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 20E
Interpretation Introduction

(a)

Interpretation:

The electron-pair geometry for the molecules, BF4, and CCl4 is to be explained and the Lewis diagram and wedge-and-dash diagram for the molecules are to be drawn.

Concept introduction:

The electron pairs in Lewis diagrams repel each other in real molecule and thus, they distribute themselves in positions around the central atoms that are as far away from one another. This arrangement of electron pairs is called electron-pair geometry. The electron pairs may be shared in covalent bond, or they may be lone pairs.

Expert Solution
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Answer to Problem 20E

The Lewis diagrams for BF4, and CCl4 are shown as below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  1 and EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  2

The wedge-and-dash diagrams for BF4, and CCl4 are shown as below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  3 and EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  4

The electron pair geometry for both molecules is tetrahedral.

Explanation of Solution

To write the Lewis diagram for a compound first the number of valence electrons is to be calculated. In the molecule, BF4, boron has three valence electrons and each fluorine has seven valence electrons. Since, BF4 has a negative molecule so, one electron is to be added in the total number of electrons. The total number of valence electron for the molecule BF4 is calculated below.

Totalnumberofvalenceelectron=(3+4×7+1)e=32e

Similarly, in the molecule, CCl4, carbon has four valence electrons and each chlorine has seven valence electron. The total number of valence electron for the molecule CCl4 is calculated below.

Totalnumberofvalenceelectron=(4+4×7)e=32e

The atom which is least electronegative is the central atom. In BF4, boron is least electronegative element. Therefore, boron is the central atom. Boron is bonded with four fluorine atoms through single bond. In Lewis diagram, each electron is placed around the atom such that the octet rule is obeyed. Therefore, the Lewis diagram of BF4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  5

Figure 1

In CCl4, carbon is least electronegative element. Therefore, carbon is the central atom. Carbon is bonded with four chlorine atoms through single bond. In Lewis diagram, each electron is placed around the atom such that the octet rule is obeyed. Therefore, the Lewis diagram of CCl4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  6

Figure 2

The electron-pair geometry depends on the number of electron pairs around the central atoms. In both the molecules BF4, and CCl4 there are four electron-pairs around the boron and carbon atom. Therefore, the electron pair geometry is tetrahedral.

The wedge-and-dash diagram for the molecules BF4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  7

Figure 3

The wedge-and-dash diagram for the molecules CCl4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  8

Figure 4

Conclusion

The Lewis and wedge-and-dash diagrams for BF4, and CCl4 are shown in the Figure 1, Figure 2, Figure 3, aand Figure 4. The electron pair geometry for both molecules is tetrahedral.

Interpretation Introduction

(b)

Interpretation:

The molecular geometry predicted by the valence shell electron-pair repulsion theory for the molecules BF4, and CCl4 is to be explained and the Lewis diagram and wedge-and-dash diagram for the molecules are to be drawn.

Concept introduction:

Molecular geometry is the precise term that is used to describe the shape of molecules and arrangement of atoms around the central atom. The molecular geometry of a molecule is predicted by valence shell electron-pair repulsion theory or in short VSEPR theory. VSEPR theory applies to substances in which a second period element is bonded to two, three, four, or other atoms.

Expert Solution
Check Mark

Answer to Problem 20E

The Lewis diagrams for BF4, and CCl4 are shown as below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  9 and EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  10

The wedge-and-dash diagrams for BF4, and CCl4 are shown as below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  11 and EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  12

The molecular geometry for both molecules is tetrahedral.

Explanation of Solution

To write the Lewis diagram for a compound first the number of valence electrons is to be calculated. In the molecule, BF4, boron has three valence electrons and each fluorine has seven valence electrons. Since, BF4 has a negative molecule so one electron is to be added in the total number of electrons. The total number of valence electron for the molecule BF4 is calculated as below.

Totalnumberofvalenceelectron=(3+4×7+1)e=32e

Similarly, in the molecule, CCl4, carbon has four valence electrons and each chlorine has seven valence electron. The total number of valence electron for the molecule CCl4 is calculated as below.

Totalnumberofvalenceelectron=(4+4×7)e=32e

The atom which is least electronegative is the central atom. In BF4 boron is least electronegative element. Therefore, boron is the central atom. Boron is bonded with four fluorine atoms through single bond. In Lewis diagram, each electron is placed around the atom such that the octet rule is obeyed. Therefore, the Lewis diagram of BF4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  13

Figure 1

In CCl4, carbon is least electronegative element. Therefore, carbon is the central atom. Carbon is bonded with four chlorine atoms through single bond. In Lewis diagram, each electron is placed around the atom such that the octet rule is obeyed. Therefore, the Lewis diagram of CCl4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  14

Figure 2

The molecular geometry depends on the number of electron pairs as well as number of unpaired electron on the central atoms. In both the molecules BF4, and CCl4 there are four electron-pairs around the boron and carbon atom and no loane pair on the central atoms. Therefore, the molecular geometry would be same as electron-pair geometry that is tetrahedral.

The wedge-and-dash diagram for the molecules BF4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  15

Figure 3

The wedge-and-dash diagram for the molecules CCl4 is shown below.

EBK INTRODUCTORY CHEMISTRY: AN ACTIVE L, Chapter 13, Problem 20E , additional homework tip  16

Figure 4

Conclusion

The Lewis and wedge-and-dash diagrams for BF4, and CCl4 are shown in the Figure 1, Figure 2, Figure 3, aand Figure 4. The molecular geometry for both molecules is tetrahedral.

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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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