Oxygen shows oxidation state of -1 in the compound -

Options

A. `NO_(2)`
B. `MnO_(2)`
C. `PbO_(2)`
D. `Na_(2)O_(2)`

Answer

To determine in which compound oxygen shows an oxidation state of -1, we will analyze each of the given options step by step.

### Step-by-Step Solution:

1. **Understanding Oxidation States**:
- The oxidation state of an element in a compound indicates the degree of oxidation or reduction of that element. Oxygen typically has an oxidation state of -2 in most compounds, but there are exceptions.

2. **Analyzing Each Option**:
- We have four options to evaluate: NO2, MnO2, PbO2, and Na2O2.

3. **Option 1: NO2**:
- In nitrogen dioxide (NO2), nitrogen is more electronegative than oxygen. The oxidation state of oxygen here is -2.
- Calculation:
- Let the oxidation state of nitrogen be x.
- The equation is: x + 2(-2) = 0 (since the compound is neutral).
- Solving gives x = +4.
- Therefore, the oxidation state of oxygen in NO2 is -2.

4. **Option 2: MnO2**:
- In manganese dioxide (MnO2), oxygen is again in the -2 oxidation state.
- Calculation:
- Let the oxidation state of manganese be y.
- The equation is: y + 2(-2) = 0.
- Solving gives y = +4.
- Thus, the oxidation state of oxygen in MnO2 is -2.

5. **Option 3: PbO2**:
- In lead dioxide (PbO2), oxygen remains in the -2 oxidation state.
- Calculation:
- Let the oxidation state of lead be z.
- The equation is: z + 2(-2) = 0.
- Solving gives z = +4.
- Therefore, the oxidation state of oxygen in PbO2 is -2.

6. **Option 4: Na2O2**:
- In sodium peroxide (Na2O2), oxygen has an oxidation state of -1.
- Calculation:
- Sodium (Na) has an oxidation state of +1.
- The equation is: 2(+1) + 2x = 0 (where x is the oxidation state of oxygen).
- Solving gives 2 + 2x = 0, hence 2x = -2, leading to x = -1.
- Thus, the oxidation state of oxygen in Na2O2 is -1.

7. **Conclusion**:
- After analyzing all options, we find that the oxidation state of oxygen is -1 in Na2O2.

### Final Answer:
Oxygen shows an oxidation state of -1 in the compound **Na2O2**.