When O_{2} is converted into O_{2}^{+},what h | vedclass

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(D) The molecular orbital configuration of $O_{2}$ is $(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}(\sigma 2p_{z})^{2}(\pi 2p_

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Paramagnetic character decreases and the bond order increases

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(D) The molecular orbital configuration of $O_{2}$ 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 $= \frac{N_{b} - N_{a}}{2} = \frac{10 - 6}{2} = 2$. It has $2$ unpaired electrons.For $O_{2}^{+}$,the configuration is $(\sigma 1s)^{2}(\sigma^{*} 1s)^{2}(\sigma 2s)^{2}(\sigma^{*} 2s)^{2}(\sigma 2p_{z})^{2}(\pi 2p_{x})^{2}(\pi 2p_{y})^{2}(\pi^{*} 2p_{x})^{1}$.Bond order $= \frac{10 - 5}{2} = 2.5$. It has $1$ unpaired electron.Thus,when $O_{2}$ is converted into $O_{2}^{+}$,the paramagnetic character decreases and the bond order increases.

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

When $O_{2}$ is converted into $O_{2}^{+}$,what happens to its paramagnetic character and bond order?

Similar Questions

According to molecular orbital theory,the paramagnetism of the $O_2$ molecule is due to the presence of

The bond order in $NO$ is $2.5$ while that in $NO^{+}$ is $3$. Which of the following statements is true for these two species?

The oxygen molecule $(O_2)$ exhibits which of the following magnetic properties?

Match the following:

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

How many ions of the following have a bond order of $2.5 ?$
$N_2^{-} ; NO^{-} ; C_2^{-} ; N_2^{+} ; C_2^{2-} ; CN^{+}$

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