The magnetic behaviour of Li$$_2$$O, Na$$_2$$O$$_2$$ and KO$$_2$$, respectively, are

We need to determine the magnetic behaviour of Li$$_2$$O, Na$$_2$$O$$_2$$, and KO$$_2$$.

Li$$_2$$O (Lithium oxide):

Contains Li$$^+$$ (1s$$^2$$, i.e., [He] configuration) and O$$^{2-}$$ (1s$$^2$$2s$$^2$$2p$$^6$$, i.e., [Ne] configuration). Both ions have all electrons paired. Li$$_2$$O is diamagnetic.

Na$$_2$$O$$_2$$ (Sodium peroxide):

Contains Na$$^+$$ ([Ne]) and the peroxide ion O$$_2^{2-}$$. The peroxide ion has a bond order of 1 with electronic configuration: $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^4$$. All electrons are paired. Na$$_2$$O$$_2$$ is diamagnetic.

KO$$_2$$ (Potassium superoxide):

Contains K$$^+$$ ([Ar]) and the superoxide ion O$$_2^-$$. The superoxide ion has electronic configuration: $$(\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p})^2(\pi_{2p})^4(\pi^*_{2p})^3$$. There is one unpaired electron in the $$\pi^*_{2p}$$ orbital. KO$$_2$$ is paramagnetic.

Therefore, the magnetic behaviours are: diamagnetic, diamagnetic, and paramagnetic.

The correct answer is Option D.