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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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  1

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  3

Figure 1

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  5

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  7

Figure 3

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  9

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  11

Figure 5

. Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  13

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 12, Problem 43E , additional homework tip  15

Figure 7

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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.

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Chapter 12 Solutions

Modified Mastering Chemistry with Pearson eText -- Standalone Access Card -- for 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
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