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 given species, we need to determine the total number of electrons: a. \(\mathrm{N}_{2}^{2-}\): N has 7 electrons, so two N atoms have 14 electrons, plus 2 more electrons due to the negative charge = 16 electrons. \(\mathrm{O}_{2}^{2-}\): O has 8 electrons, so two O atoms have 16 electrons, plus 2 more electrons due to the negative charge = 18 electrons. \(\mathrm{F}_{2}^{2-}\): F has 9 electrons, so two F atoms have 18 electrons, plus 2 more electrons due to the negative charge = 20 electrons. b. \(\mathrm{Be}_{2}\): Be has 4 electrons, so two Be atoms have 8 electrons. \(\mathrm{B}_{2}\): B has 5 electrons, so two B atoms have 10 electrons. \(\mathrm{Ne}_{2}\): Ne has 10 electrons, so two Ne atoms have 20 electrons.

Now, we'll fill the molecular orbitals for each species and determine if they are stable: a. \(\mathrm{N}_{2}^{2-}\): For 16 electrons up to molecular orbital energy level 8, 12 electrons are in bonding orbitals, and 4 electrons are in antibonding orbitals. Since there are more bonding electrons, \(\mathrm{N}_{2}^{2-}\) is stable. \(\mathrm{O}_{2}^{2-}\): For 18 electrons, 12 electrons are in bonding orbitals, and 6 electrons are in antibonding orbitals. Since there are more bonding electrons, \(\mathrm{O}_{2}^{2-}\) is stable. \(\mathrm{F}_{2}^{2-}\): For 20 electrons, 12 electrons are in bonding orbitals, and 8 electrons are in antibonding orbitals. Due to the equal number of bonding and antibonding electrons, \(\mathrm{F}_{2}^{2-}\) is unstable. b. \(\mathrm{Be}_{2}\): For 8 electrons, 4 electrons are in bonding orbitals, and 4 electrons are in antibonding orbitals. Due to the equal number of bonding and antibonding electrons, \(\mathrm{Be}_{2}\) is unstable. \(\mathrm{B}_{2}\): For 10 electrons, 6 electrons are in bonding orbitals, and 4 electrons are in antibonding orbitals. Since there are more bonding electrons, \(\mathrm{B}_{2}\) is stable. \(\mathrm{Ne}_{2}\): For 20 electrons (similar to \(\mathrm{F}_{2}^{2-}\)), 12 electrons are in bonding orbitals, and 8 electrons are in antibonding orbitals. Due to the equal number of bonding and antibonding electrons, \(\mathrm{Ne}_{2}\) is unstable.

The stable diatomic species are: a. \(\mathrm{N}_{2}^{2-}\) and \(\mathrm{O}_{2}^{2-}\) b. \(\mathrm{B}_{2}\)