Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 43

In which of the following molecules would you expect to find delocalized molecular orbitals: (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{SO}_{2} ;\) (c) \(\mathrm{H}_{2} \mathrm{CO}\) ? Explain.

Short Answer

Expert verified
Delocalized molecular orbitals are present in Ethene (\(\mathrm{C}_{2} \mathrm{H}_{4}\)) and Sulfur Dioxide (\(\mathrm{SO}_{2}\)) but not in Formaldehyde (\(\mathrm{H}_{2} \mathrm{CO}\)).

Step by step solution

01

Understanding Delocalized Orbitals

Delocalized molecular orbitals extend over three or more atoms. They contribute to the stability of molecules and are usually present in molecules that exhibit resonance.
02

Analyzing Ethene \(\mathrm{C}_{2} \mathrm{H}_{4}\)

In ethene, which is a planar and symmetrical molecule, the six electrons from the carbon atoms are localized in three sigma bonds, two with hydrogen and one with the carbon atom. The remaining electron in each carbon forms a pi bond, which is a delocalized molecular orbital
03

Analyzing Sulfur Dioxide \(\mathrm{SO}_{2}\)

In sulfur dioxide, sulfur forms one sigma bond with each oxygen atom, and one of the oxygen atoms forms a pi bond with sulfur. However, the molecule has resonance structures, which means that the pi bond electrons are spread over three atoms making them delocalized.
04

Analyzing Formaldehyde \(\mathrm{H}_{2} \mathrm{CO}\)

In formaldehyde, carbon forms sigma bonds with hydrogen and oxygen, and forms a pi bond with oxygen. The pi bond is localized between carbon and oxygen, and there are no resonating structures. Therefore, there are no delocalized electrons in formaldehyde.

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

Draw a Lewis structure(s) for the nitrite ion, \(\mathrm{NO}_{2}^{-}\) Then propose a bonding scheme to describe the \(\sigma\) and the bonding in this ion. What conclusion can you reach about the number and types of \(\pi\) molecular orbitals in this ion? Explain.

A substance in which the valence and conduction bands overlap is (a) a semiconductor; (b) a metalloid; (c) a metal; (d) an insulator.

Construct a concept map that embodies the ideas of valence bond theory.

Describe the molecular geometry of \(\mathrm{H}_{2} \mathrm{O}\) suggested by each of the following methods: (a) Lewis theory; (b) valence-bond method using simple atomic orbitals; (c) VSEPR theory; (d) valence-bond method using hybridized atomic orbitals.

Explain how it is possible to avoid the concept of resonance by using molecular orbital theory.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

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