The combustion of sodium in excess air yields a higher oxide. What is the oxidation state of the oxygen in the product ?

Explanation

To determine the oxidation state of oxygen in the product formed by the combustion of sodium in excess air, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The combustion of sodium in excess air can be represented by the reaction:
\[ 4 \text{Na} + \text{O}_2 \rightarrow 2 \text{Na}_2\text{O} \]
However, since we are dealing with excess oxygen, the sodium oxide formed can further react with oxygen to form a higher oxide.

2. **Formation of Higher Oxide**:
The higher oxide of sodium formed in the presence of excess oxygen is sodium peroxide:
\[ 2 \text{Na}_2\text{O} + \text{O}_2 \rightarrow 2 \text{Na}_2\text{O}_2 \]

3. **Identify the Chemical Formula**:
The chemical formula for sodium peroxide is \(\text{Na}_2\text{O}_2\).

4. **Determine the Oxidation State of Sodium**:
In sodium peroxide, sodium (Na) has an oxidation state of +1. This is because sodium typically forms +1 ions.

5. **Calculate the Total Charge**:
Since there are two sodium atoms, the total positive charge contributed by sodium is:
\[ 2 \times (+1) = +2 \]

6. **Determine the Charge on Oxygen**:
The overall charge of the compound must be neutral (0). Let’s denote the oxidation state of oxygen as \(x\). In sodium peroxide, there are two oxygen atoms, so we can express the total charge from oxygen as:
\[ 2x \]

7. **Set Up the Equation**:
The sum of the charges must equal zero:
\[ +2 + 2x = 0 \]

8. **Solve for \(x\)**:
Rearranging the equation gives:
\[ 2x = -2 \\ x = -1 \]

9. **Conclusion**:
The oxidation state of each oxygen atom in sodium peroxide (\(\text{Na}_2\text{O}_2\)) is \(-1\).

### Final Answer:
The oxidation state of the oxygen in the product (sodium peroxide) is \(-1\).