EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 9781305856745
Author: DECOSTE
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 13, Problem 65E

(a)

Interpretation Introduction

Interpretation: The species which have same number of atoms and same number of valence electrons in POCl3, SO42-, XeO4, PO43-, ClO4- needs to be determined.

Concept Introduction:

Lewis dot structure is the representation which shows the bonding between atoms present in a molecule. It shows lone pairs and bond pairs that exist on each bonded atom. Lewis dot structure is also known as Lewis dot formula or electron dot structure.

The sum of valence electrons must be arranged in such a way that all atoms must get octet configuration (8 electrons).

(a)

Expert Solution
Check Mark

Answer to Problem 65E

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  1EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  2EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  3EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  4EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  5

Hence here all POCl3, SO42-, XeO4, PO43-, ClO4- molecules and ions have same number of total valence electrons that is 32 electrons. Therefore they have same Lewis structure.

Explanation of Solution

The bond formation between the atoms takes place due to the sharing of valence electrons of bonded atoms while the remaining electrons present in outer shell represented as lone pair of electrons. To draw the Lewis structure, calculate the total number of valence electrons in each atom and draw the structure in such a way that each atom gets its octet configuration.

Total number of valence electrons in POCl3 :

  (1×5)+(1×6)+(3×7)=32

Hence the best Lewis structure for POCl3 must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  6

Total number of valence electrons in SO42- :

  (1×6)+(4×6)+2=32

Hence the best Lewis structure for SO42- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  7

Total number of valence electrons in XeO4 :

  (1×8)+(4×6)=32

Hence the best Lewis structure for XeO4 must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  8

Total number of valence electrons in PO43- :

  (1×5)+(4×6)+3=32

Hence the best Lewis structure for PO43- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  9

Total number of valence electrons in ClO4- :

  (1×7)+(4×6)+1=32

Hence the best Lewis structure for ClO4- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  10

Hence here all POCl3, SO42-, XeO4, PO43-, ClO4- molecules and ions have same number of total valence electrons that is 32 electrons. Therefore they have same Lewis structure.

(b)

Interpretation Introduction

Interpretation: The species which have same number of atoms and same number of valence electrons in NF3, SO32-, PO33-, ClO3- needs to be determined.

Concept Introduction:

Lewis dot structure is the representation which shows the bonding between atoms present in a molecule. It shows lone pairs and bond pairs that exist on each bonded atom. Lewis dot structure is also known as Lewis dot formula or electron dot structure.

The sum of valence electrons must be arranged in such a way that all atoms must get octet configuration (8 electrons).

(b)

Expert Solution
Check Mark

Answer to Problem 65E

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  11EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  12EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  13EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  14

Hence here all NF3, SO32-, PO33-, ClO3- molecules and ions have same number of total valence electrons that is 26 electrons. Therefore they have same Lewis structure.

Explanation of Solution

The bond formation between the atoms takes place due to the sharing of valence electrons of bonded atoms while the remaining electrons present in outer shell represented as lone pair of electrons. To draw the Lewis structure, calculate the total number of valence electrons in each atom and draw the structure in such a way that each atom gets its octet configuration.

Total number of valence electrons in NF3 :

  (1×5)+(3×7)=26

Hence the best Lewis structure for NF3 must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  15

Total number of valence electrons in SO32- :

  (1×6)+(3×6)+2=26

Hence the best Lewis structure for SO32- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  16

Total number of valence electrons in PO33- :

  (1×5)+(3×6)+3=26

Hence the best Lewis structure for PO33- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  17

Total number of valence electrons in ClO3- :

  (1×7)+(3×6)+1=26

Hence the best Lewis structure for ClO3- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  18

Hence here all NF3, SO32-, PO33-, ClO3- molecules and ions have same number of total valence electrons that is 26 electrons. Therefore they have same Lewis structure.

(c)

Interpretation Introduction

Interpretation: The species which have same number of atoms and same number of valence electrons in ClO2-, SCl2, PCl2- needs to be determined.

Concept Introduction:

Lewis dot structure is the representation which shows the bonding between atoms present in a molecule. It shows lone pairs and bond pairs that exist on each bonded atom. Lewis dot structure is also known as Lewis dot formula or electron dot structure.

The sum of valence electrons must be arranged in such a way that all atoms must get octet configuration (8 electrons).

(c)

Expert Solution
Check Mark

Answer to Problem 65E

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  19EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  20EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  21

Hence here all ClO2-, SCl2, PCl2- molecules and ions have same number of total valence electrons that is 20 electrons. Therefore they have same Lewis structure.

