Draw an MO energy diagram and predict the bond order of Be2+ and Be2-. Do you expect these molecules to exist in the gas phase?
The Correct Answer and Explanation is:
Answer:
Molecular Orbital (MO) Diagrams:
For beryllium (Be), atomic number = 4.
Each neutral Be atom has 4 electrons. So:
- Be₂ has 8 total electrons (4 from each atom).
- Be₂⁺ has 7 total electrons.
- Be₂⁻ has 9 total electrons.
The MO diagram for Be₂, Be₂⁺, and Be₂⁻ (for elements up to atomic number 7) follows this order:
scssCopyEditσ(1s) < σ*(1s) < σ(2s) < σ*(2s)
Now we fill in the electrons:
Be₂⁺ (7 electrons)
MO configuration:
σ(1s)² σ*(1s)² σ(2s)² σ*(2s)¹
Bond order = (Number of bonding electrons − Number of antibonding electrons) ÷ 2
= (4 bonding − 3 antibonding) ÷ 2 = 0.5
Be₂⁻ (9 electrons)
MO configuration:
σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² π(2p)¹ → but the π(2p) orbital is not filled until boron and higher, so stick to σ(2s)*
Thus, 9th electron goes into π(2p) for elements with p-orbitals involved (Be does not yet include p-orbitals), so the configuration remains:
σ(1s)² σ*(1s)² σ(2s)² σ*(2s)³
Bond order = (4 bonding − 5 antibonding) ÷ 2 = −0.5
Existence in the Gas Phase:
- Be₂⁺ has a bond order of 0.5. It has a weak bond and could exist briefly in the gas phase under experimental or high-energy conditions, but it is not stable.
- Be₂⁻ has a negative bond order, indicating that it is unstable and unlikely to exist under normal conditions.
Conclusion:
- Be₂⁺: Bond order = 0.5 → marginally stable, may exist briefly in gas phase.
- Be₂⁻: Bond order = −0.5 → not expected to exist, unstable.
Therefore, Be₂⁺ might be detected in specific high-energy environments, but Be₂⁻ is not likely to exist in the gas phase.
