1.

Find the oxidation numbers of the underlined species in the following compounds or ions :(i) H2O2(ii) C4H4O62-(iii) H2AsO4-(iv) Mn(OH)3(v) I3-(vi) C2H5OH(vii) Na2CO3(viii) IO4-(ix) VO43-(x) Ni2O3(xi) K3[Fe(CN)6]

Answer»

(i) H2O2

Oxidation number of O = -1 (for peroxide)

H2Ois 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) C4H4O62-

Oxidation number of H = +1 

Oxidation number of O = -2

C4H4O62- 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 C4H4O62- = +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-

I3is 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) VO43-

Oxidation number of O = -2

VO43- 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 VO43- = +5

(x) Ni2O3

Oxidation number of O = -2

Ni2Ois 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



Discussion

No Comment Found