During the disproportionation of `I_(2)` to iodide and iodate ions, the ratio of iodate and iodide ions formed in alkaline medium is

Options

A. `1:5`
B. `5:1`
C. `3:1`
D. `1:3`

Answer

To solve the problem of determining the ratio of iodate ions (IO3^-) to iodide ions (I^-) formed during the disproportionation of I2 in an alkaline medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The disproportionation of iodine (I2) in an alkaline medium can be represented by the following balanced chemical equation:
\[ 3I_2 + OH^- \rightarrow 5I^- + IO_3^- \]

2. **Analyze the Products**:
From the balanced equation, we can see that for every 3 moles of I2 that react, 5 moles of iodide ions (I^-) and 1 mole of iodate ions (IO3^-) are produced.

3. **Determine the Ratio**:
To find the ratio of iodate ions to iodide ions, we can use the coefficients from the balanced equation:
- Moles of IO3^- produced = 1
- Moles of I^- produced = 5

Therefore, the ratio of iodate ions to iodide ions is:
\[ \text{Ratio of IO3^- to I^-} = \frac{1}{5} \]

4. **Express the Ratio**:
This can be expressed as:
\[ \text{Ratio} = 1 : 5 \]

5. **Select the Correct Option**:
Based on the options given in the question, the correct answer is:
\[ 1 : 5 \]

### Final Answer:
The ratio of iodate ions to iodide ions formed during the disproportionation of I2 in an alkaline medium is **1 : 5**.

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