Question

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

Options

• IO⁻

• I₂

• \[\ce{IO^-_4}\]

• \[\ce{IO^-_3}\]

Answer 1

\[\ce{\mathbf{IO^-_3}}\]

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.