Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 7.1, Problem 6LC

(a)

Interpretation Introduction

Interpretation: The electron dot structure of the potassium atom needs to be drawn.

Concept Introduction: Electron dot structures of an element represent the number of electron/s present in its valance electron shell. In the electron dot structure, a symbol of an atom of an element is written and the number of valance electrons is arranged in pairs represented as dots around the symbol.

(a)

Expert Solution
Check Mark

Answer to Problem 6LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  1

Explanation of Solution

The given atom is potassium. The element belongs to group 1 and the atomic number is 19. The electronic configuration is represented as follows:

  1s22s22p63s23p64s1

The number of electrons present in the valance shell that is 4s is valence electrons. Therefore, the number of valence electrons is equal to 1.

Since the number of valance electrons present in potassium is 1, its electron dot structure can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  2

(b)

Interpretation Introduction

Interpretation: The electron dot structure of the carbon atom needs to be drawn.

Concept Introduction:Electron dot structures of an element represent the number of electron/s present in its valance electron shell. In the electron dot structure, a symbol of an atom of an element is written and the number of valance electrons is arranged in pairs represented as dots around the symbol.

(b)

Expert Solution
Check Mark

Answer to Problem 6LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  3

Explanation of Solution

The given atom is carbon. The element belongs to group 14 and the atomic number is 6. The electronic configuration is represented as follows:

  1s22s22p2

The number of electrons present in the valance shell that is 2s2p is valence electrons. Therefore, the number of valence electrons is equal to 4.

Since the number of valance electrons present in carbon is 4, its electron dot structure can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  4

(c)

Interpretation Introduction

Interpretation: The electron dot structure of the magnesium atom needs to be drawn.

Concept Introduction:Electron dot structures of an element represent the number of electron/s present in its valance electron shell. In the electron dot structure, a symbol of an atom of an element is written and the number of valance electrons is arranged in pairs represented as dots around the symbol.

(c)

Expert Solution
Check Mark

Answer to Problem 6LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  5

Explanation of Solution

The given atom is magnesium. The element belongs to group 2 and the atomic number is 12. The electronic configuration is represented as follows:

  1s22s22p63s2

The number of electrons present in the valance shell that is 3s is valence electrons. Therefore, the number of valence electrons is equal to 2.

Since the number of valance electrons present in magnesium is 2, its electron dot structure can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  6

(d)

Interpretation Introduction

Interpretation: The electron dot structure of the oxygen atom needs to be drawn.

Concept Introduction:Electron dot structures of an element represent the number of electron/s present in its valance electron shell. In the electron dot structure, a symbol of an atom of an element is written and the number of valance electrons is arranged in pairs represented as dots around the symbol.

(d)

Expert Solution
Check Mark

Answer to Problem 6LC

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  7

Explanation of Solution

The given atom is oxygen. The element belongs to group 16 or 6A and the atomic number is 8. The electronic configuration is represented as follows:

  1s22s22p4

The number of electrons present in the valance shell that is 2s2p is valence electrons. Therefore, the number of valence electrons is equal to 6.

Since the number of valance electrons present in oxygen is 6, its electron dot structure can be represented as follows:

  Chemistry 2012 Student Edition (hard Cover) Grade 11, Chapter 7.1, Problem 6LC , additional homework tip  8

Chapter 7 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 7.2 - Prob. 11SPCh. 7.2 - Prob. 12LCCh. 7.2 - Prob. 13LCCh. 7.2 - Prob. 14LCCh. 7.2 - Prob. 15LCCh. 7.2 - Prob. 16LCCh. 7.2 - Prob. 17LCCh. 7.2 - Prob. 18LCCh. 7.2 - Prob. 19LCCh. 7.3 - Prob. 20LCCh. 7.3 - Prob. 21LCCh. 7.3 - Prob. 22LCCh. 7.3 - Prob. 23LCCh. 7.3 - Prob. 24LCCh. 7.3 - Prob. 25LCCh. 7.3 - Prob. 26LCCh. 7 - Prob. 27ACh. 7 - Prob. 28ACh. 7 - Prob. 29ACh. 7 - Prob. 30ACh. 7 - Prob. 31ACh. 7 - Prob. 32ACh. 7 - Prob. 33ACh. 7 - Prob. 34ACh. 7 - Prob. 35ACh. 7 - Prob. 36ACh. 7 - Prob. 37ACh. 7 - Prob. 38ACh. 7 - Prob. 39ACh. 7 - Prob. 40ACh. 7 - Prob. 41ACh. 7 - Prob. 42ACh. 7 - Prob. 43ACh. 7 - Prob. 44ACh. 7 - Prob. 45ACh. 7 - Prob. 46ACh. 7 - Prob. 47ACh. 7 - Prob. 48ACh. 7 - Prob. 49ACh. 7 - Prob. 50ACh. 7 - Prob. 51ACh. 7 - Prob. 52ACh. 7 - Prob. 53ACh. 7 - Prob. 54ACh. 7 - Prob. 55ACh. 7 - Prob. 56ACh. 7 - Prob. 57ACh. 7 - Prob. 58ACh. 7 - Prob. 59ACh. 7 - Prob. 60ACh. 7 - Prob. 61ACh. 7 - Prob. 62ACh. 7 - Prob. 63ACh. 7 - Prob. 64ACh. 7 - Prob. 65ACh. 7 - Prob. 66ACh. 7 - Prob. 67ACh. 7 - Prob. 68ACh. 7 - Prob. 69ACh. 7 - Prob. 70ACh. 7 - Prob. 71ACh. 7 - Prob. 72ACh. 7 - Prob. 73ACh. 7 - Prob. 74ACh. 7 - Prob. 75ACh. 7 - Prob. 76ACh. 7 - Prob. 77ACh. 7 - Prob. 78ACh. 7 - Prob. 79ACh. 7 - Prob. 80ACh. 7 - Prob. 81ACh. 7 - Prob. 82ACh. 7 - Prob. 83ACh. 7 - Prob. 84ACh. 7 - Prob. 85ACh. 7 - Prob. 86ACh. 7 - Prob. 87ACh. 7 - Prob. 88ACh. 7 - Prob. 89ACh. 7 - Prob. 90ACh. 7 - Prob. 91ACh. 7 - Prob. 92ACh. 7 - Prob. 93ACh. 7 - Prob. 94ACh. 7 - Prob. 95ACh. 7 - Prob. 96ACh. 7 - Prob. 97ACh. 7 - Prob. 98ACh. 7 - Prob. 99ACh. 7 - Prob. 100ACh. 7 - Prob. 101ACh. 7 - Prob. 102ACh. 7 - Prob. 103ACh. 7 - Prob. 104ACh. 7 - Prob. 105ACh. 7 - Prob. 106ACh. 7 - Prob. 107ACh. 7 - Prob. 108ACh. 7 - Prob. 109ACh. 7 - Prob. 110ACh. 7 - Prob. 111ACh. 7 - Prob. 112ACh. 7 - Prob. 113ACh. 7 - Prob. 114ACh. 7 - Prob. 1STPCh. 7 - Prob. 2STPCh. 7 - Prob. 3STPCh. 7 - Prob. 4STPCh. 7 - Prob. 5STPCh. 7 - Prob. 6STPCh. 7 - Prob. 7STPCh. 7 - Prob. 8STPCh. 7 - Prob. 9STPCh. 7 - Prob. 10STPCh. 7 - Prob. 11STPCh. 7 - Prob. 12STP
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