Using MOT,compare O_{2}^{+} and O_{2}^{-} spe | vedclass

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Question

(C) Molecular orbital configurations:$O_{2}^{+} (15 \ e^{-}): \sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p

Vedclass · Accepted Answer

$O_{2}^{+}$ is diamagnetic while $O_{2}^{-}$ is paramagnetic.

Vedclass · Answer

(C) Molecular orbital configurations:$O_{2}^{+} (15 \ e^{-}): \sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{1}$Bond order $= \frac{1}{2}(10-5) = 2.5$,and it is paramagnetic due to one unpaired electron.$O_{2}^{-} (17 \ e^{-}): \sigma 1s^{2} \sigma^{*} 1s^{2} \sigma 2s^{2} \sigma^{*} 2s^{2} \sigma 2p_{z}^{2} \pi 2p_{x}^{2} = \pi 2p_{y}^{2} \pi^{*} 2p_{x}^{2} = \pi^{*} 2p_{y}^{1}$Bond order $= \frac{1}{2}(10-7) = 1.5$,and it is paramagnetic due to one unpaired electron.Since both species have unpaired electrons,both are paramagnetic.Option $C$ is incorrect because $O_{2}^{+}$ is paramagnetic,not diamagnetic.

Using $MOT$,compare $O_{2}^{+}$ and $O_{2}^{-}$ species and choose the incorrect option.

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