Question

Consider the \(\mathrm{H}_{2}^{+}\)ion. (a) Sketch the molecular orbitals of the ion and draw its energy-level diagram. (b) How many electrons are there in the \(\mathrm{H}_{2}^{+}\)ion? (c) Write the electron configuration of the ion in terms of its MOs. (d) What is the bond order in \(\mathrm{H}_{2}^{+}\)? (e) Suppose that the ion is excited by light so that an electron moves from a lower-energy to a higherenergy MO. Would you expect the excited-state \(\mathrm{H}_{2}^{+}\)ion to be stable or to fall apart? (f) Which of the following statements about part (e) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Short answer

In summary: For the \(\mathrm{H}_{2}^{+}\) ion, the molecular orbitals consist of a bonding (\(\sigma\)) and antibonding (\(\sigma^{*}\)) orbital. The energy-level diagram shows \(\sigma\) to have lower energy than the individual atomic orbitals, and \(\sigma^{*}\) to have higher energy. There is only one electron occupying the \(\sigma\) orbital, yielding an electron configuration of \(\sigma^{1}\). The bond order is \(\frac{1}{2}\), and upon excitation, the electron transitions from the bonding to the antibonding orbital, resulting in an unstable excited-state ion. Hence, the correct statement about the excited state is that the light excites an electron from a bonding orbital to an antibonding orbital (option (i)).

Step-by-step solution

(a) Sketch molecular orbitals and energy-level diagram

For \(\mathrm{H}_{2}^{+}\) ion, it has two hydrogen nuclei and only one electron. The formation of the molecular orbitals involves the combination of 1s atomic orbitals from each hydrogen atom. There are two molecular orbitals formed: the bonding orbital (denoted as \(\sigma\)) and the antibonding orbital (denoted as \(\sigma^{*}\)). For the energy-level diagram, find the energy levels of both the bonding and antibonding orbitals: 1. The bonding orbital (\(\sigma\)) has lower energy than the individual atomic orbitals. 2. The antibonding orbital (\(\sigma^{*}\)) has higher energy than the individual atomic orbitals. Draw the energy-level diagram with one level for the bonding orbital (\(\sigma\)), and another level for the antibonding orbital (\(\sigma^{*}\)), with atomic orbitals lying between them in energy.

(b) Number of electrons in \(\mathrm{H}_{2}^+\) ion

The hydrogen atom has one electron, and since the \(\mathrm{H}_{2}^{+}\) ion has lost one electron, there is only one electron present in the ion.

(c) Electron configuration of the ion

The electron configuration in terms of molecular orbitals can be written as an electron occupying the lowest energy orbital. Here, the electron occupies the bonding orbital (\(\sigma\)). So, the electron configuration is \(\sigma ^{1}\).

(d) Bond order in \(\mathrm{H}_{2}^+\) ion

The bond order can be calculated as: \(Bond\:order = \frac{1}{2}(N_b-N_a)\) Where \(N_b\) is the number of bonding electrons, and \(N_a\) is the number of antibonding electrons. In this case, there is one electron in the bonding orbital (\(\sigma\)) and none in the antibonding orbital (\(\sigma^{*}\)). So, \(N_b = 1\) and \(N_a = 0\). Therefore, the bond order is: \(Bond\:order = \frac{1}{2}(1-0) = \frac{1}{2}\).

(e) Stability of the excited-state \(\mathrm{H}_{2}^{+}\) ion

In the excited state, the electron moves from a lower-energy molecular orbital to a higher-energy molecular orbital. Here, the electron would move from the bonding orbital (\(\sigma\)) to the antibonding orbital (\(\sigma^{*}\)), leaving the bonding orbital empty. As the excited state has no electron in the bonding orbital, the excited-state \(\mathrm{H}_{2}^{+}\) ion is unstable and more likely to fall apart.

(f) Choose the correct statement about the excited state

Based on the explanation for part (e), the correct statement about the excited state is: (i) The light excites an electron from a bonding orbital to an antibonding orbital.

Key concepts

H2+ Ion

Understanding the structure and characteristics of the \(\mathrm{H}_{2}^{+}\) ion is fundamental in molecular chemistry. This dihydrogen ion, also known as the hydronium ion, is a simple cation consisting of two hydrogen nuclei sharing a single electron. Unlike a neutral hydrogen molecule (\(\mathrm{H}_2\)), the \(\mathrm{H}_{2}^{+}\) ion has lost an electron, resulting in a positive charge.

The presence of only one electron in this ion makes it an excellent subject for introductory quantum mechanical studies, as its simplicity allows for a clear illustration of molecular orbital theory without the complexity of electron-electron repulsions found in many-electron systems.

Energy-Level Diagram

An energy-level diagram visually represents the energy states available to electrons in a molecule. For the \(\mathrm{H}_{2}^{+}\) ion, its energy-level diagram includes two molecular orbitals: the lower-energy bonding orbital (\(\sigma\)) and the higher-energy antibonding orbital (\(\sigma^*\)).

The bonding orbital is a result of constructive interference between atomic orbitals, leading to an increased electron density between the nuclei, which serves to hold the two hydrogen nuclei together. Conversely, the antibonding orbital arises from destructive interference, resulting in decreased electron density between the nuclei, which weakens the bond. The electron, by default, resides in the lower-energy state, the bonding orbital, in the ground state of the ion.

Electron Configuration

Electron configuration describes the arrangement of electrons in the molecular orbitals of a molecule. For the \(\mathrm{H}_{2}^{+}\) ion, this is straightforward due to the presence of only one electron.

The electron configuration is denoted as \(\sigma^1\), indicating that the single electron occupies the bonding molecular orbital (\(\sigma\)). This configuration is what gives the \(\mathrm{H}_{2}^{+}\) ion its bonding characteristics and partially stabilizes the system, despite its overall positive charge.

Bond Order

Bond order is a concept used to predict the stability of a bond within a molecule, quantified by the difference in the number of electrons in bonding and antibonding orbitals. It is calculated with the formula: \(Bond\:order = \frac{1}{2}(N_b-N_a)\), where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.

For the \(\mathrm{H}_{2}^{+}\) ion, the bond order is \(\frac{1}{2}\), as there is one electron in a bonding orbital and none in the antibonding orbital. A bond order of \(\frac{1}{2}\) indicates the presence of a bond, albeit weaker than a single bond (which has a bond order of 1), reflecting the unstable nature of the \(\mathrm{H}_{2}^{+}\) ion.

Excited-State Stability

Stability in the excited-state refers to a molecule's ability to remain intact when an electron is promoted to a higher-energy molecular orbital. When the \(\mathrm{H}_{2}^{+}\) ion absorbs energy, typically in the form of light, the single electron transitions from the bonding orbital (\(\sigma\)) to the antibonding orbital (\(\sigma^*\)).

This transition disrupts the delicate balance of forces holding the nuclei together, as there is now no electron in the bonding orbital to maintain the bond. Consequently, the excited-state \(\mathrm{H}_{2}^{+}\) ion is considered to be unstable and likely to dissociate. The correct statement regarding this process is that the light excites an electron from a bonding orbital to an antibonding orbital, leading to a state where there are no longer bonding electrons that can effectively hold the nuclei together.