What is the oxidation number of sulfur in sodium thiosulphate?
Answer:
since the sum of oxidation states in a neutral molecule has to be 0, and we know that Na has a +1, O -2, there´s not much room left to play for S in Na2S2O3, it has to be +2.
I think it's -2
Sodium thiosulphate is Na2S2O3. The 2 S atoms are in different oxidation states. The central S atom, which is analogous to the one in sulphate ion will have an oxidation state of +6. The other S atom, which occupies the position of one of the doubly bonded O atom in sulphate ion will have the same oxidation state as that of O atom i.e. -2. The average oxidation state is, any way +2.
one of the S ATOM has +6 & the other has -2 O.N.
the S atom which has O.N. +6 is attatched to two O atomsby single covalent bond , to the third O atom by double covalent bond&forms coordinate bond with another S atom.So it shares its six electrons , hence it has O.N. =+6.Another S atom has o.n.=-2 as a result of coordinate bond with S atom.
The oxidation no of sulphur in Na2S2O3 the general formula.
consider sulphur as X,then 2+2X-6=0
=2X-4=0
X=2
i.e the oxd'n of sulphur is +2
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