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, Problem 71A
Interpretation Introduction

Interpretation: A hybridization scheme PCl3 needs to be devised. Also, the molecular shape of the molecule needs to be predicted.

Concept Introduction: VSEPR theory is used to predict the molecular shape with the help of electron pairs around the central atoms of the molecule. It is based on the assumption that the shape of the molecule is such that there is minimum electronic repulsion between the valence shell and the atom.

Expert Solution & Answer
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Explanation of Solution

The given molecule is PCl3 . Here, the phosphorus atom is the central atom and is bonded with three chlorine atoms. The number of valence electrons in the P atom is 5. It forms 3 single covalent bonds with 3 chlorine atoms. Thus, there will be 1 lone pair of electrons on the P atom. The arrangement of total valence electrons of a molecule takes place such that there will be one lone pair of electrons on the P atom, 3 bond pairs of electrons between P and Cl, and 3 lone pairs of electrons on each Cl atom.

The arrangement can be shown as follow:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 71A

The hybridization of the central atom will be sp3 . Now, the geometry of the molecule will be tetrahedral but due to the presence of one lone pair of electrons, the molecule will have a trigonal pyramidal shape.

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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