Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 8.3, Problem 24LC

(a)

Interpretation Introduction

Interpretation: The shape of the hybridization sp2 is to be classified.

Concept Introduction: In chemistry, hybridization combines two atomic orbitals to create a brand-new category of hybridized orbitals. The molecular morphologies can be well explained by the VSEPR hypothesis.

(a)

Expert Solution
Check Mark

Answer to Problem 24LC

The shape of the hybridization sp2 is a trigonal planar shape.

Explanation of Solution

Molecular bonding and shape information are both revealed by orbital hybridization.

The carbon atom with sp2 means the carbon atom has a double bond in it.

Hybridization plays a vital role in describing double covalent bonds.

Consider a simple molecule ethene where there is a double bond between carbon and carbon and a single bond between carbon and hydrogen.

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8.3, Problem 24LC , additional homework tip  1

The hybridization of sp2 is achieved by combining one 2s and two 2p atomic orbitals of carbon.

Each hybrid orbital is separated by 120 .

The shape of sp2 hybridized carbon atom is trigonal planar.

(b)

Interpretation Introduction

Interpretation: The shape of the hybridization sp3 is to be classified.

Concept Introduction: In chemistry, hybridization combines two atomic orbitals to create a brand-new category of hybridized orbitals. The molecular morphologies can be well explained by the VSEPR hypothesis.

(b)

Expert Solution
Check Mark

Answer to Problem 24LC

The shape of the hybridization sp3 is tetrahedral.

Explanation of Solution

Molecular bonding and shape information are both revealed by orbital hybridization.

The carbon atom with sp3 means the carbon atom has a single bond with its substituents.

Consider a simple molecule methane where there is carbon in the center and there are four single bonds between carbon and hydrogen.

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8.3, Problem 24LC , additional homework tip  2

The hybridization of sp3 is achieved by combining one 2s orbital and three 2p orbitals of carbon.

Each hybrid orbital is separated by 109.5 .

The shape of sp3 hybridized carbon atom is tetrahedral.

(c)

Interpretation Introduction

Interpretation: The shape of the hybridization sp2 is to be classified.

Concept Introduction: In chemistry, hybridization combines two atomic orbitals to create a brand-new category of hybridized orbitals. The molecular morphologies can be well explained by the VSEPR hypothesis.

(c)

Expert Solution
Check Mark

Answer to Problem 24LC

The shape of the hybridization sp is linear.

Explanation of Solution

Molecular bonding and shape information are both revealed by orbital hybridization.

The carbon atom with sp means the carbon atom has a triple bond in it.

Hybridization plays a vital role in describing triple covalent bonds.

Consider a single molecule of acetylene where there is a triple bond between carbon and carbon and a single bond between carbon and hydrogen.

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8.3, Problem 24LC , additional homework tip  3

The hybridization of sp is achieved by combining one 2s orbital and one of the three 2p atomic orbitals of carbon.

Each hybrid orbital is separated by 180 .

The shape of sp hybridized carbon atom is linear.

Chapter 8 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 8.2 - Prob. 11LCCh. 8.2 - Prob. 12LCCh. 8.2 - Prob. 13LCCh. 8.2 - Prob. 14LCCh. 8.2 - Prob. 15LCCh. 8.2 - Prob. 16LCCh. 8.2 - Prob. 17LCCh. 8.2 - Prob. 18LCCh. 8.2 - Prob. 19LCCh. 8.2 - Prob. 20LCCh. 8.3 - Prob. 21LCCh. 8.3 - Prob. 22LCCh. 8.3 - Prob. 23LCCh. 8.3 - Prob. 24LCCh. 8.3 - Prob. 25LCCh. 8.3 - Prob. 26LCCh. 8.3 - Prob. 27LCCh. 8.3 - Prob. 28LCCh. 8.4 - Prob. 29SPCh. 8.4 - Prob. 30SPCh. 8.4 - Prob. 31LCCh. 8.4 - Prob. 32LCCh. 8.4 - Prob. 33LCCh. 8.4 - Prob. 34LCCh. 8.4 - Prob. 35LCCh. 8.4 - Prob. 36LCCh. 8.4 - Prob. 37LCCh. 8.4 - Prob. 38LCCh. 8 - Prob. 39ACh. 8 - Prob. 40ACh. 8 - Prob. 41ACh. 8 - Prob. 42ACh. 8 - Prob. 43ACh. 8 - Prob. 44ACh. 8 - Prob. 45ACh. 8 - Prob. 46ACh. 8 - Prob. 47ACh. 8 - Prob. 48ACh. 8 - Prob. 49ACh. 8 - Prob. 50ACh. 8 - Prob. 51ACh. 8 - Prob. 52ACh. 8 - Prob. 53ACh. 8 - Prob. 54ACh. 8 - Prob. 55ACh. 8 - Prob. 56ACh. 8 - Prob. 57ACh. 8 - Prob. 58ACh. 8 - Prob. 59ACh. 8 - Prob. 60ACh. 8 - Prob. 61ACh. 8 - Prob. 62ACh. 8 - Prob. 63ACh. 8 - Prob. 64ACh. 8 - Prob. 65ACh. 8 - Prob. 66ACh. 8 - Prob. 67ACh. 8 - Prob. 68ACh. 8 - Prob. 69ACh. 8 - Prob. 70ACh. 8 - Prob. 71ACh. 8 - Prob. 72ACh. 8 - Prob. 73ACh. 8 - Prob. 74ACh. 8 - Prob. 75ACh. 8 - Prob. 76ACh. 8 - Prob. 77ACh. 8 - Prob. 78ACh. 8 - Prob. 79ACh. 8 - Prob. 80ACh. 8 - Prob. 81ACh. 8 - Prob. 82ACh. 8 - Prob. 83ACh. 8 - Prob. 84ACh. 8 - Prob. 85ACh. 8 - Prob. 86ACh. 8 - Prob. 87ACh. 8 - Prob. 88ACh. 8 - Prob. 89ACh. 8 - Prob. 90ACh. 8 - Prob. 91ACh. 8 - Prob. 92ACh. 8 - Prob. 94ACh. 8 - Prob. 95ACh. 8 - Prob. 96ACh. 8 - Prob. 97ACh. 8 - Prob. 98ACh. 8 - Prob. 99ACh. 8 - Prob. 100ACh. 8 - Prob. 101ACh. 8 - Prob. 102ACh. 8 - Prob. 103ACh. 8 - Prob. 104ACh. 8 - Prob. 105ACh. 8 - Prob. 106ACh. 8 - Prob. 107ACh. 8 - Prob. 108ACh. 8 - Prob. 109ACh. 8 - Prob. 110ACh. 8 - Prob. 111ACh. 8 - Prob. 112ACh. 8 - Prob. 113ACh. 8 - Prob. 114ACh. 8 - Prob. 115ACh. 8 - Prob. 1STPCh. 8 - Prob. 2STPCh. 8 - Prob. 3STPCh. 8 - Prob. 4STPCh. 8 - Prob. 5STPCh. 8 - Prob. 6STPCh. 8 - Prob. 7STPCh. 8 - Prob. 8STPCh. 8 - Prob. 9STPCh. 8 - Prob. 10STPCh. 8 - Prob. 11STPCh. 8 - Prob. 12STPCh. 8 - Prob. 13STPCh. 8 - Prob. 14STP
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