Which of the following species has a bond ord | vedclass

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(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$

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$O_{2}^{-}$

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(B) The bond order is calculated using the formula: $	ext{Bond Order} = \frac{N_b - N_a}{2}$,where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.For $O_{2}^{+}$ (total $15$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Bond order $= \frac{10 - 5}{2} = 2.5$.For $O_{2}^{-}$ (total $17$ electrons): Configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. Bond order $= \frac{10 - 7}{2} = 1.5$.For $O_{2}^{2-}$ (total $18$ electrons): Bond order $= \frac{10 - 8}{2} = 1.0$.For $O_{2}$ (total $16$ electrons): Bond order $= \frac{10 - 6}{2} = 2.0$.Thus,$O_{2}^{-}$ has a bond order of $1.5$.

Which of the following species has a bond order of $1.5$?

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