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.

Similar Questions

In which of the following pairs do the two molecules have identical bond order?

Which of the following molecular orbitals has two nodal planes?

Given below are two statements:
Statement $(I)$: The correct sequence of bond lengths in the following species is:
$O_2^+ < O_2 < O_2^- < O_2^{2-}$
Statement $(II)$: The correct sequence of number of unpaired electrons in the following species is:
$O_2 > O_2^+ > O_2^- > O_2^{2-}$

Arrange the following in the correct order of their bond orders.
$I. N_2, II. O_2, III. O_2^+, IV. O_2^-$

The $H.O.M.O.$ (Highest Occupied Molecular Orbital) of the $CO$ molecule is:

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