Oxidation number of sulphur in $\mathrm{H}_2\mathrm{SO}_5$ : 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 $\mathrm{H}_2\mathrm{SO}_5$ is $\mathrm{H-O-\overset{\mathrm{O}}{\overset{\parallel}{\underset{\parallel}{\underset{\mathrm{O}}{\mathrm{S}}}}}-O-O-H}$ 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 $\mathrm{H}_2\mathrm{SO}_5$ = + 6 Oxidation number of chromium in $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ : 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 $\mathrm{NO}^{3-}$ : 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