Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
bartleby

Videos

Question
Book Icon
Chapter U1.24, Problem 13E
Interpretation Introduction

(a)

Interpretation:

The element whose electronic configuration is given has to be identified.

1s2 2s2 2p63s23p64s23d4

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Chromium, Cr

Explanation of Solution

1s2 2s2 2p63s23p64s23d4 - To identify the element one has to simply look at the last subshell; 3d4. This element has valence electrons in subshell 3d. So it belongs to 4th period, d block. Total number of electrons = 24. The superscript number tells the row number of that block; here it is 4 so group is 6(2s + 4d) on the periodic table. The element is hence chromium.

Interpretation Introduction

(b)

Interpretation:

The element whose electronic configuration is given has to be identified.

1s2 2s2 2p63s23p2

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Silicon

Explanation of Solution

1s2 2s2 2p63s23p2 - To identify the element one has to simply look at the last subshell; 3p2. This element has valence electrons in subshell 3p. So it belongs to 3rdperiod, in p block. Total number of electrons = 14. The superscript number tells the row number of that block; here it is 2 so group is 14 (2s + 10d + 2p) on the periodic table. The element is hence silicon.

Interpretation Introduction

(c)

Interpretation:

The element whose electronic configuration is given has to be identified.

1s2 2s2 2p3

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Nitrogen

Explanation of Solution

1s2 2s2 2p3 - To identify the element one has to simply look at the last subshell; 2p3. This element has valence electrons in subshell 2p. So it belongs to 2ndperiod, in p block. Total number of electrons = 7. The superscript number tells the row number of that block; here it is 3 so group is 15 (2s + 10d + 3p) on the periodic table. The element is hence nitrogen.

Interpretation Introduction

(d)

Interpretation:

The element whose electronic configuration is given has to be identified.

1s2 2s2 2p63s23p6 4s23d104p65s24d10 5p66s1

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Caesium

Explanation of Solution

1s2 2s2 2p63s23p6 4s23d104p65s24d10 5p66s1-To identify the element one has to simply look at the last subshell; 6s1. This element has valence electrons in subshell 6s. So it belongs to 6thperiod, in s block. The superscript number tells the row number of that block; here it is 1 so group is 1 on the periodic table. The element is hence caesium.

Interpretation Introduction

(e)

Interpretation:

The element whose electronic configuration is given has to be identified.

1s2 2s2 2p63s23p6 4s23d104p65s24d10 5p66s24f145d106p2-

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Lead

Explanation of Solution

1s2 2s2 2p63s23p6 4s23d104p65s24d10 5p66s24f145d106p2-To identify the element one has to simply look at the last subshell; 3p2. This element has valence electrons in subshell 6p. So it belongs to 6thperiod, in p block. The superscript number tells the row number of that block; here it is 2 so group is 14 (2s + 10d + 2p) on the periodic table. The element is hence lead.

Interpretation Introduction

(f)

Interpretation:

The element whose electronic configuration is given has to be identified.

[Kr] 5s24d9

Concept introduction:

The electrons in an atom can be arranged in a shorthand notation form called electron configuration. Counting the number of electrons in the electron configuration will help us to know the atomic number of the atom. Each element has a unique atomic number.

The ending gives the exact location of the atom on the periodic table. The subscript tells the period number, the subshell letter tells the block and the superscript number tells the row number of that block.

Expert Solution
Check Mark

Answer to Problem 13E

Silver

Explanation of Solution

[Kr] 5s24d9 - The noble gas that comes before the element is krypton. The last subshell filled is5s2 and 4d9. This element has valence electrons in subshell 5. So it belongs to 5thperiod, in d block. The superscript number tells the row number of that block; here it is 9 so group is 11 (2s + 9d) on the periodic table. The element is hence silver.

Chapter U1 Solutions

Living by Chemistry

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. C1.1RECh. U1 - Prob. C1.2RECh. U1 - Prob. C1.3RECh. U1 - Prob. C1.4RECh. U1 - Prob. C1.5RECh. U1 - Prob. C1.6RECh. U1 - Prob. C2.1RECh. U1 - Prob. C2.2RECh. U1 - Prob. C2.3RECh. U1 - Prob. C2.4RECh. U1 - Prob. C3.1RECh. U1 - Prob. C3.2RECh. U1 - Prob. C3.3RECh. U1 - Prob. C3.4RECh. U1 - Prob. C4.1RECh. U1 - Prob. C4.2RECh. U1 - Prob. C4.3RECh. U1 - Prob. C4.4RECh. U1 - Prob. C5.1RECh. U1 - Prob. C5.2RECh. U1 - Prob. C5.3RECh. U1 - Prob. C5.4RECh. 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. 12RECh. U1 - Prob. 1STPCh. U1 - Prob. 2STPCh. U1 - Prob. 3STPCh. U1 - Prob. 4STPCh. U1 - Prob. 5STPCh. U1 - Prob. 6STPCh. U1 - Prob. 7STPCh. U1 - Prob. 8STPCh. U1 - Prob. 9STPCh. U1 - Prob. 10STPCh. U1 - Prob. 11STPCh. U1 - Prob. 12STPCh. U1 - Prob. 13STPCh. U1 - Prob. 14STPCh. U1 - Prob. 15STPCh. U1 - Prob. 16STPCh. U1 - Prob. 17STPCh. U1 - Prob. 18STPCh. U1 - Prob. 19STPCh. U1 - Prob. 20STPCh. U1 - Prob. 21STPCh. U1 - Prob. 22STP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Quantum Numbers, Atomic Orbitals, and Electron Configurations; Author: Professor Dave Explains;https://www.youtube.com/watch?v=Aoi4j8es4gQ;License: Standard YouTube License, CC-BY
QUANTUM MECHANICAL MODEL/Atomic Structure-21E; Author: H to O Chemistry;https://www.youtube.com/watch?v=mYHNUy5hPQE;License: Standard YouTube License, CC-BY