Problem 64 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

The stable diatomic species according to the molecular orbital model are: \(\mathrm{N}_{2}^{2-}\), \(\mathrm{O}_{2}^{2-}\), and \(\mathrm{B}_{2}\).

Step 1: Determine the total number of electrons in each species

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.

Step 2: Fill molecular orbitals and compare bonding vs antibonding 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.

Step 3: Identify stable diatomic species

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

Key Concepts

Bonding Electrons

Bonding electrons play a crucial role in the stability of molecules and can be understood better through the concept of energy levels within molecular orbitals. When atoms form molecules, their atomic orbitals combine to create molecular orbitals, which can either be bonding or antibonding.

Bonding orbitals are molecular orbitals that result from the constructive interference of overlapping atomic orbitals. This means that the atomic wave functions combine in such a way that they increase electron density between two atomic nuclei. This increased electron density leads to an attractive force between the nuclei, effectively holding them together.

• Bonding electrons lower the energy of a system making it more stable.
• Molecules tend to be stable if there are more electrons in bonding orbitals than in antibonding ones.

Thus, in the exercise, molecules like \( \text{N}_2^{2-} \) and \( \text{O}_2^{2-} \), have more bonding electrons compared to antibonding ones, making them stable. This illustrates that the presence of sufficient bonding electrons is critical for molecular stability.

Antibonding Electrons

Unlike bonding electrons, antibonding electrons occupy molecular orbitals that decrease the stability of a molecule. These orbitals are formed through destructive interference of the atomic orbitals' wave functions. Instead of increasing the electron density between the nuclei, antibonding orbitals have nodes between atoms, resulting in regions of zero electron density.

This causes:

• Repulsion between nuclei, leading to destabilization.
• Higher energy levels as compared with bonding orbitals.

A greater number of electrons in antibonding orbitals compared to bonding orbitals typically results in an unstable molecule, because the repulsive forces overcome the attraction.

For instance, in the exercise, molecules such as \( \text{F}_2^{2-} \) and \( \text{Ne}_2 \) are unstable because the numbers of electrons in antibonding orbitals either match or exceed those in bonding orbitals, reflecting the adverse effect antibonding orbitals have on molecule stability.

Diatomic Molecules

Diatomic molecules are simple molecules consisting of only two atoms. Understanding them is fundamental in molecular orbital theory because they offer a straightforward model to study bonding and antibonding interactions.

The formation of diatomic molecules involves:

• Combination of atomic orbitals from each atom forming molecular orbitals.
• Filling of these orbitals by the available electrons which will determine their stability.

In such molecules, the arrangement of electrons in these orbitals helps predict their stability.

Out of the examples given, \( \text{N}_2^{2-} \), \( \text{O}_2^{2-} \), and \( \text{B}_2 \) are stable, characterized by having an effective net number of electrons in bonding orbitals surpassing those in antibonding orbitals. Learning about such interactions helps students of molecular theory explain why certain diatomic molecules exist in nature and others do not.