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(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p

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Paramagnetic and bond order $> O_2$

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(B) The molecular orbital configuration for $O_2$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^1)$.Bond order of $O_2 = \frac{10-6}{2} = 2.0$. It has two unpaired electrons,so it is paramagnetic.The molecular orbital configuration for $O_2^+$ is $(\sigma 1s)^2(\sigma^* 1s)^2(\sigma 2s)^2(\sigma^* 2s)^2(\sigma 2p_z)^2(\pi 2p_x^2 \equiv \pi 2p_y^2)(\pi^* 2p_x^1 \equiv \pi^* 2p_y^0)$.Bond order of $O_2^+ = \frac{10-5}{2} = 2.5$.Since $O_2^+$ has one unpaired electron,it is paramagnetic.Comparing the values,$2.5 > 2.0$,so the bond order of $O_2^+ > O_2$.

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

According to molecular orbital theory,which of the following statements about the magnetic character and bond order is correct regarding $O_2^+$?

Similar Questions

Explain the structure of the $O_2$ molecule.

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

What is the total number of electrons present in bonding orbitals of $O_2$ molecule according to molecular orbital theory?

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

Based on the Molecular Orbital Theory,which of the following is the correct electronic configuration of the nitrogen molecular ion $N_2^+$?

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