Chemical Principles
Chemical Principles
8th Edition
ISBN: 9781305581982
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 18, Problem 88E
Interpretation Introduction

Interpretation: The bond angle and hybridization of the given compounds 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 shape or geometry of molecule can be predicted with the help of hybridization and VSEPR theory. It can be checked with the below formula:

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

Expert Solution & Answer
Check Mark

Answer to Problem 88E

  • XeF2 = sp3d= linear; 180°
  • XeO2F2 = sp3d= see saw; 90°,180°
  • XeO3 = sp3 = pyramide ; 103°
  • XeO4 = sp3 = tetrahedral ; 109°
  • XeF4 = sp3d= square planer ; 90°
  • XeO3F2 = sp3d= trigonal bipyramide; 90°,180°
  • XeO2F4 = sp3d2= octahedral ; 90°,180°

Explanation of Solution

Given information:

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  1

The hybridization can be checked with the help of Lewis structure and VSEPR theory which provide the geometry of molecules with lone pairs.

  • Chemical Principles, Chapter 18, Problem 88E , additional homework tip  2

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 2 + 3Hybridization = sp3d= linear; 180°

  • Chemical Principles, Chapter 18, Problem 88E , additional homework tip  3

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 4 + 1Hybridization = sp3d= see saw; 90°,180°

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  4

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 3 + 1=4Hybridization = sp3 = pyramide ; 103°

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  5

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 4 + 0 =4Hybridization = sp3 = tetrahedral ; 109°

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  6

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 4 + 2 = 6Hybridization =sp3d= square planer ; 90°

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  7

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 5 + 0 =5Hybridization = sp3d= trigonal bipyramide; 90°,180°

  Chemical Principles, Chapter 18, Problem 88E , additional homework tip  8

  Hybridization = Number of sigma bonds + Number of lone pairs on bonded atoms.Hybridization = 6 + 0 = 6Hybridization = sp3d2= octahedral ; 90°,180°

Conclusion

Thus,

  • XeF2 = sp3d= linear; 180°
  • XeO2F2 = sp3d= see saw; 90°,180°
  • XeO3 = sp3 = pyramide ; 103°
  • XeO4 = sp3 = tetrahedral ; 109°
  • XeF4 = sp3d= square planer ; 90°
  • XeO3F2 = sp3d= trigonal bipyramide; 90°,180°
  • XeO2F4 = sp3d2= octahedral ; 90°,180°

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Chapter 18 Solutions

Chemical Principles

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