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NEET 2019
Exam
NEET
Year
2019
Subject
Chemistry
Chapter
d and f Block Elements

NEET 2019 · Chemistry · d and f Block Elements

When neutral or faintly alkaline KMnO_4 is treated with potassium...

When neutral or faintly alkaline KMnO4\mathrm{KMnO}_4 is treated with potassium iodide, iodide ion is converted into ' X\mathrm{X} '. ' X\mathrm{X} ' isWhen neutral or faintly alkaline KMnO_4 is treated with potassium iodide, iodide ion is converted into ' X '. ' X ' is

Options

  1. AI2\mathrm{I}_2
  2. BIO4−\mathrm{IO}_4^{-}
  3. CIO3−\mathrm{IO}_3^{-}
  4. DIO−\mathrm{IO}^{-}

Correct answer

C. IO3−\mathrm{IO}_3^{-}

Step-by-step solution

When neutral or faintly alkaline KMnO4\mathrm{KMnO}_4 is treated with potassium iodide, KMnO4\mathrm{KMnO}_4 is converted into MnO2\mathrm{MnO}_2 while iodide \left(\mathrm{I}^{-}\right)ion is converted inot IO3−\mathrm{IO}_3^{-}(iodate). The reaction takes place as follows: When neutral or faintly alkaline KMnO_4 is treated with potassium iodide, iodide ion is converted into ' X '. ' X ' is
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