Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card
5th Edition
ISBN: 9781305367487
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 7, Problem 14QRT

(a)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for NH2Cl should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described.

Concept Introduction:

  • Lewis structures are diagrams that represent the chemical bonding of covalently bonded molecules and coordination compounds.
  • It is also known as Lewis dot structures which represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule.
  • The Lewis structure is based on the concept of the octet rule so that the electrons shared in each atom should have 8 electrons in its outer shell.

Lewis structure for any molecule is drawn by using the following steps,

First the skeletal structure for the given molecule is drawn then the total number of valence electrons for all atoms present in the molecule is determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed such that each atom contains eight electrons in its valence shell.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is NH2Cl.

The Lewis electron dot structure for given molecule can be determined by first drawing the skeletal structure. Then, the total number of valence electrons for all atoms present in the molecule is determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  1

Total number of valence electrons is given below:

  (2×1)+(1×7)+(1×5)=14.

Total number of electrons in bonds present is given below:

  (2×3)=6.

The eight electrons remaining will be distributing in such a way that each atom should have 8 electrons in its outer shell.

Therefore, the Lewis structure is given below:

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  2.

The molecular geometry will be triangular pyramidal because of the presence of three bond pairs and one lone pair around the central atom.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  3.

There will be four electron regions in the molecule and hence the electron-region geometry will be tetrahedral.

(b)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for OF2 should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described.

Concept Introduction:

Refer to (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is OF2.

The Lewis electron dot structure for given molecule can be determined by first drawing the skeletal structure. Then, the total number of valence electrons for all atoms present in the molecule is determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  4.

Total number of valence electrons is given below:

  (2×7)+(6×1)=20.

Total number of electrons in bonds present is given below:

  (2×2)=4.

The sixteen electrons remaining will be distributing in such a way that each atom should have 8 electrons in its outer shell.

Therefore, the Lewis structure is given below:

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  5.

The molecular geometry will be angular because of the presence of two bond pairs and two lone pair around the central atom.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  6.

There are four electron regions in the molecule and hence the electron-region geometry will be tetrahedral.

(c)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for SCN should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described.

Concept Introduction:

Refer to (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is SCN.

The Lewis electron dot structure for given molecule can be determined by first drawing the skeletal structure. Then, the total number of valence electrons for all atoms present in the molecule is determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  7.

Total number of valence electrons is given below:

  (4×1)+(6×1)+(5×1)+1=16.

Total number of electrons in bonds present is given below:

  (2×4)=8.

The eight electrons remaining will be distributing in such a way that each atom should have 8 electrons in its outer shell.

Therefore, the Lewis structure is given below:

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  8

The molecular geometry and electron-region geometry will be linear because of the presence of two bond pairs around the central atom.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  9.

(d)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for HOF should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described.

Concept Introduction:

Refer to (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is HOF.

The Lewis electron dot structure for given molecule can be determined by first drawing the skeletal structure. Then, the total number of valence electrons for all atoms present in the molecule is determined.

The next step is to subtract the electrons present in the total number of bonds present in the skeletal structure of the molecule with the total valence electrons such that considering each bond contains two electrons with it.

Finally, the electrons which got after subtractions have to be equally distributed considering each atom contains eight electrons in its valence shell.

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  10

Total number of valence electrons is given below:

  (1×1)(1×7)+(6×1)=14.

Total number of electrons in bonds present is given below:

  (2×2)=4.

The ten electrons remaining will be distributing in such a way that each atom should have 8 electrons in its outer shell.

Therefore, the Lewis structure is given below:

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  11.

The molecular geometry will be angular because of the presence of two bond pairs and two lone pair around the central atom.

  Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card, Chapter 7, Problem 14QRT , additional homework tip  12.

There are four electron regions in the molecule and hence the electron-region geometry will be tetrahedral.

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Chapter 7 Solutions

Bundle: Chemistry: The Molecular Science, 5th, Loose-Leaf + OWLv2 with Quick Prep 24-Months Printed Access Card

