Which of the following is not paramagnetic?

Options

A. `O_(2)`
B. `N_(2)^(+)`
C. `B_(2)`
D. `O_(2)^(2-)`

Explanation

To determine which of the given molecules is not paramagnetic, we need to analyze the molecular orbital configurations of each molecule and check for the presence of unpaired electrons. Here’s a step-by-step solution:

### Step 1: Identify the Total Number of Electrons
- **O2**: Each oxygen atom has 8 electrons, so for O2, the total is \(8 \times 2 = 16\) electrons.
- **N2+**: Each nitrogen atom has 7 electrons, so for N2, the total is \(7 \times 2 = 14\) electrons. Since it is N2+, we remove one electron, giving us \(14 - 1 = 13\) electrons.
- **B2**: Each boron atom has 5 electrons, so for B2, the total is \(5 \times 2 = 10\) electrons.
- **O2 2-**: For O2, we have 16 electrons, and the 2- charge means we add 2 more electrons, giving us \(16 + 2 = 18\) electrons.

### Step 2: Determine the Molecular Orbital Configuration
- **O2 (16 electrons)**: Since 16 is greater than 14, we use the following energy level configuration:
- \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2pz^2, \pi 2px^2, \pi 2py^2, \pi^* 2px^1, \pi^* 2py^1 \)
- This configuration has 2 unpaired electrons in the \( \pi^* \) orbitals, so O2 is **paramagnetic**.

- **N2+ (13 electrons)**: Since 13 is less than 14, we use the following configuration:
- \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^2, \pi 2py^2, \sigma 2pz^1 \)
- This configuration has 1 unpaired electron in the \( \sigma 2pz \) orbital, so N2+ is **paramagnetic**.

- **B2 (10 electrons)**: Since 10 is less than 14, we use the following configuration:
- \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2px^1, \pi 2py^1 \)
- This configuration has 2 unpaired electrons in the \( \pi \) orbitals, so B2 is **paramagnetic**.

- **O2 2- (18 electrons)**: Since 18 is greater than 14, we use the following configuration:
- \( \sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2pz^2, \pi 2px^2, \pi 2py^2, \pi^* 2px^2, \pi^* 2py^2 \)
- This configuration has no unpaired electrons, so O2 2- is **diamagnetic**.

### Step 3: Conclusion
From the analysis, we find that:
- O2: Paramagnetic
- N2+: Paramagnetic
- B2: Paramagnetic
- O2 2-: Not paramagnetic (diamagnetic)

Thus, the molecule that is **not paramagnetic** is **O2 2-**.

### Final Answer
**O2 2- is not paramagnetic.**

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