In which of the following compounds, the oxidation state of I-atom is highest -

Options

A. `KI_(3)`
B. `KIO_(4)`
C. `DIO_(3)`
D. `IF_(5)`

Answer

To determine which compound has the highest oxidation state of iodine (I), we will analyze the oxidation states of iodine in each of the given compounds: KI3, KIO4, HIO3, and IF5.

### Step-by-Step Solution:

1. **Analyze KI3:**
- Let the oxidation state of iodine be \( x \).
- Potassium (K) has an oxidation state of +1.
- The compound is neutral, so we can write the equation:
\[ +1 + 3x = 0 \]
- Solving for \( x \):
\[ 3x = -1 \implies x = -\frac{1}{3} \]
- Thus, the oxidation state of iodine in KI3 is \(-\frac{1}{3}\).

2. **Analyze KIO4:**
- Let the oxidation state of iodine be \( x \).
- Potassium (K) is +1 and oxygen (O) is -2. There are 4 oxygen atoms:
\[ +1 + x + 4(-2) = 0 \]
- Simplifying the equation:
\[ +1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7 \]
- Thus, the oxidation state of iodine in KIO4 is \( +7 \).

3. **Analyze HIO3:**
- Let the oxidation state of iodine be \( x \).
- Hydrogen (H) is +1 and there are 3 oxygen atoms:
\[ +1 + x + 3(-2) = 0 \]
- Simplifying the equation:
\[ +1 + x - 6 = 0 \implies x - 5 = 0 \implies x = +5 \]
- Thus, the oxidation state of iodine in HIO3 is \( +5 \).

4. **Analyze IF5:**
- Let the oxidation state of iodine be \( x \).
- Fluorine (F) has an oxidation state of -1 and there are 5 fluorine atoms:
\[ x + 5(-1) = 0 \]
- Simplifying the equation:
\[ x - 5 = 0 \implies x = +5 \]
- Thus, the oxidation state of iodine in IF5 is \( +5 \).

5. **Compare the oxidation states:**
- KI3: \(-\frac{1}{3}\)
- KIO4: \(+7\)
- HIO3: \(+5\)
- IF5: \(+5\)

6. **Conclusion:**
- The highest oxidation state of iodine is found in KIO4, which is \( +7 \).

### Final Answer:
The compound in which the oxidation state of the iodine atom is highest is **KIO4** (option B).