To determine which set has the same bond order, we need to calculate the bond order for each species in the given options. Bond order can be calculated using the molecular orbital theory formula:

$\text{Bond Order}=\frac{1}{2}(\text{Number of electrons in bonding MOs}-\text{Number of electrons in antibonding MOs})$

Alternatively, for diatomic molecules with up to 20 electrons, a simpler method based on the total number of electrons can be used:

• 14 electrons: Bond order = 3
• 13 or 15 electrons: Bond order = 2.5
• 12 or 16 electrons: Bond order = 2
• 11 or 17 electrons: Bond order = 1.5
• 10 or 18 electrons: Bond order = 1

Let's calculate the total number of electrons and bond order for each species:

Option A:

• ${\text{N}}_2$: Total electrons = 7 + 7 = 14. Bond order = 3.
• ${\text{O}}_2^{-}$: Total electrons = 8 + 8 + 1 = 17. Bond order = 1.5.

Bond orders are not the same (3 vs 1.5).

Option B:

• ${\text{N}}_2$: Total electrons = 14. Bond order = 3.
• ${\text{N}}_2^{+}$: Total electrons = 14 - 1 = 13. Bond order = 2.5.

Bond orders are not the same (3 vs 2.5).

Option C:

• ${\text{O}}_2^{+}$: Total electrons = 8 + 8 - 1 = 15. Bond order = 2.5.
• ${\text{O}}_2^{-}$: Total electrons = 8 + 8 + 1 = 17. Bond order = 1.5.

Bond orders are not the same (2.5 vs 1.5).

Since none of the given options have species with the same bond order, the correct answer is D.

Correct Answer: (D)