Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 78

Pentadiene, \(\mathrm{C}_{5} \mathrm{H}_{8},\) has three isomers, depending on the position of the two double bonds. Determine the shape of these isomers by using VSEPR theory. Describe the bonding in these molecules by using the valence-bond method. Do the shapes agree in the two theories? Use molecular orbital theory to decide which of these molecules has a delocalized \(\pi\) system. Sketch the molecular orbital and an energy-level diagram.

Short Answer

Expert verified
Using VSEPR theory, all isomers of pentadiene have a trigonal planar geometry. According to the valence bond method, each Carbon atom forms σ bonds with its neighboring atoms, with the fourth valence electron forming a π bond. The molecular shapes are consistent with both theories. However, only 1,3-pentadiene and 1,4-pentadiene have a delocalized pi system as per Molecular Orbital theory.

Step by step solution

01

Sketch Structures of Isomers

Identify the three isomers of pentadiene, which are 1,2-pentadiene, 1,3-pentadiene, and 1,4-pentadiene based on the different positions of two double bonds. In VSEPR model, atom with double or single bond is considered as one domain. Thus, each Carbon atom in pentadiene is surrounded by three electron domains and according to VSEPR theory, such molecules possess trigonal planar electron-domain geometry.
02

Describe Bonding Using Valence-Bond Method

For the second step, use the valence bond theory to analyze. Each Carbon atom uses three of its valence electrons to form sigma bonds with its neighboring atoms. The fourth valence electron in each Carbon atom of a double bond pairs up to form a pi bond, where the electron density lies above and below the plane of the molecule.
03

Check Agreement Between VSEPR and Valence-Bond Theory

The molecular shapes predicted by VSEPR theory for these molecules are in agreement with the molecular structure suggested by the valence-bond method.
04

Identify Delocalized Pi System Using Molecular Orbital Theory

1,2-pentadiene has two isolated double bonds. Therefore, there can be no delocalization of pi electrons. On the other hand, 1,3-pentadiene and 1,4-pentadiene both have a continuous system of parallel p orbitals. This suggests delocalization of pi-bonded electrons in them.
05

Sketch the Molecular Orbital and Energy-Level Diagram

Here we need to draw the energy level diagrams indicating the molecular orbitals of 1,3-pentadiene and 1,4-pentadiene, showing the occupied and vacant pi molecular orbitals.

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

Describe the molecular geometry of \(\mathrm{CCl}_{4}\) 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.

Use the valence molecular orbital configuration to determine which of the following species is expected to have the greatest electron affinity: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{Be}_{2}\) (c) \(\mathrm{F}_{2} ;\) (d) \(\mathrm{B}_{2}^{+}\)

The molecule formamide, \(\mathrm{HCONH}_{2}\), has the approximate bond angles \(\mathrm{H}-\mathrm{C}-\mathrm{O}, 123^{\circ} ; \mathrm{H}-\mathrm{C}-\mathrm{N}, 113^{\circ}\) \(\mathrm{N}-\mathrm{C}-\mathrm{O}, 124^{\circ} ; \mathrm{C}-\mathrm{N}-\mathrm{H}, 119^{\circ} ; \mathrm{H}-\mathrm{N}-\mathrm{H}, 119^{\circ}\) The \(\mathrm{C}-\mathrm{N}\) bond length is \(138 \mathrm{pm}\). Two Lewis structures can be written for this molecule, with the true structure being a resonance hybrid of the two. Propose a hybridization and bonding scheme for each structure.

Construct a concept map that connects the ideas of molecular orbital theory.

Explain why the molecular structure of \(\mathrm{BF}_{3}\) cannot be adequately described through overlaps involving pure \(s\) and \(p\) orbitals.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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