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(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p

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Paramagnetic and bond order $> O_2$

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(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^1)$.Bond order of $O_2 = \frac{10-6}{2} = 2.0$. It has two unpaired electrons,so it is paramagnetic.The molecular orbital configuration for $O_2^+$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^0)$.Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$.Since $O_2^+$ has one unpaired electron,it is paramagnetic.Comparing the values,$2.5 > 2.0$,so the bond order of $O_2^+ > O_2$.

According to molecular orbital theory,which of the following statements about the magnetic character and bond order is correct regarding $O_2^+$?

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