Step 1: Build the valence molecular orbital filling for neutral $O_2$.
Oxygen has 12 valence electrons per molecule (6 from each atom), and since $O_2$ has more electrons than $N_2$, the filling order is:
\[ \sigma 2s^2\, \sigma^{*}2s^2\, \sigma 2p_z^2\, \pi 2p_x^2\,\pi 2p_y^2\, \pi^{*}2p_x^1\,\pi^{*}2p_y^1 \]
Step 2: Separate bonding electrons from antibonding electrons.
Bonding: $\sigma 2s^2 + \sigma 2p_z^2 + \pi 2p_x^2 + \pi 2p_y^2 = 8$.
Antibonding: $\sigma^{*}2s^2 + \pi^{*}2p_x^1 + \pi^{*}2p_y^1 = 4$.
\[ \text{Bond order} = \frac{8-4}{2} = 2 \]
Step 3: Remove one electron for $O_2^{+}$.
Removing an electron takes it from the highest occupied, weakest-bound orbital, which is $\pi^{*}$. Antibonding count drops to $3$:
\[ \text{Bond order}(O_2^{+}) = \frac{8-3}{2} = 2.5 \]
Step 4: Add one electron for $O_2^{-}$.
The extra electron also enters $\pi^{*}$, raising antibonding count to $5$:
\[ \text{Bond order}(O_2^{-}) = \frac{8-5}{2} = 1.5 \]
Step 5: Add a second electron for $O_2^{2-}$.
Both $\pi^{*}$ orbitals are now completely filled, giving antibonding count $6$:
\[ \text{Bond order}(O_2^{2-}) = \frac{8-6}{2} = 1 \]
Step 6: Rank by bond order, since a higher bond order always means a stronger, harder-to-break bond.
\[ O_2^{+}(2.5) \gt O_2(2) \gt O_2^{-}(1.5) \gt O_2^{2-}(1) \]
Final Answer:
\[ \boxed{O_2^{+} \gt O_2 \gt O_2^{-} \gt O_2^{2-}} \]