y}^1 \\$The bond order is given by:$\\$Bond order$= \frac{1}{2} ($Number of bonding electrons$-$Number of antibonding electrons$) \\$For $O_2: \\$Number of bonding electrons$= 10 \\$Number of antibonding electrons$= 6 \\$Bond order$= \frac{1}{2} (10 - 6) = 2 \\$For $O_2^+: \\$Remove one electron from the antibonding orbital:$\\$Number of bonding electrons$= 10 \\$Number of antibonding electrons$= 5 \\$Bond order$= \frac{1}{2} (10 - 5) = 2.5 \\$For $O_2^-: \\$Add one electron to the antibonding orbital:$\\$Number of bonding electrons$= 10 \\$Number of antibonding electrons$= 7 \\$Bond order$= \frac{1}{2} (10 - 7) = 1.5 \\$Comparing the bond orders:$\\ O_2^- (1.5) < O_2 (2) < O_2^+ (2.5) \\$Therefore, the correct option is Option 1: $O_2^- < O_2 < O_2^+.$