Which of the following is paramagnetic ?

Options

A. CO
B. `NO^(+)`
C. `O_(2)^(2-)`
D. `B_(2)`

Answer

To determine which of the given molecules is paramagnetic, we need to analyze the number of electrons in each molecule and check for unpaired electrons. A species is considered paramagnetic if it has unpaired electrons, while it is diamagnetic if all electrons are paired.

### Step-by-Step Solution:

1. **Identify the Molecules**: The question provides several molecules. We will analyze each one to determine its electron configuration and whether it is paramagnetic or diamagnetic.

2. **Calculate the Total Number of Electrons**:
- For each molecule, we will sum the number of valence electrons from the constituent atoms.

3. **Analyze Each Option**:
- **Option A: CO (Carbon Monoxide)**
- Carbon (C) has 6 electrons, and Oxygen (O) has 8 electrons.
- Total electrons = 6 (C) + 8 (O) = 14 electrons.
- Since 14 is an even number and all electrons are paired, CO is **diamagnetic**.

- **Option B: NO⁺ (Nitric Oxide Cation)**
- Nitrogen (N) has 7 electrons, and Oxygen (O) has 8 electrons.
- Total electrons = 7 (N) + 8 (O) = 15 electrons. However, since it is a cation (NO⁺), we subtract 1 electron.
- Total = 15 - 1 = 14 electrons.
- Again, since 14 is even and all electrons are paired, NO⁺ is **diamagnetic**.

- **Option C: O₂²⁻ (Peroxide Ion)**
- Each Oxygen (O) has 8 electrons, and there are 2 O atoms.
- Total electrons = 8 (O) + 8 (O) = 16 electrons. However, since it has a -2 charge, we add 2 electrons.
- Total = 16 + 2 = 18 electrons.
- Since 18 is even and all electrons are paired, O₂²⁻ is **diamagnetic**.

- **Option D: B₂ (Boron Dimer)**
- Each Boron (B) has 5 electrons, and there are 2 B atoms.
- Total electrons = 5 (B) + 5 (B) = 10 electrons.
- Since 10 is an even number, we need to check for unpaired electrons. B₂ has unpaired electrons in its molecular orbital configuration.
- Thus, B₂ is **paramagnetic**.

4. **Conclusion**: After analyzing all options, we find that the only paramagnetic species is **Option D: B₂**.

### Final Answer:
The paramagnetic species is **Option D: B₂**.