Explanation of Solution

The bond formation between the atoms takes place due to the sharing of valence electrons of bonded atoms while the remaining electrons present in outer shell represented as lone pair of electrons. To draw the Lewis structure, calculate the total number of valence electrons in each atom and draw the structure in such a way that each atom gets its octet configuration.

Total number of valence electrons in ClO2- :

  (1×7)+(2×6)+1=20

Hence the best Lewis structure for ClO2- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  22

Total number of valence electrons in SCl2 :

  (1×6)+(2×7)=20

Hence the best Lewis structure for SCl2 must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  23

Total number of valence electrons in PCl2- :

  (1×5)+(2×7)+1=20

Hence the best Lewis structure for PCl2- must be:

  EBK CHEMICAL PRINCIPLES, Chapter 13, Problem 65E , additional homework tip  24

Hence here all ClO2-, SCl2, PCl2- molecules and ions have same number of total valence electrons that is 20 electrons. Therefore they have same Lewis structure.

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Chapter 13 Solutions

EBK CHEMICAL PRINCIPLES

Ch. 13 - Prob. 11DQCh. 13 - Prob. 12DQCh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - An alternative definition of electronegativity...Ch. 13 - Prob. 16ECh. 13 - Without using Fig. 13.3, predict the order of...Ch. 13 - Without using Fig. 13.3, predict which bond in...Ch. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Indicate the bond polarity (show the partial...Ch. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Give an example of an ionic compound where both...Ch. 13 - What noble gas has the same electron configuration...Ch. 13 - Which of the following ions have noble gas...Ch. 13 - Give three ions that are isoelectronic with...Ch. 13 - Prob. 36ECh. 13 - Predict the empirical formulas of the ionic...Ch. 13 - Which compound in each of the following pairs of...Ch. 13 - Use the following data to estimate Hf for...Ch. 13 - Use the following data to estimate Hf for...Ch. 13 - Consider the following:...Ch. 13 - In general, the higher the charge on the ions in...Ch. 13 - Consider the following energy changes: a....Ch. 13 - Prob. 44ECh. 13 - Prob. 45ECh. 13 - The lattice energies of FeCl3,FeCl2,andFe2O3 are...Ch. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Prob. 68ECh. 13 - Prob. 69ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Prob. 79ECh. 13 - Prob. 80ECh. 13 - Prob. 81ECh. 13 - Prob. 82ECh. 13 - Prob. 83ECh. 13 - Prob. 84ECh. 13 - Prob. 85ECh. 13 - Prob. 86ECh. 13 - Prob. 87ECh. 13 - Prob. 88ECh. 13 - Prob. 89ECh. 13 - Prob. 90ECh. 13 - Prob. 91ECh. 13 - Prob. 92ECh. 13 - Prob. 93ECh. 13 - Prob. 94ECh. 13 - Prob. 95ECh. 13 - Predict the molecular structure and the bond...Ch. 13 - Prob. 97ECh. 13 - Two variations of the octahedral geometry are...Ch. 13 - Prob. 99ECh. 13 - Predict the molecular structure and the bond...Ch. 13 - Which of the molecules in Exercise 96 have net...Ch. 13 - Prob. 102ECh. 13 - Give two requirements that should be satisfied for...Ch. 13 - What do each of the following sets of...Ch. 13 - Prob. 105ECh. 13 - Consider the following Lewis structure, where E is...Ch. 13 - Consider the following Lewis structure, where E is...Ch. 13 - Prob. 108ECh. 13 - Prob. 109ECh. 13 - Which of the following molecules have net dipole...Ch. 13 - Prob. 111AECh. 13 - Prob. 112AECh. 13 - Prob. 113AECh. 13 - Prob. 114AECh. 13 - Prob. 115AECh. 13 - There are two possible structures of XeF2Cl2 ,...Ch. 13 - Prob. 117AECh. 13 - Prob. 118AECh. 13 - Prob. 119AECh. 13 - Prob. 120AECh. 13 - Prob. 121AECh. 13 - Prob. 122AECh. 13 - Prob. 123AECh. 13 - Prob. 124AECh. 13 - Prob. 125AECh. 13 - Prob. 126AECh. 13 - Prob. 127AECh. 13 - Prob. 128AECh. 13 - Prob. 129AECh. 13 - Prob. 130AECh. 13 - Prob. 131AECh. 13 - Prob. 132AECh. 13 - Prob. 133CPCh. 13 - Prob. 134CPCh. 13 - Given the following information: Heat of...Ch. 13 - Prob. 136CPCh. 13 - A promising new material with great potential as...Ch. 13 - Think of forming an ionic compound as three steps...Ch. 13 - Prob. 139CPCh. 13 - Prob. 140CPCh. 13 - Calculate the standard heat of formation of the...Ch. 13 - Prob. 142CPCh. 13 - Prob. 143MP
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