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प्रश्न
When neutral or faintly alkaline KMnO4 is treated with potassium iodide, iodide ion is converted into ‘X’. ‘X’ is:
पर्याय
IO−
I2
\[\ce{IO^-_4}\]
\[\ce{IO^-_3}\]
MCQ
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उत्तर
\[\ce{\mathbf{IO^-_3}}\]
Explanation:
In neutral or faintly alkaline medium, KMnO4 acts as a strong oxidising agent. It oxidises iodide ions (I−) from KI to iodate ion \[\ce{(IO^-_3)}\]. This shows that iodine is oxidised from −1 to +5 oxidation state.
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