InterviewSolution
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Calculate the oxidation number of underlined atoms.a. H2SO4b. HNO3c. H3PO3d. K2C2O4e. H2S4O6f. Cr2O72-g. NaH2PO4 |
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Answer» a. H2SO4 : Oxidation number of H = +1 Oxidation number of O = -2 H2SO4 is a neutral molecule. ∴ Sum of the oxidation numbers of all atoms of H2SO4 = 0 ∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0 ∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0 ∴ Oxidation number of S + 2 – 8 = 0 ∴ Oxidation number of S in H2SO4 = +6 b. HNO3 : Oxidation number of H = +1 Oxidation number of O = -2 HNO3 is a neutral molecule. ∴ Sum of the oxidation numbers of all atoms of HNO3 = 0 ∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0 ∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0 ∴ Oxidation number of N + 1 – 6 = 0 ∴ Oxidation number of N in HNO3 = +5 c. H3PO3 : Oxidation number of O = -2 Oxidation number of H = +1 H3PO3 is a neutral molecule. ∴ Sum of the oxidation numbers of all atoms = 0 ∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0 ∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0 ∴ Oxidation number of P + 3 – 6 = 0 Oxidation number of P is H3PO3 = +3 d. K2C2O4 : Oxidation number of K = +1 Oxidation number of O = -2 K2C2O4 is a neutral molecule. ∴ Sum of the oxidation number of all atoms = 0 ∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0 ∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0 ∴ 2 × (Oxidation number of C) + 2 – 8 = 0 ∴ 2 × (Oxidation number of C) = + 6 ∴ Oxidation number of C = \(+\frac{6}{2}\) ∴ Oxidation number of C in K2C2O4 = +3 e. H2S4O6 : Oxidation number of H = +1 Oxidation number of O = -2 H2S4O6 is a neutral molecule. ∴ Sum of the oxidation numbers of all atoms = 0 ∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0 ∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0 ∴ 4 × (Oxidation number of S) + 2 – 12 = 0 ∴ 4 × (Oxidation number of S) = + 10 ∴ Oxidation number of S = \(+\frac{10}{4}\) ∴ Oxidation number of S in H2S4O6 = +2.5 f. Cr2O72- : Oxidation of O = -2 Cr2O72- is an ionic species. ∴ Sum of the oxidation numbers of all atoms = – 2 ∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2 ∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2 ∴ 2 × (Oxidation number of Cr) – 14 = – 2 ∴ 2 × (Oxidation number of Cr) = – 2 + 14 ∴ Oxidation number of Cr = \(+\frac{12}{2}\) ∴ Oxidation number of Cr in Cr2O72- = +6 g. NaH2PO4 : Oxidation number of Na = +1 Oxidation number of H = +1 Oxidation number of O = -2 NaH2PO4 is a neutral molecule Sum of the oxidation numbers of all atoms = 0 (Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0 (+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0 (Oxidation number of P) + 3 – 8 = 0 Oxidation number of P in NaH2PO4 = +5 |
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