CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM
7th Edition
ISBN: 9781319420994
Author: ATKINS
Publisher: MAC HIGHER
bartleby

Concept explainers

Formal Charges
Formal charges have an important role in organic chemistry since this concept helps us to know whether an atom in a molecule is neutral/bears a positive or negative charge. Even if some molecules are neutral, the atoms within that molecule need not be ne…
Polarity Of Water
In simple chemical terms, polarity refers to the separation of charges in a chemical species leading into formation of two polar ends which are positively charged end and negatively charged end. Polarity in any molecule occurs due to the differences in t…
Valence Bond Theory Vbt
Valence bond theory (VBT) in simple terms explains how individual atomic orbitals with an unpaired electron each, come close to each other and overlap to form a molecular orbital giving a covalent bond. It gives a quantum mechanical approach to the forma…
bartleby

Videos

Question
Book Icon
Chapter 2, Problem 2B.23E

(a)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 1

Concept Introduction:

The formal charge on each atom in the Lewis structure can be calculated from the equation written as follows:

    Formal charge=[V−L−12B]

Here,

V denotes valence electrons in free atom.

L denotes electrons present as lone pairs.

B denotes electrons present as bond pairs.

Lowest energy structure is the one that has zero or nearly zero formal charge.

(a)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

  Formal charge=[V−L−12B]        (1)

For first structure:

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

  Formal charge(O)=6−4−12(4)=0

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on O attached to H.

    Formal charge(O)=6−4−12(4)=0

Substitute 7for V, 2 for L and 10 for B in equation (1).

    Formal charge(Cl)=7−2−12(10)=0

Substitute 1 for V, 0 for L and 2 for B in equation (1).

    Formal charge(H)=1−0−12(2)=0

For first structure:

Similarly for other molecule substitute 6 for V, 6 for L and 2 for B in equation (1) to O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on O attached to H.

    Formal charge(O)=6−4−12(4)=0

Substitute 7 for V, 2 for L and 4 for B in equation (1).

    Formal charge(Cl)=7−2−12(6)=+2

Substitute 1 for V, 0 for L and 2 for B in equation (1).

    Formal charge(H)=1−0−12(2)=0

Hence formal charge in the right structure is illustrated below:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 2

Since the former hasformal charges close to zero for maximum number of atoms I is lower in energy than II.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 3

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    Formal charge=[V−L−12B]        (1)

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded O.

    Formal charge(O)=6−4−12(4)=0

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on S.

    Formal charge(S)=6−4−12(4)=0

Substitute 4for V, 0 for L and 8 for B in equation (1).

    Formal charge(C)=4−0−12(8)=0

Similarly for other Lewis structure, substitute 6 for V, 6 for L and 2 for B in equation (1) to O.

    Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on S.

    Formal charge(S)=6−2−12(6)=+1

Substitute 4 for V, 0 for L and 8 for B in equation (1).

    Formal charge(C)=4−0−12(8)=0

Hence formal charge in the two structures is illustrated below:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 4

Since the former has all formal charges equal to zero I is lower in energy than II hence I represents structure of lower energy.

(b)

Interpretation Introduction

Interpretation:

The formal charge on each atom has to be determined and structure of lower energy has to be identified.

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 5

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The formal charge on each atom in the Lewis structure can be calculated from the equation as follows:

    Formal charge=[V−L−12B]        (1)

Substitute 1 for V, 0 for L and 2 for B in equation (1).

    Formal charge(H)=1−0−12(2)=0

Substitute 4 for V, 0 for L and 8 for B in equation (1).

    Formal charge(C)=4−0−12(8)=0

Substitute 5for V, 2 for L and 6  for B in equation (1).

    Formal charge(N)=5−2−12(6)=0

Similarly for other Lewis structure, substitute 1 for V, 0 for L and 2 for B in equation (1).

    Formal charge(H)=1−0−12(2)=0

Substitute 4 for V, 2 for L and 6 for B in equation (1).

    Formal charge(C)=4−2−12(6)=−1

Substitute 5 for V, 2 for L and 4  for B in equation (1).

