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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?

Short Answer

Expert verified
\(\mathrm{B}_{2}^{-}\) and \(\mathrm{B}_{2}^{+}\) are paramagnetic while \(\mathrm{B}_{2}\) is diamagnetic. \(\mathrm{B}_{2}\) has the strongest bond.

Step by step solution

01

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.
02

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.
03

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.
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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
  • \( \mathrm{B}_{2} \): 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**.
Understanding whether a molecule is paramagnetic or not gives us insights into its electronic structure and energy arrangement.
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:
  • **\( \mathrm{B}_{2} \)**: 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 \( \mathrm{B}_{2} \).
By understanding bond order, you can determine not just the strength of bonds but also predict possible reactivity and interaction with other species.
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 \( \mathrm{B}_{2}, \mathrm{B}_{2}^{-}, \mathrm{B}_{2}^{+} \):
  • 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.
Studying diatomic molecules, therefore, helps in mastering concepts such as electron configuration, bond formation, and molecular interactions in diverse chemical environments.

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Most popular questions from this chapter

Furan, \(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O},\) is a substance derivable from oat hulls, corn cobs, and other cellulosic waste. It is a starting material for the synthesis of other chemicals used as pharmaceuticals and herbicides. The furan molecule is planar and the \(\mathrm{C}\) and \(\mathrm{O}\) atoms are bonded into a fivemembered pentagonal ring. The H atoms are attached to the C atoms. The chemical behavior of the molecule suggests that it is a resonance hybrid of several contributing structures. These structures show that the double bond character is associated with the entire ring in the form of a \(\pi\) electron cloud. (a) Draw Lewis structures for the several contributing structures to the resonance hybrid mentioned above. (b) Draw orbital diagrams to show the orbitals that are involved in the \(\sigma\) and \(\pi\) bonding in furan. [Hint: You need use only one of the contributing structures, such as the one with no formal charges.] (c) How many \(\pi\) electrons are there in the furan molecule? Show that this number of \(\pi\) electrons is the same, regardless of the contributing structure you use for this assessment.

Lewis theory is satisfactory to explain bonding in the ionic compound \(\mathrm{K}_{2} \mathrm{O},\) but it does not readily explain formation of the ionic compounds potassium superoxide, \(\mathrm{KO}_{2}\), and potassium peroxide, \(\mathrm{K}_{2} \mathrm{O}_{2}\) (a) Show that molecular orbital theory can provide this explanation. (b) Write Lewis structures consistent with the molecular orbital explanation.

A conjugated hydrocarbon has an alternation of double and single bonds. Draw the molecular orbitals of the \(\pi\) system of 1,3,5 -hexatriene. If the energy required to excite an electron from the HOMO to the LUMO corresponds to a wavelength of \(256 \mathrm{nm},\) do you expect the wavelength for the corresponding excitation in 1,3,5,7 -octatetraene to be a longer or shorter wavelength?

The bond angle in \(\mathrm{H}_{2} \mathrm{Se}\) is best described as (a) between \(109^{\circ}\) and \(120^{\circ} ;\) (b) less than in \(\mathrm{H}_{2} \mathrm{S} ;\) (c) less than in \(\mathrm{H}_{2} \mathrm{S},\) but not less than \(90^{\circ} ;(\mathrm{d})\) less than \(90^{\circ}\)

Use the valence molecular orbital configuration to determine which of the following species is expected to have the greatest electron affinity: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{Be}_{2}\) (c) \(\mathrm{F}_{2} ;\) (d) \(\mathrm{B}_{2}^{+}\)

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