General, Organic, and Biochemistry
General, Organic, and Biochemistry
9th Edition
ISBN: 9780078021541
Author: Katherine J Denniston, Joseph J Topping, Dr Danae Quirk Dorr
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3.4, Problem 3.9Q

(a)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given molecule is CS2.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  SkeletalStructureCountValenceElectrons1C×4e-=4e-S  CS2S×6e-=12e-------------------16e-

Distribute valence electron:

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  1

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of each electronegative sulfur atom.  Since the dipole moment cancels each other, the molecule is not polar.

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  2

(b)

Interpretation Introduction

Interpretation:

The molecule NF3 is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

(b)

Expert Solution
Check Mark

Explanation of Solution

The given molecule is NF3.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  SkeletalStructureCountValenceElectrons1N×5e-=5e-FNF3F×7e-=21e-F------------------26e-

Distribute valence electron:

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  3

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of electronegative fluorine atom.  The given compound is polar.

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  4

(c)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

(c)

Expert Solution
Check Mark

Explanation of Solution

The given molecule is HCl.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  SkeletalStructureCountValenceElectrons1H×1e-=1e-HCl1Cl×7e-=7e-------------------8e-

Distribute valence electron:

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  5

The given molecule is polar because there is an electronegativity difference between two atoms.

General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  6

(d)

Interpretation Introduction

Interpretation:

The molecule given is polar or not has to be predicted, if it is polar by using a vector the direction in the electrons are pulled has to be indicated.

(d)

Expert Solution
Check Mark

Explanation of Solution

The given molecule is SiCl4.

The geometry of the molecule has to be determined first.  To determine the polarity, draw the Lewis structure.

  SkeletalStructureCountValenceElectronsCl1S×6e-=6e-ClSi  Cl4Cl×7e-=28e-Cl------------------34e-

Distribute valence electron:

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  7

Vectors can be used to determine the polarity of this compound.  The vectors should point in the direction of each electronegative chlorine atom.  Since the dipole moment cancels each other, the molecule is non-polar.

    General, Organic, and Biochemistry, Chapter 3.4, Problem 3.9Q , additional homework tip  8

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!
Students have asked these similar questions
X 5 Check the box under each molecule that has a total of five ẞ hydrogens. If none of the molecules fit this description, check the box underneath the table. CI Br Br Br 0 None of these molecules have a total of five ẞ hydrogens. Explanation Check esc F1 F2 tab caps lock fn Q @2 A W # 3 OH O OH HO © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility IK F7 F7 F8 TA F9 F10 & 6 28 * ( > 7 8 9 0 80 F3 O F4 KKO F5 F6 S 64 $ D % 25 R T Y U பட F G H O J K L Z X C V B N M H control option command P H F11 F12 + || { [ command option
An open vessel containing water stands in a laboratory measuring 5.0 m x 5.0 m x 3.0 m at 25 °C ; the vapor pressure (vp) of water at this temperature is 3.2 kPa. When the system has come to equilibrium, what mass of water will be found in the air if there is no ventilation? Repeat the calculation for open vessels containing benzene (vp = 13.1 kPa) and mercury (vp = 0.23 Pa)
Every chemist knows to ‘add acid to water with constant stirring’ when diluting a concentrated acid in order to keep the solution from spewing boiling acid all over the place.  Explain how this one fact is enough to prove that strong acids and water do not form ideal solutions.

