/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Using molecular orbital theory, ... [FREE SOLUTION] | 91影视

91影视

Using molecular orbital theory, explain why the removal of one electron in \(\mathrm{O}_{2}\) strengthens bonding, while the removal of one electron in \(\mathrm{N}_{2}\) weakens bonding.

Short Answer

Expert verified
When one electron is removed from O鈧, it comes from a 蟺*(2p) antibonding orbital, which results in strengthened overall bonding. In contrast, when one electron is removed from N鈧, it comes from a 蟺(2p) bonding orbital, leading to weakened overall bonding. This difference is explained by the molecular orbital theory and the energy diagrams of the molecular orbitals formed from the combination of atomic orbitals in both O鈧 and N鈧 molecules.

Step by step solution

01

Determine the Atomic Orbitals

For this exercise, we will focus on the valence electrons, since they're the ones involved in bonding. For both O鈧 and N鈧, we will look at the 2s and 2p atomic orbitals.
02

Form Molecular Orbitals

Using molecular orbital theory, atomic orbitals can form molecular orbitals through linear combinations. For diatomic molecules, the resulting molecular orbitals are either bonding or antibonding. We will focus on the sigma and pi orbitals formed by the combination of the 2s and 2p atomic orbitals.
03

Arrange Molecular Orbitals in Energy Diagrams

Next, arrange the molecular orbitals formed in Step 2 in the energy diagrams. For O鈧 and N鈧 molecules, the energy order of molecular orbitals is: 1. 蟽(2s) 2. 蟽*(2s) 3. 蟽(2p) 4. 蟺(2p) = 蟺(2p) 5. 蟺*(2p) = 蟺*(2p) 6. 蟽*(2p) The star (*) denotes an antibonding orbital.
04

Fill in Electrons in Molecular Orbitals

Oxygen has 6 valence electrons, giving O鈧 a total of 12 valence electrons, while nitrogen has 5 valence electrons, giving N鈧 a total of 10 valence electrons. Fill in the electrons in molecular orbitals, starting from the lowest energy orbital, taking into account Hund's rule and the Pauli exclusion principle. For O鈧: 1. 蟽(2s) - 2e鈦 2. 蟽*(2s) - 2e鈦 3. 蟽(2p) - 2e鈦 4. 蟺(2p) = 蟺(2p) - 4e鈦 (2 in each 蟺 orbital) 5. 蟺*(2p) = 蟺*(2p) - 2e鈦 (1 in each 蟺* orbital) For N鈧: 1. 蟽(2s) - 2e鈦 2. 蟽*(2s) - 2e鈦 3. 蟽(2p) - 2e鈦 4. 蟺(2p) = 蟺(2p) - 4e鈦 (2 in each 蟺 orbital)
05

Analyze the Effect of Removing an Electron

Now, let's analyze the effect of removing one electron from each molecule. For O鈧: When you remove one electron, it comes from one of the 蟺*(2p) antibonding orbitals. With one electron less in an antibonding orbital, the overall bonding in O鈧 is strengthened. For N鈧: When you remove one electron, it comes from one of the 蟺(2p) bonding orbitals. With one electron less in a bonding orbital, the overall bonding in N鈧 is weakened.
06

Conclusion

Using molecular orbital theory, we can explain the different effects of removing one electron from O鈧 and N鈧. In O鈧, electron removal strengthens bonding as it removes an electron from an antibonding orbital. In N鈧, electron removal weakens bonding because it removes an electron from a bonding orbital.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Bonding in Diatomic Molecules
Understanding the nature of the bond between atoms in diatomic molecules is crucial for grasping how these entities hold together. The Molecular Orbital (MO) Theory offers a clear picture of how electrons are shared and organized between the two atoms. In MO Theory, individual atomic orbitals (AO) of the atoms in the molecule mix to form molecular orbitals, which extend over both atoms and can thus house the shared electrons of the bond.

These new orbitals come in two types: bonding molecular orbitals, which are lower in energy than the original atomic orbitals, and antibonding molecular orbitals, which are higher in energy. Electrons placed in bonding orbitals stabilize the molecule because they are in an energy state lower than they would be in separate atoms. In essence, for diatomic molecules like O鈧 and N鈧, electrons inhabiting bonding orbitals are the glue holding these molecules together.
Antibonding Orbitals
Antibonding orbitals play an intriguing, albeit less intuitive role in molecular stability. These orbitals, designated with an asterisk (e.g., 蟽* or 蟺*), arise from the same AO interactions that create bonding orbitals, but they have a node between the nuclei where electron probability density is zero. Consequently, placing electrons in antibonding orbitals actually serves to increase the energy of the system, making the molecule less stable.

It's important to notice that the presence of electrons in antibonding orbitals does not always result in a molecular breakup. Molecules can and do exist with electrons in antibonding orbitals; the key is the balance between the number of electrons in bonding versus antibonding orbitals. For a stable molecule, there must be more electrons in bonding than in antibonding orbitals. This balance, known as bond order, helps us understand why removing an electron from an antibonding orbital in O鈧 strengthens the bond, whereas removing one from a bonding orbital in N鈧 weakens it.
Electronic Configuration of Diatomic Molecules
The electronic configuration of diatomic molecules is the arrangement of electrons in the molecular orbitals. According to MO theory, this configuration informs us about the bonding nature and stability of the molecule. To determine the electronic configuration, we follow a process similar to the Aufbau principle for atomic orbitals, with a focus on the energy levels specific to molecular orbitals.

In our exercise dealing with O鈧 and N鈧, the valence electrons from each atom are distributed across the resulting sigma (蟽) and pi (蟺) orbitals. For O鈧, with 12 valence electrons, the configuration ends up with two electrons in a high-energy antibonding orbital. For N鈧, with 10 valence electrons, all electrons are paired in lower-energy bonding orbitals. This electron allocation in diatomic molecules is crucial to predict molecular properties like magnetism and reactivity, and to understand changes that occur when an electron is added or removed.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibonding orbitals which is lower in energy? Explain.

Show how two \(2 p\) atomic orbitals can combine to form a \(\sigma\) or a \(\pi\) molecular orbital.

Draw the Lewis structures for \(\mathrm{TeCl}_{4}, \mathrm{ICl}_{5}, \mathrm{PCl}_{5}, \mathrm{KrCl}_{4}\), and \(\mathrm{XeCl}_{2}\). Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit \(d^{2} s p^{3}\) hybridization? Which of the compounds have a square planar molecular structure? Which of the compounds are polar?

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the \(\mathrm{OH}\) molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lowerenergy \(p_{z}\) orbital from oxygen with the higher-energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for \(\mathrm{OH}\). Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of OH. Explain.

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-}\), and NO using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.