In the following the correct bond order sequence is:

For molecules and ions of oxygen we have to use the Molecular Orbital (MO) theory. According to MO theory, the electronic energy-level sequence for the 2p orbitals in O2 and its ions is

$$\sigma_{2p_z}\,\lt \,\pi_{2p_x}=\pi_{2p_y}\,\lt \,\pi^{*}_{2p_x}=\pi^{*}_{2p_y}\,\lt \,\sigma^{*}_{2p_z}$$

All lower-energy 1s and 2s orbitals are completely filled in every species and will be included in the counting. The bond order of any species is obtained from the formula that MO theory gives:

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

where $$N_b$$ is the total number of electrons in bonding MOs and $$N_a$$ is the total number of electrons in antibonding MOs.

Now we shall count electrons one by one for each species, write the values of $$N_b$$ and $$N_a$$ explicitly, and then apply the formula.

1. The neutral molecule $$O_2$$ (16 electrons)

The filling is

$$\sigma_{1s}^2\,\sigma_{1s}^{*\,2}\,\sigma_{2s}^2\,\sigma_{2s}^{*\,2}\, \sigma_{2p_z}^2\,\pi_{2p_x}^2\,\pi_{2p_y}^2\,\pi^{*1}_{2p_x}\,\pi^{*1}_{2p_y}$$

Counting electrons:

Bonding electrons $$N_b=2+2+2+2+2=10$$ Antibonding electrons $$N_a=2+2+1+1=6$$

Hence

$$\text{B.O.}(O_2)=\dfrac{10-6}{2}=2$$

2. The ion $$O_2^{+}$$ (15 electrons)

Removal of one electron occurs from the highest occupied antibonding orbital, so

$$\pi^{*1}_{2p_x}\,\pi^{*0}_{2p_y}$$

Now

Antibonding electrons $$N_a=2+2+1=5$$ (Bonding electrons remain $$N_b=10$$.)

Thus

$$\text{B.O.}(O_2^{+})=\dfrac{10-5}{2}=2.5$$

3. The ion $$O_2^{-}$$ (17 electrons)

An extra electron goes into the still singly occupied antibonding orbital, giving

$$\pi^{*1}_{2p_x}\,\pi^{*2}_{2p_y}$$

Antibonding electrons $$N_a=2+2+1+2=7$$ (Bonding electrons are still $$N_b=10$$.)

Therefore

$$\text{B.O.}(O_2^{-})=\dfrac{10-7}{2}=1.5$$

4. The ion $$O_2^{2-}$$ (18 electrons)

The added electron pairs up in the remaining antibonding orbital, making both $$\pi^{*}$$ orbitals completely filled:

$$\pi^{*2}_{2p_x}\,\pi^{*2}_{2p_y}$$

Antibonding electrons $$N_a=2+2+2+2=8$$ (Bonding electrons again $$N_b=10$$.)

So

$$\text{B.O.}(O_2^{2-})=\dfrac{10-8}{2}=1$$

We now have all four bond orders:

$$O_2^{+}:2.5,\; O_2:2,\; O_2^{-}:1.5,\; O_2^{2-}:1$$

Clearly the descending order of bond order is

$$O_2^{+} \;\gt \; O_2 \;\gt \; O_2^{-} \;\gt \; O_2^{2-}$$

This matches Option C in the list provided.

Hence, the correct answer is Option C.