The pairs of species of oxygen and their magnetic behaviours are given below. Which of the following represents the correct description?

To determine the correct description, we analyze the magnetic behavior of each oxygen species based on the number of unpaired electrons in their electronic configurations:

1. $O_{2}$ (Dioxygen): Has 16 electrons. Its molecular orbital configuration ends in $(\pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1})$, meaning it has two unpaired electrons and is paramagnetic .
2. $O_{2}^{+}$: Has 15 electrons. It is formed by removing one electron from a $\pi^{*}$ orbital of $O_{2}$, leaving one unpaired electron. Thus, it is paramagnetic.
3. $O_{2}^{-}$ (Superoxide): Has 17 electrons. It has one additional electron in the $\pi^{*}$ orbitals compared to $O_{2}$, leaving one unpaired electron. Thus, it is paramagnetic.
4. $O_{2}^{2-}$ (Peroxide): Has 18 electrons. Both $\pi^{*}$ orbitals are fully occupied $(\pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{2})$, meaning all electrons are paired. Thus, it is diamagnetic.
5. $O$ (Oxygen atom): Has 8 electrons. Its atomic configuration is $1s^{2} 2s^{2} 2p^{4}$. According to Hund's rule, it has two unpaired electrons in the $2p$ orbitals and is paramagnetic.
6. $O^{+}$ (Oxygen atom cation): Has 7 electrons. Its configuration is $1s^{2} 2s^{2} 2p^{3}$, with three unpaired electrons in the $2p$ orbitals, making it paramagnetic.

Evaluating the options: Option A is incorrect because $O_{2}^{2-}$ is diamagnetic. Option B is incorrect because $O_{2}^{-}$ is paramagnetic. Option C is incorrect because $O_{2}^{2-}$ is diamagnetic. Option D is correct because both $O_{2}^{+}$ and $O_{2}$ are paramagnetic.