Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781337247269
Author: Steven S. Zumdahl; Donald J. DeCoste
Publisher: Cengage Learning US
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Chapter 13, Problem 109E

(a)

Interpretation Introduction

Interpretation: The Lewis structure and the molecular structure with polarity of molecule needs to be determined.

  OCl2, KrF2, BeH2, SO2

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 hybridization and molecular structure of molecule determine the molecular structure of molecule.

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.

For a molecule to be polar, it should have polar bond with no symmetry in the molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 109E

  OCl2 is a polar molecule as it has two O-F polar bonds with bent geometry.

  KrF2 is a non-polar molecule as it has two Kr-F polar bonds are arranged in linear geometry.

  BeH2 is a non-polar molecule as it has linear geometry.

  SO2 is a polar molecule as it has two O-S polar bonds with bent geometry.

Explanation of Solution

In OCl2 molecule, the total number of valence electrons; (1×6)+(2×7)=20

Hence the best Lewis structure for OCl2 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  1

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 2  = 4 = sp 3= bent 

It is a polar molecule as it has two O-F polar bonds with bent geometry.

In KrF2 molecule, the total number of valence electrons; (1×8)+(2×7)=22

Hence the best Lewis structure for KrF2 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  2

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 3  = 5 = spd3 = linear

  KrF2 is a non-polar molecule as it has two Kr-F polar bonds are arranged in linear geometry.

In BeH2 molecule, the total number of valence electrons; (1×2)+(2×1)=4

Hence the best Lewis structure for BeH2 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  3

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 0  = 2 = sp = linear

  BeH2 is a non-polar molecule as it has linear geometry.

In SO2 molecule, the total number of valence electrons; (1×6)+(2×6)=18

Hence the best Lewis structure for SO2 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  4

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 1  = 3 = sp 2= bent 

  SO2 is a polar molecule as it has two O-S polar bonds with bent geometry.

(c)

Interpretation Introduction

Interpretation: The Lewis structure and the molecular structure with polarity of molecule needs to be determined.

  CF4, SeF4, KrF4 

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 hybridization and molecular structure of molecule determine the molecular structure of molecule.

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.

For a molecule to be polar, it should have polar bond with no symmetry in the molecule.

(c)

Expert Solution
Check Mark

Answer to Problem 109E

  CF4 is a non-polar molecule as it has 4 C-F polar bonds with tetrahedral geometry.

  KrF4 has square planer structure therefore it is non-polar.

  BeH2 is a non-polar molecule as it has linear geometry.

  SO2 is a polar molecule as it has two O-S polar bonds with bent geometry.

Explanation of Solution

In CF4 molecule, the total number of valence electrons; (1×4)+(4×7)=32

Hence the best Lewis structure for CF4 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  5

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 2  = 4 = sp 3= tetrahedral

  CF4 is a non-polar molecule as it has 4 C-F polar bonds with tetrahedral geometry.

In KrF4 molecule, the total number of valence electrons; (1×8)+(4×7)=36

Hence the best Lewis structure for KrF4 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  6

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 4 + 2 = 6= spd32 = square planer

  KrF4 hassquare planer structure therefore it is non-polar.

In SeF4 molecule, the total number of valence electrons; (1×6)+(4×7)=34

Hence the best Lewis structure for SeF4 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  7

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 4 + 1  = 5= sp3d = see-saw

  SeF4 is a non-polar molecule as it has see-saw geometry.

(d)

Interpretation Introduction

Interpretation: The Lewis structure and the molecular structure with polarity of molecule needs to be determined.

  IF5,AsF5

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 hybridization and molecular structure of molecule determine the molecular structure of molecule.

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.

For a molecule to be polar, it should have polar bond with no symmetry in the molecule.

(d)

Expert Solution
Check Mark

Answer to Problem 109E

  ICl5 is a polar molecule due to square pyramid geometry.

  AsF5 is a non-polar molecule due to trigonal bipyramid geometry.

Explanation of Solution

According to the given molecular formula, the structure of ICl5 must be:

  (1×7)+(5×7)=42

Hence the best Lewis structure for ICl5 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  8

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 5 + 1  = 6 = spd32 = square pyramid

  ICl5 is a polar molecule due to square pyramid geometry.

According to the given molecular formula, the structure of AsF5 must be:

  (1×5)+(5×7)=40

Hence the best Lewis structure for AsF5 must be:

  Chemical Principles, Chapter 13, Problem 109E , additional homework tip  9

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 5 +0  = 5  = sp3d = trigonal bipyramidal

  AsF5 is a non-polar molecule due to trigonal bipyramidgeometry.

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Chapter 13 Solutions

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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