To determine which of the given species does not exist under normal conditions, we will analyze the bond order of each species using molecular orbital theory. The bond order is 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. **Identify the Species**: We need to analyze the bond order for the given species. Let's assume the species are BE2, BE2+, and others.
2. **Calculate for BE2**:
- Beryllium (Be) has an atomic number of 4, which means it has 4 electrons.
- For BE2, there are 2 beryllium atoms, giving a total of 8 electrons.
- The electron configuration is:
- 1s² (2 electrons)
- 2s² (2 electrons)
- 2p (4 electrons) - we will fill these according to the molecular orbital theory.
- Filling the molecular orbitals:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)²
- Bonding Electrons = 4 (2 from 1s and 2 from 2s)
- Antibonding Electrons = 2 (from 1s*)
- Bond Order = (4 - 2) / 2 = 1
- Since the bond order is 1, BE2 exists.
3. **Calculate for BE2+**:
- BE2+ means we remove one electron from BE2.
- The total electrons now = 7 (8 - 1).
- The electron configuration will be:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)²
- Bonding Electrons = 4 (2 from 1s and 2 from 2s)
- Antibonding Electrons = 3 (2 from 1s* and 1 from 2s*)
- Bond Order = (4 - 3) / 2 = 0.5
- Since the bond order is 0.5, BE2+ exists.
4. **Calculate for Barium (Ba)**:
- Barium has an atomic number of 56, and for Ba2+, we consider 2 barium atoms.
- Total electrons = 4 (2 from each Ba).
- The electron configuration will be:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p)²
- Bonding Electrons = 2 (from 1s)
- Antibonding Electrons = 2 (from 1s*)
- Bond Order = (2 - 2) / 2 = 0
- Since the bond order is 0, Ba2+ does not exist.
### Conclusion:
The species that does not exist under normal conditions is **Ba2+**.
