EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L
EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L
6th Edition
ISBN: 9780100547506
Author: CRACOLICE
Publisher: YUZU
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Chapter 13, Problem 9E
Interpretation Introduction

Interpretation:

The Lewis diagram for C2H2Cl4,C2H2Cl2F2 and C3H4Br3I are to be drawn.

Concept introduction:

The Lewis structure shows the connectivity between atoms by identifying the lone pairs of electrons in a compound. Lewis structures are also known as Lewis dot structures. The valence electrons around an atom are shown by dots. Bonds between atoms are shown by lines and the lone pair of electrons is shown by a pair of dots.

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Answer to Problem 9E

The Lewis diagram for C2H2Cl4,C2H2Cl2F2 and C3H4Br3I are shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 9E , additional homework tip  1

Explanation of Solution

The number of valence electrons in the carbon atom is 4. The number of valence electrons in the hydrogen atom is 1. Chlorine, fluorine, bromine and iodine atoms belong to the same group and have 7 valence electrons.

In C2H2Cl4, the number of carbon atoms is 2 which gives the valence electrons from carbon atom to be (4×2)8. The number of hydrogen atoms in C2H2Cl4 is 2. This gives the valence electrons from hydrogen atoms to be (2×1)2. The number of chlorine atoms in C2H2Cl4 is 4. This gives the valence electrons from chlorine atoms to be (4×7)28.

Therefore, the total number of valence electrons in C2H2Cl4 is (8+2+28)38.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains 2 electrons. This uses total 14 electrons in making seven bonds. The remaining electrons are distributed around the chlorine atoms in order to complete their octet. Therefore, the Lewis diagram for C2H2Cl4 is drawn as shown in Figure 1.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 9E , additional homework tip  2

Figure 1

In C2H2Cl2F2, the number of carbon atoms is two which gives the valence electrons from carbon atom to be (4×2)8. The number of hydrogen atom in C2H2Cl2F2 is 2. This gives the valence electrons from hydrogen atoms to be (2×1)2. The number of fluorine atoms in C2H2Cl2F2 is 2. This gives the valence electrons from fluorine atoms to be (2×7)14. The number of chlorine atoms in C2H2Cl2F2 is 2. This gives the valence electrons from chlorine atoms to be (2×7)14.

Therefore, the total number of valence electrons in C2H2Cl2F2 is (8+2+14+14)38.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains 2 electrons. This uses total 14 electrons in making seven bonds. The remaining electrons are distributed around the chlorine and fluorine atoms in order to complete their octet. Therefore, the Lewis diagram for C2H2Cl2F2 is drawn as shown in Figure 2.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 9E , additional homework tip  3

Figure 2

In C3H4Br3I, the number of carbon atoms is three which give the valence electrons from carbon atoms to be (4×3)12. The number of hydrogen atoms in C3H4Br3I is 4. This gives the valence electrons from hydrogen atoms to be (4×1)4. The number of bromine atoms in C3H4Br3I is 3. This gives the valence electrons from bromine atoms to be (7×3)21. The number of iodine atoms in C3H4Br3I is 1. This gives the valence electrons from iodine atom to be (1×7)7.

Therefore, the total number of valence electrons in C3H4Br3I is (12+4+7+21)44.

These valence electrons are distributed in the Lewis diagram using bonds and lone pair of electrons.

At first, the single bonds between the carbon atoms and other atoms are drawn. Each bond contains 2 electrons. This uses total 20 electrons in making ten bonds. The remaining electrons are distributed around the bromine and iodine atoms in order to complete their octet. Therefore, the Lewis diagram for C3H4Br3I is drawn as shown in Figure 3.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 9E , additional homework tip  4

Figure 3

Conclusion

The Lewis diagram for C2H2Cl4,C2H2Cl2F2 and C3H4Br3I are shown in Figure 1, 2 and 3 respectively.

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Chapter 13 Solutions

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L

Ch. 13 - Prob. 11ECh. 13 - Prob. 12ECh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - Prob. 15ECh. 13 - Prob. 16ECh. 13 - Prob. 17ECh. 13 - Prob. 18ECh. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Prob. 21ECh. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Prob. 32ECh. 13 - Prob. 33ECh. 13 - Prob. 34ECh. 13 - Prob. 35ECh. 13 - Prob. 36ECh. 13 - Prob. 37ECh. 13 - Prob. 38ECh. 13 - Prob. 39ECh. 13 - Prob. 40ECh. 13 - Prob. 41ECh. 13 - Prob. 42ECh. 13 - Prob. 43ECh. 13 - Prob. 44ECh. 13 - Is the carbon tetrachloride molecule, CCl4, which...Ch. 13 - Prob. 46ECh. 13 - Describe the shapes and compare the polarities of...Ch. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Classify each of the following statements as true...Ch. 13 - Prob. 69ECh. 13 - Draw Lewis diagrams for these five acids of...Ch. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Describe the shapes of C2H6 and C2H4. In doing so,...Ch. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - C4H10O is the formula of diethyl ether. The same...Ch. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Draw Lewis diagrams for water and dihydrogen...Ch. 13 - Prob. 2PECh. 13 - Prob. 3PECh. 13 - Prob. 4PECh. 13 - Prob. 5PECh. 13 - What is the Lewis diagram of butane, C4H10?Ch. 13 - Prob. 7PECh. 13 - Prob. 8PECh. 13 - Prob. 9PECh. 13 - Prob. 10PECh. 13 - In the gas phase, tin (II) chloride is a...Ch. 13 - Prob. 12PECh. 13 - Determine the molecular geometry around each...Ch. 13 - Describe the molecular geometry around each carbon...Ch. 13 - Is the difluoromethane molecule polar or nonpolar?...Ch. 13 - Prob. 1LDRECh. 13 - Prob. 2LDRECh. 13 - Prob. 3LDRECh. 13 - Prob. 4LDRECh. 13 - Prob. 5LDRECh. 13 - Prob. 6LDRECh. 13 - Prob. 7LDRECh. 13 - Prob. 8LDRECh. 13 - Prob. 9LDRECh. 13 - Prob. 10LDRE
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