Question

Use molecular orbital theory to explain why the \(\mathrm{Be}_{2}\) molecule does not exist.

Short answer

\(\mathrm{Be}_{2}\) does not exist because it has a bond order of 0.

Step-by-step solution

Identify Valence Electrons

Molecular orbital theory starts by examining the valence electrons of each atom. Beryllium (\(\mathrm{Be}\)) has an atomic number of 4, therefore a single \(\mathrm{Be}\) atom has 2 electrons in the valence shell (the 2s orbital). Thus, \(\mathrm{Be}_{2}\) would have a total of 4 valence electrons.

Construct Molecular Orbital Diagram

Next, we construct the molecular orbital (MO) diagram for \(\mathrm{Be}_{2}\). The molecular orbitals form by combining the atomic orbitals: the two 2s atomic orbitals of beryllium combine to form one bonding sigma (\(\sigma_{2s}\)) orbital and one antibonding sigma star (\(\sigma_{2s}^{*}\)) orbital.

Fill Molecular Orbitals with Electrons

Following the Aufbau principle, fill these molecular orbitals starting from the lowest energy level. The 4 electrons will fill as follows: 2 electrons fill the \(\sigma_{2s}\) orbital (bonding), and the remaining 2 electrons will fill the \(\sigma_{2s}^{*}\) orbital (antibonding).

Calculate Bond Order

Bond order is calculated using the formula: \[ \text{Bond Order} = \frac{(\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals})}{2} \] For \(\mathrm{Be}_{2}\), both the \(\sigma_{2s}\) and \(\sigma_{2s}^{*}\) orbitals have 2 electrons each, resulting in a bond order of: \[ \text{Bond Order} = \frac{2 - 2}{2} = 0 \]

Interpret Bond Order

A bond order of 0 indicates that there are no net stabilizing bonds formed, hence there is no force holding the \(\mathrm{Be}_{2}\) molecule together. As a result, \(\mathrm{Be}_{2}\) does not exist in a stable form.

Key concepts

Valence Electrons

In the context of molecular orbital theory, valence electrons are of particular importance as they participate in forming bonds between atoms. For beryllium (\(\mathrm{Be}\)), understanding its valence electrons is the key to analyzing why the \(\mathrm{Be}_{2}\) molecule does not exist. Beryllium is an element with the atomic number 4. This means it has a total of four electrons, with two of them residing in the valence shell, specifically in the 2s orbital.
When we consider the hypothetical \(\mathrm{Be}_{2}\) molecule, it would incorporate two beryllium atoms, thus possessing a total of four valence electrons.
These valence electrons are utilized in forming molecular orbitals when the atoms combine, setting the stage for subsequent analysis of molecular stability.

Molecular Orbital Diagram

A molecular orbital (MO) diagram visualizes how atomic orbitals overlap to form molecular orbitals when atoms interact to form a molecule. In our exploration of the \(\mathrm{Be}_{2}\) molecule, this diagram is central to understanding the molecule's potential stability or instability. Here's how we construct an MO diagram for \(\mathrm{Be}_{2}\):

• The two 2s orbitals from each beryllium atom merge to create two distinct types of molecular orbitals:
  • A bonding sigma (\(\sigma_{2s}\)) orbital.
  • An antibonding sigma star (\(\sigma_{2s}^{*}\)) orbital.
• Bonding orbitals are lower in energy and stabilize the molecule, while antibonding orbitals, denoted by a "*", are higher in energy and destabilize the molecule.

For \(\mathrm{Be}_{2}\), all 4 valence electrons fill these molecular orbitals according to the Aufbau principle, where they occupy the lowest available energy orbitals first. The first two electrons fill the bonding \(\sigma_{2s}\) orbital, and the remaining two electrons fill the antibonding \(\sigma_{2s}^{*}\) orbital. This electron configuration is pivotal in determining the \(\mathrm{Be}_{2}\) molecule's bond order and overall stability.

Bond Order

Bond order is a crucial concept in understanding the stability of a molecule. It represents the net number of chemical bonds between a pair of atoms. In molecular orbital theory, it is calculated using the formula:\[\text{Bond Order} = \frac{(\text{Number of electrons in bonding orbitals} - \text{Number of electrons in antibonding orbitals})}{2}\]For the \(\mathrm{Be}_{2}\) molecule, all electrons in valence orbitals are taken into account:

• There are 2 electrons in the bonding \(\sigma_{2s}\) orbital.
• There are also 2 electrons in the antibonding \(\sigma_{2s}^{*}\) orbital.

Substituting these values into the formula results in:\[\text{Bond Order} = \frac{2 - 2}{2} = 0\]A bond order of 0 indicates that the \(\mathrm{Be}_{2}\) molecule does not have any effective bonding interactions. Simply put, there is no net stabilization provided by the molecular orbitals, which means that there is no force holding \(\mathrm{Be}_{2}\) together in a stable form. Consequently, \(\mathrm{Be}_{2}\) does not exist naturally as a stable molecule due to zero bond order.