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: Write the formula and assign variables
Let the oxidation number of phosphorus be \( x \). In the ion \( HP_2O_7^{-} \):
- There is 1 hydrogen (H) atom.
- There are 2 phosphorus (P) atoms.
- There are 7 oxygen (O) atoms.

### Step 2: Assign known oxidation numbers
- The oxidation number of hydrogen (H) is +1.
- The oxidation number of oxygen (O) is -2.

### Step 3: Set up the equation
The overall charge of the ion is -1. Therefore, we can set up the equation based on the sum of the oxidation numbers:
\[ \text{Total charge} = (\text{Oxidation number of H}) + 2 \times (\text{Oxidation number of P}) + 7 \times (\text{Oxidation number of O}) \]
Substituting the known values:
\[ -1 = (+1) + 2x + 7(-2) \]

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

### Step 5: Solve for \( x \)
Now, isolate \( x \):
\[ 2x - 13 = -1 \]
\[ 2x = -1 + 13 \]
\[ 2x = 12 \]
\[ x = 6 \]

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

### Final Answer
The oxidation number of P in \( HP_2O_7^{-} \) is +6.

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