The oxidation state of I in `IPO_4` is

Options

A. `+1`
B. `+3`
C. `+5`
D. `+7`

Explanation

To determine the oxidation state of iodine (I) in the compound \( \text{IPO}_4 \), we can follow these steps:

### Step 1: Identify the oxidation states of other elements
In the compound \( \text{IPO}_4 \):
- The oxidation state of phosphorus (P) is typically +5.
- The oxidation state of oxygen (O) is typically -2.

### Step 2: Set up the equation
Let the oxidation state of iodine (I) be \( x \). The compound is neutral, so the sum of the oxidation states must equal zero. Therefore, we can set up the following equation:

\[ x + \text{(oxidation state of P)} + 4 \times \text{(oxidation state of O)} = 0 \]

Substituting the known oxidation states:

\[ x + 5 + 4(-2) = 0 \]

### Step 3: Simplify the equation
Now, simplify the equation:

\[ x + 5 - 8 = 0 \]

This simplifies to:

\[ x - 3 = 0 \]

### Step 4: Solve for \( x \)
Now, solve for \( x \):

\[ x = 3 \]

### Conclusion
The oxidation state of iodine (I) in \( \text{IPO}_4 \) is +3.

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