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

Permanganate ion $$\left(MnO_4^- \right)$$ is a strong oxidising agent. The species to which it gets reduced and the product obtained from the reducing agent (here $$I^-$$) depend on the pH of the solution.

Useful facts to remember:
  • In acidic medium: $$MnO_4^- \rightarrow Mn^{2+}$$ and $$I^- \rightarrow I_2$$.
  • In neutral (or weakly basic) medium: $$MnO_4^- \rightarrow MnO_2$$ and $$I^- \rightarrow IO_3^-$$.
  • In strongly alkaline medium: $$MnO_4^- \rightarrow MnO_4^{2-}$$ and $$I^- \rightarrow IO_4^-$$.

Because the question specifically mentions a neutral aqueous medium, we focus on the second line above.

Half-reactions in neutral solution:
Reduction half-reaction for permanganate:
$$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$$ $$-(1)$$
Oxidation half-reaction for iodide:
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$ $$-(2)$$

To combine $$-(1)$$ and $$-(2)$$, multiply $$-(1)$$ by 2 so that the electrons cancel (total 6 e$$^-$$):
$$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$ $$-(3)$$

Add $$-(2)$$ and $$-(3)$$:

$$2MnO_4^- + I^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$$
$$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$$

Electrons cancel. Combine, then simplify water and $$H^+/OH^-$$ to obtain the overall balanced equation in neutral medium:

$$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$$

The iodine species formed is $$IO_3^-$$ (iodate ion).

Hence, in a neutral aqueous medium, the reaction of permanganate ion with iodide ion produces iodate ion.

Option B which is: IO$$_3^-$$