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 7STP
Interpretation Introduction

Interpretation: The tetrahedral arrangement of electron pairs in each molecule needs to be explained.

Concept Introduction: An electron dot structure represents the arrangement of total valence electrons in a molecule. Here, electrons are represented as dots in pairs to show the lone pairs as well as the bond pairs.

Expert Solution & Answer
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Explanation of Solution

The molecules in the given example are as follows:

  CH4 , NCl3 , H2S and, HF

In CH4 molecule, C atom is bonded with 4 H atoms via a single bond. There is no lone pair of electrons on C and H atoms; thus, all the electrons will be bond pairs.

The arrangement of electrons can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 7STP , additional homework tip  1

It has tetrahedral geometry because all 4 bond pairs are arranged tetrahedrally around the central C atom.

In NCl3 , N atom is bonded with 3 Cl atoms via a single bond. There is one lone pair of electrons on N atom and three lone pairs of electrons on the Cl atom.

The arrangement of electrons can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 7STP , additional homework tip  2

The above geometry is tetrahedral because 3 N-Cl bond pairs and 1 lone pair of N atoms are tetrahedrally arranged.

In H2S molecule, 2 H atoms are bonded with 1 S atom via single bonds. The valence electrons of the H and S atom is 1 and 6 respectively.

The arrangement of electrons can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 7STP , additional homework tip  3

The above geometry is tetrahedral because 2 bond pairs of H-S bonds and 2 lone pairs of electrons are tetrahedrally arranged.

In the HF molecule, 1 H and 1 F atoms are bonded together via a single bond. The valence electrons of the H and F atom is 1 and 7 respectively. There are 3 lone pairs of electrons on the F atom.

The arrangement of electrons can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 7STP , additional homework tip  4

The above geometry is also tetrahedral because 1 bond pair of H-F bond and 3 lone pairs of electrons on the F atom are tetrahedrally arranged.

Also, this can be checked from the total number of bonding pairs of electrons and non-bonding pairs of electrons (lone pairs). Here, in all the cases, the total number is 4.

According to VSEPR theory, the following table shows the relation between the total number of bonding and non-bonding pair of electrons and the geometry of a molecule.

    Total number of bonding and non-bonding pairsGeometry
    6Octahedral
    5Trigonal pyramidal
    4Tetrahedral
    3Trigonal planar
    2Linear

Since the total number is 4, the geometry will be tetrahedral.

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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