Step-by-step Solution:

1. Identify Reactants and Reaction Conditions: The reaction involves permanganate ion ($MnO_4^-$) and iodide ion ($I^-$) in a neutral aqueous medium.

2. Identify Oxidizing and Reducing Agents:

   • Permanganate ion ($MnO_4^-$) is a strong oxidizing agent. The oxidation state of manganese (Mn) is +7.
   • Iodide ion ($I^-$) is a reducing agent. The oxidation state of iodine (I) is -1. A redox reaction will occur between these two ions.

3. Determine the Half-Reactions in a Neutral Medium: The products of a redox reaction involving permanganate depend on the pH of the medium.

   • Reduction of Permanganate ($MnO_4^-$): In a neutral or weakly alkaline aqueous medium, permanganate ion is reduced to manganese dioxide ($MnO_2$), where the oxidation state of Mn is +4. The unbalanced reduction half-reaction is: $MnO_4^- \to MnO_2$ To balance this half-reaction: a. The oxidation state of Mn changes from +7 to +4, a gain of 3 electrons. $MnO_4^- + 3e^- \to MnO_2$ b. Balance the oxygen atoms by adding water ($H_2O$) to the right side. $MnO_4^- + 3e^- \to MnO_2 + 2H_2O$ c. Balance the hydrogen atoms by adding H$^+$ to the left side. $MnO_4^- + 4H^+ + 3e^- \to MnO_2 + 2H_2O$ d. Since the reaction is in a neutral medium, we neutralize the H$^+$ ions by adding an equal number of OH$^-$ ions to both sides. $MnO_4^- + 4H^+ + 4OH^- + 3e^- \to MnO_2 + 2H_2O + 4OH^-$ This simplifies to: $MnO_4^- + 4H_2O + 3e^- \to MnO_2 + 2H_2O + 4OH^-$ e. Canceling water molecules from both sides gives the final balanced reduction half-reaction: $MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$

   • Oxidation of Iodide ($I^-$): In a neutral or alkaline medium, the strong oxidizing agent permanganate oxidizes iodide ion ($I^-$) to iodate ion ($IO_3^-$), where the oxidation state of I is +5. The unbalanced oxidation half-reaction is: $I^- \to IO_3^-$ To balance this half-reaction: a. The oxidation state of I changes from -1 to +5, a loss of 6 electrons. $I^- \to IO_3^- + 6e^-$ b. Balance the oxygen atoms by adding OH$^-$ ions to the left side (since the reduction half-reaction produces OH$^-$, the medium will become basic). $I^- + 6OH^- \to IO_3^- + 6e^-$ c. Balance the hydrogen atoms by adding water ($H_2O$) to the right side. $I^- + 6OH^- \to IO_3^- + 3H_2O + 6e^-$ This is the balanced oxidation half-reaction.

4. Combine the Half-Reactions: To get the overall balanced equation, the number of electrons lost in oxidation must equal the number of electrons gained in reduction.

   • Reduction: $MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-$
   • Oxidation: $I^- + 6OH^- \to IO_3^- + 3H_2O + 6e^-$

   Multiply the reduction half-reaction by 2 to balance the electrons (6 electrons in both). $2(MnO_4^- + 2H_2O + 3e^- \to MnO_2 + 4OH^-) \implies 2MnO_4^- + 4H_2O + 6e^- \to 2MnO_2 + 8OH^-$

   Now add the two balanced half-reactions: $(2MnO_4^- + 4H_2O + 6e^-) + (I^- + 6OH^-) \to (2MnO_2 + 8OH^-) + (IO_3^- + 3H_2O + 6e^-)$

5. Simplify the Overall Equation: Cancel the electrons ($6e^-$) and simplify the water ($H_2O$) and hydroxide ($OH^-$) ions from both sides. $2MnO_4^- + 4H_2O + I^- + 6OH^- \to 2MnO_2 + 8OH^- + IO_3^- + 3H_2O$ $2MnO_4^- + H_2O + I^- \to 2MnO_2 + 2OH^- + IO_3^-$

6. Identify the Products and Compare with Options: The final products of the reaction are manganese dioxide ($MnO_2$), hydroxide ions ($OH^-$), and iodate ion ($IO_3^-$). Looking at the given options: A: $I_2$ B: $IO_3^-$ C: $IO_4^-$ D: $IO_2^-$

   One of the products is $IO_3^-$, which corresponds to option B.

Conclusion

In a neutral aqueous medium, permanganate ion oxidizes iodide ion to iodate ion ($IO_3^-$). Therefore, $IO_3^-$ is one of the products formed. The correct option is B.