EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 11, Problem 88CE

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound H2SO4 has to be drawn showing the full charges if charges exist.

Concept Introduction:

Lewis structure:

The representation of valence shell electrons around the atom is known as Lewis structure or Lewis dot structure.  Electrons are represented as a dot in Lewis structures, a single dot represents unpaired electron and paired of dots represents paired electrons.

(a)

Expert Solution
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Explanation of Solution

Given compound,

        H2SO4

The total number of valence electrons in H2SO4 is 32. 1 electron each from two hydrogen atoms, 6 valence electrons from one sulfur atom and 6 valence electrons each from four oxygen atom.

The skeletal structure of H2SO4 can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  1

The total number of 12 electrons are used for bonding. Remaining 22 electrons can be distributed over oxygen atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  2

The Lewis structure of compound showing charges can be given as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  3

(b)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound NaNO3 has to be drawn showing the full charges if charges exist.

Concept Introduction:

Refer part (a)

(b)

Expert Solution
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Explanation of Solution

Given compound,

        NaNO3

The total number of valence electrons in NaNO3 is 24.  1 electron from sodium atom, 5 valence electrons from one nitrogen atom and 6 valence electrons each from three oxygen atom.

The skeletal structure of NaNO3 can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  4

The total number of 8 electrons are used for bonding. Remaining 16 electrons can be distributed over oxygen atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  5

The Lewis structure of compound showing charges can be given as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  6

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound K2CO3 has to be drawn showing the full charges if charges exist.

Concept Introduction:

Refer part (a)

(c)

Expert Solution
Check Mark

Explanation of Solution

Given compound,

        K2CO3

The total number of valence electrons in K2CO3 is 24. 1 electron from each potassium atom, 4 valence electrons from one carbon atom and 6 valence electrons each from three oxygen atom.

The skeletal structure of K2CO3 can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  7

The total number of 6 electrons are used for bonding. Remaining 18 electrons can be distributed over oxygen atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  8

The Lewis structure of compound showing charges can be given as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  9

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound HCN has to be drawn showing the full charges if charges exist.

Concept Introduction:

Refer part (a)

(d)

Expert Solution
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Explanation of Solution

Given compound,

        HCN

The total number of valence electrons in HCN is 10. 1 electron from hydrogen atom, 4 valence electrons from one carbon atom and 5 valence electrons each nitrogen atom.

The skeletal structure of HCN can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  10

The total number of four electrons are used for bonding. Remaining 6 electrons can be distributed over nitrogen and carbon atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  11

The Lewis structure of compound showing charges can be given as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  12

(e)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound Al2(SO4)3 has to be drawn showing the full charges if charges exist.

Concept Introduction:

Refer part (a)

(e)

Expert Solution
Check Mark

Explanation of Solution

Given compound,

        Al2(SO4)3

The total number of valence electrons in Al2(SO4)3 is 96. 3 electrons each from two aluminum atoms, 6 valence electrons from three sulfur atoms and 6 valence electrons each from 12 oxygen atoms.

The skeletal structure of Al2(SO4)3 can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  13

The total number of 32 electrons are used for bonding. Remaining 64 electrons can be distributed over oxygen atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  14

The Lewis structure of compound showing charges can be also be represented as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  15

(f)

Interpretation Introduction

Interpretation:

The Lewis structure of the given compound CH3COOH has to be drawn showing the full charges if charges exist.

Concept Introduction:

Refer part (a)

(f)

Expert Solution
Check Mark

Explanation of Solution

Given compound,

        CH3COOH

The total number of valence electrons in CH3COOH is 24. 1 electron each from  four hydrogen atoms, 4 valence electrons from one carbon atom and 6 valence electrons each oxygen atom.

The skeletal structure of CH3COOH can be drawn as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  16

The total number of 14 electrons are used for bonding. Remaining 10 electrons can be distributed over oxygen and carbon atoms to get the octet.

