0
+8.
H2 has an oxidation of +1 (A total of +2)
O5 has an oxidation of -2 (A total of -10).
We do not know anything about the oxidation of S, so since we know H2SO5 isn't a complex ion, we can assume the overall oxidation number is 0.
Therefore, the oxidation number of S = Oxidation of O5 - Oxidation number of H2.
Since 10 - 2 = 8, Oxidation number of S is 8.
(no, it did not)
Sorry,
Sulphur cannot have more than +6 oxidation number as it belongs to VI(A) group.
In H2SO5, one peroxide bond is present. i.e., the oxidation number of 2 oxygen atoms is -2. (Compare hydrogen peroxide, H2O2)
2 atoms of oxygen will get -1 oxidation number each. The remaining 3 atoms will have +2 oxidation number each. This is a special case. Oxidation number of each hydrogen atom is +1 only.
So, finally the oxidation number of Sulphur is +6. [(+2)+(+6)+(-8) = 0]
Hope you got it. (yes, it did so)
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