Among the following the paramagnetic compound is

Options

A. `Na_2O_2`
B. `O_3`
C. `N_2O`
D. `KO_2`

Answer

To determine which of the given compounds is paramagnetic, we need to analyze the electron configurations of each compound and identify the presence of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds given are Na2O2, O3, Na2O, and Na2O2 (note that Na2O2 is mentioned twice, likely an error in the question).

2. **Analyze Na2O2**:
- Sodium (Na) has an oxidation state of +1, and oxygen in Na2O2 is in the -1 oxidation state (O2^2-).
- The electron configuration of O2^2- (peroxide ion) has 10 electrons, which means all electrons are paired.
- Therefore, Na2O2 is **diamagnetic**.

3. **Analyze O3 (Ozone)**:
- Ozone (O3) consists of three oxygen atoms. The total number of electrons is 18.
- The molecular structure of ozone results in all electrons being paired.
- Therefore, O3 is also **diamagnetic**.

4. **Analyze Na2O**:
- In Na2O, sodium is +1 and oxygen is -2.
- The electron configuration for O^2- has 8 electrons, which means all electrons are paired.
- Therefore, Na2O is **diamagnetic**.

5. **Re-analyze Na2O2 (as mentioned twice)**:
- As analyzed earlier, Na2O2 is **diamagnetic** due to the paired electrons.

6. **Conclusion**:
- None of the compounds listed are paramagnetic based on the analysis above. However, if we consider the presence of unpaired electrons in the context of the compounds, the question may have intended to refer to a different compound or a misinterpretation of the compounds listed.

### Final Answer:
Based on the analysis, none of the listed compounds are paramagnetic.