Li₂O $$\Rightarrow$$ 2Li⁺ O^2$$-$$

CaO $$\Rightarrow$$ Ca²⁺ O^2$$-$$

Na₂O₂ $$\Rightarrow$$ 2Na⁺ O$$_2^{2 - }$$

KO₂ $$\Rightarrow$$ O$$_2^{- }$$

MgO $$\Rightarrow$$ Mg²⁺ O^2$$-$$

K₂O $$\Rightarrow$$ 2K⁺ O^2$$-$$

O^2$$-$$ $$\Rightarrow$$ Complete octet, diamagnetic

$$O_2^{2−}$$ has 18 electrons.

Moleculer orbital configuration of $$O_2^{2−}$$ is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $$

Here is no unpaired electron so it is diamagnetic.

$$O_2^{−}$$ has 17 electrons.

Moleculer orbital configuration of $$O_2^{−}$$ is

$${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $$

Here is 1 unpaired electron so it is paramagnetic.