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.

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

Structure of N${{O}