`KMnO_(4)` acts as an oxidising agent in alkaline medium when alkaline `KMn_(4)` is treated with KI iodide ion is oxidised to

Options

A. `I_(2)`
B. `IO^(-)`
C. `IO_(3)^(-)`
D. `IO_(4)^(-)`

Explanation

To solve the question, we need to determine what the iodide ion (I⁻) is oxidized to when treated with alkaline potassium permanganate (KMnO₄). Here’s a step-by-step solution:

### Step 1: Understand the Role of KMnO₄
KMnO₄ is a strong oxidizing agent, especially in alkaline medium. In this reaction, it will facilitate the oxidation of iodide ions (I⁻) to a higher oxidation state.

**Hint:** Remember that oxidizing agents gain electrons while the substance being oxidized loses electrons.

### Step 2: Identify the Oxidation Process
In alkaline medium, iodide ions (I⁻) can be oxidized to iodate ions (IO₃⁻). The half-reaction for the oxidation of iodide can be represented as follows:

\[ 2 I^- \rightarrow IO_3^- + 6 e^- + 6 OH^- \]

This shows that iodide ions lose electrons and are converted into iodate ions.

**Hint:** Look for the changes in oxidation states to identify what species is being oxidized.

### Step 3: Write the Overall Reaction
The overall balanced reaction when KMnO₄ reacts with KI in alkaline medium can be written as:

\[ 2 KMnO_4 + 6 KOH + 3 I^- \rightarrow 2 MnO_2 + 3 IO_3^- + 3 H_2O + 6 K^+ \]

In this reaction, the KMnO₄ is reduced to manganese dioxide (MnO₂), while the iodide ions are oxidized to iodate ions (IO₃⁻).

**Hint:** Ensure that the number of atoms and charges are balanced on both sides of the equation.

### Step 4: Conclusion
From the reaction, we conclude that the iodide ion (I⁻) is oxidized to iodate ion (IO₃⁻) when treated with alkaline KMnO₄.

**Final Answer:** The iodide ion (I⁻) is oxidized to iodate ion (IO₃⁻).

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