Problem 95 - Chemistry Principles and Practice

Question

Identify two homonuclear diatomic molecules or ions with each of the following molecular orbital electron configurations. Are these species stable? (a) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\left(\pi_{2 p}^{*}\right)^{3}\) (b) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\) (c) \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\)

Step-by-Step Solution

Answer

(a) \(\text{N}_2^{+}\) and \(\text{O}_2^{2+}\) are stable with bond order 1.5. (b) \(\text{N}_2\) is stable with bond order 3. (c) \(\text{He}_2\) is not stable with bond order 0.

Step 1: Understand the Problem

We need to identify two homonuclear diatomic molecules or ions for each given molecular orbital electron configuration, and determine their stability based on the bond order.

Step 2: Analyze Configuration (a)

The configuration \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\left(\pi_{2 p}^{*}\right)^{3}\) corresponds to a 13-electron system. Possible species include \(\text{N}_2^{+}\) and \(\text{O}_2^{2+}\).

Step 3: Calculate Bond Order for Configuration (a)

Bond order is calculated as \((\text{number of bonding electrons} - \text{number of antibonding electrons})/2\). For configuration (a), bond order is \((10 - 7)/2 = 1.5\). The species \(\text{N}_2^{+}\) and \(\text{O}_2^{2+}\) are comparatively stable, due to positive bond orders.

Step 4: Analyze and Calculate for Configuration (b)

The configuration \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\pi_{2 p}\right)^{4}\left(\sigma_{2 p}\right)^{2}\) corresponds to a 12-electron system, like \(\text{N}_2\). Its bond order is calculated as \((10 - 4)/2 = 3\). This means \(\text{N}_2\), with a bond order of 3, is very stable.

Step 5: Analyze and Calculate for Configuration (c)

Configuration \(\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\) corresponds to a 4-electron system, like \(\text{He}_2\). Its bond order is \((2 - 2)/2 = 0\). Thus, \(\text{He}_2\) is not stable, as it has a bond order of 0.

Key Concepts

Homonuclear Diatomic Molecules

A homonuclear diatomic molecule is a molecule composed of two identical atoms. These types of molecules can be found in elements such as hydrogen (\(\text{H}_2\)), oxygen (\(\text{O}_2\)), nitrogen (\(\text{N}_2\)), and the halogens like fluorine (\(\text{F}_2\)).

Homodiatomic molecules are crucial in understanding Molecular Orbital Theory because they simplify the study of bonding between atoms. By comparing these molecules, students can better comprehend how bonding electrons are shared equally between identical atoms, making them an excellent introduction to more complicated molecular structures.

Electron Configuration

Electron configuration in molecular orbitals refers to the arrangement of electrons in a molecule's orbitals. For homonuclear diatomic molecules, this configuration allows us to determine how electrons fill available molecular orbitals from lower to higher energy levels.

For example, in a molecule like \(\text{N}_2\), the electrons fill molecular orbitals based on their energy across both atoms, following the order: \(\sigma_{1s}\), \(\sigma_{1s}^{*}\), \(\sigma_{2s}\), \(\sigma_{2s}^{*}\), \(\pi_{2p}\), and \(\sigma_{2p}\). This sequence leads to a unique electron configuration that greatly influences the molecule's properties and its stability.

Bond Order

Bond order is a concept in molecular orbital theory that indicates the strength and stability of a bond between two atoms. It is calculated as half the difference between the number of bonding electrons and antibonding electrons:\[\text{Bond order} = \frac{(\text{number of bonding electrons} - \text{number of antibonding electrons})}{2}\]A higher bond order usually implies a stronger, more stable bond. For instance, \(\text{N}_2\) has a bond order of 3, which makes it exceptionally stable, as the triple bond holds the nitrogen atoms tightly together. In contrast, a bond order of 0 suggests that the molecule, like \(\text{He}_2\), doesn't have enough bonding interaction to hold it together.

Stability of Molecules

The stability of a molecule is closely linked to its bond order. A positive bond order indicates that a molecule is stable because there are more bonding electrons than antibonding electrons. This means the electron clouds are effectively holding the nuclei together.

For example, \(\text{N}_2\)'s high bond order of 3 reflects its high stability due to strong covalent bonds. On the contrary, molecules with a bond order of 0, like helium molecules (\(\text{He}_2\)), are not stable since the electrons do not provide any net bonding energy to keep the atoms together. When comparing different molecular species, a higher bond order usually suggests better stability, leading to more robust chemical bonds and lower reactivity with other chemical species.