Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11.44P

(a)

Interpretation Introduction

Interpretation:

The hybridization change of boron in the following reaction is to be determined.

  BF3+NaFNa+BF4

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

(a)

Expert Solution
Check Mark

Answer to Problem 11.44P

The hybridization of boron changes from sp2 to sp3.

Explanation of Solution

The Lewis structure of BF3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  1

Boron forms three single bonds with three fluorine atoms so three hybrid orbitals are required and therefore the hybridization of boron in BF3 is sp2.

The Lewis structure of BF4 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  2

Boron forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of boron in BF4 is sp3.

The hybridization of boron changes from sp2 to sp3.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(b)

Interpretation Introduction

Interpretation:

The hybridization change for phosphorus in the following reaction is to be determined.

  PCl3+Cl2PCl5

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

(b)

Expert Solution
Check Mark

Answer to Problem 11.44P

The hybridization of phosphorus changes from sp3 to sp3d.

Explanation of Solution

The Lewis structure of PCl3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  3

Phosphorus forms three single bonds with three chlorine atoms and has a lone pair on it so four hybrid orbitals are required and therefore the hybridization of phosphorus in PCl3 is sp3.

The Lewis structure of PCl5 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  4

Phosphorus forms five single bonds with five chlorine atoms so five hybrid orbitals are required and therefore the hybridization of phosphorus in PCl5 is sp3d.

The hybridization of phosphorus changes from sp3 to sp3d.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(c)

Interpretation Introduction

Interpretation:

The hybridization change for carbon in the following reaction is to be determined.

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  5

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

(c)

Expert Solution
Check Mark

Answer to Problem 11.44P

The hybridization of carbon changes from sp to sp2.

Explanation of Solution

The Lewis structure of acetylene is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  6

Carbon forms one triple bond with other carbon atom and one single bond with hydrogen atom so two hybrid orbitals are required and therefore the hybridization of carbon in acetylene is sp.

The Lewis structure of ethene is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  7

Carbon forms two single bonds with two hydrogen atoms and one double bond with another carbon atom so three hybrid orbitals are required and therefore the hybridization of carbon in ethene is sp2.

The hybridization of carbon changes from sp to sp2.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(d)

Interpretation Introduction

Interpretation:

The hybridization change for silicon in the following reaction is to be determined.

  SiF4+2FSiF62

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

(d)

Expert Solution
Check Mark

Answer to Problem 11.44P

The hybridization of silicon changes from sp3 to sp3d2.

Explanation of Solution

The Lewis structure of SiF4 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  8

Silicon forms four single bonds with four fluorine atoms so four hybrid orbitals are required and therefore the hybridization of silicon in SiF4 is sp3.

The Lewis structure of SiF62 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  9

Silicon forms six single bonds with six fluorine atoms so six hybrid orbitals are required and therefore the hybridization of silicon in SiF62 is sp3d2.

The hybridization of silicon changes from sp3 to sp3d2.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

(e)

Interpretation Introduction

Interpretation:

The hybridization of sulphur in the following reaction is to be determined.

  SO2+12O2SO3

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

Hybridization of the central atom can be determined from the number of electron groups around the central atom in the Lewis structure of the molecule. Single bond, double bond, triple bond and lone pair all are considered as single electron group.

(e)

Expert Solution
Check Mark

Answer to Problem 11.44P

There is no change in the hybridization of sulphur.

Explanation of Solution

The Lewis structure of SO2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  10

Sulphur forms one single, one double bond with two oxygen atoms separately. It has a lone pair of electrons on it. So three hybrid orbitals are required and therefore the hybridization of sulphur in SO2 is sp2.

The Lewis structure of SO3 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.44P , additional homework tip  11

Sulphur forms two single bonds and one double bond with three oxygen atoms separately so three hybrid orbitals are required and therefore the hybridization of sulphur in SO3 is sp2.

The hybridization of sulphur remains the same in going from SO2 to SO3.

Conclusion

Hybridization is determined from the number of electron groups around the central atom in the Lewis structure of the molecule.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 11 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Give the number and type of hybrid orbital that...Ch. 11 - What is the hybridization of nitrogen in each of...Ch. 11 - What is the hybridization of carbon in each of the...Ch. 11 - Prob. 11.9PCh. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Phosphine (PH3) reacts with borane (BH3) as...Ch. 11 - The illustrations below depict differences in...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Methyl isocyanate, , is an intermediate in the...Ch. 11 - Are these statements true or false? Correct any...Ch. 11 - Prob. 11.21PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.23PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Certain atomic orbitals on two atoms were combined...Ch. 11 - Prob. 11.28PCh. 11 - Antibonding MOs always have at least one node. Can...Ch. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - The molecular orbitals depicted are derived from...Ch. 11 - The molecular orbitals depicted below are derived...Ch. 11 - Prob. 11.34PCh. 11 - Use an MO diagram and the bond order you obtain...Ch. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Epinephrine (or adrenaline; below) is a naturally...Ch. 11 - Prob. 11.41PCh. 11 - Isoniazid (below) is an antibacterial agent that...Ch. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Tryptophan is one of the amino acids found in...Ch. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Sulfur forms oxides, oxoanions, and halides. What...Ch. 11 - Prob. 11.54PCh. 11 - Use an MO diagram to find the bond order and...Ch. 11 - Acetylsalicylic acid (aspirin), the most widely...Ch. 11 - Prob. 11.57P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY