When sodium reacts with excess of oxygen, oxidation number of oxygen changes from

Options

A. 0 to -1
B. 0 to -2
C. `-1` to `-2`
D. No change.

Explanation

To determine how the oxidation number of oxygen changes when sodium reacts with excess oxygen, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
When sodium (Na) reacts with excess oxygen (O2), it forms sodium peroxide (Na2O2). The balanced chemical equation is:
\[ 2 \text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2 \]

2. **Determine the Oxidation State of Oxygen in Reactants**:
In the reactants, oxygen is in its elemental form (O2), where the oxidation state of oxygen is 0.

3. **Determine the Oxidation State of Oxygen in Products**:
In sodium peroxide (Na2O2), we need to find the oxidation state of oxygen. The oxidation state of sodium (Na) is +1.

4. **Set Up the Equation for Oxidation State**:
Let the oxidation state of oxygen be \( x \). In Na2O2, we have:
\[ 2(\text{oxidation state of Na}) + 2(\text{oxidation state of O}) = 0 \]
This can be expressed as:
\[ 2(+1) + 2x = 0 \]
Simplifying this gives:
\[ 2 + 2x = 0 \]
\[ 2x = -2 \]
\[ x = -1 \]

5. **Conclusion on Change in Oxidation State**:
The oxidation state of oxygen changes from 0 (in O2) to -1 (in Na2O2). Therefore, the oxidation number of oxygen changes from 0 to -1.

6. **Identify the Type of Reaction**:
Since the oxidation state of oxygen decreases, we can conclude that reduction has taken place.

### Final Answer:
The oxidation number of oxygen changes from 0 to -1.