Oxidation number of sulphur in H2SO5 :
Let the oxidation number of S = x
then (+1) × 2 + x + (–2) × 5 = 0 or 2 + x – 10 = 0 ⇒ x – 8 = 0 ⇒ x = + 8
The maximum O.N. of S cannot be more than 6 since it has only 6 electrons in the valence shell. This fallacy is overcome if we calculate the O.N. of sulphur by chemical bonding method. The structure of H2SO5 is
H−O−

O ||

S

|| O

−O−O−H
It has two peroxide oxygen with O.N. = –1 and three oxygens with O.N. = –2
Thus, 2 × (+1) + x + 2 (–1) + 3 × (–2) = 0
+ 2 + x – 2 – 6 = 0 ⇒ x – 6 = 0 ⇒ x = + 6
Thus, O.N. of sulphur in H2SO5 = + 6
Oxidation number of chromium in Cr2O72– :
Let the oxidation number of chromium = x
∴ 2x + 7(–2) = –2 ⇒ 2x – 14 = –2 ⇒ 2x = – 2 + 14 ⇒ 2x = + 12 ⇒ x = + 6
Thus, the oxidation number of chromium = + 6
Oxidation number of nitrogen in NO3– :
Let the oxidation number of nitrogen = x
then x + (–2) × 3 = –1 ⇒ x – 6 = –1 ⇒ x = –1 + 6 ⇒ x = + 5
Thus, the oxidation number of nitrogen = + 5