Neutral or weakly alkaline \(KMnO_4\) solution will oxidize iodide ion into:

iodate ion

The correct answer is option 1. iodate ion.

Let us delve into the detailed process of how potassium permanganate (\(KMnO_4\)) oxidizes iodide ion (\(I^-\)) in neutral or weakly alkaline conditions.

Potassium permanganate (\(KMnO_4\)) is a strong oxidizing agent. Its oxidation state of manganese changes from +7 in \(KMnO_4\) to +2 in \(Mn^{2+}\) during the reaction. The iodide ion (\(I^-\)) undergoes oxidation, where iodine's oxidation state increases.

Steps of the Reaction

Reduction of Potassium Permanganate:

In neutral or weakly alkaline solutions, \(KMnO_4\) is reduced to manganese(II) ion (\(Mn^{2+}\)). The balanced half-reaction for this reduction process is:

\(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)

However, in neutral or weakly alkaline solutions, the reaction is adjusted to include hydroxide ions (\(OH^-\)):

\(MnO_4^- + 4H_2O + 3e^- \rightarrow Mn^{2+} + 8OH^-\)

Oxidation of Iodide Ion:

The iodide ion (\(I^-\)) is oxidized to iodine species with a higher oxidation state. In neutral or weakly alkaline solutions, the iodide ion (\(I^-\)) is typically oxidized to iodate ion (\(IO_3^-\)). The balanced oxidation half-reaction in alkaline conditions is:

\(6I^- + 3MnO_4^- + 6H_2O \rightarrow 6IO_3^- + 3Mn^{2+} + 12OH^-\)

Detailed Explanation of the Products

Iodate Ion (\(IO_3^-\)): This is the main product of oxidation in neutral or weakly alkaline conditions. Iodide (\(I^-\)) is oxidized from an oxidation state of -1 to +5. The formation of iodate ion (\(IO_3^-\)) occurs through the following steps:

Oxidation State Change: The iodine in iodide (\(I^-\)) changes from an oxidation state of -1 to +5 in iodate (\(IO_3^-\)).

Reaction Conditions: The reaction occurs efficiently in alkaline conditions where iodide is converted to iodate ion.

Other Possible Products:

Periodate Ion (\(IO_4^-\)): This is typically formed in strongly alkaline solutions or under different conditions, and is not the primary product in neutral or weakly alkaline conditions.

Iodite Ion (\(IO_2^-\)): This can be formed in specific conditions but is less common in neutral or weakly alkaline solutions.

Hypoiodite Ion (\(IO^-\)): Generally forms in acidic conditions, not in neutral or weakly alkaline solutions.

Summary of the Reaction

Oxidizing Agent: Potassium permanganate (\(KMnO_4\))

Oxidation of Iodide Ion: Converts to iodate ion (\(IO_3^-\))

Reduction of \(KMnO_4\): Reduces to manganese(II) ion (\(Mn^{2+}\))

Thus, the oxidation of iodide ion by potassium permanganate in neutral or weakly alkaline conditions leads to the formation of iodate ion (\(IO_3^-\)).

So, the correct answer is: 1. iodate ion