Create a Molecular Orbital Diagram for the B2 Plus Ion and O2 2 Plus Ion

To create a molecular orbital (MO) diagram for the B₂⁺ ion and the O₂²⁺ ion, we first need to understand the electron configurations and the arrangement of molecular orbitals for these species.

B₂⁺ Ion:

The B₂ molecule has a total of 10 electrons (5 from each boron atom). In the B₂⁺ ion, we remove one electron, resulting in 9 electrons. The molecular orbital configuration for B₂ is:

• σ_1s²
• σ_1s^*2
• σ_2s²
• σ_2s^*2
• σ_2p²
• π_2p²
• π_2p^*

The important point here is that for B₂⁺, the electron configuration becomes:

• σ_1s²
• σ_1s^*2
• σ_2s²
• σ_2s^*2
• σ_2p²
• π_2p²

This means there are 3 electrons in the molecular orbitals above the level of the σ_2s, making B₂⁺ a paramagnetic species.

O₂²⁺ Ion:

The O₂ molecule has a total of 16 electrons (8 from each oxygen atom). In the O₂²⁺ ion, we remove two electrons, resulting in 14 electrons. The molecular orbital configuration for O₂ is:

• σ_1s²
• σ_1s^*2
• σ_2s²
• σ_2s^*2
• σ_2p²
• π_2p⁴
• π_2p^*2

So, for O₂²⁺, we can describe the molecular orbital configuration as:

• σ_1s²
• σ_1s^*2
• σ_2s²
• σ_2s^*2
• σ_2p²
• π_2p⁴

With 14 electrons, all molecular orbitals are filled, resulting in O₂²⁺ being a diamagnetic species.

Conclusion:

In summary, the B₂⁺ ion has unpaired electrons and exhibits paramagnetism, while the O₂²⁺ ion has all electrons paired, making it diamagnetic. Reflexively, the molecular orbital diagrams illustrate these properties, emphasizing the arrangement and occupation of the electrons in their respective molecular orbitals.