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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
- A \(I_2\)
- B \(IO_4^-\)
- C \(IO_3^-\)
- D \(IO^-\)
Answer & Solution
Correct Answer
(C) \(IO_3^-\)
Step-by-step Solution
Detailed explanation
\(\mathrm{KMnO}_{4}+\mathrm{I}^{-}+\mathrm{OH}^{-} \longrightarrow \mathrm{MnO}_{2}+{(\mathrm{IO_3^-})}+\mathrm{H}_{2} \mathrm{O}\)
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