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 U1.20, Problem 6E
Interpretation Introduction

(a)

Interpretation:

Total numbers of valence electrons of all the atoms needs to be determined in the compound Potassium bromide (KBr).

Concept introduction:

Valence electrons in a molecule, is the sum of all the outer most orbital electrons of each atom present in it. In the periodic table, depending upon the number of valence electrons, elements are placed in specific groups.

Expert Solution
Check Mark

Answer to Problem 6E

8 valence electrons.

Explanation of Solution

Potassium bromide consists of potassium and bromine atom.

Electronic configuration of K atom is [Ar]4s1

Thus, the valence orbital is 4s and the number of valence electron is 1.

Similarly, electronic configuration of Br atom is [Ar] 3d104s² 4p5

Thus, the valence orbital is 4s and 4p and the number of valence electrons are 7.

Therefore, total number of valence electron in KBr molecule = (1 + 7) = 8.

Interpretation Introduction

(b)

Interpretation:

Total number of valence electrons of all the atoms need to be determined in the compound Calcium Oxide (CaO).

Concept introduction:

Valence electrons in a molecule, is the sum of all the outer most orbital electrons of each atom present in it. In the periodic table, depending upon the number of valence electrons, elements are placed in specific groups.

Expert Solution
Check Mark

Answer to Problem 6E

8 valence electrons.

Explanation of Solution

Electronic configuration of Ca atom is [Ar]4s2, so valence orbital is 4s and the number of valence electron is 2.

Electronic configuration of O atom is [He]2s22p4 so valence orbital is 2s, 2p and the number of valence electron is 6.

Therefore total number of valence electron in CaO molecule = (2 + 6) = 8.

Interpretation Introduction

(c)

Interpretation:

Total numbers of valence electrons of all the atoms needs to be determined in the compound Lithium oxide (Li2O).

Concept introduction:

Valence electrons in a molecule, is the sum of all the outer most orbital electrons of each atom present in it. In the periodic table, depending upon the number of valence electrons, elements are placed in specific groups.

Expert Solution
Check Mark

Answer to Problem 6E

8 valence electrons.

Explanation of Solution

Lithium oxide consists of Lithium and Oxygen atoms.

Electronic configuration of Li atom: [He]2s1 so valence orbital is 2s and the number of valence electron is 1.

Electronic configuration of O atom:  [He] 2s22p4, so valence orbital is 2s, 2p and the number of valence electron is 6.

Therefore, total number of valence electron in Li2O molecule =(2×1 + 6)=8.

Interpretation Introduction

(d)

Interpretation:

Total numbers of valence electrons of all the atoms needs to be determined in the compound Calcium chloride (CaCl2).

Concept introduction:

Valence electrons in a molecule, is the sum of all the outer most orbital electrons of each atom present in it. In the periodic table, depending upon the number of valence electrons, elements are placed in specific groups.

Expert Solution
Check Mark

Answer to Problem 6E

16 valence electrons.

Explanation of Solution

Calcium chloride is consists of Calcium and Chlorine atom.

Electronic configuration of Ca atom: [Ar]4s2 so valence orbital is 4s and the number of valence electron is 2.

Electronic configuration of Cl atom: [Ne] 3s² 3p5, so valence orbital is 3s and 3p and the number of valence electrons are 7.

Therefore, total number of valence electron in CaCl2 is =2 + (2×7)=16.

Interpretation Introduction

(e)

Interpretation:

Total numbers of valence electrons of all the atoms needs to be determined in the compound Aluminum chloride (AlCl3).

Concept introduction:

Valence electrons in a molecule, is the sum of all the outer most orbital electrons of each atom present in it. In the periodic table, depending upon the number of valence electrons, elements are placed in specific groups.

Expert Solution
Check Mark

Answer to Problem 6E

24 valence electrons.

Explanation of Solution

Aluminum chloride is consists of Aluminum and chlorine atom.

Electronic configuration of Al atom: [Ne]3s23p1 so valence orbital is 3s and 3p and the number of valence electrons are 3.

Electronic configuration of Cl atom: [Ne]3s²3p5, so valence orbital is 3s and 3p and the number of valence electrons are 7.

Therefore, total number of valence electron in AlCl3 is =3 + (3×7)=24.

