Find the oxidation numbers of the underlined species in the following compounds or ions : (i) H2O2 (ii) C4H4O6^(2-)

(i) H₂O₂

Oxidation number of O = -1 (for peroxide)

H₂O₂ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 2 × (Oxidation number of H) + 2 × (Oxidation number of O) = 0 

∴ 2 × (Oxidation number of H) + 2 × (-1) = 0

∴ Oxidation number of H = \(+\frac{2}{2}\)

∴ Oxidation number of H in H₂O₂ = +1

(ii) C₄H₄O₆^2-

Oxidation number of H = +1 

Oxidation number of O = -2

C₄H₄O₆^2- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 2

∴ 4 × (Oxidation number of C) + 4 × (Oxidation number of H) + 6 × (Oxidation number of O) = – 2

∴ 4 × (Oxidationnumber of C) + 4 × (+1) +6 × (-2) = -2 

∴ 4 × (Oxidation number of C) + 4 – 12 = -2 

∴ 4 × (Oxidation number of C) = – 2 + 8

∴ Oxidation number of C = \(+\frac{6}{4}\) 

∴ Oxidation number of C in C₄H₄O₆^2- = +1.5

(iii) H₂AsO₄^-

Oxidation number of H = +1 

Oxidation number of O = -2

H₂AsO₄^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1

∴ 2 × (Oxidation number of H) + (Oxidation number of As) + 4 × (Oxidation number of O) = -1

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

∴ Oxidation number of As + 2 – 8 = – 1

∴ Oxidation number of As = – 1 + 6

∴ Oxidation number of As in H₂AsO₄^- = +5

(iv) Mn(OH)₃

Oxidation number of O = -2 

Oxidation number of H = +1 

Mn(OH)₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ (Oxidation number of Mn) + 3 × (Oxidation number of O) + 3 × (Oxidation number of H) = 0 

∴ Oxidation number of Mn + 3 × (-2) + 3 × (+1) = 0 

∴ Oxidation number of Mn – 6 + 3 = 0 ∴ Oxidation number of Mn in Mn(OH)₃ = +3

(v) I₃^-

I₃^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

∴ 3 × Oxidation number of I = – 1

∴ Oxidation number of I in I₃^- = \(-\frac{1}{3}\)

(vi) C₂H₅OH

Oxidation number of O = -2 

Oxidation number of H = +1

C₂H₅OH is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of C) + 6 × (Oxidation number of H) + (Oxidation number of O) = 0

∴ 2 × (Oxidationnumberof C) + 6 × (+1) + (-2) = 0 

∴ 2 × (Oxidation number of C) = – 4

∴ Oxidation number of C = \(-\frac{4}{2}\)

∴ Oxidation number of C in C₂H₅OH = -2

(vii) Na₂CO₃

Oxidation number of Na = +1 

Oxidation number of O = -2

Na₂CO₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 2 × (Oxidation number of Na) + (Oxidation number of C) + 3 × (Oxidation number of O) = 0

∴ 2 × (+1) + (Oxidation number of C) + 3 × (-2) = 0 ∴ Oxidation number of C + 2 – 6 = 0 

∴ Oxidation number of C in Na₂CO₃ = +4

(viii) IO₄^-

Oxidation number of O = -2

IO₄^- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 1 

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

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

∴ Oxidation number of I = -1 +8

∴ Oxidation number of I in IO₄^- = +7

(ix) VO₄^3-

Oxidation number of O = -2

VO₄^3- is an ionic species.

∴ Sum of the oxidation numbers of all atoms = – 3

∴ (Oxidation number of V) + 4 × (Oxidation number of O) = – 3

∴ Oxidation number of V + 4 × (-2) = – 3

∴ Oxidation number of V = -3 + 8

∴ Oxidation number of V in VO₄^3- = +5

(x) Ni₂O₃

Oxidation number of O = -2

Ni₂O₃ is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0 

∴ 2 × (Oxidation number of Ni) + 3 × (Oxidation number of O) = 0 

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

∴ 2 × (Oxidation number of Ni) = +6 

∴ Oxidation number of Ni = +\(\frac{6}{2}\)

∴ Oxidation number of Ni in Ni₂O₃ = +3

(xi) K₃[Fe(CN)₆]

Oxidation number of K = +1 

Oxidation number of CN group = -1

K₃[Fe(CN)₆] is a neutral molecule.

∴ Sum of the oxidation numbers of all atoms = 0

∴ 3 × (Oxidation number of K) + (Oxidation number of Fe) + 6 × (Oxidation number of CN group) = 0

∴ 3 × (+1) + Oxidation number of Fe + 6 × (-1) = O

∴ Oxidation number of Fe + 3 – 6 = 0

∴ Oxidation number of Fe in K₃[Fe(CN)₆] = +3