Determine the oxidation number of the element as indicated
(i) P in `NaH_(2)PO_(4)`

Answer

To determine the oxidation number of phosphorus (P) in sodium dihydrogen phosphate, \( \text{NaH}_2\text{PO}_4 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states of the other elements in the compound:**
- Sodium (Na) has an oxidation number of +1.
- Hydrogen (H) has an oxidation number of +1.
- Oxygen (O) has an oxidation number of -2.

2. **Set up the equation based on the overall charge of the compound:**
- The compound \( \text{NaH}_2\text{PO}_4 \) is neutral, meaning the sum of the oxidation numbers must equal 0.

3. **Write the equation:**
- Let the oxidation number of phosphorus (P) be \( x \).
- The total contribution from sodium is \( +1 \) (from 1 Na).
- The total contribution from hydrogen is \( +2 \) (from 2 H).
- The total contribution from oxygen is \( -8 \) (from 4 O, since \( 4 \times -2 = -8 \)).
- Therefore, the equation can be set up as:
\[ +1 + 2 + x - 8 = 0 \]

4. **Simplify the equation:**
- Combine the constants:
\[ x - 5 = 0 \]

5. **Solve for \( x \):**
- Rearranging gives:
\[ x = +5 \]

6. **Conclusion:**
- The oxidation number of phosphorus (P) in \( \text{NaH}_2\text{PO}_4 \) is \( +5 \).

### Final Answer:
The oxidation number of phosphorus in \( \text{NaH}_2\text{PO}_4 \) is \( +5 \).