Compare the relative stability of the following species and indicate the magnetic properties: $O_2$,$O_2^+$,$O_2^-$ (superoxide),and $O_2^{2-}$ (peroxide).

The stability of a species is directly proportional to its bond order $(BO)$. The bond order is calculated using the formula: $BO = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
$1$. $O_2$ $(16 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1(\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 6) = 2.0$. It has unpaired electrons,so it is paramagnetic.
$2$. $O_2^+$ $(15 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10 - 5) = 2.5$. It has an unpaired electron,so it is paramagnetic.
$3$. $O_2^-$ $(17 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^1$. $BO = \frac{1}{2}(10 - 7) = 1.5$. It has an unpaired electron,so it is paramagnetic.
$4$. $O_2^{2-}$ $(18 \ e^-)$: Configuration: $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x)^2(\pi 2p_y)^2(\pi^* 2p_x)^2(\pi^* 2p_y)^2$. $BO = \frac{1}{2}(10 - 8) = 1.0$. All electrons are paired,so it is diamagnetic.
Stability order: $O_2^+ (2.5) > O_2 (2.0) > O_2^- (1.5) > O_2^{2-} (1.0)$.