Question

(a) Use MO theory to determine the bond order in each of \(\left[\mathrm{He}_{2}\right]^{+}\) and \(\left[\mathrm{He}_{2}\right]^{2+}\) (b) Does the MO picture of the bonding in these ions suggest that they are viable species?

Short answer

(a) \\([\mathrm{He}_2]^{+}\\) has bond order 0.5; \\([\mathrm{He}_2]^{2+}\\) has bond order 0. (b) \\([\mathrm{He}_2]^{+}\\) might be weakly stable; \\([\mathrm{He}_2]^{2+}\\) is not viable.

Step-by-step solution

Write Molecular Electron Configuration

For a diatomic molecule like \([\mathrm{He}_2]^{+}\) and \([\mathrm{He}_2]^{2+}\), determine the number of electrons each species has. - \([\mathrm{He}_2]^{+}\) has 5 electrons in total (2 from each He minus 1 due to the positive charge).- \([\mathrm{He}_2]^{2+}\) has 4 electrons in total (2 from each He minus 2 due to the double positive charge).Label the molecular orbitals as \(\sigma_{1s}\), \(\sigma_{1s}^*\) (bonding and antibonding orbitals for 1s atomic orbitals).

Fill Molecular Orbitals for \\([\mathrm{He}_2]^{+}\\)

For the \[\mathrm{He}_2]^{+}\] ion, we need to fill 5 electrons into the available molecular orbitals.- Place 2 electrons in the bonding orbital \(\sigma_{1s}\).- Place 2 electrons in the antibonding orbital \(\sigma_{1s}^*\).- Place 1 electron in the bonding orbital \(\sigma_{1s}\) again (means 3 electrons in total for the bonding orbital).

Calculate Bond Order for \\([\mathrm{He}_2]^{+}\\)

Bond order is calculated as \text{Bond Order} = \frac{1}{2}\left(N_b - N_a\right)\, where \(N_b\) is the number of electrons in bonding orbitals and \(N_a\) is the number in antibonding orbitals.- For \([\mathrm{He}_2]^{+}\), \(N_b = 3\) and \(N_a = 2\) giving a bond order of: \text{Bond Order} = \frac{1}{2}(3 - 2) = 0.5\.

Fill Molecular Orbitals for \\([\mathrm{He}_2]^{2+}\\)

For the \[\mathrm{He}_2]^{2+}\] ion, we need to fill 4 electrons into the available molecular orbitals.- Place 2 electrons in the bonding orbital \(\sigma_{1s}\).- Place 2 electrons in the antibonding orbital \(\sigma_{1s}^*\).

Calculate Bond Order for \\([\mathrm{He}_2]^{2+}\\)

Using the bond order formula \text{Bond Order} = \frac{1}{2}\left(N_b - N_a\right)\, calculate the bond order:- For \([\mathrm{He}_2]^{2+}\), \(N_b = 2\) and \(N_a = 2\), giving \text{Bond Order} = \frac{1}{2}(2-2) = 0\.

Evaluate Viability of the Species

A positive bond order indicates a stable bond. \([\mathrm{He}_2]^{+}\) with a bond order of 0.5 suggests some degree of stability, although it is weakly bonded. \([\mathrm{He}_2]^{2+}\) with a bond order of 0 indicates no bond, meaning it is not a viable species.

Key concepts

Bond Order

Bond order is a crucial concept in the Molecular Orbital Theory. It provides an insight into the stability of a molecule. To calculate the bond order, you use the formula:
\[ \text{Bond Order} = \frac{1}{2}(N_b - N_a) \]
Where:\[N_b\] is the number of electrons in bonding orbitals and \(N_a\) is the number of electrons in antibonding orbitals.
The bond order tells you about the strength and stability of a bond:

• A higher bond order usually indicates a stronger and more stable bond.
• A bond order of zero means that there are no bonds between the atoms, suggesting instability.
• For example, in the molecules \( ext{He}_2^{+}\) and \( ext{He}_2^{2+}\), \( ext{He}_2^{+}\) has a bond order of \(0.5\) which indicates some weak stability. Whereas \( ext{He}_2^{2+}\) has a bond order of zero, indicating it is not likely to exist in nature.

Understanding bond order helps us predict which molecules are likely to be stable and thus, exist.

Diatomic Molecules

Diatomic molecules consist of just two atoms. These atoms may either be the same, as in oxygen (O\(_2\)) or different, as in carbon monoxide (CO). In the context of Molecular Orbital Theory, it is important to understand how these molecules form and interact.

For diatomic molecules, the combination of atomic orbitals from each atom results in the formation of molecular orbitals. These can be either bonding or antibonding orbitals. Each occupied molecular orbital will receive electrons from both atoms.

Some characteristics of diatomic molecules include:

• Diatomic molecules are highly significant in nature, as they encompass elements like hydrogen (H\(_2\)), nitrogen (N\(_2\)), and many more.
• The simplicity of the diatomic species makes them excellent examples for studying basic chemistry principles.
• They help illustrate fundamental concepts such as bonding, antibonding, and molecular configuration.

Looking at \( ext{He}_2^{+}\)and \( ext{He}_2^{2+}\),we see how molecular orbital theory applies to even these simple molecules, providing insights into their design and stability.

Bonding and Antibonding Orbitals

In the framework of Molecular Orbital Theory, atomic orbitals combine to form molecular orbitals. These are categorized as either bonding orbitals or antibonding orbitals.

Bonding orbitals are formed when atomic orbitals from two atoms combine constructively.They lower the energy of the system making it more stable. Antibonding orbitals, denoted with a \(*\) (e.g., \(\sigma^*\)), form when atomic orbitals combine destructively, raising the system's energy and causing instability.

Here are some key points about bonding and antibonding orbitals:

• Bonding orbitals help hold atoms together in a molecule, by effectively overlapping and increasing electron density between atoms.
• Antibonding orbitals do the opposite by decreasing electron density, thereby making bonds less stable.
• The position of electrons in these orbitals directly impacts the bond order and, consequently, the stability of a molecule.For example, in \( ext{He}_2^{+}\),3 electrons occupy \( ext{bonding orbitals}\)and 2 occupy\( ext{antibonding orbitals}\),resulting in a bond order of \(0.5\),an indicator of slight stability.

Understanding these concepts aids in determining molecular structure and potential chemical reactions.