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 79A

(a)

Interpretation Introduction

Interpretation: The electron dot structure of ethanoic acid needs to be drawn.

Concept Introduction: An electron dot structure represents the arrangement of total valence electrons in a molecule. Here, electrons are represented as dots in pairs around the symbol of atoms of the molecule.

(a)

Expert Solution
Check Mark

Explanation of Solution

The molecular formula of ethanoic acid is CH3COOH . In ethanoic acid, there are carbon-carbon both and two carbon-oxygen bonds. Here, one oxygen is bonded to each carbon atom.

The number of valence electrons in C, H, and O atoms is 4, 1, and 6 respectively. Thus, the total number of valence electrons in ethanoic acid will be:

  42+14+26=8+4+12=24

The total number of valence electrons is distributed among atoms such that there are 8 bonding-pair of electrons and 4 lone-pair of electrons.

The electron dot structure can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 79A , additional homework tip  1

(b)

Interpretation Introduction

Interpretation: Whether the bonding between each oxygen atom and carbon is the same or not 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 around the symbol of atoms of the molecule.

(b)

Expert Solution
Check Mark

Explanation of Solution

In the given molecule, there are two carbon and two oxygen atoms. The two carbon atoms are bonded together via a single covalent bond. Here, one oxygen atom is bonded with a carbon atom via one double bond, and the other oxygen atom is bonded via a single bond. The oxygen atom singly bonded with the carbon atom is also bonded with one hydrogen atom.

Therefore, the bonding between each oxygen atom and the carbon atom is not the same.

(c)

Interpretation Introduction

Interpretation: Whether the bonding between the carbon and each oxygen atom is polar or nonpolar needs to be explained.

Concept Introduction: For a covalent bond to be polar, the difference in electronegativity should be between 0.5 to 2.0. Here, one atom has more electronegativity than another; thus, the electronegative atom has a partial negative charge and the less electronegative atom has a partial positive charge.

(c)

Expert Solution
Check Mark

Explanation of Solution

A covalent bond is said to be polar in nature if the difference between the electronegativity value of two atoms is between 0.5 to 2.0. Here, the bond formed between carbon and oxygen needs to be analyzed. The electronegativity value of carbon and oxygen is 2.55 and 3.44 respectively. The difference in electronegativity can be calculated as follows:

  d=3.442.55=0.89

Since the calculated value is between 0.5 to 2.0, the carbon-oxygen bond will be polar in nature.

(d)

Interpretation Introduction

Interpretation: Whether ethanoic acid is a polar molecule or not needs to be determined.

Concept Introduction: A molecule is said to be polar if there are polar bonds and the molecule has some net dipole moment.

(d)

Expert Solution
Check Mark

Explanation of Solution

A molecule is said to be polar in nature if it has polar bonds and it has an overall dipole moment.

The structure of ethanoic acid is as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 8, Problem 79A , additional homework tip  2

Here, dipole moments of two C-O bonds are not get canceled out; thus, the overall molecule is polar in nature.

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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