Oxidation number if iodine in `IO_(3)^(-), IO_(4)^(-),KI and I_(2)` respectively are

Options

A. `-2, -5, -1, 0`
B. `+5, +7, -1, 0`
C. `+2, +5, +1, 0`
D. `-1, +1, 0, +1`

Explanation

To find the oxidation number of iodine in the compounds \( IO_3^- \), \( IO_4^- \), \( KI \), and \( I_2 \), we will follow a systematic approach for each compound.

### Step-by-Step Solution:

1. **Finding the oxidation number of iodine in \( IO_3^- \)**:
- Let the oxidation number of iodine be \( x \).
- Oxygen has an oxidation number of \(-2\).
- There are 3 oxygen atoms, so their total contribution is \( 3 \times (-2) = -6 \).
- The overall charge of the ion is \(-1\).
- Thus, we can set up the equation:
\[ x + (-6) = -1 \]
- Simplifying this gives:
\[ x - 6 = -1 \implies x = +5 \]

2. **Finding the oxidation number of iodine in \( IO_4^- \)**:
- Again, let the oxidation number of iodine be \( x \).
- There are 4 oxygen atoms, contributing \( 4 \times (-2) = -8 \).
- The overall charge is \(-1\).
- Setting up the equation:
\[ x + (-8) = -1 \]
- Simplifying this gives:
\[ x - 8 = -1 \implies x = +7 \]

3. **Finding the oxidation number of iodine in \( KI \)**:
- In \( KI \), potassium (K) has an oxidation number of \( +1 \).
- Let the oxidation number of iodine be \( x \).
- The compound is neutral, so:
\[ +1 + x = 0 \]
- Solving for \( x \):
\[ x = -1 \]

4. **Finding the oxidation number of iodine in \( I_2 \)**:
- In \( I_2 \), iodine is in its elemental form.
- The oxidation number of any element in its free state is \( 0 \).
- Therefore, the oxidation number of iodine in \( I_2 \) is \( 0 \).

### Summary of Results:
- The oxidation number of iodine in \( IO_3^- \) is \( +5 \).
- The oxidation number of iodine in \( IO_4^- \) is \( +7 \).
- The oxidation number of iodine in \( KI \) is \( -1 \).
- The oxidation number of iodine in \( I_2 \) is \( 0 \).

### Final Answer:
- \( IO_3^- \): +5
- \( IO_4^- \): +7
- \( KI \): -1
- \( I_2 \): 0