In the of the following pairs of molecules /ions both the species are not likely to exist?

Options

A. `H_(2)^(-),He_(2)^(2+)`
B. `H_(2)^(+),He_(2)^(2-)`
C. `H_(2)^(-),He_(2)^(2-)`
D. `H_(2)^(2+),He_(2)`

Explanation

To determine which pairs of molecules or ions are not likely to exist, we need to analyze the bond order of each species. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2} \]

If the bond order is zero, the species does not exist.

### Step-by-Step Solution:

1. **Analyze H2-**:
- **Number of Electrons**: 3 (2 from H2 and 1 extra for the negative charge)
- **Electron Configuration**: \(\sigma_{1s}^2 \sigma^*_{1s}^1\)
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{(2 - 1)}{2} = \frac{1}{2} = 0.5 \]
- **Conclusion**: H2- exists.

2. **Analyze H2+**:
- **Number of Electrons**: 1 (1 from H2 and 1 less due to the positive charge)
- **Electron Configuration**: \(\sigma_{1s}^1\)
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{(1 - 0)}{2} = \frac{1}{2} = 0.5 \]
- **Conclusion**: H2+ exists.

3. **Analyze H2 2+**:
- **Number of Electrons**: 0 (2 from H2 and 2 less due to the positive charge)
- **Electron Configuration**: No electrons
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{(0 - 0)}{2} = 0 \]
- **Conclusion**: H2 2+ does not exist.

4. **Analyze He2**:
- **Number of Electrons**: 4 (2 from each He atom)
- **Electron Configuration**: \(\sigma_{1s}^2 \sigma^*_{1s}^2\)
- **Bond Order Calculation**:
\[ \text{Bond Order} = \frac{(2 - 2)}{2} = 0 \]
- **Conclusion**: He2 does not exist.

### Final Conclusion:
The pairs of molecules/ions that are not likely to exist are **H2 2+** and **He2**.

### Answer:
**H2 2+ and He2 do not exist.**