Chapter U1 Solutions

Living By Chemistry: First Edition Textbook

Ch. U1.2 - Prob. 1ECh. U1.2 - Prob. 2ECh. U1.2 - Prob. 5ECh. U1.3 - Prob. 1TAICh. U1.3 - Prob. 1ECh. U1.3 - Prob. 2ECh. U1.3 - Prob. 3ECh. U1.3 - Prob. 4ECh. U1.3 - Prob. 5ECh. U1.3 - Prob. 6ECh. U1.4 - Prob. 1TAICh. U1.4 - Prob. 1ECh. U1.4 - Prob. 2ECh. U1.4 - Prob. 3ECh. U1.4 - Prob. 4ECh. U1.4 - Prob. 5ECh. U1.4 - Prob. 6ECh. U1.4 - Prob. 7ECh. U1.4 - Prob. 8ECh. U1.4 - Prob. 9ECh. U1.4 - Prob. 10ECh. U1.4 - Prob. 11ECh. U1.5 - Prob. 1TAICh. U1.5 - Prob. 1ECh. U1.5 - Prob. 2ECh. U1.5 - Prob. 3ECh. U1.5 - Prob. 4ECh. U1.5 - Prob. 5ECh. U1.5 - Prob. 6ECh. U1.5 - Prob. 7ECh. U1.6 - Prob. 1TAICh. U1.6 - Prob. 1ECh. U1.6 - Prob. 2ECh. U1.6 - Prob. 3ECh. U1.6 - Prob. 4ECh. U1.6 - Prob. 5ECh. U1.6 - Prob. 6ECh. U1.7 - Prob. 1TAICh. U1.7 - Prob. 1ECh. U1.7 - Prob. 2ECh. U1.7 - Prob. 3ECh. U1.7 - Prob. 4ECh. U1.7 - Prob. 5ECh. U1.8 - Prob. 1TAICh. U1.8 - Prob. 1ECh. U1.8 - Prob. 2ECh. U1.8 - Prob. 4ECh. U1.8 - Prob. 5ECh. U1.8 - Prob. 6ECh. U1.8 - Prob. 7ECh. U1.9 - Prob. 1TAICh. U1.9 - Prob. 1ECh. U1.9 - Prob. 2ECh. U1.9 - Prob. 5ECh. U1.9 - Prob. 7ECh. U1.10 - Prob. 1TAICh. U1.10 - Prob. 1ECh. U1.10 - Prob. 2ECh. U1.10 - Prob. 3ECh. U1.10 - Prob. 4ECh. U1.10 - Prob. 5ECh. U1.10 - Prob. 6ECh. U1.10 - Prob. 7ECh. U1.10 - Prob. 8ECh. U1.11 - Prob. 1TAICh. U1.11 - Prob. 1ECh. U1.11 - Prob. 2ECh. U1.11 - Prob. 3ECh. U1.11 - Prob. 4ECh. U1.11 - Prob. 5ECh. U1.11 - Prob. 6ECh. U1.11 - Prob. 7ECh. U1.11 - Prob. 9ECh. U1.11 - Prob. 11ECh. U1.11 - Prob. 12ECh. U1.12 - Prob. 1TAICh. U1.12 - Prob. 1ECh. U1.12 - Prob. 2ECh. U1.12 - Prob. 3ECh. U1.12 - Prob. 4ECh. U1.12 - Prob. 5ECh. U1.12 - Prob. 6ECh. U1.12 - Prob. 7ECh. U1.12 - Prob. 8ECh. U1.13 - Prob. 1TAICh. U1.13 - Prob. 1ECh. U1.13 - Prob. 2ECh. U1.13 - Prob. 3ECh. U1.13 - Prob. 4ECh. U1.13 - Prob. 5ECh. U1.13 - Prob. 6ECh. U1.13 - Prob. 7ECh. U1.13 - Prob. 8ECh. U1.13 - Prob. 9ECh. U1.14 - Prob. 1TAICh. U1.14 - Prob. 1ECh. U1.14 - Prob. 2ECh. U1.14 - Prob. 3ECh. U1.14 - Prob. 4ECh. U1.14 - Prob. 5ECh. U1.14 - Prob. 6ECh. U1.14 - Prob. 7ECh. U1.14 - Prob. 8ECh. U1.14 - Prob. 9ECh. U1.14 - Prob. 10ECh. U1.14 - Prob. 11ECh. U1.14 - Prob. 12ECh. U1.14 - Prob. 13ECh. U1.14 - Prob. 14ECh. U1.15 - Prob. 1TAICh. U1.15 - Prob. 1ECh. U1.15 - Prob. 2ECh. U1.15 - Prob. 3ECh. U1.15 - Prob. 4ECh. U1.15 - Prob. 5ECh. U1.15 - Prob. 6ECh. U1.15 - Prob. 7ECh. U1.15 - Prob. 8ECh. U1.15 - Prob. 9ECh. U1.15 - Prob. 10ECh. U1.15 - Prob. 11ECh. U1.15 - Prob. 12ECh. U1.16 - Prob. 1TAICh. U1.16 - Prob. 1ECh. U1.16 - Prob. 2ECh. U1.16 - Prob. 3ECh. U1.16 - Prob. 4ECh. U1.16 - Prob. 5ECh. U1.16 - Prob. 6ECh. U1.17 - Prob. 1TAICh. U1.17 - Prob. 1ECh. U1.17 - Prob. 2ECh. U1.17 - Prob. 3ECh. U1.17 - Prob. 4ECh. U1.17 - Prob. 5ECh. U1.17 - Prob. 6ECh. U1.17 - Prob. 7ECh. U1.17 - Prob. 8ECh. U1.17 - Prob. 