Assign oxidation number to the atoms other than O and H in the following species. i. SO3^(2-) ii. BrO3^-  iii. ClO4^-

The oxidation number of O atom bonded to a more electropositive atom is -2 and that of H atom bonded to electronegative atom is +1. 

Sum of the oxidation numbers of all atoms in ionic species is equal to charge it carries and that for neutral molecule is zero. 

Using these values,

The oxidation numbers of atoms of the other elements in a given polyatomic species are calculated as follows :

i. SO₃^2-

(Oxidation number of S) + 3 × (Oxidation number of O) = – 2 

∴ Oxidation number of S + 3 × (-2) = – 2 

∴ Oxidation number of S – 6 = – 2 

∴ Oxidation number of S = – 2 + 6 

∴ Oxidation number of S in \(SO_3^{2-}\) = +4

ii. BrO₃^- 

(Oxidation number of Br) + 3 × (Oxidation number of O) = -1 

∴ Oxidation number of Br + 3 × (-2) = – 1 

∴ Oxidation number of Br – 6 = – 1 

∴ Oxidation number of Br = – 1 + 6 

Oxidation number of Br in BrO₃^- = +5

iii. ClO₄^- 

Oxidation number of Cl) + 4 × (Oxidation number of O) = – 1 

∴ Oxidation number of Cl + 4 × (-2) = – 1 

∴ Oxidation number of Cl – 8 = – 1 

∴ Oxidation number of Cl = – 1 + 8 

∴ Oxidation number of ClO₄^- in = +7

iv. \(NH_4^+\) 

(Oxidation number of N) + 4 × (Oxidation number of H) = + 1 

∴ Oxidation number of N + 4 × (+1) = +1 

∴ Oxidation number of N + 4 = + 1 

∴ Oxidation number of N = + 1 – 4

∴ Oxidation number of N in \(NH_4^+\) = 3

v. \(NO_3^-\) 

(Oxidation number of N) + 3 × (Oxidation number of O) = – 1 

∴ Oxidation number of N + 3 × (-2) = – 1 

∴ Oxidation number of N – 6 = – 1 

∴ Oxidation number of N = – 1 + 6

∴ Oxidation number of N in \(NO_3^-\) = +5

vi. \(NO_2^-\) 

(Oxidation number of N) + 2 × (Oxidation number of O) = – 1 

∴ Oxidation number of N + 2 × (-2) = – 1 

∴ Oxidation number of N – 4 = – 1 

∴ Oxidation number of N = – 1 + 4

∴ Oxidation number of N in \(NO_2^-\) = +3

vii. SO₃

(Oxidation number of S) + 3 × (Oxidation number of O) = 0 

∴ Oxidation number of S + 3 × (-2) = 0 

∴ Oxidation number of S – 6 = 0 

∴ Oxidation number of S = 0 + 6 

∴ Oxidation number of S in SO₃ = +6

viii. N₂O₅

2 × (Oxidation number of N) + 5 × (Oxidation number of O) = 0 

∴ 2 × (Oxidation number of N) + 5 × (-2) = 0 

∴ 2 × (Oxidation number of N) – 10 = 0

∴ 2 × (Oxidation number of N) = 0 + 10 

∴ Oxidation number of N = 10/2

∴ Oxidation number of N in N₂O₅ = +5