The product of oxidation of `I^(-)` with `MnO_(4)^(-)` in alkaline medium is:

Options

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

Explanation

To solve the question regarding the product of oxidation of \( I^- \) with \( MnO_4^- \) in alkaline medium, we can follow these steps:

### Step 1: Identify the reactants and the medium
We have the reactants:
- \( I^- \) (iodide ion)
- \( MnO_4^- \) (permanganate ion)
The reaction occurs in an alkaline medium (presence of \( OH^- \)).

**Hint:** Always start by identifying the reactants and the conditions of the reaction.

### Step 2: Determine the oxidation states
In this reaction:
- The oxidation state of iodine in \( I^- \) is -1.
- The oxidation state of manganese in \( MnO_4^- \) is +7.

**Hint:** Knowing the oxidation states helps in identifying which species is oxidized and which is reduced.

### Step 3: Identify the changes in oxidation states
During the reaction:
- \( I^- \) is oxidized to \( IO_3^- \) (iodate ion), where the oxidation state of iodine becomes +5.
- \( MnO_4^- \) is reduced to \( MnO_2 \), where the oxidation state of manganese becomes +4.

**Hint:** Write down the half-reactions to clearly see the changes in oxidation states.

### Step 4: Write the balanced reaction
The balanced reaction in alkaline medium can be written as:
\[ 2 MnO_4^- + 3 I^- + 4 OH^- \rightarrow 2 MnO_2 + 3 IO_3^- + 2 H_2O \]

**Hint:** Balancing reactions often requires adjusting the coefficients in front of the compounds.

### Step 5: Identify the products
From the balanced reaction, we can see that the products formed are:
- \( MnO_2 \) (manganese dioxide)
- \( IO_3^- \) (iodate ion)
- \( H_2O \) (water)

**Hint:** Always check the balanced equation to identify all products formed.

### Conclusion
The product of the oxidation of \( I^- \) with \( MnO_4^- \) in alkaline medium is \( IO_3^- \).

**Final Answer:** \( IO_3^- \)

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