Consider the following statements.I. In the c | vedclass

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(C) 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

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$III$ & $IV$ only

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(C) 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 = $(10-6)/2 = 2$. It is paramagnetic.The electronic configuration of $O_2^{2+}$ ($14$ electrons) is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2$. Bond order = $(10-4)/2 = 3$. It is diamagnetic.Statement $I$ is incorrect because bond order increases from $2$ to $3$.Statement $II$ is incorrect because $O_2$ is paramagnetic and $O_2^{2+}$ is diamagnetic.Statement $III$ is correct because bond order increases,so bond length decreases.For $O_2^{2-}$ ($18$ electrons),bond order = $(10-8)/2 = 1$. For $B_2$ ($10$ electrons),bond order = $(6-4)/2 = 1$.Statement $IV$ is correct because both have a bond order of $1$.

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