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(B) To determine the magnetic behavior,we analyze the molecular orbital configuration of each species.$1$. $NO$ has $15$ electrons: $\sigma_{1s}^2, \s

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$O_2^{2-}$

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(B) To determine the magnetic behavior,we analyze the molecular orbital configuration of each species.$1$. $NO$ has $15$ electrons: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*1}$. It is paramagnetic due to the unpaired electron.$2$. $O_2^{2-}$ has $18$ electrons. Its configuration is: $\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p_z}^2, \pi_{2p_x}^2 = \pi_{2p_y}^2, \pi_{2p_x}^{*2} = \pi_{2p_y}^{*2}$. Since all electrons are paired,it is diamagnetic.$3$. $O_2^{+}$ has $15$ electrons and is paramagnetic.$4$. $O_2$ has $16$ electrons and is paramagnetic due to two unpaired electrons in the $\pi^*$ orbitals.

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 exhibits the diamagnetic behaviour?

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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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