Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 54

Compare the Lewis and molecular orbital treatments of the oxygen molecule.

Short Answer

Expert verified
While both theories are integral to our understanding of molecular structure and behavior, Lewis' theory is much simpler and considers that electrons are localized between atoms; it unfortunately falls short when describing paramagnetism and other phenomena. Molecular Orbital Theory is more complex but also more comprehensive, explaining phenomena such as paramagnetism through the concept of electron sharing in molecular orbitals.

Step by step solution

01

Lewis Structure of Oxygen Molecule

First, let's talk about the Lewis structure of an oxygen molecule, \( O_2 \). Oxygen is in Group 6 on the periodic table, so an atom of oxygen has 6 electrons in its outer level. With two oxygen atoms, there are a total of 12 valence electrons. In \( O_2 \), you can see that each oxygen atom is connected to the other by a double bond which accounts for 4 of the 12 valence electrons. The remaining 8 electrons complete the octets of both the oxygen atoms. Hence, the Lewis structure shows us that oxygen molecule is a double bonded compound.
02

Molecular Orbital Theory of Oxygen Molecule

Let's now discuss the molecular orbital theory for \( O_2 \). According to this theory, when atomic orbitals combine, the number of molecular orbitals remains the same. For example, oxygen has two unpaired electrons in separate orbitals according to its molecular orbital diagram. Therefore, oxygen is paramagnetic. These properties cannot be revealed with Lewis structures which is considered as a limitation.
03

Comparing Both Treatments

Lewis structures provide a simple and easy-to-understand picture of bonding. However, some phenomena like the paramagnetism of \( O_2 \), cannot be explained with it. On the other hand, Molecular Orbital Theory, although mathematically intense, provides a more comprehensive description of molecular structure including the distribution of electrons in molecules.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider a \(\mathrm{N}_{2}\) molecule in its first excited electronic state; that is, when an electron in the highest occupied molecular orbital is promoted to the lowest empty molecular obital. (a) Identify the molecular orbitals involved and sketch a diagram to show the transition. (b) Compare the bond order and bond length of \(\mathrm{N}_{2}{ }^{*}\) with \(\mathrm{N}_{2}\), where the asterisk denotes the excited molecule. (c) Is \(\mathrm{N}_{2}{ }^{*}\) diamagnetic or paramagnetic? (d) When \(\mathrm{N}_{2} *\) loses its excess energy and converts to the ground state \(\mathrm{N}_{2}\), it emits a photon of wavelength \(470 \mathrm{nm}\), which makes up part of the auroras lights. Calculate the energy difference between these levels.

The geometry of \(\mathrm{CH}_{4}\) could be square planar, with the four \(\mathrm{H}\) atoms at the corners of a square and the \(\mathrm{C}\) atom at the center of the square. Sketch this geometry and compare its stability with that of a tetrahedral \(\mathrm{CH}_{4}\) molecule.

What is the angle between these two hybrid orbitals on the same atom? (a) \(s p\) and \(s p\) hybrid orbitals, (b) \(s p^{2}\) and \(s p^{2}\) hybrid orbitals, (c) \(s p^{3}\) and \(s p^{3}\) hybrid orbitals.

Write the ground-state electron configuration for \(\mathrm{B}_{2}\). Is the molecule diamagnetic or paramagnetic?

Explain why the bond order of \(\mathrm{N}_{2}\) is greater than that of \(\mathrm{N}_{2}^{+},\) but the bond order of \(\mathrm{O}_{2}\) is less than that of \(\mathrm{O}_{2}^{+}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation
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.

Sign-up for free