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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.

Which of the following species exhibits the diamagnetic behaviour?

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