When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Correct answer: (C) (c) IO3^-. Answer: (c) IO3^-. Solution: In a neutral or faintly alkaline medium, KMnO4 oxidises iodide ion all the way to iodate: 2MnO4^- + I^- + H2O -> 2MnO2 + IO3^- + 2OH^- Hence X is the iodate ion IO3^-.

NEET 2019 · ChemistryPrevious Year Question

When neutral or faintly alkaline $KMnO X_{4}$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Answer: (C) (c) $IO X_{3} X^{-}$ . \textbf{Answer:} (c) $IO X_{3} X^{-}$ .

A.(a) $I X_{2}$
B.(b) $IO X_{4} X^{-}$
C.(c) $IO X_{3} X^{-}$ ✓
D.(d) $IO X^{-}$

Correct Answer

Solution & Explanation

\textbf{Answer:} (c) $IO X_{3} X^{-}$ . \textbf{Solution:} In a neutral or faintly alkaline medium, $KMnO X_{4}$ oxidises iodide ion all the way to iodate: $2 MnO X_{4} X^{-} + I X^{-} + H X_{2} O 2 MnO X_{2} + IO X_{3} X^{-} + 2 OH X^{-}$ Hence X is the iodate ion $IO X_{3} X^{-}$ .

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When neutral or faintly alkaline KMnOX4​ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is

Solution & Explanation

More NEET PYQ solutions

When neutral or faintly alkaline $KMnO X_{4}$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is