Answer:

Correct Answer: B

Solution:

Bond order of $ O_2^{-} $

$O_2^{-}=\sigma 1s^{2},\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2}$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma {2p_z}^{2}(\pi {2p_x}^{2}=\pi 2p_y^{2}) ({{\pi }^*}{2p_x}^2$ =${{\pi }^*}{2p_y}^1) $

Bond order $ =\frac{\text{number of electrons in BMO}-\text{number of elections ABMO}}{2} $

$ =\frac{10-7}{2}=\frac{3}{2}=1.5 $

$ O_2^{+}=\sigma 1s^{2},\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2},$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma 2p_z^{2} $ $ (\pi 2p_x^{2}=\pi {2p_y}^{2})({{\pi }^{*}}2p_x^{1}$ =${{\pi }^{*}}{2p_y}^{0}) $

$ BO=\frac{10-5}{2}=\frac{5}{2}=2.5 $

$ O _2=\sigma 1s^{2}\overset{*}{\mathop{\sigma }},1s^{2},\sigma 2s^{2},$ $\overset{*}{\mathop{\sigma }},2s^{2}\sigma 2p_z^{2}(\pi 2p_x^{2}=\pi 2p_y^{2}) $ $ (\overset{*}{\mathop{\pi }},{2p_x}^{1}$ =$\overset{*}{\mathop{\pi }},{2p_y}^{1}) $

$ BO=\frac{10-6}{2}=\frac{4}{2}=2 $ So, the correct sequence is $ O_2^{-}<O _2<O_2^{+} $