9ECh. U1.17 - Prob. 10ECh. U1.17 - Prob. 11ECh. U1.18 - Prob. 1TAICh. U1.18 - Prob. 1ECh. U1.18 - Prob. 2ECh. U1.18 - Prob. 3ECh. U1.18 - Prob. 4ECh. U1.18 - Prob. 5ECh. U1.18 - Prob. 6ECh. U1.18 - Prob. 7ECh. U1.18 - Prob. 8ECh. U1.18 - Prob. 9ECh. U1.18 - Prob. 10ECh. U1.19 - Prob. 1TAICh. U1.19 - Prob. 1ECh. U1.19 - Prob. 2ECh. U1.19 - Prob. 3ECh. U1.19 - Prob. 4ECh. U1.19 - Prob. 5ECh. U1.19 - Prob. 6ECh. U1.19 - Prob. 7ECh. U1.19 - Prob. 8ECh. U1.19 - Prob. 9ECh. U1.19 - Prob. 10ECh. U1.19 - Prob. 11ECh. U1.19 - Prob. 12ECh. U1.19 - Prob. 13ECh. U1.19 - Prob. 14ECh. U1.19 - Prob. 15ECh. U1.19 - Prob. 16ECh. U1.20 - Prob. 1TAICh. U1.20 - Prob. 1ECh. U1.20 - Prob. 2ECh. U1.20 - Prob. 3ECh. U1.20 - Prob. 4ECh. U1.20 - Prob. 5ECh. U1.20 - Prob. 6ECh. U1.20 - Prob. 7ECh. U1.21 - Prob. 1TAICh. U1.21 - Prob. 1ECh. U1.21 - Prob. 2ECh. U1.21 - Prob. 3ECh. U1.21 - Prob. 4ECh. U1.21 - Prob. 5ECh. U1.21 - Prob. 6ECh. U1.21 - Prob. 7ECh. U1.21 - Prob. 8ECh. U1.22 - Prob. 1TAICh. U1.22 - Prob. 1ECh. U1.22 - Prob. 2ECh. U1.22 - Prob. 3ECh. U1.22 - Prob. 4ECh. U1.22 - Prob. 5ECh. U1.22 - Prob. 6ECh. U1.22 - Prob. 7ECh. U1.23 - Prob. 1TAICh. U1.23 - Prob. 1ECh. U1.23 - Prob. 2ECh. U1.23 - Prob. 3ECh. U1.23 - Prob. 4ECh. U1.23 - Prob. 5ECh. U1.24 - Prob. 1TAICh. U1.24 - Prob. 1ECh. U1.24 - Prob. 2ECh. U1.24 - Prob. 3ECh. U1.24 - Prob. 4ECh. U1.24 - Prob. 5ECh. U1.24 - Prob. 6ECh. U1.24 - Prob. 7ECh. U1.24 - Prob. 8ECh. U1.24 - Prob. 9ECh. U1.24 - Prob. 10ECh. U1.24 - Prob. 11ECh. U1.24 - Prob. 12ECh. U1.24 - Prob. 13ECh. U1.25 - Prob. 1TAICh. U1.25 - Prob. 1ECh. U1.25 - Prob. 2ECh. U1.25 - Prob. 3ECh. U1.25 - Prob. 4ECh. U1.25 - Prob. 5ECh. U1.25 - Prob. 6ECh. U1.26 - Prob. 1TAICh. U1.26 - Prob. 1ECh. U1.26 - Prob. 2ECh. U1.26 - Prob. 3ECh. U1.26 - Prob. 4ECh. U1.26 - Prob. 5ECh. U1.26 - Prob. 6ECh. U1.26 - Prob. 7ECh. U1.26 - Prob. 8ECh. U1.26 - Prob. 9ECh. U1.26 - Prob. 10ECh. U1.27 - Prob. 1TAICh. U1.27 - Prob. 1ECh. U1.27 - Prob. 2ECh. U1.27 - Prob. 3ECh. U1.27 - Prob. 4ECh. U1.27 - Prob. 5ECh. U1.27 - Prob. 6ECh. U1.27 - Prob. 7ECh. U1 - Prob. SI1RECh. U1 - Prob. SI2RECh. U1 - Prob. SI3RECh. U1 - Prob. SI4RECh. U1 - Prob. SI5RECh. U1 - Prob. SI6RECh. U1 - Prob. SII1RQCh. U1 - Prob. SII2RQCh. U1 - Prob. SII3RQCh. U1 - Prob. SII4RQCh. U1 - Prob. SIII1RQCh. U1 - Prob. SIII2RQCh. U1 - Prob. SIII3RQCh. U1 - Prob. SIII4RQCh. U1 - Prob. SIV1RQCh. U1 - Prob. SIV2RQCh. U1 - Prob. SIV3RQCh. U1 - Prob. SIV4RQCh. U1 - Prob. SIV5RQCh. U1 - Prob. SIV6RQCh. U1 - Prob. SV1RQCh. U1 - Prob. SV2RQCh. U1 - Prob. SV3RQCh. U1 - Prob. SV4RQCh. U1 - Prob. 1RECh. U1 - Prob. 2RECh. U1 - Prob. 3RECh. U1 - Prob. 4RECh. U1 - Prob. 5RECh. U1 - Prob. 6RECh. U1 - Prob. 7RECh. U1 - Prob. 8RECh. U1 - Prob. 9RECh. U1 - Prob. 10RECh. U1 - Prob. 11RECh. U1 - Prob. 12RE
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