The bond order of O_2^{+} is x. The bond orde | vedclass

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(B) $	ext{B.O.} = \frac{N_b - N_a}{2}$ For $O_2^{+}$: $\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^*

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$\frac{3}{5} x, \frac{6}{5} x$

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(B) $	ext{B.O.} = \frac{N_b - N_a}{2}$ For $O_2^{+}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^1$ $\Rightarrow 	ext{B.O.} = \frac{10 - 5}{2} = 2.5 = x$ For $O_2^{-}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2 \pi^* 2p_x^2 \pi^* 2p_y^1$ $\Rightarrow 	ext{B.O.} = \frac{10 - 7}{2} = 1.5$ For $O_2^{2+}$: $\sigma 1s^2 \sigma^* 1s^2 \sigma 2s^2 \sigma^* 2s^2 \sigma 2p_z^2 \pi 2p_x^2 \pi 2p_y^2$ $\Rightarrow 	ext{B.O.} = \frac{10 - 4}{2} = 3$ Now,express $1.5$ and $3$ in terms of $x = 2.5$: $\frac{1.5}{2.5} x = \frac{3}{5} x$ $\frac{3}{2.5} x = \frac{6}{5} x$ Thus,the bond orders are $\frac{3}{5} x$ and $\frac{6}{5} x$.

The bond order of $O_2^{+}$ is $x$. The bond orders of $O_2^{-}$ and $O_2^{2+}$ are respectively

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