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. \(\mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{Ne}_{2}\)

First, we need to find the atomic electron configuration for each atom involved in the diatomic species. 1. N (7): 1s², 2s², 2p³ 2. O (8): 1s², 2s², 2p⁴ 3. F (9): 1s², 2s², 2p⁵ 4. Be (4): 1s², 2s² 5. B (5): 1s², 2s², 2p¹ 6. Ne (10): 1s², 2s², 2p⁶

Use the atomic electron configuration and knowledge of molecular orbitals to draw the molecular orbital diagrams for each diatomic species. a. For \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{F}_{2}^{2-}\), we need to fill in the electrons and consider the extra two electrons due to the negative charge: 1. \(\mathrm{N}_{2}{ }^{2-}\): N has 7 electrons + 2 from the charge = 9 electrons per atom in the molecule 2. \(\mathrm{O}_{2}^{2-}\): O has 8 electrons + 2 from the charge = 10 electrons per atom in the molecule 3. \(\mathrm{F}_{2}^{2-}\): F has 9 electrons + 2 from the charge = 11 electrons per atom in the molecule b. For \(\mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{Ne}_{2}\), we need to fill in the electrons as they are in their neutral state: 1. \(\mathrm{Be}_{2}\): Be has 4 electrons per atom in the molecule 2. \(\mathrm{B}_{2}\): B has 5 electrons per atom in the molecule 3. \(\mathrm{Ne}_{2}\): Ne has 10 electrons per atom in the molecule

Compare the number of bonding and antibonding electrons for each diatomic species to determine their stability: a. 1. \(\mathrm{N}_{2}{ }^{2-}\): 10 bonding electrons - 8 antibonding electrons = 2 (Stable) 2. \(\mathrm{O}_{2}^{2-}\): 10 bonding electrons - 10 antibonding electrons = 0 (Unstable) 3. \(\mathrm{F}_{2}^{2-}\): 12 bonding electrons - 10 antibonding electrons = 2 (Stable) b. 1. \(\mathrm{Be}_{2}\): 4 bonding electrons - 4 antibonding electrons = 0 (Unstable) 2. \(\mathrm{B}_{2}\): 6 bonding electrons - 4 antibonding electrons = 2 (Stable) 3. \(\mathrm{Ne}_{2}\): 10 bonding electrons - 10 antibonding electrons = 0 (Unstable)

According to the molecular orbital model, the stable diatomic species among the given options are: a. \(\mathrm{N}_{2}{ }^{2-}\) and \(\mathrm{F}_{2}^{2-}\) b. \(\mathrm{B}_{2}\)