The reaction of neutral MnO_4 and I^- gives:

The correct answer is option 3. \(MnO_2\) and \(IO_3^-\).

Let us break down the reaction between neutral \(MnO_4^-\) (permanganate) and \(I^-\) (iodide) and why it leads to the formation of \(MnO_2\) (manganese dioxide) and \(IO_3^-\) (iodate):

1. Properties of Permanganate (\(MnO_4^-\)):

Permanganate (\(MnO_4^-\)) is a strong oxidizing agent because manganese in this ion is in the +7 oxidation state. It readily undergoes reduction reactions in various environments.

2. Neutral Medium:

In a neutral environment, permanganate (\(MnO_4^-\)) tends to get reduced to manganese dioxide (\(MnO_2\)). The reduction involves a decrease in the oxidation state of manganese from +7 to +4.

3. Properties of Iodide (\(I^-\)):

Iodide (\(I^-\)) is capable of being oxidized by strong oxidizing agents like permanganate. It readily undergoes oxidation reactions, where iodine is converted from its -1 oxidation state to higher oxidation states.

4. Expected Products:

When neutral \(MnO_4^-\) reacts with \(I^-\), the reaction is a redox reaction where \(MnO_4^-\) is reduced and \(I^-\) is oxidized. The most likely products are:

\(MnO_2\) (Manganese Dioxide): This is formed from the reduction of \(MnO_4^-\) to \(MnO_2\).

\(IO_3^-\) (Iodate): Iodide (\(I^-\)) is oxidized to iodate (\(IO_3^-\)), where iodine goes from a -1 oxidation state to a +5 oxidation state.

5. Explanation of Excluded Options:

\(Mn^{2+}\) and \(I_2\): While iodide (\(I^-\)) can be oxidized to \(I_2\) by strong oxidizing agents, this reaction typically occurs in acidic conditions, not in a neutral medium. Additionally, \(Mn^{2+}\) is a lower oxidation state for manganese compared to the +4 state achieved in \(MnO_2\).

\(Mn^{2+}\) and \(IO_3^-\): Similar to the previous option, \(Mn^{2+}\) is less likely to be formed than \(MnO_2\) in this context.

6. Conclusion:

The reaction between neutral \(MnO_4^-\) and \(I^-\) in a neutral medium most likely produces \(MnO_2\) (manganese dioxide) and \(IO_3^-\) (iodate).

This explanation clarifies why option 3, \(MnO_2\) and \(IO_3^-\), is the correct answer.