InterviewSolution
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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] |
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Answer» (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) 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- 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) 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 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 |
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