    Formal charge(N)=5−2−12(4)=+1

Hence formal charge in the two structures is illustrated below:

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM, Chapter 2, Problem 2B.23E , additional homework tip 6

Since the former has all formal charges equal to zero I is lower in energy than II hence I represents structure of lower energy.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Previous
Chapter 2, Problem 2B.22E
Next
Chapter 2, Problem 2B.24E
Students have asked these similar questions
What would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 3 2. n-BuLi • Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Click and drag to start drawing a structure.
Identify the missing organic reactants in the following reaction: X + Y H+ two steps Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H2O) are not shown. In the drawing area below, draw the skeletal ("line") structures of the missing organic reactants X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. Х :
Draw the mechanism of friedel-crafts acylation using acetyl chloride of m-Xylene

Chapter 2 Solutions

CHEM PRINCIPLES LL W/ACHIEVE ONE-SEM

Ch. 2 - Prob. 2A.1ASTCh. 2 - Prob. 2A.1BSTCh. 2 - Prob. 2A.2ASTCh. 2 - Prob. 2A.2BSTCh. 2 - Prob. 2A.3ASTCh. 2 - Prob. 2A.3BSTCh. 2 - Prob. 2A.4ASTCh. 2 - Prob. 2A.4BSTCh. 2 - Prob. 2A.1ECh. 2 - Prob. 2A.2E
Ch. 2 - Prob. 2A.3ECh. 2 - Prob. 2A.4ECh. 2 - Prob. 2A.5ECh. 2 - Prob. 2A.6ECh. 2 - Prob. 2A.7ECh. 2 - Prob. 2A.8ECh. 2 - Prob. 2A.9ECh. 2 - Prob. 2A.10ECh. 2 - Prob. 2A.11ECh. 2 - Prob. 2A.12ECh. 2 - Prob. 2A.13ECh. 2 - Prob. 2A.14ECh. 2 - Prob. 2A.15ECh. 2 - Prob. 2A.16ECh. 2 - Prob. 2A.17ECh. 2 - Prob. 2A.18ECh. 2 - Prob. 2A.19ECh. 2 - Prob. 2A.20ECh. 2 - Prob. 2A.21ECh. 2 - Prob. 2A.22ECh. 2 - Prob. 2A.23ECh. 2 - Prob. 2A.24ECh. 2 - Prob. 2A.25ECh. 2 - Prob. 2A.26ECh. 2 - Prob. 2A.27ECh. 2 - Prob. 2A.28ECh. 2 - Prob. 2A.29ECh. 2 - Prob. 2A.30ECh. 2 - Prob. 2B.1ASTCh. 2 - Prob. 2B.1BSTCh. 2 - Prob. 2B.2ASTCh. 2 - Prob. 2B.2BSTCh. 2 - Prob. 2B.3ASTCh. 2 - Prob. 2B.3BSTCh. 2 - Prob. 2B.4ASTCh. 2 - Prob. 2B.4BSTCh. 2 - Prob. 2B.5ASTCh. 2 - Prob. 2B.5BSTCh. 