Consider the following changes I and II as sh | vedclass

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(B) The electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \p

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$A$ & $D$ only

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(B) The electronic configuration of $O_2$ ($16$ electrons) is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2.0$. It is paramagnetic.Change $(I)$: $O_2 \rightarrow O_2^+$ ($15$ electrons). Configuration: $\dots \pi^* 2p_x^1, \pi^* 2p_y^0$. Bond order = $(10-5)/2 = 2.5$. Increase in bond order = $2.5 - 2.0 = 0.5$. It is paramagnetic. Magnetic property remains unchanged.Change $(II)$: $O_2 \rightarrow O_2^-$ ($17$ electrons). Configuration: $\dots \pi^* 2p_x^2, \pi^* 2p_y^1$. Bond order = $(10-7)/2 = 1.5$. Decrease in bond order = $2.0 - 1.5 = 0.5$. It is paramagnetic. Magnetic property remains unchanged.Thus,statement $A$ is correct (bond order increases by $0.5$ in $I$).Statement $D$ is correct (magnetic property is not changed in both $I$ and $II$ as all species are paramagnetic).Therefore,the correct option is $B$ ($A$ & $D$ only).

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