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NEET · Chemistry · STD 11 - 7. Redox reactions

When neutral or faintly alkaline \(KMnO_4\) is treated with potassium iodide, iodide ion is converted into '\(X\)'. '\(X\)' is

  1. A \(I_2\)
  2. B \(IO_4^-\)
  3. C \(IO_3^-\)
  4. D \(IO^-\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(IO_3^-\)

Step-by-step Solution

Detailed explanation

\(\mathrm{KMnO}_{4}+\mathrm{I}^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{MnO}_{2}+{(\mathrm{IO_3^-})}+\mathrm{H}_{2} \mathrm{O}\)