Ch. 7.5 - Prob. 7.5ECh. 7.6 - Prob. 7.8PSPCh. 7.6 - Prob. 7.7CECh. 7.6 - Prob. 7.9PSPCh. 7.7 - Prob. 7.8CECh. 7.7 - Prob. 7.9CECh. 7 - Write the Lewis structures and give the...Ch. 7 - The structural formula for the open-chain form of...Ch. 7 - Describe the VSEPR model. How is the model used to...Ch. 7 - What is the difference between the electron-region...Ch. 7 - Prob. 3QRTCh. 7 - Prob. 4QRTCh. 7 - If you have three electron regions around a...Ch. 7 - Prob. 6QRTCh. 7 - Prob. 7QRTCh. 7 - Prob. 8QRTCh. 7 - Prob. 9QRTCh. 7 - Prob. 10QRTCh. 7 - Prob. 11QRTCh. 7 - Prob. 12QRTCh. 7 - Prob. 13QRTCh. 7 - Prob. 14QRTCh. 7 - Prob. 15QRTCh. 7 - Prob. 16QRTCh. 7 - Write Lewis structures for XeOF2 and ClOF3. Use...Ch. 7 - Write Lewis structures for HCP and [IOF4]. Use...Ch. 7 - Prob. 19QRTCh. 7 - Prob. 20QRTCh. 7 - Explain why (I3)+ is bent, but (I3) is linear.Ch. 7 - Prob. 22QRTCh. 7 - Prob. 23QRTCh. 7 - Give approximate values for the indicated bond...Ch. 7 - Give approximate values for the indicated bond...Ch. 7 - Prob. 26QRTCh. 7 - Compare the FClF angles in ClF2+ and ClF2. From...Ch. 7 - Prob. 28QRTCh. 7 - Prob. 29QRTCh. 7 - Prob. 30QRTCh. 7 - Prob. 31QRTCh. 7 - Describe the geometry and hybridization of carbon...Ch. 7 - Describe the geometry and hybridization for each C...Ch. 7 - Describe the hybridization around the central atom...Ch. 7 - The hybridization of the two carbon atoms differs...Ch. 7 - The hybridization of the two nitrogen atoms...Ch. 7 - Identify the type of hybridization, approximate...Ch. 7 - Prob. 38QRTCh. 7 - Prob. 39QRTCh. 7 - Prob. 40QRTCh. 7 - Prob. 41QRTCh. 7 - Methylcyanoacrylate is the active ingredient in...Ch. 7 - Prob. 43QRTCh. 7 - Prob. 44QRTCh. 7 - Prob. 45QRTCh. 7 - Prob. 46QRTCh. 7 - Which of these molecules has a net dipole moment?...Ch. 7 - Prob. 48QRTCh. 7 - Use molecular structures and noncovalent...Ch. 7 - Prob. 50QRTCh. 7 - Explain why water “beads up” on a freshly waxed...Ch. 7 - Explain why water will not remove tar from your...Ch. 7 - Prob. 53QRTCh. 7 - Prob. 54QRTCh. 7 - Prob. 55QRTCh. 7 - Prob. 56QRTCh. 7 - The structural formula for vitamin C is Give a...Ch. 7 - Prob. 58QRTCh. 7 - Prob. 59QRTCh. 7 - Prob. 60QRTCh. 7 - Prob. 61QRTCh. 7 - Prob. 62QRTCh. 7 - Prob. 63QRTCh. 7 - Prob. 64QRTCh. 7 - Prob. 65QRTCh. 7 - Prob. 66QRTCh. 7 - Methylcyanoacrylate is the active ingredient in...Ch. 7 - Prob. 68QRTCh. 7 - Prob. 69QRTCh. 7 - Use Lewis structures and VSEPR theory to predict...Ch. 7 - In addition to CO, CO2, and C3O2, there is another...Ch. 7 - Prob. 72QRTCh. 7 - Prob. 73QRTCh. 7 - Prob. 74QRTCh. 7 - Prob. 75QRTCh. 7 - In the gas phase, positive and negative ions form...Ch. 7 - Prob. 77QRTCh. 7 - Prob. 78QRTCh. 7 - Prob. 79QRTCh. 7 - Prob. 80QRTCh. 7 - Prob. 81QRTCh. 7 - Prob. 82QRTCh. 7 - Prob. 83QRTCh. 7 - Prob. 84QRTCh. 7 - Prob. 85QRTCh. 7 - Prob. 86QRTCh. 7 - Prob. 87QRTCh. 7 - Prob. 88QRTCh. 7 - Prob. 89QRTCh. 7 - Prob. 90QRTCh. 7 - Prob. 91QRTCh. 7 - Prob. 92QRTCh. 7 - Prob. 93QRTCh. 7 - Prob. 94QRTCh. 7 - Which of these are examples of hydrogen bonding?Ch. 7 - Prob. 96QRTCh. 7 - Prob. 97QRTCh. 7 - Prob. 98QRTCh. 7 - Halothane, which had been used as an anesthetic,...Ch. 7 - Ketene, C2H2O, is a reactant for synthesizing...Ch. 7 - Gamma hydroxybutyric acid, GHB, infamous as a date...Ch. 7 - There are two compounds with the molecular formula...Ch. 7 - Piperine, the active ingredient in black pepper,...Ch. 7 - Prob. 105QRTCh. 7 - Two compounds have the molecular formula N3H3. One...Ch. 7 - Prob. 108QRTCh. 7 - Prob. 109QRTCh. 7 - Prob. 110QRTCh. 7 - Prob. 111QRTCh. 7 - Prob. 7.ACPCh. 7 - Prob. 7.BCPCh. 7 - Prob. 7.CCPCh. 7 - Prob. 7.DCP
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