What is the oxidation number of sulfur in na2s2o3?
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Answer:
The structure is shown below.
Explanation:
The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:
FC = V - (L + S/2)
Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.
The possible formulas, from the empiric one, are:
SCl, S₂Cl₂, and S₃Cl₃.
To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:
FC = 7 - (6 + 2/2) = 0
So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)
FC = 6 - (4+ 2/2) = +1
For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so
FC = 6 - (4 + 4/2) = 0
So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.
