To determine which pairs of species have the same bond order, we need to calculate the bond order for each species.$\\$The bond order can be calculated using the molecular orbital theory formula:$\\$Bond Order$= \frac{(N_b - N_a)}{2} \\$where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.$\\$Let's calculate the bond order for each species:$\\$1. $O_2: \\$• Electron configuration: $( \sigma_{1s})^2 ( \sigma_{1s}^*)^2 ( \sigma_{2s})^2 ( \sigma_{2s}^*)^2 ( \sigma_{2p_z})^2 ( \pi_{2p_x})^2 ( \pi_{2p_y})^2 ( \pi_{2p_x}^*)^1 ( \pi_{2p_y}^*)^1 \\$• Total electrons: 16$\\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 6 \\$• Bond Order$= \frac{10 - 6}{2} = 2 \\$2. $NO^+: \\$• Total electrons: 14$\\$• Electron configuration similar to $N_2$ due to isoelectronic nature$\\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 4 \\$• Bond Order$= \frac{10 - 4}{2} = 3 \\$3. $CN^-: \\$• Total electrons: 14$\\$• Electron configuration similar to $N_2 \\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 4 \\$• Bond Order$= \frac{10 - 4}{2} = 3 \\$4. $CO: \\$• Total electrons: 14$\\$• Electron configuration similar to $N_2 \\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 4 \\$• Bond Order$= \frac{10 - 4}{2} = 3 \\$5. $N_2: \\$• Total electrons: 14$\\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 4 \\$• Bond Order$= \frac{10 - 4}{2} = 3 \\$6. $O_2^-: \\$• Total electrons: 17$\\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 7 \\$• Bond Order$= \frac{10 - 7}{2} = 1.5 \\$7. $NO: \\$• Total electrons: 15$\\$• Bonding electrons $(N_b) = 10 \\$• Antibonding electrons $(N_a) = 5 \\$• Bond Order$= \frac{10 - 5}{2} = 2.5 \\$From the calculations, the pair of species with the same bond order is:$\\$Option 2: $CN^-$ and $CO$ both have a bond order of 3.$\\$Therefore, the correct option is Option 2.