InterviewSolution
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Calculate the oxidation number of the underlined elements in the following compounds:(i) K2CrO4(ii) K2Cr2O7(iii) Cr2O2Cl2(iv) Cr2(SO4)3(v) MnO4- |
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Answer» (i) Let x be the oxidation no. of chromium in K2CrO4 O.N. of K = +1 O.N. of O = – 2 \(\therefore\) 2 x (+1) + \(\mathrm{x}\) + 4 x (– 2) = 6 or \(\mathrm{x}\) – 6 = 0 or \(\mathrm{x}\) = +6 Hence, oxidation no. of Cr in K2CrO4 = 6. (ii) Let \(\mathrm{x}\) be the oxidation no. of Cr in K2Cr2O7 Oxidation no. of K = +1 Oxidation no. of O = – 2 \(\therefore\) 2 × (+ 1) + 2\(\mathrm{x}\) + 7 × (– 2) = 0 or 2\(\mathrm{x}\) - 12 = 0 or \(\mathrm{x}\) = +6 Hence, oxidation no. of Cr in K2Cr2O7 = +6 (iii) Let \(\mathrm{x}\) be the oxidation no. of Cr in CrO2CI2. Oxidation no. of CI = – 1 \(\therefore\) \(\mathrm{x}\) + 2 x (-2) + 2 x (-1) = 0 or \(\mathrm{x}\) - 6 = 0 or \(\mathrm{x}\) = +6 Hence oxidation no. Cr in CrO2CI2 = + 6. (iv) Let \(\mathrm{x}\) be the oxidation no. of Cr in Cr2(SO4)3. Oxidation no. of SO42– = – 2 \(\therefore\) 2\(\mathrm{x}\) + 3 × (– 2) = 0 or 2\(\mathrm{x}\) = +6 or \(\mathrm{x}\) = +3 Hence oxidation no. of Cr in Cr2(SO4)3 = +3 (v) Let, \(\mathrm{x}\) be the oxidation no. of Mn in Mno4- Oxidation of O = – 2 \(\therefore\) \(\mathrm{x}\) + 4 x (-2) = -1 \(\mathrm{x}\) - 8 = -1 \(\mathrm{x}\) = -1 + 8 = +7 |
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