ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM
7th Edition
ISBN: 9781319403959
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 2.13E

(a)

Interpretation Introduction

Interpretation:

The most important Lewis structure of HONCO along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Lewis structures represent covalent bonds and describe valence electrons configuration of atoms. The covalent bonds are depicted by lines, and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

  • The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
  • Total valence electrons is estimated.
  • single bond is first placed between each atom pair.
  • The electrons left can be allocated as unshared electron pairs or as multiple bonds around the right symbol of the element to satisfy the octet (or duplet) for each atom.
  • Add charge on the overall structure in case of polyatomic cation or anion.

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.

(a)

Expert Solution
Check Mark

Explanation of Solution

Lewis structure possible for HONCO is illustrated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 1

Since the double and triple bonds can conjugate therefore delocalization occurs that results in various equivalent resonance structures as indicated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 2

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

  Formal charge=[V−L−12B]        (1)

Substitute 5 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on doubly bonded nitrogen in structure I.

  Formal charge(N)=5−2−12(6)=0

Substitute 5 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on triply bonded nitrogen in structure II.

  Formal charge(N)=5−0−12(8)=+1

Substitute 4 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on central doubly- bonded carbon in structure I.

  Formal charge(C)=4−0−12(4)=0

Substitute 5 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on singly bonded nitrogen in structure II.

  Formal charge(N)=5−2−12(6)=0

Substitute 4 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on carbon in structure III.

  Formal charge(C)=4−2−12(6)=−1

Substitute 6 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on terminal doubly- bonded oxygen.

  Formal charge(O)=6−4−12(4)=0

Substitute 6 for V, 6 for L and 2 for B in equation (1) to calculate formal charge on terminal oxygen in structure II.

  Formal charge(O)=6−6−12(2)=−1

Substitute 6 for V, 6 for L and 2 for B in equation (1) to calculate formal charge on central oxygen in structure II.

  Formal charge(O)=6−2−12(6)=+1

Substitute 1 for V, 0 for L and 2 for B in equation (1) to calculate formal charge on terminal hydrogen.

  Formal charge(H)=1−0−12(2)=0

Therefore the non-zero formal charges can be assigned as follows:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 3

(b)

Interpretation Introduction

Interpretation:

The most important Lewis structure of H2CSO along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

Lewis structure possible for H2CSO is illustrated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 4

Since the double and triple bonds can conjugate therefore delocalization occurs that results in various equivalent resonance structures as indicated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 5

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

  Formal charge=[V−L−12B]        (1)

Substitute 6 for V, 2 for L and 8 for B in equation (1) to calculate formal charge on sulphur in I.

  Formal charge(S)=6−2−12(8)=0

Substitute 6 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on sulphur in II.

  Formal charge(S)=6−2−12(6)=+1

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

  Formal charge(O)=6−4−12(4)=0

Substitute 6 for V, 6 for L and 2 for B in equation (1) to calculate formal charge on oxygen in II.

  Formal charge(O)=6−6−12(2)=−1

Substitute 1 for V, 0 for L and 2 for B in equation (1) to calculate formal charge on terminal hydrogen.

  Formal charge(H)=1−0−12(2)=0

Therefore the formal charges can be assigned as follows:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 6

(c)

Interpretation Introduction

Interpretation:

The most important Lewis structure of H2CNN along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

Lewis structure for ONCN is illustrated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 7

Since the double and triple bonds can conjugate therefore delocalization occurs that results in two equivalent resonance structures as indicated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 8

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

  Formal charge=[V−L−12B]        (1)

Substitute 5 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on doubly bonded central nitrogen in structure I.

  Formal charge(N)=5−0−12(8)=+1

Substitute 5 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded terminal nitrogen in structure I.

  Formal charge(N)=5−4−12(4)=−1

Substitute 5 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on triply bonded central nitrogen in structure II.

  Formal charge(N)=5−0−12(8)=+1

Substitute 5 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on triply bonded terminal nitrogen in structure II.

  Formal charge(N)=5−2−12(6)=0

Substitute 4 for V, 2 for L and 6 for B in equation (1) to calculate formal charge on carbon in structure II.

Formal charge(C)=4−2−12(6)=−1

Substitute 1 for V, 0 for L and 2 for B in equation (1) to calculate formal charge on terminal hydrogen.

  Formal charge(H)=1−0−12(2)=0

Therefore the formal charges in H2CNN can be assigned as follows:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 9

(d)

Interpretation Introduction

Interpretation:

The most important Lewis structure of ONCN along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

Lewis structure for ONCN is illustrated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 10

Since the double and triple bonds can conjugate therefore delocalization occurs that results in two equivalent resonance structures as indicated below:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 11

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

  Formal charge=[V−L−12B]        (1)

Substitute 5 for V, 4 for L and 4 for B in equation (1) to calculate formal charge on doubly bonded terminal nitrogen in structure II.

  Formal charge(N)=5−4−12(4)=−1

Substitute 5 for V, 0 for L and 8 for B in equation (1) to calculate formal charge on doubly bonded non-terminal nitrogen in structure II.

  Formal charge(N)=5−0−12(8)=+1

Therefore the non-zero formal charges in ONCN can be assigned as follows:

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM, Chapter 2, Problem 2.13E , additional homework tip 12

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 2.12E
Next
Chapter 2, Problem 2.14E
Students have asked these similar questions
What is the pH at the cathode for the following cell written in line notation at 25.0 oC with a Ecell = -0.2749 V? Ni(s)|Ni+2(aq, 1.00 M)||H+1(aq, ?M)|H2(g, 1.00 atm)|Pt(s)
Calculate Ecell for a hydrogen fuel cell at 95.0 oC using the following half-reactions with PH2 = 25.0 atm and PO2 = 25.0 atm. O2(g) + 4H+1(aq) + 4e-1 → 2H2O(l)  Eo = 1.229 V 2H2(g) → 4H+1(aq) + 4e-1   Eo = 0.00 V
Calculate Ecell at 25.0 oC using the following half-reactions with [Ag+1] = 0.0100 M and [Sn+2] = 0.0200 M. Ag+1(aq) + 1e-1 Ag(s) Sn+2(aq) + 2e-1 Sn(s)

Chapter 2 Solutions

ACHIEVE/CHEMICAL PRINCIPLES ACCESS 2TERM

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
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
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
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