Which among the following in a paramagnetic compound?

Options

A. `N_2O`
B. `NO`
C. `N_2O_3`
D. `N_2O_4`

Answer

To determine which among the given compounds is paramagnetic, we need to analyze the number of unpaired electrons in each compound. A compound is considered paramagnetic if it has unpaired electrons, while it is diamagnetic if all electrons are paired.

### Step-by-Step Solution:

1. **Identify the Compounds**:
We need to analyze the compounds mentioned in the question. Let's assume we have the following compounds to analyze: N2, O2, and N2O3.

2. **Determine the Valence Electrons**:
- **For N2**:
Each nitrogen atom has 5 valence electrons. Therefore, for N2:
\[ \text{Total valence electrons} = 5 \times 2 = 10 \]
Since 10 is an even number, we will check the electron configuration to see if there are unpaired electrons.
- The molecular orbital configuration for N2 is:
\[ (σ_{1s})^2 (σ^*_{1s})^2 (σ_{2s})^2 (σ^*_{2s})^2 (σ_{2p_z})^2 (π_{2p_x})^2 (π_{2p_y})^2 \]
All electrons are paired, hence N2 is **diamagnetic**.

- **For O2**:
Each oxygen atom has 6 valence electrons. Therefore, for O2:
\[ \text{Total valence electrons} = 6 \times 2 = 12 \]
The molecular orbital configuration for O2 is:
\[ (σ_{1s})^2 (σ^*_{1s})^2 (σ_{2s})^2 (σ^*_{2s})^2 (σ_{2p_z})^2 (π_{2p_x})^2 (π_{2p_y})^1 \]
Here, there are 2 unpaired electrons in the π orbitals, hence O2 is **paramagnetic**.

- **For N2O3**:
Nitrogen has 5 valence electrons and oxygen has 6. The total valence electrons for N2O3:
\[ \text{Total valence electrons} = (5 \times 2) + (6 \times 3) = 10 + 18 = 28 \]
Since 28 is even, we need to check the electron configuration. However, typically N2O3 has unpaired electrons due to its structure, making it **paramagnetic**.

3. **Conclusion**:
Among the compounds analyzed, O2 and N2O3 exhibit paramagnetic behavior due to the presence of unpaired electrons.

### Final Answer:
The paramagnetic compounds among the options are O2 and N2O3.