Which of the following species is paramagneti | vedclass

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(A) To determine the magnetic nature,we use Molecular Orbital Theory $(MOT)$. $1$. $O_2^+$: Total electrons = $8 + 8 - 1 = 15$. The configuration is $

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$O_2^+$

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(A) To determine the magnetic nature,we use Molecular Orbital Theory $(MOT)$. $1$. $O_2^+$: Total electrons = $8 + 8 - 1 = 15$. The configuration is $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. Since it has one unpaired electron,it is paramagnetic. $2$. $CN^-$: Total electrons = $6 + 7 + 1 = 14$. All electrons are paired,so it is diamagnetic. $3$. $CO$: Total electrons = $6 + 8 = 14$. All electrons are paired,so it is diamagnetic. $4$. $N_2$: Total electrons = $7 + 7 = 14$. All electrons are paired,so it is diamagnetic. Therefore,the correct option is $A$.

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

Which of the following species is paramagnetic?

Similar Questions

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

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