Oxygen exhibits positive oxidation state in

Options

A. CO
B. `F_2O`
C. `NO`
D. `N_2O`

Answer

To determine in which compound oxygen exhibits a positive oxidation state, we will analyze the oxidation states of oxygen in each of the given compounds: carbon monoxide (CO), difluorooxide (F2O), nitric oxide (NO), and nitrous oxide (N2O).

### Step 1: Analyze Carbon Monoxide (CO)
1. **Structure**: Carbon monoxide consists of one carbon atom and one oxygen atom.
2. **Electronegativity**: Oxygen is more electronegative than carbon.
3. **Oxidation State Calculation**:
- Oxygen will take both electrons from the bond.
- Therefore, the oxidation state of oxygen in CO = -2.

### Step 2: Analyze Difluorooxide (F2O)
1. **Structure**: Difluorooxide consists of two fluorine atoms and one oxygen atom.
2. **Electronegativity**: Fluorine is more electronegative than oxygen.
3. **Oxidation State Calculation**:
- Each fluorine takes one electron from the bond.
- Therefore, the oxidation state of oxygen in F2O = +2 (since it loses two electrons to fluorine).

### Step 3: Analyze Nitric Oxide (NO)
1. **Structure**: Nitric oxide consists of one nitrogen atom and one oxygen atom.
2. **Electronegativity**: Oxygen is more electronegative than nitrogen.
3. **Oxidation State Calculation**:
- Oxygen takes both electrons from the bond.
- Therefore, the oxidation state of oxygen in NO = -2.

### Step 4: Analyze Nitrous Oxide (N2O)
1. **Structure**: Nitrous oxide consists of two nitrogen atoms and one oxygen atom.
2. **Electronegativity**: Oxygen is more electronegative than nitrogen.
3. **Oxidation State Calculation**:
- Oxygen takes both electrons from the bond.
- Therefore, the oxidation state of oxygen in N2O = -2.

### Conclusion
After analyzing all four compounds, we find that:
- In CO, the oxidation state of oxygen is -2.
- In F2O, the oxidation state of oxygen is +2.
- In NO, the oxidation state of oxygen is -2.
- In N2O, the oxidation state of oxygen is -2.

Thus, **oxygen exhibits a positive oxidation state in F2O**.