Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}{ }^{2-}, \mathrm{F}_{2}{ }^{2-}\) b. \(\mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{Ne}_{2}\)

For each element in the given list, determine its electron configuration based on its atomic number and the aufbau principle. Recall that for diatomic molecules, the electrons are shared between the two atoms, and the given oxidation states need to be considered: a. \(\mathrm{N}_{2}{ }^{2-}\): N has 7 electrons; N₂ will have 14 electrons; N₂²⁻ will have 14 + 2 = 16 electrons \(\mathrm{O}_{2}{ }^{2-}\): O has 8 electrons; O₂ will have 16 electrons; O₂²⁻ will have 16 + 2 = 18 electrons \(\mathrm{F}_{2}{ }^{2-}\): F has 9 electrons; F₂ will have 18 electrons; F₂²⁻ will have 18 + 2 = 20 electrons b. \(\mathrm{Be}_{2}\): Be has 4 electrons, so Be₂ will have 4 + 4 = 8 electrons \(\mathrm{B}_{2}\): B has 5 electrons, so B₂ will have 5 + 5 = 10 electrons \(\mathrm{Ne}_{2}\): Ne has 10 electrons, so Ne₂ will have 10 + 10 = 20 electrons

Analyze the molecular orbital diagrams for each element, considering the number of bonding and antibonding electrons. a. For N₂²⁻, we have 16 electrons. The molecular orbital diagram for nitrogen shows that it fills the 1σ bonding, 2σ bonding, 1π bonding, and 3σ antibonding orbitals completely, leaving two extra electrons to occupy the 1π antibonding orbitals. This overall combination yields a bond order of 3 - 1 = 2. For O₂²⁻, we have 18 electrons. The molecular orbital diagram for oxygen shows that it fills the 1σ bonding, 2σ bonding, 1π bonding, and 4σ antibonding orbitals completely, leaving two extra electrons to occupy the 1π antibonding orbitals. This overall combination results in a bond order of 2 - 2 = 0. For F₂²⁻, we have 20 electrons. The molecular orbital diagram for fluorine shows that it fills the 1σ bonding, 2σ bonding, 1π bonding, 4σ antibonding, and 1π antibonding orbitals completely. This overall combination results in a bond order of 1 - 3 = -2. b. For Be₂, we have 8 electrons. The molecular orbital diagram for beryllium indicates that it fills the 1σ bonding, 2σ bonding, and 1σ antibonding orbitals completely. This overall combination results in a bond order of 1 - 1 = 0. For B₂, we have 10 electrons. The molecular orbital diagram for boron shows that it fills the 1σ bonding, 2σ bonding, 1π bonding, and 1σ antibonding orbitals completely. This overall combination results in a bond order of 2 - 1 = 1. For Ne₂, we have 20 electrons. The molecular orbital diagram for neon does not offer any stable diatomic configuration as Ne atoms have filled orbitals and any added bonding electrons would be canceled out by antibonding ones.

Based on the calculated bond orders in step 2, determine the stability of the diatomic species. If the bond order > 0, the diatomic species is stable; if the bond order ≤ 0, the diatomic species is not stable: a. N₂²⁻: Bond order = 2, stable O₂²⁻: Bond order = 0, not stable F₂²⁻: Bond order = -2, not stable b. Be₂: Bond order = 0, not stable B₂: Bond order = 1, stable Ne₂: No stable diatomic configuration Hence, the stable diatomic species are N₂²⁻ and B₂.