Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
bartleby

Videos

Question
Book Icon
Chapter 12, Problem 43E
Interpretation Introduction

(a)

Interpretation:

The electron dot formula and structural formula of H3O+ are to be drawn.

Concept introduction:

An electron dot formula is a way of representing the molecular structure in which electrons are represented by a dot. Structural formula is a way in which atoms are linked together through a solid line. This solid line represents the covalent bond. An electron dot structure is known as Lewis structure. Electron dot structure indicates the valence electrons of an atom which are involved in bonding.

Expert Solution
Check Mark

Answer to Problem 43E

Electron dot structure and structural formula of H3O+ are shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  1

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  2

Explanation of Solution

In molecule H3O+ oxygen is present as the central atom and hydrogen is present as the surrounding atom of oxygen. Oxygen has 6 valence electrons and hydrogen has 1 valence electron. Positive charge present on the molecule indicates that one electron is lost by the molecule. Total number of electron pairs is calculated by adding the valence electrons and subtracting the positive charge present on the molecule. So, number of electrons is (6+31) which is 8. Total electrons present in each bond pair is 2. Rest of the electrons are present as the lone pairs on oxygen atoms. An electron dot structure and structural formula of H3O+ is shown below in Figure 1 and 2 respectively.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  3

Figure 1

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  4

Figure 2

Solid line, in Figure 2, between the oxygen atom and the hydrogen atom is the covalent bond which is made up of two electrons. This bond is formed by sharing of electrons between the atoms present in that bond.

Conclusion

An electron dot structure and structural formula of H3O+ are shown above in Figure 1 and Figure 2.

Interpretation Introduction

(b)

Interpretation:

The electron dot formula and structural formula of OH are to be drawn.

Concept introduction:

An electron dot formula is a way of representing the molecular structure in which electrons are represented by a dot. Structural formula is a way in which atoms are linked together through a solid line. This solid line represents the covalent bond. An electron dot structure is known as Lewis structure. Electron dot structure indicates the valence electrons of an atom which are involved in bonding.

Expert Solution
Check Mark

Answer to Problem 43E

Electron dot structure and structural formula of OH are shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  5

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  6

Explanation of Solution

In molecule OH, Oxygen and hydrogen is present as the bonding atoms. Oxygen has 6 valence electrons and hydrogen has 1 valence electron. Negative charge present on the molecule indicates that one electron is gained by the molecule. Total number of electron pairs is calculated by adding the valence electrons and the negative charge present on the molecule. So, number of electrons is (6+1+1) which is 8. This shows that total electrons must be 8 or 4 pair of electrons. Total electrons present in each bond pair is 2. Rest of the electrons are present as the lone pair on oxygen atom. An electron dot structure and structural formula of OH is shown below in Figure 3 and 4 respectively.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  7

Figure 3

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  8

Figure 4

Solid line, in Figure 4, between the oxygen atom and the hydrogen atom shows the covalent bond which is made up of two electrons. This bond is formed by sharing of electrons between the atoms present in that bond.

Conclusion

An electron dot structure and structural formula of OH are shown above in Figure 3 and Figure 4.

Interpretation Introduction

(c)

Interpretation:

The electron dot formula and structural formula of HS are to be drawn.

Concept introduction:

An electron dot formula is a way of representing the molecular structure in which electrons are represented by a dot. Structural formula is a way in which atoms are linked together through a solid line. This solid line represents the covalent bond. An electron dot structure is known as Lewis structure. Electron dot structure indicates the valence electrons of an atom which are involved in bonding.

Expert Solution
Check Mark

Answer to Problem 43E

Electron dot structure and structural formula of HS is shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  9

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  10

Explanation of Solution

In molecule HS, sulfur and hydrogen is present as the bonding atoms. sulfur has 6 valence electrons and hydrogen has 1 valence electron. Negative charge present on the molecule indicates that one electron is gained by the the molecule. Total number of electron pairs is calculated by adding the valence electrons and the negative charge present on the molecule. So, number of electrons is (6+1+1) which is 8. This shows that total electrons must be 8 or 4 pair of electrons. Total electrons present in each bond pair is 2. Rest of the electrons are present as the lone pair on sulfur atom. An electron dot structure and structural formula of HS is shown below in Figure 5 and 6 respectively.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  11

Figure 5

. Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  12

Figure 6

Solid line, in Figure 6, between the sulfur and hydrogen atom is the covalent bond which is made up of two electrons. This bond is formed by sharing of electrons between the atoms present in that bond.

Conclusion

An electron dot structure and structural formula of HS are shown above in Figure 5 and Figure 6.

Interpretation Introduction

(d)

Interpretation:

The electron dot formula and structural formula of CN are to be drawn.

Concept introduction:

An electron dot formula is a way of representing the molecular structure in which electrons are represented by a dot. Structural formula is a way in which atoms are linked together through a solid line. This solid line represents the covalent bond. An electron dot structure is known as Lewis structure. Electron dot structure indicates the valence electrons of an atom which are involved in bonding.

Expert Solution
Check Mark

Answer to Problem 43E

Electron dot structure and structural formula of CN are shown below.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  13

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  14

Explanation of Solution

In molecule CN, carbon and nitrogen are present as the bonding atoms. Carbon has 4 valence electrons and nitrogen has 5 valence electrons. Negative charge present on the molecule indicates that one electron is gained by the molecule. Total number of electron pairs is calculated by adding the valence electrons and the negative charge present on the molecule. So, number of electrons is (4+5+1) which is 10. This shows that total electrons must be 10 or 5 pair of electrons. Total electrons present in each bond pair is 2. Rest of the electrons are present as the lone pair on carbon and nitrogen atom. An electron dot structure and structural formula of CN is shown below in Figure 7 and 8 respectively.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  15

Figure 7

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  16

Figure 8

Each solid line, in Figure 8, between the carbon and nitrogen atom is the covalent bond which is made up of two electrons. This bond is formed by sharing of electrons between the atoms present in that bond.

