Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U2.4, Problem 6E

(a)

Interpretation Introduction

Interpretation: The number of lone pairs in TeCl2 molecule needs to be determined.

Concept Introduction:

Lewis dot structure or electron dot structure can be defined as the structure of a molecule in which all the valence electrons and lone pairs on each atom is represented with the help of line and dot or cross.

The Lewis structure is used to predict the hybridization and molecular geometry of the molecule as it gives complete idea about the bonding between bonded atoms in a molecule.

(a)

Expert Solution
Check Mark

Answer to Problem 6E

8 lone pairs; Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  1

Explanation of Solution

The number of lone pairs in any molecule can be determined with the help of Lewis structure. In the Lewis structure of TeCl2 molecule;

Number of valence electrons in Te = 6 electrons

Number of valence electrons in Cl = 7

Total number of valence electrons = 6 + (2 x 7) = 20 electrons.

Thus the Lewis structure must be:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  2

Hence in TeCl2 molecule there are 8 lone pairs.

(b)

Interpretation Introduction

Interpretation: The number of lone pairs in HI molecule needs to be determined.

Concept Introduction:

Lewis dot structure or electron dot structure can be defined as the structure of a molecule in which all the valence electrons and lone pairs on each atom is represented with the help of line and dot or cross.

The Lewis structure is used to predict the hybridization and molecular geometry of the molecule as it gives complete idea about the bonding between bonded atoms in a molecule.

(b)

Expert Solution
Check Mark

Answer to Problem 6E

3 lone pairs; Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  3

Explanation of Solution

The number of lone pairs in any molecule can be determined with the help of Lewis structure. In the Lewis structure of HI molecule;

Number of valence electrons in I = 7 electrons

Number of valence electrons in H= 1

Total number of valence electrons = 1 + (1 x 7) = 8 electrons.

Thus the Lewis structure must be:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  4

Hence in HI molecule there are 3 lone pairs.

(c)

Interpretation Introduction

Interpretation: The number of lone pairs in AsBr3molecule needs to be determined.

Concept Introduction:

Lewis dot structure or electron dot structure can be defined as the structure of a molecule in which all the valence electrons and lone pairs on each atom is represented with the help of line and dot or cross.

The Lewis structure is used to predict the hybridization and molecular geometry of the molecule as it gives complete idea about the bonding between bonded atoms in a molecule.

(c)

Expert Solution
Check Mark

Answer to Problem 6E

10 lone pairs; Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  5

Explanation of Solution

The number of lone pairs in any molecule can be determined with the help of Lewis structure. In the Lewis structure of AsBr3 molecule;

Number of valence electrons in As = 5 electrons

Number of valence electrons in Br = 7

Total number of valence electrons = 5 + (3 x 7) = 26 electrons.

Thus the Lewis structure must be:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  6

Hence in AsBr3 molecule there are 10 lone pairs.

(d)

Interpretation Introduction

Interpretation: The number of lone pairs in SiF4 molecule needs to be determined.

Concept Introduction:

Lewis dot structure or electron dot structure can be defined as the structure of a molecule in which all the valence electrons and lone pairs on each atom is represented with the help of line and dot or cross.

The Lewis structure is used to predict the hybridization and molecular geometry of the molecule as it gives complete idea about the bonding between bonded atoms in a molecule.

(d)

Expert Solution
Check Mark

Answer to Problem 6E

12 lone pairs; Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  7

Explanation of Solution

The number of lone pairs in any molecule can be determined with the help of Lewis structure. In the Lewis structure of SiF4 molecule;

Number of valence electrons in Si = 4 electrons

Number of valence electrons in F = 7

Total number of valence electrons = 4 + (4 x 7) = 32 electrons.

Thus the Lewis structure must be:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  8

Hence in SiF4 moleculethere are 12 lone pairs.

(e)

Interpretation Introduction

Interpretation: The number of lone pairs in F2 molecule needs to be determined.

Concept Introduction:

Lewis dot structure or electron dot structure can be defined as the structure of a molecule in which all the valence electrons and lone pairs on each atom is represented with the help of line and dot or cross.

