Consider the ions/molecule $$O_2^+, O_2, O_2^-, O_2^{2-}$$. For increasing bond order the correct option is

We need to arrange $$O_2^+$$, $$O_2$$, $$O_2^-$$, and $$O_2^{2-}$$ in increasing order of bond order.

To begin, recall that the molecular orbital configuration of $$O_2$$, which has 16 electrons, is:

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

Using the formula for bond order, $$\frac{N_b - N_a}{2}$$, where $$N_b$$ represents bonding electrons and $$N_a$$ represents antibonding electrons, we can determine the bond orders for each species.

In $$O_2$$, with $$N_b = 10$$ and $$N_a = 6$$, the bond order is $$\frac{10 - 6}{2} = 2$$.

When one electron is removed from the antibonding orbitals to form $$O_2^+$$ (15 electrons), we have $$N_b = 10$$ and $$N_a = 5$$, giving a bond order of $$\frac{10 - 5}{2} = 2.5$$.

Adding one extra electron to the antibonding orbitals yields $$O_2^-$$ with $$N_b = 10$$ and $$N_a = 7$$, resulting in a bond order of $$\frac{10 - 7}{2} = 1.5$$.

Similarly, $$O_2^{2-}$$ incorporates two additional electrons in the antibonding orbitals, giving $$N_b = 10$$ and $$N_a = 8$$ and a bond order of $$\frac{10 - 8}{2} = 1$$.

Arranging these bond orders in ascending order gives $$O_2^{2-}(1) < O_2^-(1.5) < O_2(2) < O_2^+(2.5)$$. Therefore, the correct sequence is $$O_2^{2-} < O_2^- < O_2 < O_2^+$$, which corresponds to Option D.