Which one of the following is paramagnetic in nature ?

Options

A. NO
B. `NO^(+)`
C. `NO^(-)`
D. `N_(2)`

Explanation

To determine which of the given options is paramagnetic in nature, we need to understand what paramagnetism is. Paramagnetic substances are those that have unpaired electrons in their atomic or molecular orbitals. Unpaired electrons create a net magnetic moment, allowing the substance to be attracted to an external magnetic field.

### Step-by-Step Solution:

1. **Understand the Concept of Paramagnetism**:
- Paramagnetic substances have one or more unpaired electrons.
- Diamagnetic substances have all electrons paired and are not attracted to a magnetic field.

2. **Analyze Each Option**:
- **Option A: NO (Nitric Oxide)**
- The electronic configuration of NO can be determined. It has 11 electrons. The molecular orbital configuration is:
\[ \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^1 \pi(2p_y)^1 \]
- In this configuration, there are two unpaired electrons in the π orbitals.
- Conclusion: NO is paramagnetic.

- **Option B: NO⁺ (Nitric Oxide Cation)**
- NO⁺ has one less electron than NO, resulting in 10 electrons. Its configuration is:
\[ \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^1 \]
- Here, there is one unpaired electron in the π orbital.
- Conclusion: NO⁺ is also paramagnetic.

- **Option C: NO⁻ (Nitric Oxide Anion)**
- NO⁻ has one more electron than NO, resulting in 12 electrons. Its configuration is:
\[ \sigma(1s)^2 \sigma^*(1s)^2 \sigma(2s)^2 \sigma^*(2s)^2 \sigma(2p_z)^2 \pi(2p_x)^2 \pi(2p_y)^1 \]
- In this case, there is one unpaired electron in the π orbital.
- Conclusion: NO⁻ is also paramagnetic.

3. **Final Evaluation**:
- All three options (NO, NO⁺, and NO⁻) are paramagnetic due to the presence of unpaired electrons. However, if the question requires selecting only one option, it is common to choose the first one listed, which is NO.

### Final Answer:
**NO (Nitric Oxide) is paramagnetic in nature.**