The Lewis structure is used to predict the hybridization and molecular geometry of the molecule as it gives complete idea about the bonding between bonded atoms in a molecule.

(e)

Expert Solution
Check Mark

Answer to Problem 6E

6 lone pairs; Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  9

Explanation of Solution

The number of lone pairs in any molecule can be determined with the help of Lewis structure. In the Lewis structure of F2 molecule;

Number of valence electrons in F = 7

Total number of valence electrons = (2x 7) = 14 electrons.

Thus the Lewis structure must be:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 6E , additional homework tip  10

Hence in F2 molecule there are 6 lone pairs.

Chapter U2 Solutions

Living By Chemistry: First Edition Textbook

Ch. U2.2 - Prob. 3ECh. U2.2 - Prob. 4ECh. U2.2 - Prob. 5ECh. U2.2 - Prob. 6ECh. U2.2 - Prob. 7ECh. U2.3 - Prob. 1TAICh. U2.3 - Prob. 1ECh. U2.3 - Prob. 2ECh. U2.3 - Prob. 3ECh. U2.3 - Prob. 4ECh. U2.3 - Prob. 5ECh. U2.4 - Prob. 1TAICh. U2.4 - Prob. 1ECh. U2.4 - Prob. 2ECh. U2.4 - Prob. 3ECh. U2.4 - Prob. 4ECh. U2.4 - Prob. 5ECh. U2.4 - Prob. 6ECh. U2.4 - Prob. 7ECh. U2.4 - Prob. 8ECh. U2.5 - Prob. 1TAICh. U2.5 - Prob. 1ECh. U2.5 - Prob. 2ECh. U2.5 - Prob. 3ECh. U2.5 - Prob. 4ECh. U2.5 - Prob. 5ECh. U2.5 - Prob. 6ECh. U2.5 - Prob. 7ECh. U2.5 - Prob. 8ECh. U2.5 - Prob. 9ECh. U2.6 - Prob. 1TAICh. U2.6 - Prob. 1ECh. U2.6 - Prob. 2ECh. U2.6 - Prob. 3ECh. U2.6 - Prob. 4ECh. U2.6 - Prob. 5ECh. U2.6 - Prob. 6ECh. U2.6 - Prob. 7ECh. U2.6 - Prob. 8ECh. U2.6 - Prob. 9ECh. U2.6 - Prob. 10ECh. U2.6 - Prob. 11ECh. U2.7 - Prob. 1TAICh. U2.7 - Prob. 1ECh. U2.7 - Prob. 2ECh. U2.7 - Prob. 3ECh. U2.7 - Prob. 4ECh. U2.7 - Prob. 5ECh. U2.8 - Prob. 1TAICh. U2.8 - Prob. 1ECh. U2.8 - Prob. 2ECh. U2.8 - Prob. 3ECh. U2.8 - Prob. 4ECh. U2.8 - Prob. 5ECh. U2.8 - Prob. 6ECh. U2.9 - Prob. 1TAICh. U2.9 - Prob. 1ECh. U2.9 - Prob. 2ECh. U2.9 - Prob. 3ECh. U2.9 - Prob. 4ECh. U2.9 - Prob. 5ECh. U2.9 - Prob. 6ECh. U2.9 - Prob. 7ECh. U2.10 - Prob. 1TAICh. U2.10 - Prob. 1ECh. U2.10 - Prob. 2ECh. U2.10 - Prob. 3ECh. U2.10 - Prob. 4ECh. U2.10 - Prob. 5ECh. U2.10 - Prob. 6ECh. U2.10 - Prob. 7ECh. U2.10 - Prob. 8ECh. U2.10 - Prob. 9ECh. U2.11 - Prob. 1TAICh. U2.11 - Prob. 1ECh. U2.11 - Prob. 