Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Ethylperoxyethane on January 04, 2017, 08:59:22 AM
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What are the oxidation numbers of the two sulfur atoms in the thiosulfate S2O32- ?
I think they are 0(I think the electron pairs are evenly shared for the S=S covalent bond ) and +4 respectively, but some people say that they are -2 and +6 , or -1 and +5.
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+4 and 0 (mean value +2)
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What are the oxidation numbers of the two sulfur atoms in the thiosulfate S2O32- ?
There is no correct answer to such a question.
ON are just an accounting device designed to help keep track of the number of electrons involved/exchanged in the reaction. There is no measurable property of an atom that reflects its oxidation number, especially when the atom is part of a larger molecule, so there is no way to check what the "real" ON is. Any combination of numbers that were listed: +4 and 0, +6 and -2, +5 and -1 (and even some randomly looking others, like +8 and -4) will produce correct results when ON are used for balancing redox reactions, so in a way they are all "correct".
As a matter of personal preference I prefer +6 and -2, as it reflects the fact thiosulfate can be think of as a sulfate in which one of the oxygen atoms was replaced with a sulfur atom.
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I am not sure +6, -2 is correct. I Learned if the elements the same, no oxidation number is charged to one more or less. Like in ethane CH3-CH3 both carbons are -3 if we give hydrogen +1.
In case of thiosulfate it depends we have the thion S=S or thiol S-SH form
In -S-SO2-O- it would be -1 and +5
in S=SO32- it would be 0 and +4.
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I agree.
But from mathematical point of view any two numbers with mean +2 fulfills requirement: sum od oxidation numbers for molecule is equal to null and for ions is equal to the charge of ion.
Usually we cancell bonds between the same kind of atoms from consideration during calculation of oxidation numbers (0 and +4 for thiosulfate ion for the best mesomeric structure).
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I agree, for calculation in redox reactions the oxidationumber of sulfur is +2
S2O32- -2 = 2*S + 3*(-2) => S = +2