Which of the following is paramagnetic?

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Question

(A) Paramagnetism is due to the presence of unpaired electrons.In $O_2[AsF_6]$,the $O_2^+$ ion has a bond order of $2.5$ and contains one unpaired ele

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$O_2[AsF_6]$

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(A) Paramagnetism is due to the presence of unpaired electrons.In $O_2[AsF_6]$,the $O_2^+$ ion has a bond order of $2.5$ and contains one unpaired electron in its molecular orbitals,making it paramagnetic.$Al_2O_3$ is an ionic solid where $Al^{3+}$ ($[Ne]$ configuration) and $O^{2-}$ ($[Ne]$ configuration) have no unpaired electrons,making it diamagnetic.For $C_2$,the molecular orbital configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired,so it is diamagnetic.For $Be_2$,the configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2$. All electrons are paired,so it is diamagnetic.

List-$I$ (Molecule / ion)	List-$II$ (Bond order)
$1$. $NO$	$i$. $1.5$
$2$. $CO$	$ii$. $2.0$
$3$. $O_2^{-}$	$iii$. $2.5$
$4$. $O_2$	$iv$. $3.0$

Similar Questions

Match List-$I$ with List-$II$ and select the correct option:
List-$I$ (Molecule / ion) List-$II$ (Bond order)
$1$. $NO$ $i$. $1.5$
$2$. $CO$ $ii$. $2.0$
$3$. $O_2^{-}$ $iii$. $2.5$
$4$. $O_2$ $iv$. $3.0$

Which of the following is correctly matched?

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

In the conversion of $N_2$ into $N_2^+$,the electron will be lost from which of the following molecular orbitals?

"Isoelectronic molecules and ions have identical bond orders." Explain by examples.

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