Problem 76 - Chemistry The Central Science In Si Units

Question

Explain the following: \((\mathbf{a})\) The peroxide ion, \(\mathrm{O}_{2}^{2-},\) has a longer bond length than the superoxide ion, \(\mathrm{O}_{2}^{-}\). (b) The magnetic properties of \(\mathrm{B}_{2}\) are consistent with the \(\pi_{2 p}\) MOs being lower in energy than the \(\sigma_{2 p}\) MO. (c) The \(\mathrm{O}_{2}^{2+}\) ion has a stronger \(\mathrm{O}-\mathrm{O}\) bond than \(\mathrm{O}_{2}\) itself.

Step-by-Step Solution

Answer

(a) The peroxide ion \(\mathrm{O}_{2}^{2-}\) has a longer bond due to lower bond order. (b) \(\mathrm{B}_{2}\) is paramagnetic, confirming \(\pi_{2p}\) is lower than \(\sigma_{2p}\). (c) \(\mathrm{O}_{2}^{2+}\) has a stronger bond due to higher bond order.

Step 1: Analyze Bond Order

The bond order for an ion is calculated by the formula: \[(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})/2\]. The bond order is inversely related to bond length; higher bond order implies a shorter bond. For \(\mathrm{O}_{2}^{2-}\), there are more anti-bonding electrons compared to \(\mathrm{O}_{2}^{-}\), leading to a bond order of 1 for \(\mathrm{O}_{2}^{2-}\) versus 1.5 for \(\mathrm{O}_{2}^{-}\). Thus, \(\mathrm{O}_{2}^{2-}\) has a longer bond length.

Step 2: Explore Molecular Orbital (MO) Theory for Magnetic Properties

For \(\mathrm{B}_{2}\), the molecular orbital diagram shows how electrons fill the orbitals. The \(\pi_{2p}\) molecular orbitals (MOs) being lower in energy than \(\sigma_{2p}\) means that electrons occupy the \(\pi_{2p}\) before the \(\sigma_{2p}\). \(\mathrm{B}_{2}\) has two unpaired electrons in the \(\pi_{2p}\) MOs, making it paramagnetic; this confirms the energy ordering where \(\pi_{2p}\) is indeed lower than \(\sigma_{2p}\).

Step 3: Compare Bond Strengths Using Bond Order

For \(\mathrm{O}_{2}^{2+}\), the bond order increases compared to \(\mathrm{O}_{2}\). \(\mathrm{O}_{2}\) has a bond order of 2, whereas \(\mathrm{O}_{2}^{2+}\) has a bond order of 3. This higher bond order in \(\mathrm{O}_{2}^{2+}\) indicates a stronger bond between the oxygen atoms compared to \(\mathrm{O}_{2}\) itself.

Key Concepts

Peroxide ion bond length

In understanding the peroxide ion, \( \mathrm{O}_{2}^{2-} \), bond length, we need to dive into the concept of molecular orbital theory. This theory provides insight into how electrons are distributed in a molecule's structure. The main takeaway is that bond order is inversely related to bond length.

• A lower bond order means a bond is longer.
• The bond order is calculated as \((\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})/2\).

For \( \mathrm{O}_{2}^{2-} \), there are more electrons filling anti-bonding orbitals compared to \( \mathrm{O}_{2}^{-} \). This results in a lower bond order of 1 for \( \mathrm{O}_{2}^{2-} \), making the bond longer. This comparison with \( \mathrm{O}_{2}^{-} \), which has a bond order of 1.5, explains why \( \mathrm{O}_{2}^{-} \) has a shorter bond length.

Superoxide ion bond length

The superoxide ion, \( \mathrm{O}_{2}^{-} \), exhibits a unique bond length due to its electronic configuration. By understanding the superoxide ion’s molecular orbital configuration, we can understand why its bond is relatively shorter.

• The presence of fewer anti-bonding electrons than in \( \mathrm{O}_{2}^{2-} \) leads to a higher bond order.
• It has a bond order of 1.5, reflecting a stronger interaction compared to the peroxide ion.

This stronger bond in \( \mathrm{O}_{2}^{-} \) results in a shorter bond length compared to \( \mathrm{O}_{2}^{2-} \). Thus, the better bonding efficiency due to the reduced number of anti-bonding electron contributions highlights the effect of higher bond order on the ionic bond lengths.

Bond order calculation

The bond order is a simple yet significant concept in molecular orbital theory, serving as an indicator of bond strength and length. To compute bond order:1. Count the total number of electrons in bonding orbitals.2. Count the total number of electrons in anti-bonding orbitals.3. Use the formula: \( \text{Bond order} = \frac{(\text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \).The bond order also provides insights into the stability of a molecule and the strength of the bond.

• Higher bond orders indicate stronger bonds and shorter bond lengths.
• Lower bond orders indicate weaker bonds and longer bond lengths.

By applying this calculation to different ions and molecules, one can predict their relative strength and stability, as demonstrated in the cases of \( \mathrm{O}_{2}^{2-} \), \( \mathrm{O}_{2}^{-} \), and \( \mathrm{O}_{2} \).

Magnetic properties of B2

Boron in its diatomic form, \( \mathrm{B}_{2} \), displays intriguing magnetic properties, attributed to its distinct molecular orbital configuration. The peculiarity arises mainly from the arrangement of electrons in its molecular orbitals.

• The \( \pi_{2p} \) orbitals are filled before the \( \sigma_{2p} \) orbitals.
• \( \mathrm{B}_{2} \) has unpaired electrons in the \( \pi_{2p} \) orbitals.

These unpaired electrons imbue \( \mathrm{B}_{2} \) with paramagnetic properties, meaning that it is attracted to magnetic fields. This phenomenon confirms that the \( \pi_{2p} \) orbitals are indeed lower in energy than the \( \sigma_{2p} \) orbital in the energy arrangement of \( \mathrm{B}_{2} \). This is contrary to the typical energy ordering seen in more common diatomic molecules but provides an exceptional example of the variance in molecular orbital energy levels across different elements.