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

Answer

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**:
When sodium (Na) combusts in excess air (O2), it forms sodium peroxide (Na2O2).

2. **Write the Chemical Formula**:
The product of the reaction is sodium peroxide, which has the formula Na2O2.

3. **Assign Oxidation States**:
In sodium peroxide (Na2O2), we need to find the oxidation state of oxygen (O). Let’s denote the oxidation state of oxygen as \( x \).

4. **Set Up the Equation**:
The overall charge of the compound must equal zero. Sodium has an oxidation state of +1 (since it is an alkali metal). Therefore, for Na2O2:
\[ 2(+1) + 2(x) = 0 \]
This simplifies to:
\[ 2 + 2x = 0 \]

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

6. **Conclusion**:
The oxidation state of oxygen in sodium peroxide (Na2O2) is -1. Since the question asks to neglect the negative sign, the final answer is:
\[ \text{Oxidation state of oxygen (neglecting the negative sign)} = 1 \]