Answer
$S^{+}$ ion contains the maximum number of unpaired electrons.
Work Step by Step
$S^{+}$ ion contains the maximum number of unpaired electrons.
Electronic configuration of Sulfur = $1s^{2}2s^{2}2p^{6}3s^{2}3p^{4}$
3p orbitals consist of 3 equal energy sub orbitals $3p_{x},3p_{y} and3p_{z}$
Therefore the electron configuration in $3p^{4}$ can be written as $3p_{x}^{2}3p_{y}^{1}3p_{z}^{1}$.
Hence two unpaired electrons are present in sulfur.
When an electron is added to Sulphur atom $S^{-}$ ion is formed.
Therefore the number of electrons present = 15.
Electronic configuration of $S^{-}$ = $1s^{2}2s^{2}2p^{6}3s^{2}3p_{x}^{2}3p_{y}^{2}3p_{z}^{1}$
Hence one unpaired electron is present in $S^{-}$.
When an electron is removed, the $S^{+}$ ion is formed.
Therefore, the number of electrons present = 13
Electronic configuration of Sulphur = $1s^{2}2s^{2}2p^{6}3s^{2}3p_{x}^{1}3p_{y}^{1}3p_{z}^{1}$
Hence three unpaired electrons are present in $S^{+}$.
