1. Use MO theory for the oxygen species

For $O_2$ and its ions, the relevant molecular orbital filling is:

$\sigma(2s)^2\,\sigma^*(2s)^2\,\sigma(2p_z)^2\,\pi(2p_x)^2=\pi(2p_y)^2\,\pi^*(2p_x)^1\,\pi^*(2p_y)^1$

For $O_2$, the bond order is:

$\text{Bond order} = \frac{N_b - N_a}{2}$

where $N_b$ = bonding electrons and $N_a$ = antibonding electrons.

For $O_2$:

• Bonding electrons = $8$
• Antibonding electrons = $4$

So,

$\text{B.O.}(O_2)=\frac{8-4}{2}=2$

2. Find bond orders of each species

(i) $O_2^+$

One electron is removed from antibonding $\pi^*$ orbital.

$\text{B.O.}(O_2^+) = \frac{8-3}{2} = 2.5$

(ii) $O_2$

$\text{B.O.}(O_2)=2$

(iii) $O_2^-$

One electron is added to antibonding $\pi^*$ orbital.

$\text{B.O.}(O_2^-) = \frac{8-5}{2} = 1.5$

(iv) $O_2^{2-}$

Two electrons are added to antibonding $\pi^*$ orbitals.

$\text{B.O.}(O_2^{2-}) = \frac{8-6}{2} = 1$

3. Arrange in increasing bond order

Increasing bond order means lowest to highest:

$O_2^{2-} < O_2^- < O_2 < O_2^+$

4. Match with the options

Option A corresponds to the correct increasing trend.

So, the correct answer is:

$\boxed{A}$

5. Comparison with stored correct answer

Stored correct answer: $A$

My derived answer is also $A$, so they agree.