2 - Prob. 2B.1ECh. 2 - Prob. 2B.2ECh. 2 - Prob. 2B.3ECh. 2 - Prob. 2B.4ECh. 2 - Prob. 2B.5ECh. 2 - Prob. 2B.6ECh. 2 - Prob. 2B.7ECh. 2 - Prob. 2B.8ECh. 2 - Prob. 2B.9ECh. 2 - Prob. 2B.10ECh. 2 - Prob. 2B.11ECh. 2 - Prob. 2B.12ECh. 2 - Prob. 2B.13ECh. 2 - Prob. 2B.14ECh. 2 - Prob. 2B.15ECh. 2 - Prob. 2B.16ECh. 2 - Prob. 2B.17ECh. 2 - Prob. 2B.18ECh. 2 - Prob. 2B.19ECh. 2 - Prob. 2B.20ECh. 2 - Prob. 2B.21ECh. 2 - Prob. 2B.22ECh. 2 - Prob. 2B.23ECh. 2 - Prob. 2B.24ECh. 2 - Prob. 2C.1ASTCh. 2 - Prob. 2C.1BSTCh. 2 - Prob. 2C.2ASTCh. 2 - Prob. 2C.2BSTCh. 2 - Prob. 2C.3ASTCh. 2 - Prob. 2C.3BSTCh. 2 - Prob. 2C.1ECh. 2 - Prob. 2C.2ECh. 2 - Prob. 2C.3ECh. 2 - Prob. 2C.4ECh. 2 - Prob. 2C.5ECh. 2 - Prob. 2C.6ECh. 2 - Prob. 2C.7ECh. 2 - Prob. 2C.8ECh. 2 - Prob. 2C.9ECh. 2 - Prob. 2C.10ECh. 2 - Prob. 2C.11ECh. 2 - Prob. 2C.12ECh. 2 - Prob. 2C.13ECh. 2 - Prob. 2C.14ECh. 2 - Prob. 2C.15ECh. 2 - Prob. 2C.16ECh. 2 - Prob. 2C.17ECh. 2 - Prob. 2C.18ECh. 2 - Prob. 2D.1ASTCh. 2 - Prob. 2D.1BSTCh. 2 - Prob. 2D.2ASTCh. 2 - Prob. 2D.2BSTCh. 2 - Prob. 2D.1ECh. 2 - Prob. 2D.2ECh. 2 - Prob. 2D.3ECh. 2 - Prob. 2D.4ECh. 2 - Prob. 2D.5ECh. 2 - Prob. 2D.6ECh. 2 - Prob. 2D.7ECh. 2 - Prob. 2D.8ECh. 2 - Prob. 2D.9ECh. 2 - Prob. 2D.10ECh. 2 - Prob. 2D.11ECh. 2 - Prob. 2D.12ECh. 2 - Prob. 2D.13ECh. 2 - Prob. 2D.14ECh. 2 - Prob. 2D.15ECh. 2 - Prob. 2D.16ECh. 2 - Prob. 2D.17ECh. 2 - Prob. 2D.18ECh. 2 - Prob. 2D.19ECh. 2 - Prob. 2D.20ECh. 2 - Prob. 2E.1ASTCh. 2 - Prob. 2E.1BSTCh. 2 - Prob. 2E.2ASTCh. 2 - Prob. 2E.2BSTCh. 2 - Prob. 2E.3ASTCh. 2 - Prob. 2E.3BSTCh. 2 - Prob. 2E.4ASTCh. 2 - Prob. 2E.4BSTCh. 2 - Prob. 2E.5ASTCh. 2 - Prob. 2E.5BSTCh. 2 - Prob. 2E.1ECh. 2 - Prob. 2E.2ECh. 2 - Prob. 2E.3ECh. 2 - Prob. 2E.4ECh. 2 - Prob. 2E.5ECh. 2 - Prob. 2E.6ECh. 2 - Prob. 2E.7ECh. 2 - Prob. 2E.8ECh. 2 - Prob. 2E.9ECh. 2 - Prob. 2E.10ECh. 2 - Prob. 2E.11ECh. 2 - Prob. 2E.12ECh. 2 - Prob. 2E.13ECh. 2 - Prob. 2E.14ECh. 2 - Prob. 2E.15ECh. 2 - Prob. 2E.16ECh. 2 - Prob. 2E.17ECh. 2 - Prob. 2E.18ECh. 2 - Prob. 2E.19ECh. 2 - Prob. 2E.20ECh. 2 - Prob. 2E.21ECh. 2 - Prob. 2E.22ECh. 2 - Prob. 2E.23ECh. 2 - Prob. 2E.24ECh. 2 - Prob. 2E.25ECh. 2 - Prob. 2E.26ECh. 2 - Prob. 2E.27ECh. 2 - Prob. 2E.28ECh. 2 - Prob. 2E.29ECh. 2 - Prob. 2E.30ECh. 2 - Prob. 2F.1ASTCh. 2 - Prob. 2F.1BSTCh. 2 - Prob. 2F.2ASTCh. 2 - Prob. 2F.2BSTCh. 2 - Prob. 2F.3ASTCh. 2 - Prob. 2F.3BSTCh. 2 - Prob. 2F.4ASTCh. 2 - Prob. 2F.4BSTCh. 