The correct order of bond orders of $$C_2^{2-}$$, $$N_2^{2-}$$ and $$O_2^{2-}$$ is

We need to find the correct order of bond orders of $$C_2^{2-}$$, $$N_2^{2-}$$, and $$O_2^{2-}$$.

Since each carbon atom contributes 6 electrons, the total electron count in $$C_2^{2-}$$ is:

$$2(6) + 2 = 14$$ electrons.

Similarly, nitrogen contributes 7 electrons each, so for $$N_2^{2-}$$:

$$2(7) + 2 = 16$$ electrons,

and oxygen contributes 8 electrons each, giving for $$O_2^{2-}$$:

$$2(8) + 2 = 18$$ electrons.

For $$C_2^{2-}$$ (14 electrons), which is isoelectronic with $$N_2$$, the molecular orbital configuration is:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\pi_{2p})^4 (\sigma_{2p})^2$$

This configuration gives bonding electrons = $$2 + 2 + 4 + 2 = 10$$ and anti-bonding electrons = $$2 + 2 = 4$$. From the above,

$$\text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3$$

Now, for $$N_2^{2-}$$ (16 electrons), which is isoelectronic with $$O_2$$, the configuration becomes:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^2$$

Here, bonding electrons = $$2 + 2 + 2 + 4 = 10$$ and anti-bonding electrons = $$2 + 2 + 2 = 6$$. Hence,

$$\text{Bond Order} = \frac{10 - 6}{2} = \frac{4}{2} = 2$$

Finally, for $$O_2^{2-}$$ (18 electrons), the peroxide ion, the configuration is:

$$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 (\sigma_{2p})^2 (\pi_{2p})^4 (\pi_{2p}^*)^4$$

This yields bonding electrons = $$2 + 2 + 2 + 4 = 10$$ and anti-bonding electrons = $$2 + 2 + 4 = 8$$, so

$$\text{Bond Order} = \frac{10 - 8}{2} = \frac{2}{2} = 1$$

From the above calculations, $$C_2^{2-}$$ has bond order 3, $$N_2^{2-}$$ has bond order 2, and $$O_2^{2-}$$ has bond order 1, giving the order:

$$C_2^{2-} > N_2^{2-} > O_2^{2-}$$

Therefore, the correct option is Option A: $$C_2^{2-} > N_2^{2-} > O_2^{2-}$$.