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 84A

(a)

Interpretation Introduction

Interpretation: If there are two bonding pairs and two non-bonding pairs of electrons of a central atom in a molecule, its shape 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.

(a)

Expert Solution
Check Mark

Explanation of Solution

The total number of pairs of electrons will be the sum of bonding and non-bonding pair of electrons. Thus, it will be 4 for the given molecule. For 4 electrons, the hybridization will be sp3 that is tetrahedral geometry. Now, due to two lone pairs of electrons, there will be bond pair-lone pair repulsion between two atoms, and the shape of the molecule will be bent.

(b)

Interpretation Introduction

Interpretation: If there are four bonding pairs and zero non-bonding pairs of electrons of a central atom in the molecule, its shape 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.

(b)

Expert Solution
Check Mark

Explanation of Solution

The total number of pairs of electrons will be the sum of bonding and non-bonding pair of electrons. Thus, it will be 4 for the given molecule. For 4 electrons, the hybridization will be sp3 that is tetrahedral geometry. Since there are no lone pairs of electrons the shape will also be tetrahedral.

(c)

Interpretation Introduction

Interpretation: If there are three bonding pairs and one non-bonding pair of electrons of a central atom in the molecule, its shape 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.

(c)

Expert Solution
Check Mark

Explanation of Solution

The total number of pairs of electrons will be the sum of bonding and non-bonding pair of electrons. Thus, it will be 4 for the given molecule. For 4 electrons, the hybridization will be sp3 that is tetrahedral geometry. Since there is one lone pair of electrons the shape will also be trigonal pyramidal.

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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