Problem 102
Question
102\. Which of the following species are paramagnetic? (a) \(\mathrm{B}_{2} ;\) (b) \(\mathrm{B}_{2}^{-} ;\) (c) \(\mathrm{B}_{2}^{+}\). Which species has the strongest bond?
Step-by-Step Solution
Verified Answer
\(\mathrm{B}_{2}^{-}\) and \(\mathrm{B}_{2}^{+}\) are paramagnetic while \(\mathrm{B}_{2}\) is diamagnetic. \(\mathrm{B}_{2}\) has the strongest bond.
1Step 1: Understand the Molecules
Identify the particles under review, here \(\mathrm{B}_{2}, \mathrm{B}_{2}^{-},\) and \(\mathrm{B}_{2}^{+}\). All three are variations of a diatomic boron molecule, which differ in their electron number due to charge.
2Step 2: Apply the Molecular Orbital Theory
Use the rules of molecular orbital theory to fill up the molecular orbitals. For boron molecules, the energy order of molecular orbitals is \(\sigma_{1s}<\sigma_{1s}^*<\sigma_{2s}<\sigma_{2s}^*<\pi_{2p}\). As each boron atom contributes 3 valence electrons, \(\mathrm{B}_{2}\) has a total of 6, \(\mathrm{B}_{2}^{-}\) has 7, and \(\mathrm{B}_{2}^{+}\) has 5 electrons to distributed in the molecular orbitals.
3Step 3: Identify the Paramagnetic Species
A paramagnetic species has unpaired electrons. Hence, distribute the electrons in the molecular orbitals. For \(\mathrm{B}_{2},\) the six electrons can be paired in the molecular orbitals, so it is diamagnetic. For \(\mathrm{B}_{2}^{-},\) with seven electrons, one electron will remain unpaired, so it is paramagnetic. For \(\mathrm{B}_{2}^{+}\) with five electrons, one electron will also remain unpaired, so it is paramagnetic.
4Step 4: Determine the Strongest Bond
For \(\mathrm{B}_{2}, \mathrm{B}_{2}^{-},\) and \(\mathrm{B}_{2}^{+}\), the bond order can be calculated as \([Number\:of\:bonding\:electrons - Number\:of\:antibonding\:electrons] / 2\). The \(\mathrm{B}_{2}\) molecule has the highest bond order, indicating the strongest bond among the three.
Key Concepts
ParamagnetismBond OrderDiatomic Molecules
Paramagnetism
When we talk about **paramagnetism**, we're referring to a property of certain materials that have unpaired electrons. These unpaired electrons align with an external magnetic field, causing the material to be attracted to the field. In the context of molecular orbital theory, we can identify paramagnetic molecules by examining their molecular orbital diagrams after filling in the electrons. More electrons that are unpaired means the molecule is paramagnetic. In contrast, when all electrons are paired, the molecule exhibits **diamagnetism** and won't be attracted to magnetic fields.
For instance, in the exercise, when examining the species
For instance, in the exercise, when examining the species
- \( \mathrm{B}_{2} \): All electrons are paired, thus it is **diamagnetic**.
- \( \mathrm{B}_{2}^{-} \): With one unpaired electron, it is **paramagnetic**.
- \( \mathrm{B}_{2}^{+} \): Also has an unpaired electron, making it **paramagnetic**.
Bond Order
**Bond order** is a concept that helps us predict the stability and strength of a chemical bond in a molecule. The higher the bond order, the stronger and more stable the bond. It is calculated using the formula: \[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} \]This simple formula gives great insight into molecular stability. In molecular orbital theory, bonding electrons are situated in lower-energy orbitals, while antibonding electrons occupy higher-energy orbitals.
For the given diatomic molecules:
For the given diatomic molecules:
- **\( \mathrm{B}_{2} \)**: Contains a bond order of 1, indicating a relatively strong bond within the group.
- **\( \mathrm{B}_{2}^{-} \)**: Has a slightly decreased bond order due to additional antibonding electrons, weakening the bond.
- **\( \mathrm{B}_{2}^{+} \)**: With a decreased electron count, its bond order is also less than that of \( \mathrm{B}_{2} \).
Diatomic Molecules
**Diatomic molecules** consist of only two atoms, which can be the same or different elements. For example, oxygen (\( \mathrm{O}_2 \)) and hydrogen chloride (\( \mathrm{HCl} \)) are both diatomic molecules. In the context of molecular orbital theory, diatomic molecules are particularly interesting because they allow for a relatively simple yet insightful analysis of bonding and electronic properties.
For the diatomic boron species \( \mathrm{B}_{2}, \mathrm{B}_{2}^{-}, \mathrm{B}_{2}^{+} \):
For the diatomic boron species \( \mathrm{B}_{2}, \mathrm{B}_{2}^{-}, \mathrm{B}_{2}^{+} \):
- Each boron atom contributes three valence electrons, affecting how electrons fill the molecular orbitals.
- Differences emerge based on additional or missing electrons, leading to variations in paramagnetic and diamagnetic properties.
Other exercises in this chapter
Problem 100
Why does the hybridization \(s p^{3} d\) not account for bonding in the molecule BrF \(_{5} ?\) What hybridization scheme does work? Explain.
View solution Problem 101
What is the total number of (a) \(\sigma\) bonds and (b) \(\pi\) bonds in the molecule \(\mathrm{CH}_{3} \mathrm{NCO}\) ?
View solution Problem 103
Use the valence molecular orbital configuration to determine which of the following species is expected to have the lowest ionization energy: (a) \(\mathrm{C}_{
View solution Problem 104
Use the valence molecular orbital configuration to determine which of the following species is expected to have the greatest electron affinity: (a) \(\mathrm{C}
View solution