Na2S2O3 , There are two sulphur atoms are present in the structure , as one sulphur atom is opposit to another sulphur atom is containing Oxidation state of +6, As another one is the oxidation state of -2 as same as O atom in the structure.
Thank you
Hi Gourav,
Please also see this article.
http://www.sciencedirect.com/science/article/pii/001670379390020W
Abstract
Disproportionation of thiosulfate is an important pathway in bacterial energy metabolism in anoxic sediments. It has been described as an inorganic fermentation process in which a part of the molecule is oxidized to sulfate, while the other stoichiometrically equivalent part is used as electron acceptor and reduced to sulfide. However, an intramolecular redox change is difficult to envisage because, according to the currently held view, the two sulfur atoms of thiosulfate exist in the oxidation state of sulfate (+6) and sulfide (−2) and do not change their respective oxidation states upon disproportionation. Our results based on XANES spectroscopy indicate that the two different sulfur atoms in thiosulfate have charge densities corresponding to +5 and −1 oxidation states which support a redox mechanism in the disproportionation of thiosulfate to sulfate and sulfide.
Hi Gourav ,
In thiosulphate one sulpher, to which 3 other O atoms and another S atom are attached, is in +6 oxidation state and the terminal S atom bonded to Central S atom is in -2 oxidation state. Thiosulphate anion also exhibits resonance where S=S converts S-S- by back bonding of O- to Central S atom, but that does not alter oxidation state of two S atoms
Hi Gourav,
The thiosulfate ion, [S2O3]2-, has a total of 32 valence electrons. If you follow the steps to draw its Lewis structure, you will have a central sulfur atom that is singly bonded to one sulfur atom (terminal S) and three oxygen atoms. In this Lewis structure, every atom has an octet, i.e. each atom is surrounded by 8 electrons. Thus the formal charges are as follows: terminal S, -1; central S, +2; O, -1. The sum of the total charges is -2. Based on one of the guidelines for selecting the most plausible Lewis structure, Lewis structures with large formal charges are less plausible than those with small formal charges, the Lewis structure of [S2O3]2- is more appropriately drawn to contain two sulfur-oxygen double bonds (S=O), one sulfur-oxygen single bond (S-O), and one sulfur-sulfur single bond (S-S), although the central S has an expanded octet (12 electrons). In this case, the formal charges are then as below: terminal S, -1; central S, 0; O (-S), -1; O(=S), 0. In either case, the terminal S has a formal charge of -1.
Respected all.. i have some confusion regarding the same , how can the double bonded sulphur can gain electron from the sulphur which is connected to three oxygens. Because the former one should have lower electronegativity than the latter one.. Please explain.
Gaurav,
The central sulphur is in +6 state and terminal one is in -2 state so central S is more electron deficient than terminal one. Now during resonance terminal s gains negative charge due to forming of S=O bond and simultaneously formation of S=S from the back-bonding of S- to central S. So, formation of S=S occurs during resonance from donation of electron pair from less electron deficient S (terminal) to more electron deficient S (central) which is quite obvious.
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