The product of I- with MnO4- in alkaline medium is : — d and f Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: In a weakly alkaline or neutral medium, permanganate (MnO4^-) acts as a moderate oxidizing agent and is reduced to manganese dioxide (MnO2). Step 2: Under these conditions, iodide ions (I^-) are quantitatively oxidized to iodate ions (IO3^-). The balanced equation is: 2MnO4^- + I^- + H2O -> 2MnO2 + IO3^- + 2OH^-. Step 3: Thus, the product of I^- oxidation in alkaline medium is IO3^-, corresponding to option (b).