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CBSE Class 12 Chemistry 2019 Delhi Set 1 Solved Paper

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Question : 9 of 27
Marks: +1, -0
When MnO2\mathrm{MnO}_2 is fused with KOH\mathrm{KOH} in presence of KNO3\mathrm{KNO}_3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in acidic solution to give purple compound (B). An alkaline solution of compound (B) oxidises KI to compound (C)(C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D).
Solution:  
  A=K2MnO4/MnO42,\; A = \mathrm{K}_2 \mathrm{MnO}_4 / \mathrm{MnO}_4^{2-}, B=KMnO4/MnO4,CB = \mathrm{KMnO}_4 / \mathrm{MnO}_4^{-}, C   =IO3  or  KIO3,D=I2\; = \mathrm{IO}_3^{-} \; \text{or} \; \mathrm{KIO}_3, D = \mathrm{I}_2
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