The oxidation nunber of P in `HP_(2)O_(7)^(-)` ion is

Options

A. `+5`
B. `+6`
C. `+7`
D. `+3`

Explanation

To find the oxidation number of phosphorus (P) in the ion \( HP_2O_7^{-} \), we can follow these steps:

### Step 1: Identify the known oxidation states
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) is -2.

### Step 2: Set up the equation
Let the oxidation number of phosphorus (P) be \( x \). In the ion \( HP_2O_7^{-} \), we have:
- 1 hydrogen atom contributing \( +1 \)
- 2 phosphorus atoms contributing \( 2x \)
- 7 oxygen atoms contributing \( 7 \times (-2) = -14 \)

### Step 3: Write the equation based on the charge of the ion
The overall charge of the ion is -1. Therefore, we can set up the equation:
\[ 1 + 2x - 14 = -1 \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 1 + 2x - 14 = -1 \]
\[ 2x - 13 = -1 \]

### Step 5: Solve for \( x \)
Add 13 to both sides:
\[ 2x = 12 \]
Now, divide by 2:
\[ x = 6 \]

### Conclusion
The oxidation number of phosphorus (P) in the ion \( HP_2O_7^{-} \) is +6.

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