Which of the following electronic configurati | vedclass

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(A) According to Molecular Orbital Theory $(MOT)$,for homonuclear and heteronuclear diatomic molecules with $14$ or fewer electrons,the order of energ

Vedclass · Accepted Answer

$CO : \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^2$

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$,for homonuclear and heteronuclear diatomic molecules with $14$ or fewer electrons,the order of energy is $\sigma 1s For molecules with more than $14$ electrons,the order is $\sigma 1s $CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^2$.$CN^-$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^2$.$NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2pz^2, \pi 2px^2, \pi 2py^2, \pi^* 2px^1$.$N_2^+$ ($13$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^1$.All given options represent correct configurations based on the $MOT$ energy order.

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 electronic configurations is correct for the given species?

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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,the paramagnetism of the $O_2$ molecule is due to the presence of

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