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Chapter 9: Problem 50

Draw a molecular orbital energy level diagram for each of the following species: \(\mathrm{He}_{2}, \mathrm{HHe}, \mathrm{He}_{2}^{+}\). Compare their relative stabilities in terms of bond orders. (Treat HHe as a diatomic molecule with three electrons.)

Short Answer

Expert verified
He鈧� is unstable; HHe and He鈧傗伜 are equally stable with bond order 0.5.

Step by step solution

01

Understand the Basics of Molecular Orbital Theory

Molecular Orbital (MO) theory explains how atomic orbitals combine to form molecular orbitals, which belong to the entire molecule. Relative energy levels of these orbitals help in determining the stability of the molecule through bond order, where bond order is defined as \( \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \).
02

Draw the Molecular Orbital Diagram for He鈧�

For \( \mathrm{He}_{2} \), we place the electrons of each helium atom (2 each, so 4 electrons total) in the molecular orbitals. The lowest energy levels are the bonding \( \sigma_{1s} \) and the antibonding \( \sigma_{1s}^* \) orbitals. With 4 electrons available, 2 electrons fill the \( \sigma_{1s} \) and 2 the \( \sigma_{1s}^* \). Therefore, the bond order is \( \frac{(2 - 2)}{2} = 0 \).
03

Draw the Molecular Orbital Diagram for HHe

HHe has 3 electrons: 1 from H and 2 from He. Electrons fill the lowest energy states available. The \( \sigma_{1s} \) orbital is filled with 2 electrons, and the remaining 1 electron goes into the \( \sigma_{1s}^* \) orbital. The bond order is \( \frac{(2 - 1)}{2} = 0.5 \).
04

Draw the Molecular Orbital Diagram for He鈧傗伜

\( \mathrm{He}_{2}^{+} \) contains 3 electrons due to the positive charge (removing one electron from \( \mathrm{He}_{2} \)). 2 electrons fill the \( \sigma_{1s} \) and 1 electron fills the \( \sigma_{1s}^* \). The bond order here is \( \frac{(2 - 1)}{2} = 0.5 \).
05

Compare the Relative Stabilities of He鈧�, HHe, and He鈧傗伜

A higher bond order generally indicates greater stability. \( \mathrm{He}_{2} \) has a bond order of 0, meaning it is not stable. Both \( \mathrm{HHe} \) and \( \mathrm{He}_{2}^{+} \) have a bond order of 0.5, making them equally more stable than \( \mathrm{He}_{2} \).

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Key Concepts

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

Bond Order
In the realm of molecular orbital theory, bond order is an essential concept used to predict the stability of a molecule. The bond order tells us how many chemical bonds exist between a pair of atoms. It is calculated using the formula:
\[\text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2}.\]Key insights about bond order include:
  • A higher bond order usually correlates with a stronger, more stable bond.
  • A bond order of zero indicates that no stable bond exists between the atoms.
  • In molecules with unpaired electrons in the antibonding orbital, a fractional bond order can occur, reflecting partial bonding.
Understanding the bond order helps in assessing molecular stability and predicting chemical reactivity, which is crucial for students studying chemistry. In our exercise, both HHe and He鈧傗伜 exhibit a bond order of 0.5, while He鈧� has a bond order of 0 showing no stability.
Molecular Stability
Molecular stability is directly related to the bond order you鈥檝e just learned about. The stability of a molecule refers to its ability to remain intact without breaking apart into individual atoms or smaller molecules.
Here鈥檚 how bond order influences molecular stability:
  • Molecules with a bond order greater than zero are more stable, indicating a bonding interaction between atoms.
  • Higher bond orders suggest that a molecule will require more energy to break its bonds, and hence, is more stable.
  • Bond order of zero, as seen in He鈧�, means that the molecule lacks stability and would not typically exist under normal conditions.
In the exercise, we've compared He鈧�, HHe, and He鈧傗伜. The bond order of He鈧� at 0 suggests it is not stable enough to exist. However, HHe and He鈧傗伜 with bond orders of 0.5 imply they have some degree of stable bonding, but still less stable compared to molecules with higher bond orders.
Diatomic Molecules
Diatomic molecules are composed of only two atoms, which may or may not be the same chemical element. They are a great way to explore basic molecular concepts such as molecular orbital theory and bond order.
Here's what to know about diatomic molecules:
  • Diatomic molecules composed of similar atoms, like O鈧� or N鈧�, are often used as simple models to describe molecular bonding.
  • Heteronuclear diatomic molecules, such as HCl, involve atoms of different elements, providing more complex bonding scenarios.
  • For a diatomic molecule, the number of electron pairs shared between the two atoms can be analyzed using their bond order, helping to predict their chemical behavior and reactivity.
In the context of the exercise, HHe is treated as a diatomic molecule, even though it is less conventional due to its unique combination of hydrogen and helium atoms, influencing its bond order and stability in interesting ways. He鈧� and He鈧傗伜 also fit under diatomic fuels, providing clear examples of how bond order affects stability.

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Most popular questions from this chapter

Assume that the third-period element phosphorus forms a diatomic molecule, \(\mathrm{P}_{2}\), in an analogous way as nitrogen does to form \(\mathrm{N}_{2}\). (a) Write the electronic configuration for \(\mathrm{P}_{2}\). Use \(\left[\mathrm{Ne}_{2}\right]\) to represent the electron configuration for the first two periods. (b) Calculate its bond order. (c) What are its magnetic properties (diamagnetic or paramagnetic)?

Sketch the shapes of the following molecular orbitals: \(\sigma_{1 s}, \sigma_{1 s}^{*}, \pi_{2 p}, \pi_{2 p}^{*} .\) How do their energies compare?

Use molecular orbital theory to explain the bonding in the azide ion \(\left(\mathrm{N}_{3}^{-}\right)\). (The arrangement of atoms is NNN.)

Determine whether (a) \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and (b) \(\mathrm{XeF}_{4}\) are polar.

Describe the hybridization of phosphorus in \(\mathrm{PF}_{5}\).
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