Which of the following is not paramagnetic?

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Question

(A) To determine paramagnetism,we look for the presence of unpaired electrons in the molecular orbitals.$A$. $S^{2-}$: The sulfur atom has an atomic n

Vedclass · Accepted Answer

$S^{2-}$

Vedclass · Answer

(A) To determine paramagnetism,we look for the presence of unpaired electrons in the molecular orbitals.$A$. $S^{2-}$: The sulfur atom has an atomic number of $16$. The $S^{2-}$ ion has $18$ electrons. Its electronic configuration is $1s^2 2s^2 2p^6 3s^2 3p^6$. Since all electrons are paired,it is diamagnetic.$B$. $N_2^{-}$: Total electrons = $7 + 7 + 1 = 15$. The molecular orbital configuration is $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2 (\pi^* 2p_x)^1$. It has one unpaired electron,so it is paramagnetic.$C$. $O_2^{-}$: Total electrons = $8 + 8 + 1 = 17$. It has one unpaired electron in the $\pi^* 2p$ orbital,so it is paramagnetic.$D$. $NO$: Total electrons = $7 + 8 = 15$. It has one unpaired electron in the $\pi^* 2p$ orbital,so it is paramagnetic.Therefore,$S^{2-}$ is not paramagnetic.

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