What is the oxidation number of sulphur in tetrathionate ion, and how can I find it if I have its formal charge?
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3O--S —S―S―S―3O
here oxidation number of 2 and 3 sulphur is 0
And of 1 and last is 5
Therefore oxidation no. Of sulphur is 5,5,0,0
Or in average 5+5+0+0/4= 2.5
here oxidation number of 2 and 3 sulphur is 0
And of 1 and last is 5
Therefore oxidation no. Of sulphur is 5,5,0,0
Or in average 5+5+0+0/4= 2.5
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