Using MOT, compare \( \mathrm{O}_{2}^{+} \)and \( \mathrm{O}_{2}^{-} \)species and choose the incorrect option.

Molecular orbital configurations:
\( O_{2}^{+}\left(15 e^{-} s\right): \) \( \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma * 2 s^{2} \sigma 2 p_{z}^{2} \Pi 2 p_{x}^{2}=\Pi 2 p_{y}^{2} \Pi^{*} 2 p_{x}{ }^{1} \chi \)
Bond order \( =\frac{1}{2}(10-5)=2.5 \), and paramagnetic due to one unpaired electron.
\( \mathrm{O}_{2}^{-}\left(17 e^{-} s\right): \)
\( 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma * 2 s^{2} \sigma 2 p_{z}^{2} \Pi 2 p_{x}^{2}=\Pi 2 p_{y}^{2} \Pi^{*} 2 p_{x}^{2}=\Pi^{*} 2 p_{y}^{1} \)
Bond order \( =\frac{1}{2}(10-7)=1.5 \) and paramagnetic due to one unpaired electron.
Therefore, both \( \mathrm{O}_{2}^{+} \)and \( \mathrm{O}_{2}^{-} \)are paramagnetic and \( \mathrm{O}_{2}^{+} \)have higher bond order than \( \mathrm{O}_{2}^{-} \)Also, \( \mathrm{O}_{2}^{-} \)is less stable.