Which of the following is paramagnetic with bond order 0.5?
A. F2
B. H2+
C. N2
D. O2-

To determine which of the given molecules is paramagnetic with a bond order of \(0.5\), we need to analyze each option based on their electronic configuration, bond order calculation, and magnetic properties.

### 1. Bond Order Calculation

The bond order can be calculated using the formula:

\[
\text{Bond Order} = \frac{(n_b - n_a)}{2}
\]

Where:
- \(n_b\) = number of bonding electrons
- \(n_a\) = number of antibonding electrons

### Analyzing Each Option

#### A. \(F_2\) (Fluorine molecule)
- Electronic configuration: \(1s^2 2s^2 2p^6\) for each fluorine, leading to \(F_2\) having:
\[
\text{Total electrons} = 2 \times 9 = 18
\]
- Molecular orbital configuration:
\[
\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2
\]
- All electrons are paired.
- **Bond order**:
\[
\text{Bond Order} = \frac{(10 - 0)}{2} = 5
\]
- **Magnetic property**: Diamagnetic (no unpaired electrons).

#### B. \(H_2^+\) (Hydrogen ion)
- Electronic configuration: Has 1 electron.
- Molecular orbital configuration:
\[
\sigma_{1s}^1
\]
- **Bond order**:
\[
\text{Bond Order} = \frac{(1 - 0)}{2} = 0.5
\]
- **Magnetic property**: Paramagnetic (1 unpaired electron).

#### C. \(N_2\) (Nitrogen molecule)
- Electronic configuration: \(1s^2 2s^2 2p^3\) for each nitrogen, leading to \(N_2\) having:
\[
\text{Total electrons} = 2 \times 7 = 14
\]
- Molecular orbital configuration:
\[
\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2
\]
- **Bond order**:
\[
\text{Bond Order} = \frac{(10 - 4)}{2} = 3
\]
- **Magnetic property**: Diamagnetic (no unpaired electrons).

#### D. \(O_2^-\) (Superoxide ion)
- Electronic configuration: For \(O_2\), the molecular configuration is:
\[
\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1
\]
- The additional electron in \(O_2^-\) gives:
\[
\text{Total electrons} = 2 \times 8 + 1 = 17
\]
- **Bond order**:
\[
\text{Bond Order} = \frac{(10 - 5)}{2} = 2.5
\]
- **Magnetic property**: Paramagnetic (1 unpaired electron).

### Conclusion

Only **B. \(H_2^+\)** has a bond order of \(0.5\) and is paramagnetic.

So, the answer is **B. \(H_2^+\)**.