Problem 23 - Chemistry and Chemical Reactivity

Question

When sodium and oxygen react, one of the products obtained is sodium peroxide, \(\mathrm{Na}_{2} \mathrm{O}_{2}\). The anion in this compound is the peroxide ion, \(\mathrm{O}_{2}^{2-}\) Write the electron configuration for this ion in molecular orbital terms, and draw the electron dot structure. (a) Compare the ion with the \(\mathrm{O}_{2}\) molecule with respect to the following: magnetic character, net number of \(\sigma\) and \(\pi\) bonds, bond order, and oxygen-oxygen bond length. (b) Compare the valence bond and MO pictures with regard to the number of \(\sigma\) and \(\pi\) bonds and the bond order.

Step-by-Step Solution

Answer

\(\mathrm{O}_2\) is paramagnetic with bond order 2; \(\mathrm{O}_2^{2-}\) is diamagnetic with bond order 1 and longer bond length.

Step 1: Identify the Molecular Orbitals

The molecular orbitals for both the peroxide ion \(\mathrm{O}_2^{2-}\) and molecular oxygen \(\mathrm{O}_2\) include the \(\sigma_{1s}\), \(\sigma^*_{1s}\), \(\sigma_{2s}\), \(\sigma^*_{2s}\), \(\sigma_{2p_z}\), \(\pi_{2p_x}\), \(\pi_{2p_y}\), \(\pi^*_{2p_x}\), and \(\pi^*_{2p_y}\). The allocation of electrons differ for each species.

Step 2: Assign Electrons to Molecular Orbitals

For \(\mathrm{O}_2\) (16 electrons): \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\sigma_{2p_z}^2\pi_{2p_x}^2\pi_{2p_y}^2\pi^*_{2p_x}^1\pi^*_{2p_y}^1\). For \(\mathrm{O}_2^{2-}\) (18 electrons): \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\sigma_{2p_z}^2\pi_{2p_x}^2\pi_{2p_y}^2\pi^*_{2p_x}^2\pi^*_{2p_y}^2\).

Step 3: Determine Magnetic Character

\(\mathrm{O}_2\) has unpaired electrons in its \(\pi^*\) orbitals, making it paramagnetic. \(\mathrm{O}_2^{2-}\) has all paired electrons, making it diamagnetic.

Step 4: Calculate Bond Order

The bond order is calculated as: \((\text{No. of bonding electrons} - \text{No. of antibonding electrons})/2\). For \(\mathrm{O}_2\), the bond order is \((10-6)/2 = 2\). For \(\mathrm{O}_2^{2-}\), the bond order is \((10-8)/2 = 1\).

Step 5: Compare Bond Lengths

Bond order is inversely proportional to bond length; thus, \(\mathrm{O}_2^{2-}\) with a bond order of 1 will have a longer bond length compared to \(\mathrm{O}_2\), which has a bond order of 2.

Step 6: Valence Bond vs MO Picture in Terms of Bonds

In valence bond theory, \(\mathrm{O}_2\) is depicted as having two \(\pi\)-bonds with a bond order of 2, while \(\mathrm{O}_2^{2-}\) has one \(\pi\)-bond with a bond order of 1. MO theory's predictions align with this observation of bond order.

Key Concepts

Electron Configuration

When tackling electron configuration using molecular orbital theory, we consider the sequence of filling molecular orbitals based on their energy levels. For the peroxide ion, \(\mathrm{O}_2^{2-}\), we begin with the

• \(\sigma_{1s}\)
• \(\sigma^*_{1s}\)
• \(\sigma_{2s}\)
• \(\sigma^*_{2s}\)
• \(\sigma_{2p_z}\)
• \(\pi_{2p_x}\)
• \(\pi_{2p_y}\)
• \(\pi^*_{2p_x}\)
• \(\pi^*_{2p_y}\)

with a total of 18 electrons to distribute. The electron configuration of \(\mathrm{O}_2^{2-}\) is: \(\sigma_{1s}^2\sigma^*_{1s}^2\sigma_{2s}^2\sigma^*_{2s}^2\sigma_{2p_z}^2\pi_{2p_x}^2\pi_{2p_y}^2\pi^*_{2p_x}^2\pi^*_{2p_y}^2\). Each molecule has a unique
electron configuration that directly influences its properties, such as magnetic behavior and bond order.

Magnetic Character

Magnetic character arises from the presence of unpaired electrons in a molecule. In molecular orbital terms, a molecule is paramagnetic if it has unpaired electrons.
For \(\mathrm{O}_2\), the molecular oxygen has two unpaired electrons in the \(\pi^*\) orbitals, making it paramagnetic. On the other hand, \(\mathrm{O}_2^{2-}\) has all its electrons paired, so it is diamagnetic. This difference stems directly from their electron configurations, leading to clear contrasts in magnetic behavior. Recognizing whether a molecule has unpaired electrons is essential to understanding its magnetic properties.

Bond Order

Bond order is a central concept in molecular orbital theory, offering insight into a molecule's stability and bond strength. Calculate bond order using the formula:\[\text{Bond Order} = \frac{(\text{number of bonding electrons} - \text{number of antibonding electrons})}{2}\]For molecular oxygen \(\mathrm{O}_2\), the bond order is \((10-6)/2 = 2\), indicating stronger bonding with double bonds. Meanwhile, for peroxide ions \(\mathrm{O}_2^{2-}\), the bond order is \((10-8)/2 = 1\), suggesting weaker single bonds.
This calculation informs us about bond lengths too. Generally, a lower bond order corresponds to a longer bond length, as observed here. Bond order also provides insight into the energetic stability of a molecule.

Valence Bond Theory

While molecular orbital theory offers a detailed electron configuration, valence bond theory provides a different perspective on bonding. This theory illustrates bonds as overlapping atomic orbitals, where valence electrons form shared pairs.
For \(\mathrm{O}_2\), valence bond theory accounts for two \(\pi\)-bonds, maintaining a bond order of 2. This aligns closely with what we expect from molecular orbital theory. Whereas in \(\mathrm{O}_2^{2-}\), it's depicted as having just one \(\pi\)-bond, corresponding to a bond order of 1, consistent with MO predictions. Both theories reach similar conclusions about bond order, but they approach the concept utilizing different angles, offering complementary views.