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Chapter 10: Problem 82

The ionization energy of \(\mathrm{O}_{2}\) is smaller than the ionization energy of atomic \(\mathrm{O} ;\) the opposite is true for the ionization energies of \(\mathrm{N}_{2}\) and atomic \(\mathrm{N}\). Explain this behavior in terms of the molecular orbital energy diagrams of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\).

Short Answer

Expert verified
The \\(\mathrm{O}_2\\) molecule has a lower ionization energy due to its unstable antibonding orbitals, while \\(\mathrm{N}_2\\) has a higher ionization energy due to its stable bonding orbitals.

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or molecule in the gaseous state. When comparing the ionization energy of diatomic molecules to their atomic counterparts, we consider the energy levels and stability of their molecular orbitals.
02

Molecular Orbitals of O2

The molecular orbital (MO) configuration of \(\mathrm{O}_2\) is \((\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p})^2(\pi_{2p})^4(\pi_{2p}^*)^2\). The last filled orbitals are \(\pi_{2p}^*\), which are antibonding orbitals. Since the antibonding electrons are higher in energy and less stable, removing one requires less energy compared to an atomic \(\mathrm{O}\) with its stable half-filled p orbitals.
03

Molecular Orbitals of N2

The molecular orbital configuration for \(\mathrm{N}_2\) is \((\sigma_{2s})^2(\sigma_{2s}^*)^2(\pi_{2p})^4(\sigma_{2p})^2\). The last filled orbital is \(\sigma_{2p}\), which is bonding and thus very stable. Removing an electron from \(\mathrm{N}_2\) requires breaking a stable bonding orbital, which is more demanding energetically than removing an electron from an atomic \(\mathrm{N}\) with comparably filled p orbitals.
04

Comparative Analysis

In \(\mathrm{O}_2\), the ionization energy is lower due to the removal of an unstable antibonding electron. Conversely, in \(\mathrm{N}_2\), ionization involves removing an electron from a very stable bonding orbital, resulting in a higher ionization energy compared to the atomic form.

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Key Concepts

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

Ionization Energy
Ionization energy is a fundamental concept in chemistry.
  • It refers to the amount of energy needed to detach an electron from an atom or molecule in its gaseous state.
  • Ionization energy helps us understand the reactivity and stability of atoms and molecules.
  • A higher ionization energy means more energy is required, indicating greater stability and lower reactivity.
In the context of molecular orbital theory, the ionization energy of diatomic molecules can differ from their atomic counterparts.For example, the ionization energy of molecular oxygen (\( \mathrm{O}_2 \) ) is less than that of a single oxygen atom.This reflects how electrons are arranged in molecular orbitals, influencing their stability and energy levels.
Comparing this to molecular nitrogen (\( \mathrm{N}_2 \) ) shows opposite behavior due to different molecular orbital arrangements.
Antibonding Orbitals
Antibonding orbitals are a key concept in molecular orbital theory.
  • These orbitals are formed when atomic orbitals come together and interact destructively.
  • Electrons in antibonding orbitals destabilize a molecule because they raise the energy state.
  • The presence of electrons in these orbitals weakens or negates the bond between atoms.
When examining \( \mathrm{O}_2 \) , the electrons in the antibonding \( \pi_{2p}^* \) orbitals make it easier to remove an electron.Thus, \( \mathrm{O}_2 \) requires less ionization energy compared to the stable individual atomic form,because the antibonding electrons are already at a higher energy level.
Bonding Orbitals
Bonding orbitals play a pivotal role in the stability of diatomic molecules.
  • These orbitals result from the constructive interaction of atomic orbitals, providing energy advantages and increased stability.
  • Electrons within bonding orbitals help hold the atoms together by lowering the molecule's overall energy.
  • Molecules with electrons in bonding orbitals are typically stronger and more stable.
In \( \mathrm{N}_2 \) , the electrons found in the \( \sigma_{2p} \) bonding orbital contribute to its high ionization energy.This high ionization energy reflects the difficulty of removing an electron from such a stable orbital,indicating that the molecule is significantly more stable than an isolated nitrogen atom.
Diatomic Molecules
Diatomic molecules consist of exactly two atoms bonded together.
  • These molecules, such as \( \mathrm{O}_2 \) and \( \mathrm{N}_2 \) , are prevalent in nature and have unique properties.
  • Their stability and reactivity differ significantly from their constituent atoms due to molecular orbital interactions.
  • Diatomic molecules exhibit distinct differences in properties, such as ionization energy, based on how their atomic orbitals overlap.
For example, while both \( \mathrm{O}_2 \) and \( \mathrm{N}_2 \) are composed of the same atom types,the stable nature of their bonding and antibonding orbitals results in different ionization energies.This makes understanding diatomic molecules crucial for predicting their chemical behavior.
Molecular Orbitals
Molecular orbitals are formed when atomic orbitals at close proximity combine through either constructive or destructive interference.
  • They directly influence the distribution of electrons and the energy stability of a molecule.
  • Constructive interactions produce bonding molecular orbitals, which stabilize the molecule.
  • Destructive interactions create antibonding molecular orbitals, which can destabilize the molecule.
The arrangement of electrons in these molecular orbitals determines many chemical properties,including ionization energies of molecules like \( \mathrm{O}_2 \) and \( \mathrm{N}_2 \) .By analyzing the molecular orbitals, we can better understand why certain diatomic molecules behave differently from their atomic forms.

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Most popular questions from this chapter

Describe the bonding in \(\mathrm{O}_{3}\), using molecular orbital theory. Compare this with the resonance description. 10.17 According to the VSEPR model, the \(\mathrm{H}-\mathrm{P}-\mathrm{H}\) bond angle in \(\mathrm{PH}_{3}\) is a. \(120^{\circ}\) b. \(109.5^{\circ}\) c. \(90^{\circ}\) d. a little less than \(120^{\circ}\) e. a little less than \(109.5^{\circ}\)

An atom in a molecule has two bonds to other atoms and one lone pair. What kind of hybrid orbitals do you expect for this atom? Describe how you arrived at your answer.

Each of the following compounds has a nitrogen-nitrogen bond: \(\mathrm{N}_{2}, \mathrm{~N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{~F}_{2} .\) Match each compound with one of the following bond lengths: \(110 \mathrm{pm}, 122 \mathrm{pm}, 145 \mathrm{pm} .\) Describe the geometry about one of the \(\mathrm{N}\) atoms in each compound. What hybrid orbitals are needed to describe the bonding in valence bond theory?

Explain how the dipole moment could be used to distinguish between the cis and trans isomers of 1,2 -dibromoethene:

Predict the geometry of the following ions, using the electron-pair repulsion model. a. \(\mathrm{ClO}_{3}^{-}\) b. \(\mathrm{PO}_{4}{ }^{3-}\) c. \(\mathrm{SCN}^{-}\) d. \(\mathrm{H}_{3} \mathrm{O}^{+}\)
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