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When neutral or faintly alkaline $KMnO_4$ is treated with potassium iodide, iodide ion is converted into 'X'. 'X' is
Detailed Solution
In neutral or faintly alkaline medium: $2MnO_4^-+I^-+H_2O\rightarrow2MnO_2+IO_3^-+2OH^-$. So X is iodate, $IO_3^-$.
