The combustion of sodium is excess air yeilds a higher oxide. What is the oxidation state of the oxygen in the product? Neglect the negative sign.

Answer

To find the oxidation state of oxygen in the higher oxide produced from the combustion of sodium in excess air, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
Sodium (Na) reacts with oxygen (O2) in excess air to form sodium peroxide (Na2O2), which is the higher oxide mentioned in the question.

2. **Write the Chemical Formula**:
The chemical formula for sodium peroxide is Na2O2.

3. **Assign Oxidation States**:
- Sodium (Na) typically has an oxidation state of +1.
- Let the oxidation state of oxygen in Na2O2 be represented as \( x \).

4. **Set Up the Equation**:
In the compound Na2O2, there are 2 sodium atoms and 2 oxygen atoms. The overall charge of the compound is neutral (0). Therefore, we can set up the equation based on the oxidation states:
\[ 2(+1) + 2(x) = 0 \]
This simplifies to:
\[ 2 + 2x = 0 \]

5. **Solve for \( x \)**:
Rearranging the equation gives:
\[ 2x = -2 \]
Dividing both sides by 2 results in:
\[ x = -1 \]

6. **Neglect the Negative Sign**:
The question asks to neglect the negative sign. Therefore, we report the oxidation state of oxygen as:
\[ 1 \]

### Final Answer:
The oxidation state of oxygen in the product (sodium peroxide, Na2O2) is **1**.