To determine which pair of molecules/ions does not exist, we need to calculate the bond order for each pair using the molecular orbital theory. The bond order can be calculated using the formula:
\[
\text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2}
\]
Let's analyze each pair step by step.
### Step 1: Analyze the first pair - H2²⁺ and He2
1. **H2²⁺**:
- Hydrogen has 1 electron. For H2²⁺, 2 electrons are removed from H2, which originally has 2 electrons (1 from each H atom).
- Therefore, H2²⁺ has 0 electrons.
- Bond order = \(\frac{(0 - 0)}{2} = 0\)
2. **He2**:
- Helium has 2 electrons. For He2, there are 2 He atoms, so total electrons = 4.
- Configuration: \(\sigma 1s^2 \sigma 1s^*\)
- Bond order = \(\frac{(2 - 2)}{2} = 0\)
**Conclusion for first pair**: Both H2²⁺ and He2 have a bond order of 0, meaning neither exists.
### Step 2: Analyze the second pair - H2⁻ and He2²⁺
1. **H2⁻**:
- H2 has 2 electrons, and H2⁻ gains 1 electron, resulting in 3 electrons.
- Configuration: \(\sigma 1s^2 \sigma 1s^*\)
- Bond order = \(\frac{(2 - 1)}{2} = 0.5\)
2. **He2²⁺**:
- He2 has 4 electrons, and 2 are removed, resulting in 2 electrons.
- Configuration: \(\sigma 1s^2\)
- Bond order = \(\frac{(2 - 0)}{2} = 1\)
**Conclusion for second pair**: H2⁻ exists, and He2²⁺ exists.
### Step 3: Analyze the third pair - H2⁺ and He2²⁻
1. **H2⁺**:
- H2 has 2 electrons, and H2⁺ loses 1 electron, resulting in 1 electron.
- Configuration: \(\sigma 1s^1\)
- Bond order = \(\frac{(1 - 0)}{2} = 0.5\)
2. **He2²⁻**:
- He2 has 4 electrons, and 2 are added, resulting in 6 electrons.
- Configuration: \(\sigma 1s^2 \sigma 2s^2 \sigma 2s^*\)
- Bond order = \(\frac{(4 - 2)}{2} = 1\)
**Conclusion for third pair**: Both H2⁺ and He2²⁻ exist.
### Step 4: Analyze the fourth pair - H2⁻ and He2²⁻
1. **H2⁻**:
- As calculated before, bond order = 0.5.
2. **He2²⁻**:
- Helium has 2 electrons, and for He2²⁻, 2 electrons are added, resulting in 6 electrons.
- Configuration: \(\sigma 1s^2 \sigma 2s^2 \sigma 2s^*\)
- Bond order = \(\frac{(4 - 2)}{2} = 1\)
**Conclusion for fourth pair**: Both H2⁻ and He2²⁻ exist.
### Final Conclusion
The only pair in which both species do not exist is the first pair: **H2²⁺ and He2**.
### Answer
The answer is the first option: **H2²⁺ and He2**.
