To determine the bond order of the given species, we will use the molecular orbital (MO) theory.$\\$The bond order is calculated using the formula:$\\$Bond order$= \frac{1}{2} ($Number of bonding electrons$-$Number of antibonding electrons$) \\$Let's analyze each species:$\\$1. $O_2^{2+}: \\$• The neutral $O_2$ molecule has 16 electrons.$\\$• $O_2^{2+}$ has 14 electrons (16 - 2 = 14).$\\$• Electron configuration: $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \sigma_{2p_z})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 \\$• Bond order$= \frac{1}{2} (10 - 4) = 3 \\$2. $O_2^+: \\$• $O_2^+$ has 15 electrons (16 - 1 = 15).$\\$• Electron configuration: $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \sigma_{2p_z})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^1 \\$• Bond order$= \frac{1}{2} (10 - 5) = 2.5 \\$3. $O_2^-: \\$• $O_2^-$ has 17 electrons (16 + 1 = 17).$\\$• Electron configuration: $( \sigma_{1s})^2 ( \sigma^*_{1s})^2 ( \sigma_{2s})^2 ( \sigma^*_{2s})^2 ( \sigma_{2p_z})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \pi^*_{2p_x})^1 \\$• Bond order$= \frac{1}{2} (10 - 7) = 1.5 \\$Comparing the bond orders:$\\ O_2^- < O_2^+ < O_2^{2+} \\$Therefore, the correct option is Option 3.