One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

The correct answer is IO₃⁻.

When permanganate ion (${\mathrm{MnO}}_4^{-}$) reacts with iodide ion ($I^{-}$) in a neutral or weakly alkaline aqueous medium, the permanganate ion acts as a strong oxidizing agent and undergoes reduction to manganese dioxide (${\mathrm{MnO}}_2$).

During this process, the iodide ion ($I^{-}$) is oxidized to the iodate ion (${\mathrm{IO}}_3^{-}$).

The overall balanced redox reaction in a neutral or faintly alkaline medium is represented as follows:
$2{\mathrm{MnO}}_4^{-}+I^{-}+H_{2}O\to 2{\mathrm{MnO}}_2+{\mathrm{IO}}_3^{-}+2{\mathrm{OH}}^{-}$

We can break this down by analyzing the change in oxidation states:
1. Reduction half-reaction: The manganese in permanganate (${\mathrm{MnO}}_4^{-}$) starts in the +7 oxidation state and is reduced to the +4 oxidation state in manganese dioxide (${\mathrm{MnO}}_2$).
2. Oxidation half-reaction: The iodine in iodide ($I^{-}$) starts in the -1 oxidation state and is oxidized to the +5 oxidation state in iodate (${\mathrm{IO}}_3^{-}$).

Note: In an acidic medium, iodide is oxidized to iodine ($I_2$), but in a neutral or weakly alkaline medium, the oxidation proceeds further to form the iodate ion (${\mathrm{IO}}_3^{-}$).