2ECh. U2.11 - Prob. 3ECh. U2.11 - Prob. 4ECh. U2.11 - Prob. 5ECh. U2.11 - Prob. 6ECh. U2.11 - Prob. 7ECh. U2.12 - Prob. 1TAICh. U2.12 - Prob. 1ECh. U2.12 - Prob. 2ECh. U2.12 - Prob. 3ECh. U2.12 - Prob. 4ECh. U2.12 - Prob. 5ECh. U2.12 - Prob. 6ECh. U2.12 - Prob. 7ECh. U2.13 - Prob. 1TAICh. U2.13 - Prob. 1ECh. U2.13 - Prob. 2ECh. U2.13 - Prob. 3ECh. U2.13 - Prob. 4ECh. U2.13 - Prob. 5ECh. U2.14 - Prob. 1TAICh. U2.14 - Prob. 1ECh. U2.14 - Prob. 2ECh. U2.14 - Prob. 3ECh. U2.14 - Prob. 4ECh. U2.14 - Prob. 5ECh. U2.14 - Prob. 6ECh. U2.14 - Prob. 7ECh. U2.14 - Prob. 8ECh. U2.15 - Prob. 1TAICh. U2.15 - Prob. 1ECh. U2.15 - Prob. 2ECh. U2.15 - Prob. 4ECh. U2.15 - Prob. 5ECh. U2.15 - Prob. 6ECh. U2.16 - Prob. 1TAICh. U2.16 - Prob. 1ECh. U2.16 - Prob. 2ECh. U2.16 - Prob. 3ECh. U2.16 - Prob. 4ECh. U2.16 - Prob. 6ECh. U2.16 - Prob. 7ECh. U2.17 - Prob. 1TAICh. U2.17 - Prob. 1ECh. U2.17 - Prob. 2ECh. U2.17 - Prob. 3ECh. U2.17 - Prob. 4ECh. U2.17 - Prob. 5ECh. U2.17 - Prob. 6ECh. U2.17 - Prob. 7ECh. U2.17 - Prob. 8ECh. U2.17 - Prob. 9ECh. U2.18 - Prob. 1TAICh. U2.18 - Prob. 1ECh. U2.18 - Prob. 2ECh. U2.18 - Prob. 3ECh. U2.18 - Prob. 4ECh. U2.18 - Prob. 5ECh. U2.18 - Prob. 6ECh. U2.18 - Prob. 7ECh. U2.19 - Prob. 1TAICh. U2.19 - Prob. 1ECh. U2.19 - Prob. 2ECh. U2.19 - Prob. 3ECh. U2.19 - Prob. 4ECh. U2.19 - Prob. 5ECh. U2.19 - Prob. 6ECh. U2.19 - Prob. 7ECh. U2.19 - Prob. 8ECh. U2.19 - Prob. 9ECh. U2.20 - Prob. 1TAICh. U2.20 - Prob. 1ECh. U2.20 - Prob. 2ECh. U2.20 - Prob. 3ECh. U2.20 - Prob. 4ECh. U2.20 - Prob. 5ECh. U2.20 - Prob. 6ECh. U2.20 - Prob. 7ECh. U2.20 - Prob. 8ECh. U2.20 - Prob. 9ECh. U2.21 - Prob. 1TAICh. U2.21 - Prob. 1ECh. U2.21 - Prob. 2ECh. U2.21 - Prob. 3ECh. U2.21 - Prob. 4ECh. U2.21 - Prob. 6ECh. U2 - Prob. SI1RECh. U2 - Prob. SI2RECh. U2 - Prob. SI3RECh. U2 - Prob. SI4RECh. U2 - Prob. SI5RECh. U2 - Prob. SII1RECh. U2 - Prob. SII2RECh. U2 - Prob. SII3RECh. U2 - Prob. SII4RECh. U2 - Prob. SII5RECh. U2 - Prob. SIII1RECh. U2 - Prob. SIII2RECh. U2 - Prob. SIII3RECh. U2 - Prob. SIII4RECh. U2 - Prob. SIII5RECh. U2 - Prob. SIII6RECh. U2 - Prob. SIV1ECh. U2 - Prob. SIV2ECh. U2 - Prob. SIV3ECh. U2 - Prob. SIV4ECh. U2 - Prob. SIV5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RE

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