To create a molecular orbital (MO) diagram for the B2+ ion, we first need to understand the molecular orbitals formed from the atomic orbitals of the two boron atoms. Boron has an atomic number of 5, which means it has 5 electrons. In B2, there are a total of 10 electrons (5 from each boron atom). However, since we are dealing with the B2+ ion, we must remove one electron, leaving us with 9 electrons to place in our diagram.
The molecular orbital energy levels for B2 are as follows:
- σ(1s)
- σ*(1s)
- σ(2s)
- σ*(2s)
- σ(2px)
- π(2py) = π(2pz)
- π*(2py) = π*(2pz)
- σ*(2px)
For B2, the electron configuration in the molecular orbitals would be:
- σ(1s)2
- σ*(1s)2
- σ(2s)2
- σ*(2s)2
- σ(2px)2
- π(2py)2
- π(2pz)0
- π*(2py)0
- π*(2pz)0
- σ*(2px)0
This configuration corresponds to 2 electrons in the σ(1s), 2 in σ*(1s), 2 in σ(2s), 2 in σ*(2s), and 2 in σ(2px) and π(2py), competing for the remaining available orbitals. Removing one electron from the configuration for B2 leads us to consider where the missing electron would come from. Based on our arrangement, it’s logical for it to be removed from the π orbitals, so we would have:
- σ(1s)2
- σ*(1s)2
- σ(2s)2
- σ*(2s)2
- σ(2px)2
- π(2py)1
- π(2pz)0
This electronic configuration means that B2+ has one unpaired electron located in the π(2py) molecular orbital, suggesting that B2+ would exhibit paramagnetic properties.
In summary, the molecular orbital diagram for B2+ will reflect these arrangements, with most of the MOs filled, and it will indicate that this ion is likely to be stable due to the bond order being greater than zero as a result of the configuration, which can be calculated as follows:
Bond Order = (Number of bonding electrons – Number of antibonding electrons) / 2
For B2+: Bond Order = (2 + 2 + 2 + 1 – 2 – 2)/2 = 1, indicating a single bond.