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  17

Here octet of carbon attached to oxygen is not completed. Therefore, the Lewis structure of compound can be given as follows:

EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 11, Problem 88CE , additional homework tip  18

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Chapter 11 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 11.9 - Prob. 11.11PCh. 11.10 - Prob. 11.12PCh. 11 - Prob. 1RQCh. 11 - Prob. 2RQCh. 11 - Prob. 3RQCh. 11 - Prob. 4RQCh. 11 - Prob. 5RQCh. 11 - Prob. 6RQCh. 11 - Prob. 7RQCh. 11 - Prob. 8RQCh. 11 - Prob. 9RQCh. 11 - Prob. 10RQCh. 11 - Prob. 11RQCh. 11 - Prob. 12RQCh. 11 - Prob. 13RQCh. 11 - Prob. 14RQCh. 11 - Prob. 15RQCh. 11 - Prob. 16RQCh. 11 - Prob. 17RQCh. 11 - Prob. 18RQCh. 11 - Prob. 19RQCh. 11 - Prob. 20RQCh. 11 - Prob. 21RQCh. 11 - Prob. 22RQCh. 11 - Prob. 23RQCh. 11 - Prob. 24RQCh. 11 - Prob. 25RQCh. 11 - Prob. 26RQCh. 11 - Prob. 28RQCh. 11 - Prob. 30RQCh. 11 - Prob. 31RQCh. 11 - Prob. 33RQCh. 11 - Prob. 36RQCh. 11 - Prob. 1PECh. 11 - Prob. 2PECh. 11 - Prob. 3PECh. 11 - Prob. 4PECh. 11 - Prob. 5PECh. 11 - Prob. 6PECh. 11 - Prob. 7PECh. 11 - Prob. 8PECh. 11 - Prob. 9PECh. 11 - Prob. 10PECh. 11 - Prob. 11PECh. 11 - Prob. 12PECh. 11 - Prob. 13PECh. 11 - Prob. 14PECh. 11 - Prob. 15PECh. 11 - Prob. 16PECh. 11 - Prob. 17PECh. 11 - Prob. 18PECh. 11 - Prob. 19PECh. 11 - Prob. 20PECh. 11 - Prob. 21PECh. 11 - Prob. 22PECh. 11 - Prob. 23PECh. 11 - Prob. 24PECh. 11 - Prob. 25PECh. 11 - Prob. 26PECh. 11 - Prob. 27PECh. 11 - Prob. 28PECh. 11 - Prob. 29PECh. 11 - Prob. 30PECh. 11 - Prob. 31PECh. 11 - Prob. 32PECh. 11 - Prob. 33PECh. 11 - Prob. 34PECh. 11 - Prob. 35PECh. 11 - Prob. 36PECh. 11 - Prob. 37PECh. 11 - Prob. 38PECh. 11 - Prob. 39PECh. 11 - Prob. 40PECh. 11 - Prob. 47PECh. 11 - Prob. 48PECh. 11 - Prob. 49PECh. 11 - Prob. 50PECh. 11 - Prob. 51PECh. 11 - Prob. 52PECh. 11 - Prob. 55AECh. 11 - Prob. 56AECh. 11 - Prob. 57AECh. 11 - Prob. 58AECh. 11 - Prob. 59AECh. 11 - Prob. 63AECh. 11 - Prob. 64AECh. 11 - Prob. 65AECh. 11 - Prob. 66AECh. 11 - Prob. 67AECh. 11 - Prob. 68AECh. 11 - Prob. 76AECh. 11 - Prob. 77AECh. 11 - Prob. 78AECh. 11 - Prob. 81AECh. 11 - Prob. 82AECh. 11 - Prob. 83AECh. 11 - Prob. 84AECh. 11 - Prob. 85AECh. 11 - Prob. 86AECh. 11 - Prob. 87AECh. 11 - Prob. 88CECh. 11 - Prob. 89CECh. 11 - Prob. 90CECh. 11 - Prob. 92CECh. 11 - Prob. 93CECh. 11 - Prob. 94CECh. 11 - Prob. 95CE
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