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

To calculate bond order, we first need to find the number of valence electrons for each atom in the diatomic species provided. For Group 1(a): - N has 5 valence electrons, O has 6, and F has 7. - Since all species have 2 extra electrons, the total number of valence electrons for each will be increased accordingly, to N: 5+2=7, O: 6+2=8, and F: 7+2=9. For Group 1(b): - Be has 2 valence electrons, B has 3, and Ne has 8.

Based on the molecular orbital theory, for each valence electron count, follow to populate orbitals in the order of increasing energy levels: 1. \(\sigma_{1s}\) 2. \(\sigma_{1s}^{*}\) 3. \(\sigma_{2s}\) 4. \(\sigma_{2s}^{*}\) 5. \(\sigma_{2p}\) 6. \(\pi_{2p}\) 7. \(\pi_{2p}^{*}\) 8. \(\sigma_{2p}^{*}\) For Group 1(a): - \(\mathrm{N}_{2}^{2-}\): 7 electrons, the order is \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^1\) - \(\mathrm{O}_{2}^{2-}\): 8 electrons, the order is \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\) - \(\mathrm{F}_{2}^{2-}\): 9 electrons, the order is \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^1\) For Group 1(b): - \(\mathrm{Be}_{2}\): 2 electrons, the order is \(\sigma_{1s}^2\) - \(\mathrm{B}_{2}\): 3 electrons, the order is \(\sigma_{1s}^2\sigma_{1s}^{*1}\) - \(\mathrm{Ne}_{2}\): 8 electrons, the order is \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\)

To calculate the bond order, subtract the number of electrons in antibonding orbitals from the number in bonding orbitals, and divide by 2. For Group 1(a): - \(\mathrm{N}_{2}^{2-}\): Bonding electrons: 7, Antibonding electrons: 4, Bond order: \(\frac{7-4}{2}=1.5\) - \(\mathrm{O}_{2}^{2-}\): Bonding electrons: 8, Antibonding electrons: 4, Bond order: \(\frac{8-4}{2}=2\) - \(\mathrm{F}_{2}^{2-}\): Bonding electrons: 9, Antibonding electrons: 5, Bond order: \(\frac{9-5}{2}=2\) For Group 1(b): - \(\mathrm{Be}_{2}\): Bonding electrons: 2, Antibonding electrons: 0, Bond order: \(\frac{2-0}{2}=1\) - \(\mathrm{B}_{2}\): Bonding electrons: 2, Antibonding electrons: 1, Bond order: \(\frac{2-1}{2}=0.5\) - \(\mathrm{Ne}_{2}\): Bonding electrons: 4, Antibonding electrons: 4, Bond order: \(\frac{4-4}{2}=0\)

A diatomic species is considered stable if its bond order is greater than 0. For Group 1(a): - \(\mathrm{N}_{2}^{2-}\): Stable - \(\mathrm{O}_{2}^{2-}\): Stable - \(\mathrm{F}_{2}^{2-}\): Stable For Group 1(b): - \(\mathrm{Be}_{2}\): Stable - \(\mathrm{B}_{2}\): Stable - \(\mathrm{Ne}_{2}\): Not Stable (Bond order: 0) In conclusion, the stable diatomic species predicted by the molecular orbital model are \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{F}_{2}^{2-}, \mathrm{Be}_{2}\), and \(\mathrm{B}_{2}\).