What are the oxidation states of phosphorous in the following:
`Na_3PO_4` ​

Answer

To determine the oxidation state of phosphorus in the compound \( Na_3PO_4 \) (sodium phosphate), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- Sodium (Na) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write down the formula and assign oxidation states
In the compound \( Na_3PO_4 \):
- There are 3 sodium atoms, so the total contribution from sodium is:
\[ 3 \times (+1) = +3 \]
- There are 4 oxygen atoms, so the total contribution from oxygen is:
\[ 4 \times (-2) = -8 \]

### Step 3: Set up the equation for the overall charge
The overall charge of the compound is neutral (0). Therefore, we can set up the equation:
\[ \text{Total oxidation state} = \text{Oxidation state of Na} + \text{Oxidation state of P} + \text{Oxidation state of O} \]
Substituting the known values:
\[ +3 + x - 8 = 0 \]
where \( x \) is the oxidation state of phosphorus (P).

### Step 4: Solve for the oxidation state of phosphorus
Rearranging the equation gives:
\[ x - 5 = 0 \]
Thus,
\[ x = +5 \]

### Conclusion
The oxidation state of phosphorus in \( Na_3PO_4 \) is +5.

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