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 species is paramagnetic, we need to analyze the electronic configurations of each species and check for the presence of unpaired electrons. A species is considered paramagnetic if it has one or more unpaired electrons in its electronic configuration.

### Step-by-Step Solution:

1. **Identify the Species**: The question asks us to identify which of the given species is paramagnetic. The species to analyze are CO, NO+, O2^2-, and Boron (B).

2. **Calculate Total Electrons**:
- **CO (Carbon Monoxide)**:
- Carbon (C) has 6 electrons, Oxygen (O) has 8 electrons.
- Total = 6 + 8 = 14 electrons.

- **NO+ (Nitric Oxide Cation)**:
- Nitrogen (N) has 7 electrons, Oxygen (O) has 8 electrons.
- Since it has a positive charge, we subtract 1 electron.
- Total = 7 + 8 - 1 = 14 electrons.

- **O2^2- (Dioxygen Dioxide)**:
- Each Oxygen (O) has 8 electrons, so for two O atoms, it’s 8 + 8 = 16 electrons.
- With a -2 charge, we add 2 electrons.
- Total = 16 + 2 = 18 electrons.

- **Boron (B)**:
- Boron has 5 electrons.
- Total = 5 electrons.

3. **Determine Electronic Configuration**:
- **For CO (14 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² = π2p_y² (All paired)
- Result: No unpaired electrons → Diamagnetic.

- **For NO+ (14 electrons)**:
- Configuration: Same as CO (since it has the same number of electrons).
- Result: No unpaired electrons → Diamagnetic.

- **For O2^2- (18 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² = π2p_y² π*2p_x² = π*2p_y² (All paired)
- Result: No unpaired electrons → Diamagnetic.

- **For Boron (5 electrons)**:
- Configuration: σ1s² σ*1s² σ2s² π2p_x¹ π2p_y¹ (2 unpaired electrons)
- Result: 2 unpaired electrons → Paramagnetic.

4. **Conclusion**: The only species that is paramagnetic is Boron (B) because it has unpaired electrons.

### Final Answer:
**Boron (B) is paramagnetic.**