Conclusion

An electron dot structure and structural formula of CN are shown above in Figure 7 and Figure 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
(a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³
fcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transfer
34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10

Chapter 12 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 12 - Prob. 11CECh. 12 - Prob. 12CECh. 12 - Prob. 1KTCh. 12 - Prob. 2KTCh. 12 - Prob. 3KTCh. 12 - Prob. 4KTCh. 12 - Prob. 5KTCh. 12 - Prob. 6KTCh. 12 - Prob. 7KTCh. 12 - Prob. 8KTCh. 12 - Prob. 9KTCh. 12 - Prob. 10KTCh. 12 - Prob. 11KTCh. 12 - Prob. 12KTCh. 12 - Prob. 13KTCh. 12 - Prob. 14KTCh. 12 - Prob. 15KTCh. 12 - Prob. 16KTCh. 12 - Prob. 17KTCh. 12 - Prob. 18KTCh. 12 - Prob. 19KTCh. 12 - Prob. 20KTCh. 12 - Prob. 21KTCh. 12 - Prob. 22KTCh. 12 - Prob. 23KTCh. 12 - Prob. 24KTCh. 12 - Prob. 25KTCh. 12 - Prob. 26KTCh. 12 - Prob. 27KTCh. 12 - Prob. 28KTCh. 12 - Prob. 29KTCh. 12 - Prob. 1ECh. 12 - Prob. 2ECh. 12 - Prob. 3ECh. 12 - Prob. 4ECh. 12 - Prob. 5ECh. 12 - Prob. 6ECh. 12 - Prob. 7ECh. 12 - Prob. 8ECh. 12 - Prob. 9ECh. 12 - Prob. 10ECh. 12 - Prob. 11ECh. 12 - Prob. 12ECh. 12 - Prob. 13ECh. 12 - Prob. 14ECh. 12 - Prob. 15ECh. 12 - Prob. 16ECh. 12 - Prob. 17ECh. 12 - Prob. 18ECh. 12 - Prob. 19ECh. 12 - Prob. 20ECh. 12 - Prob. 21ECh. 12 - Prob. 22ECh. 12 - Prob. 23ECh. 12 - Prob. 24ECh. 12 - Prob. 25ECh. 12 - Prob. 26ECh. 12 - Prob. 27ECh. 12 - Prob. 28ECh. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Prob. 34ECh. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - Prob. 37ECh. 12 - Prob. 38ECh. 12 - Prob. 39ECh. 12 - Prob. 40ECh. 12 - Prob. 41ECh. 12 - Prob. 42ECh. 12 - Prob. 43ECh. 12 - Prob. 44ECh. 12 - Prob. 45ECh. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Prob. 50ECh. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - Prob. 54ECh. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - Prob. 58ECh. 12 - Prob. 59ECh. 12 - Prob. 60ECh. 12 - Prob. 61ECh. 12 - Prob. 62ECh. 12 - Prob. 63ECh. 12 - Prob. 64ECh. 12 - Prob. 65ECh. 12 - Prob. 66ECh. 12 - Prob. 67ECh. 12 - Prob. 68ECh. 12 - Prob. 69ECh. 12 - Prob. 70ECh. 12 - Prob. 71ECh. 12 - Prob. 72ECh. 12 - Prob. 73ECh. 12 - Prob. 74ECh. 12 - Prob. 75ECh. 12 - Prob. 76ECh. 12 - Prob. 77ECh. 12 - Prob. 78ECh. 12 - Prob. 79ECh. 12 - Prob. 80ECh. 12 - Prob. 81ECh. 12 - Prob. 82ECh. 12 - Prob. 83ECh. 12 - Prob. 84ECh. 12 - Prob. 85ECh. 12 - Prob. 86ECh. 12 - Prob. 87ECh. 12 - Prob. 88ECh. 12 - Prob. 89ECh. 12 - Prob. 90ECh. 12 - Prob. 91ECh. 12 - Prob. 92ECh. 12 - Prob. 93ECh. 12 - Prob. 94ECh. 12 - Prob. 95ECh. 12 - Prob. 96ECh. 12 - Prob. 97ECh. 12 - Prob. 98ECh. 12 - Prob. 99ECh. 12 - Prob. 100ECh. 12 - Prob. 1STCh. 12 - Prob. 2STCh. 12 - Prob. 3STCh. 12 - Prob. 4STCh. 12 - Prob. 5STCh. 12 - Prob. 6STCh. 12 - Prob. 7STCh. 12 - Prob. 8STCh. 12 - Prob. 9STCh. 12 - Prob. 10STCh. 12 - Prob. 11STCh. 12 - Prob. 12STCh. 12 - Prob. 13STCh. 12 - Prob. 14STCh. 12 - Prob. 15STCh. 12 - Prob. 16STCh. 12 - Prob. 17STCh. 12 - Prob. 18ST
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
Introduction to General, Organic and Biochemistry
Chemistry
ISBN:9781285869759
Author:Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher:Cengage Learning
Periodic Properties of Elements | Chemistry | IIT-JEE | NEET | CBSE | Misostudy; Author: Misostudy;https://www.youtube.com/watch?v=L26rRWz4_AI;License: Standard YouTube License, CC-BY
Periodic Trends: Electronegativity, Ionization Energy, Atomic Radius - TUTOR HOTLINE; Author: Melissa Maribel;https://www.youtube.com/watch?v=0h8q1GIQ-H4;License: Standard YouTube License, CC-BY