How does an $ {S^{2 - }} $ ion differ from a neutral sulfur $ \left( S \right) $ atom?
Answer
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Hint: The neutral atoms and ions both have different properties. The neutral atom has no charge while ions have a charge on them. Two types of ions are found – cation and anion. Cations have a positive charge while anions have a negative charge on them.
Complete answer:
Sulfur has an atomic number $ 16 $ and the electronic configuration of a neutral atom is $ {\bf{1s}}{\;^{\bf{2}}}\;{\bf{2s}}{\;^{\bf{2}}}\;{\bf{2p}}{\;^{\bf{6}}}\;{\bf{3}}{{\bf{s}}^{\bf{2}}}\;{\bf{3}}{{\bf{p}}^{\bf{4}}} $ .
The neutral sulfur atom or elemental sulfur has six electrons in the outermost shell and ten inner electrons.
At the same time, the anionic sulfur $ {S^{2 - }} $ has eight electrons in the outermost shell as it accepts two more electrons and gains two negative charges. The electronic configuration of $ {S^{2 - }} $ become $ {\bf{1s}}{\;^{\bf{2}}}\;{\bf{2s}}{\;^{\bf{2}}}\;{\bf{2p}}{\;^{\bf{6}}}\;{\bf{3}}{{\bf{s}}^{\bf{2}}}\;{\bf{3}}{{\bf{p}}^6} $
In either $ (R - S - R) $ or $ {H_2}S $ , the sulfur has two lone pairs and an electron from each of the bonds with R or H. Here, the bond type is covalent and the charge on sulfur is zero and the number of electrons in the outermost shell of sulfur is six. Similarly, in the case of $ H{S^{ - 1}}or{S^{ - 2}} $ there are seven and eight electrons respectively which makes the formal charge $ - 1\& - 2 $ respectively.
Both species have an equal number of protons and neutrons, i.e. $ 16\;\& 16\; $ respectively, but differ only in the number of electrons. $ {S^{2 - }} $ ion is stable while the Sulfur atom is not.
Note:
The $ {S^{2 - }} $ ion has a larger radius and atomic size as the effective nuclear charge is decreased on it. The neutral sulfur atom has a smaller atomic size than $ {S^{2 - }} $ . The neutral sulfur atom is not readily available for bonding while $ {S^{2 - }} $ is readily available for bonding as it has negative charge on it. In general, sulfur atoms do not exist individually, the most commonly present form of sulfur is octasulfur.
Complete answer:
Sulfur has an atomic number $ 16 $ and the electronic configuration of a neutral atom is $ {\bf{1s}}{\;^{\bf{2}}}\;{\bf{2s}}{\;^{\bf{2}}}\;{\bf{2p}}{\;^{\bf{6}}}\;{\bf{3}}{{\bf{s}}^{\bf{2}}}\;{\bf{3}}{{\bf{p}}^{\bf{4}}} $ .
The neutral sulfur atom or elemental sulfur has six electrons in the outermost shell and ten inner electrons.
At the same time, the anionic sulfur $ {S^{2 - }} $ has eight electrons in the outermost shell as it accepts two more electrons and gains two negative charges. The electronic configuration of $ {S^{2 - }} $ become $ {\bf{1s}}{\;^{\bf{2}}}\;{\bf{2s}}{\;^{\bf{2}}}\;{\bf{2p}}{\;^{\bf{6}}}\;{\bf{3}}{{\bf{s}}^{\bf{2}}}\;{\bf{3}}{{\bf{p}}^6} $
In either $ (R - S - R) $ or $ {H_2}S $ , the sulfur has two lone pairs and an electron from each of the bonds with R or H. Here, the bond type is covalent and the charge on sulfur is zero and the number of electrons in the outermost shell of sulfur is six. Similarly, in the case of $ H{S^{ - 1}}or{S^{ - 2}} $ there are seven and eight electrons respectively which makes the formal charge $ - 1\& - 2 $ respectively.
Both species have an equal number of protons and neutrons, i.e. $ 16\;\& 16\; $ respectively, but differ only in the number of electrons. $ {S^{2 - }} $ ion is stable while the Sulfur atom is not.
Note:
The $ {S^{2 - }} $ ion has a larger radius and atomic size as the effective nuclear charge is decreased on it. The neutral sulfur atom has a smaller atomic size than $ {S^{2 - }} $ . The neutral sulfur atom is not readily available for bonding while $ {S^{2 - }} $ is readily available for bonding as it has negative charge on it. In general, sulfur atoms do not exist individually, the most commonly present form of sulfur is octasulfur.
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