During the change of O_{2} to O_{2}^{-},the i | vedclass

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Question

(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} \p

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$\pi^{*} 2 p_{x}$

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(C) The electronic configuration of $O_{2}$ ($16$ electrons) is: $\sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1} = \pi^{*} 2p_{y}^{1}$.When an electron is added to $O_{2}$ to form $O_{2}^{-}$,the incoming electron enters the lowest energy vacant or half-filled molecular orbital,which is the $\pi^{*}$ antibonding orbital.Thus,the electron enters the $\pi^{*} 2p_{x}$ or $\pi^{*} 2p_{y}$ orbital.Therefore,option $(C)$ is correct.

Pair of species	Identical property
$A$. $B_2 \& O_2$	$P$. Bond order $= 2.5$
$B$. $Be_2 \& H_2^{2-}$	$Q$. Paramagnetic nature
$C$. $N_2^{+} \& N_2^{-}$	$R$. Diamagnetic nature
$D$. $O_2^{+} \& O_2^{-}$	$S$. Doesn't exist

During the change of $O_{2}$ to $O_{2}^{-}$,the incoming electron goes to the orbital:

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Pair of species Identical property

$A$. $B_2 \& O_2$ $P$. Bond order $= 2.5$

$B$. $Be_2 \& H_2^{2-}$ $Q$. Paramagnetic nature

$C$. $N_2^{+} \& N_2^{-}$ $R$. Diamagnetic nature

$D$. $O_2^{+} \& O_2^{-}$ $S$. Doesn't exist

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