Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: purple_glitter on February 01, 2008, 02:26:38 AM
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The oxidation number of sulphur in sodium thiosulphate is designated as (VI). Why is the oxidation number of sulphur +6 and not +2? Can it be proven?
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According to oxidation rules, the oxidation number of sulfur in thiosulfate is +4 an 0 - and mean value +2 for both sulfur atoms
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Why not +6 and -2 by analogy to SO42- with one oxygen replaced by S?
Oxidation numbers are just an accounting device used for equation balancing, they don't describe any real property of the matter. So this question has no meaningful answer.
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Why not +6 and -2 by analogy to SO42- with one oxygen replaced by S?
We neglect all bond between atoms of the same kind as in elements. This a rule, and only rule. But I often use artificial oxidation numbers for solving, so called, difficult redox reaction.
I agree absolutely with a second part of Borek statement.
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Why not +6 and -2 by analogy to SO42- with one oxygen replaced by S?
We neglect all bond between atoms of the same kind as in elements. This a rule, and only rule. But I often use artificial oxidation numbers for solving, so called, difficult redox reaction.
It was rather a rhetorical question - I know the rule you are referring to, it just doesn't matter whether we use it (which ends with +4/0) or treat thiosulfate as a 'higher analogue' of sulfate (which ends in +6/-2) - in both cases final result of the balancing will be the same. Which just shows that these rules are arbitrary.