Provide the molecular orbital (MO) diagram for Be2 . b) Is this molecule paramagnetic or diamagnetic? Why? c) What is the bond order of Be2 ?
The Correct Answer and Explanation is:
) Molecular Orbital (MO) Diagram for Be₂:
For beryllium (Be), the atomic number is 4, so each atom has 4 electrons. In a Be₂ molecule, the total number of electrons is 8.
The molecular orbital configuration for molecules with atomic numbers up to 7 (including Be) follows this order:
- σ(1s), σ(1s), σ(2s), σ(2s)**
The 1s orbitals combine to form σ(1s) and σ*(1s), and the 2s orbitals combine to form σ(2s) and σ*(2s).
So, filling the MOs with 8 electrons:
- σ(1s) → 2 electrons
- σ(1s)* → 2 electrons
- σ(2s) → 2 electrons
- σ(2s)* → 2 electrons
MO diagram summary:
scssCopyEditσ*(2s) ↑↓
σ(2s) ↑↓
σ*(1s) ↑↓
σ(1s) ↑↓
b) Is Be₂ Paramagnetic or Diamagnetic?
Be₂ is diamagnetic. This is because all the electrons in the molecular orbitals are paired. There are no unpaired electrons in the configuration, so it does not have a net magnetic moment.
A molecule is paramagnetic if it has at least one unpaired electron. Diamagnetism arises when all electrons are paired, causing the molecule to be repelled by a magnetic field.
c) What Is the Bond Order of Be₂?
Bond order is calculated using the formula:Bond Order=12×(number of bonding electrons−number of antibonding electrons)\text{Bond Order} = \frac{1}{2} \times (\text{number of bonding electrons} – \text{number of antibonding electrons})Bond Order=21×(number of bonding electrons−number of antibonding electrons)
From the MO configuration:
- Bonding electrons: 2 in σ(1s) + 2 in σ(2s) = 4
- Antibonding electrons: 2 in σ*(1s) + 2 in σ*(2s) = 4
Bond Order=12×(4−4)=0\text{Bond Order} = \frac{1}{2} \times (4 – 4) = 0Bond Order=21×(4−4)=0
So, bond order = 0, meaning Be₂ is not a stable molecule under normal conditions.
