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 16QRT

(a)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for CO2 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 CO2.

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 16QRT , additional homework tip  1.

Total number of valence electrons is given below:

  (4×1)+(2×6)=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 16QRT , additional homework tip  2.

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

There will be two electron regions in the molecule and hence the electron-region geometry will also be linear.

(b)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for NO2 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 NO2.

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 16QRT , additional homework tip  3.

Total number of valence electrons is given below:

  (5×1)+(2×6)+(1)=18.

Total number of electrons in bonds present is given below:

  (2×3)=6.

The twelve 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 16QRT , additional homework tip  4.

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

There will be three electron regions in the molecule and hence the electron-region geometry will also be triangular planar.

(c)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for SO2 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 SO2.

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 16QRT , additional homework tip  5.

Total number of valence electrons is given below:

  (6×1)+(2×6)=18.

Total number of electrons in bonds present is given below:

  (2×3)=6.

The twelve 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 16QRT , additional homework tip  6

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

There will be three electron regions in the molecule and hence the electron-region geometry will also be triangular planar.

(d)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for O3 should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described. The similarities and differences in the series have to be given.

Concept Introduction:

Refer to (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is O3.

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 16QRT , additional homework tip  7.

Total number of valence electrons is given below:

  (3×6)=18.

Total number of electrons in bonds present is given below:

  (2×3)=6.

The twelve 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 16QRT , additional homework tip  8.

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

There will be three electron regions in the molecule and hence the electron-region geometry will also be triangular planar.

(e)

Interpretation Introduction

Interpretation:

The Lewis electron dot structure for ClO2 should be drawn and the electron-region geometry and the molecular geometry of the molecule has to be described. The similarities and differences in the series have to be given.

Concept Introduction:

Refer to (a).

(e)

Expert Solution
Check Mark

Explanation of Solution

Given molecule is ClO2.

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 16QRT , additional homework tip  9.

Total number of valence electrons is given below:

  (7×1)+(2×6)+(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 16QRT , additional homework tip  10.

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

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

The similarity is that two oxygen atoms are bonded to the central atom in all these molecules. But, there are differences in the geometry of the molecules because of the differences in lone pair of electrons around the central atom.

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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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