Chapter 3 Solutions

General, Organic, and Biochemistry

Ch. 3.2 - Prob. 3.9PPCh. 3.4 - Prob. 3.10PPCh. 3.4 - Prob. 3.11PPCh. 3.4 - Prob. 3.12PPCh. 3.4 - Prob. 3.13PPCh. 3.4 - Prob. 3.3QCh. 3.4 - Prob. 3.4QCh. 3.4 - Prob. 3.14PPCh. 3.4 - Prob. 3.5QCh. 3.4 - Prob. 3.6QCh. 3.4 - Prob. 3.16PPCh. 3.4 - Prob. 3.7QCh. 3.4 - Prob. 3.8QCh. 3.4 - Prob. 3.9QCh. 3 - Prob. 3.13QPCh. 3 - Draw the appropriate Lewis symbol for each of the...Ch. 3 - Draw the appropriate Lewis symbol for each of the...Ch. 3 - Prob. 3.16QPCh. 3 - Describe the differences between covalent bonding...Ch. 3 - Describe the difference between nonpolar covalent...Ch. 3 - What is the periodic trend of electronegativity? Ch. 3 - What role does electronegativity play in...Ch. 3 - Use electronegativity values to classify the bonds...Ch. 3 - Use electronegativity values to classify the bonds...Ch. 3 - When there is a reaction between each of these...Ch. 3 - Prob. 3.24QPCh. 3 - Explain, using Lewis symbols and the octet rule,...Ch. 3 - Explain, using Lewis symbols and the octet rule,...Ch. 3 - Prob. 3.27QPCh. 3 - Prob. 3.28QPCh. 3 - Name each of the following ions: Na+ Cu+ Mg2+ Ch. 3 - Name each of the following ions: Cu2+ Fe2+ Fe3+ Ch. 3 - Name each of the following ions: HCO3– H3O+ CO32− Ch. 3 - Prob. 3.32QPCh. 3 - Prob. 3.33QPCh. 3 - Write the formula for each of the following...Ch. 3 - Prob. 3.35QPCh. 3 - Prob. 3.36QPCh. 3 - Prob. 3.37QPCh. 3 - Predict the formula of a compound formed...Ch. 3 - Prob. 3.39QPCh. 3 - Prob. 3.40QPCh. 3 - Prob. 3.41QPCh. 3 - Write the correct formula for each of the...Ch. 3 - Prob. 3.43QPCh. 3 - Write the correct formula for each of the...Ch. 3 - Prob. 3.45QPCh. 3 - Write the correct formula for each of the...Ch. 3 - Write a suitable formula for: sodium...Ch. 3 - Write a suitable formula for: aluminum...Ch. 3 - Prob. 3.49QPCh. 3 - Prob. 3.50QPCh. 3 - Prob. 3.51QPCh. 3 - Prob. 3.52QPCh. 3 - Write a suitable formula for: silicon...Ch. 3 - Prob. 3.54QPCh. 3 - Contrast ionic and covalent compounds with respect...Ch. 3 - Prob. 3.56QPCh. 3 - Prob. 3.57QPCh. 3 - Prob. 3.58QPCh. 3 - Prob. 3.59QPCh. 3 - Prob. 3.60QPCh. 3 - Prob. 3.61QPCh. 3 - Prob. 3.62QPCh. 3 - Prob. 3.63QPCh. 3 - Prob. 3.64QPCh. 3 - Prob. 3.65QPCh. 3 - Prob. 3.66QPCh. 3 - How is the positive charge of a polyatomic cation...Ch. 3 - Prob. 3.68QPCh. 3 - Prob. 3.69QPCh. 3 - Prob. 3.70QPCh. 3 - Prob. 3.71QPCh. 3 - Prob. 3.72QPCh. 3 - Prob. 3.73QPCh. 3 - Prob. 3.74QPCh. 3 - Prob. 3.75QPCh. 3 - Prob. 3.76QPCh. 3 - Prob. 3.77QPCh. 3 - Prob. 3.78QPCh. 3 - Prob. 3.79QPCh. 3 - Prob. 3.80QPCh. 3 - Prob. 3.81QPCh. 3 - Prob. 3.82QPCh. 3 - Prob. 3.83QPCh. 3 - Prob. 3.84QPCh. 3 - Prob. 3.85QPCh. 3 - Prob. 3.86QPCh. 3 - Prob. 3.87QPCh. 3 - Prob. 3.88QPCh. 3 - Prob. 3.89QPCh. 3 - Prob. 3.90QPCh. 3 - Prob. 3.91QPCh. 3 - Prob. 3.93QPCh. 3 - Prob. 3.94QPCh. 3 - Prob. 3.95QPCh. 3 - Prob. 3.96QPCh. 3 - Prob. 3.97QPCh. 3 - Prob. 3.98QPCh. 3 - Prob. 3.99QPCh. 3 - Prob. 3.100QPCh. 3 - Prob. 3.101QPCh. 3 - Prob. 3.102QPCh. 3 - Prob. 3.103QPCh. 3 - Prob. 3.104QPCh. 3 - Prob. 3.107QPCh. 3 - Prob. 3.108QPCh. 3 - Prob. 3.109QPCh. 3 - Prob. 3.110QPCh. 3 - Prob. 3.112QPCh. 3 - Predict differences in our global environment that...Ch. 3 - Prob. 2CPCh. 3 - Prob. 3CPCh. 3 - Prob. 4CPCh. 3 - Prob. 5CP
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.
Similar questions
SEE MORE QUESTIONS
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