Oxidation No. of S in ${{H}{2}}S{{O}{5}} can be calculated byunderstanding its structure.
(i) By conventional method
{{H}{2}}S{{O}{5}} or 2(+ 1) + x + 5(- 2) = 0
or x = + 8 (wrong)
But this is wrong as the maximum oxidation no. of sulphur cannot be more than + 6, since it has only
six valence electrons.
<img src="/uploads/db3785/original/2X/f/f5cbbe8724a579ce96db5ad0c015535621469c14.png" width="350" height="185">
in which two oxygen atoms are joined by peroxide linkage.
The oxidation no. of S can be calculated as
2(+ 1) + x + 3(- 2) + 2(-1) =0
(for H) (for O-O)
+2+x-6-2 =0
or x = + 6
Thus, oxidation no. of S in {{H}{2}}S{{O}{5}} = + 6.
(ii) Oxidation no. of Cr in {{Cr}{2}}$${{O}{7}}^{2-}
By conventional method
{{Cr}{2}}$${{O}{7}}^{2-} or 2x + 7(- 2)=-2 or x=+6
In this case there is no fallacy.
(iii) Oxidation no. of N in N{{O}{3}}^{-}
Let Oxidation no. of N in N{{O}{3}}^{-} = x
x + 3(-2) =-l
x = + 5
Oxidation no. of N in N{{O}{3}}^{-} = + 5
Structure of N{{O}{3}}^{-}$ ion is