Question

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion, and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\) (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Short answer

In short, the molecular orbitals for H₂⁻ consist of a bonding MO (σ₁s) and an antibonding MO (σ₁s*), with the electron configuration being σ₁s² σ₁s*¹. The bond order is 1/2, indicating a weak bond. When excited by light, an electron moves from σ₁s to σ₁s*, creating a new unstable configuration (σ₁s¹ σ₁s*²) with a negative bond order of -1/2. Therefore, the excited-state H₂⁻ ion is not stable.

Step-by-step solution

Draw the atomic orbitals of hydrogen atoms

Each hydrogen atom has one electron in the 1s orbital. To determine the MOs of H₂⁻, we will need to sketch the 1s atomic orbitals of two hydrogen atoms.

Combine the atomic orbitals to form MOs

When two 1s atomic orbitals combine, they form two MOs: a bonding MO (σ₁s) and an antibonding MO (σ₁s*). The bonding MO has a lower energy than the original atomic orbitals, while the antibonding MO has a higher energy level. Now we can sketch the MO diagram by putting σ₁s at a lower energy level and σ₁s* at a higher energy level.

Find the electron configuration of H₂⁻

Typically, a hydrogen molecule (H₂) has two electrons, with both occupying the bonding MO (σ₁s). However, the hydrogen ion (H₂⁻) has an additional electron. Therefore, the electron configuration of the ion in terms of its MOs will be: σ₁s² σ₁s*¹.

Calculate the bond order

The bond order is defined as the difference between the number of electrons in bonding MOs and antibonding MOs, divided by 2: \[Bond\,order = \frac{(number\,of\,bonding\,MO\,electrons) - (number\,of\,antibonding\,MO\,electrons)}{2}\] For H₂⁻: Bond order = \(\frac{2 - 1}{2} = \frac{1}{2}\)

Predict the stability of the excited state

When the ion is excited, an electron moves from the lower-energy MO (σ₁s) to the higher-energy MO (σ₁s*). The new electron configuration will be: σ₁s¹ σ₁s*². Calculating the new bond order will give us: Bond order = \(\frac{1 - 2}{2} = -\frac{1}{2}\) Since the excited state has a negative bond order, it is not stable. The negative bond order implies that there are more electrons in the antibonding MO than in the bonding MO, which leads to a net repulsion between the hydrogen atoms and an unstable H₂⁻ ion.

Key concepts

Bond Order

Bond order is a concept used to determine the stability of a molecule. It refers to the number of chemical bonds between a pair of atoms. In simple terms, bond order can tell us how strong a chemical bond is. The higher the bond order, the stronger the bond.

To calculate bond order, use the formula: \[Bond\,order = \frac{(\text{number of bonding MO electrons}) - (\text{number of antibonding MO electrons})}{2}\]
For \(\mathrm{H}_{2}^{-}\), we have 2 electrons in bonding molecular orbitals (MOs) and 1 electron in antibonding MOs. By using the formula, the bond order is \[Bond\,order = \frac{2-1}{2} = 0.5\]
A bond order of 0.5 suggests that there is a bond, but it is weaker than a single bond, indicating partial bond strength. If the bond order is zero or negative, it often suggests the molecule, or ion, may not be stable under normal conditions.

Electron Configuration

Electron configuration describes how electrons are arranged in an atom or molecule. For molecules, we use molecular orbitals (MOs) which are formed when atomic orbitals combine.

In \(\mathrm{H}_{2}^{-}\), two hydrogen atoms each with one electron in the 1s orbital combine to form molecular orbitals. Typically, the hydrogen molecule, \(\mathrm{H}_{2}\), has a configuration of \(\sigma_{1s}^{2}\) with both electrons in a bonding molecular orbital. However, for \(\mathrm{H}_{2}^{-}\), an additional electron makes its electronic configuration \(\sigma_{1s}^{2}\sigma^{*}_{1s}^{1}\).

The notation \(\sigma_{1s}^{2}\sigma^{*}_{1s}^{1}\) shows two electrons in the bonding orbital \(\sigma_{1s}\) and one electron in the antibonding orbital \(\sigma^{*}_{1s}\). This configuration provides a glimpse into the electron behavior and helps predict the stability of the molecule.

Excited State

In the context of molecular orbitals, an excited state occurs when an electron absorbs energy and moves to a higher energy orbital.

For \(\mathrm{H}_{2}^{-}\), when it is excited, an electron from the lower energy bonding orbital moves to the higher energy antibonding orbital, changing the electron configuration to \(\sigma_{1s}^{1}\sigma^{*}_{1s}^{2}\).

Calculating the bond order in this excited state:- Bond order = \(\frac{1 - 2}{2} = -0.5\)A negative bond order implies that the excited state has more electrons in antibonding orbitals than bonding orbitals. This results in repulsion rather than attraction between the atoms, suggesting this excited state is unstable for \(\mathrm{H}_{2}^{-}\).

Energy-Level Diagram

An energy-level diagram visually represents the energy of various orbitals in a molecule.

For \(\mathrm{H}_{2}^{-}\), this diagram includes both bonding and antibonding molecular orbitals:- **Bonding Orbital (\(\sigma_{1s}\))**: Located at a lower energy level than equivalent atomic orbitals. Electrons in this orbital contribute to bond strength.- **Antibonding Orbital (\(\sigma^{*}_{1s}\))**: Found at a higher energy level than atomic orbitals. Electrons here reduce bond strength.
When constructing an energy-level diagram, the bonding orbital (\(\sigma_{1s}\)) typically appears below the antibonding orbital (\(\sigma^{*}_{1s}\)).
This means electrons will first fill the \(\sigma_{1s}\) before entering the \(\sigma^{*}_{1s}\). Such diagrams are crucial for predicting molecular properties, such as stability and reactivity.