2 - Prob. 2F.1ECh. 2 - Prob. 2F.2ECh. 2 - Prob. 2F.3ECh. 2 - Prob. 2F.4ECh. 2 - Prob. 2F.5ECh. 2 - Prob. 2F.6ECh. 2 - Prob. 2F.7ECh. 2 - Prob. 2F.8ECh. 2 - Prob. 2F.9ECh. 2 - Prob. 2F.10ECh. 2 - Prob. 2F.11ECh. 2 - Prob. 2F.12ECh. 2 - Prob. 2F.13ECh. 2 - Prob. 2F.14ECh. 2 - Prob. 2F.15ECh. 2 - Prob. 2F.16ECh. 2 - Prob. 2F.17ECh. 2 - Prob. 2F.18ECh. 2 - Prob. 2F.19ECh. 2 - Prob. 2F.20ECh. 2 - Prob. 2F.21ECh. 2 - Prob. 2G.1ASTCh. 2 - Prob. 2G.1BSTCh. 2 - Prob. 2G.2ASTCh. 2 - Prob. 2G.2BSTCh. 2 - Prob. 2G.1ECh. 2 - Prob. 2G.2ECh. 2 - Prob. 2G.3ECh. 2 - Prob. 2G.4ECh. 2 - Prob. 2G.5ECh. 2 - Prob. 2G.6ECh. 2 - Prob. 2G.7ECh. 2 - Prob. 2G.8ECh. 2 - Prob. 2G.9ECh. 2 - Prob. 2G.11ECh. 2 - Prob. 2G.12ECh. 2 - Prob. 2G.13ECh. 2 - Prob. 2G.14ECh. 2 - Prob. 2G.15ECh. 2 - Prob. 2G.16ECh. 2 - Prob. 2G.17ECh. 2 - Prob. 2G.18ECh. 2 - Prob. 2G.19ECh. 2 - Prob. 2G.20ECh. 2 - Prob. 2G.21ECh. 2 - Prob. 2G.22ECh. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 2.4ECh. 2 - Prob. 2.5ECh. 2 - Prob. 2.6ECh. 2 - Prob. 2.7ECh. 2 - Prob. 2.8ECh. 2 - Prob. 2.9ECh. 2 - Prob. 2.10ECh. 2 - Prob. 2.11ECh. 2 - Prob. 2.12ECh. 2 - Prob. 2.13ECh. 2 - Prob. 2.14ECh. 2 - Prob. 2.17ECh. 2 - Prob. 2.19ECh. 2 - Prob. 2.22ECh. 2 - Prob. 2.23ECh. 2 - Prob. 2.24ECh. 2 - Prob. 2.25ECh. 2 - Prob. 2.26ECh. 2 - Prob. 2.27ECh. 2 - Prob. 2.28ECh. 2 - Prob. 2.29ECh. 2 - Prob. 2.30ECh. 2 - Prob. 2.31ECh. 2 - Prob. 2.32ECh. 2 - Prob. 2.33ECh. 2 - Prob. 2.34ECh. 2 - Prob. 2.35ECh. 2 - Prob. 2.36ECh. 2 - Prob. 2.37ECh. 2 - Prob. 2.39ECh. 2 - Prob. 2.40ECh. 2 - Prob. 2.41ECh. 2 - Prob. 2.42ECh. 2 - Prob. 2.43ECh. 2 - Prob. 2.44ECh. 2 - Prob. 2.45ECh. 2 - Prob. 2.46ECh. 2 - Prob. 2.47ECh. 2 - Prob. 2.48ECh. 2 - Prob. 2.49ECh. 2 - Prob. 2.50ECh. 2 - Prob. 2.51ECh. 2 - Prob. 2.52ECh. 2 - Prob. 2.53ECh. 2 - Prob. 2.54ECh. 2 - Prob. 2.55ECh. 2 - Prob. 2.56ECh. 2 - Prob. 2.57ECh. 2 - Prob. 2.58ECh. 2 - Prob. 2.59ECh. 2 - Prob. 2.60ECh. 2 - Prob. 2.61ECh. 2 - Prob. 2.62ECh. 2 - Prob. 2.63ECh. 2 - Prob. 2.64E
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.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Text book image
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Pushing Electrons
Chemistry
ISBN:9781133951889
Author:Weeks, Daniel P.
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
SEE MORE TEXTBOOKS
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
General Chemistry 1A. Lecture 12. Two Theories of Bonding.; Author: UCI Open;https://www.youtube.com/watch?v=dLTlL9Z1bh0;License: CC-BY