Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 8, Problem 8.92QP
Interpretation Introduction

Interpretation: The preferred resonance form among the given ions is to be stated on the basis of formal charges.

Concept introduction: Formal charges play an important role in choosing between the possible molecular structures. The preferred structure is the one in which formal charges are zero.

To determine: The preferred resonance form among the given ions on the basis of formal charges.

Expert Solution & Answer
Check Mark

Answer to Problem 8.92QP

Solution

The preferred resonance form in the CNO ion is resonating structure (a).

The preferred resonance form in the NCO ion is resonating structure (b).

The preferred resonance form in the CON ion is resonating structure (b).

Explanation of Solution

Explanation

The given ion is CNO . Number of valence electrons in oxygen is 6 , number of valence electrons in carbon is 4 and number of valence electrons in nitrogen is 5 . There is one nitrogen atom, one carbon atom and one oxygen atom present in CNO . There is an additional valence electron due to 1 charge on CNO .

Therefore the total valence electrons are =6+4+5+1=16

The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structures. The lewis structures for CNO are,

Chemistry, Chapter 8, Problem 8.92QP , additional homework tip  1

Figure 1

The formal charge on each atom of resonating structure (a) of CNO is calculated as,

Formal charge is calculated as,

Formalcharge=Valenceelectrons(Lonepairelectrons+12Bondpairelectrons) (1)

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 0 .

Number of bond pair electrons in nitrogen is 10 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(0+12(10))=0

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 4 .

Number of bond pair electrons in oxygen is 4 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(4+12(4))=0

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 0 .

Number of bond pair electrons in carbon is 6 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(0+12(6))=1

The formal charge on each atom of resonating structure (b) of CNO is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 0 .

Number of bond pair electrons in nitrogen is 10 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(0+12(10))=0

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 2 .

Number of bond pair electrons in oxygen is 6 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(2+12(6))=+1

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 2 .

Number of bond pair electrons in carbon is 4 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(2+12(4))=0

According to formal charge calculations, there is a formal charge of +1 on oxygen atom in the resonanting structure (b). Oxygen is more electronegative than nitrogen. Formal positive charge is not favorable on most electronegative atom. Hence the resonating structure (a) is preffered.

The given ion is NCO . Number of valence electrons in oxygen is 6 , number of valence electrons in carbon is 4 and number of valence electrons in nitrogen is 5 . There is one nitrogen atom, one carbon atom and one oxygen atom present in NCO . There is an additional valence electron due to 1 charge on NCO .

Therefore the total valence electrons are =6+4+5+1=16

Since carbon is least electronegative, it will act as central atom. The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structures. The lewis structures for NCO are,

Chemistry, Chapter 8, Problem 8.92QP , additional homework tip  2

Figure 2

The formal charge on each atom of resonating structure (a) of NCO is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 4 .

Number of bond pair electrons in nitrogen is 4 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(4+12(4))=1

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 4 .

Number of bond pair electrons in oxygen is 4 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(4+12(4))=0

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 0 .

Number of bond pair electrons in carbon is 8 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(0+12(8))=0

The formal charge on each atom of resonating structure (b) of NCO is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 2 .

Number of bond pair electrons in nitrogen is 6 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(2+12(6))=0

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 6 .

Number of bond pair electrons in oxygen is 2 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(6+12(2))=1

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 0 .

Number of bond pair electrons in carbon is 8 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(0+12(8))=0

The formal charge on each atom of resonating structure (c) of NCO is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 6 .

Number of bond pair electrons in nitrogen is 2 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(6+12(2))=2

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 2 .

Number of bond pair electrons in oxygen is 6 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(2+12(6))=+1

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 0 .

Number of bond pair electrons in carbon is 8 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(0+12(8))=0

According to formal charge calculations, there is a formal charge of 1 on oxygen atom in the resonanting structure (b). Oxygen is more electronegative than nitrogen. Formal negative charge is favorable on most electronegative atom. Hence the resonating structure (b) is preferred.

The given ion is CON . Number of valence electrons in oxygen is 6 , number of valence electrons in carbon is 4 and number of valence electrons in nitrogen is 5 . There is one nitrogen atom, one carbon atom and one oxygen atom present in CON . There is an additional valence electron due to 1 charge on CON . Therefore the total valence electrons are 6+4+5+1=16 . The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structuresThe lewis structures for CON are,

Chemistry, Chapter 8, Problem 8.92QP , additional homework tip  3

Figure 3

The formal charge on each atom of resonating structure (a) of CON is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 4 .

Number of bond pair electrons in nitrogen is 2 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(4+12(2))=0

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 2 .

Number of bond pair electrons in oxygen is 8 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(2+12(8))=0

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 2 .

Number of bond pair electrons in carbon is 6 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(2+12(6))=1

The formal charge on each atom of resonating structure (b) of CON is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 4 .

Number of bond pair electrons in nitrogen is 4 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(4+12(4))=1

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 2 .

Number of bond pair electrons in oxygen is 8 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(2+12(8))=0

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 2 .

Number of bond pair electrons in carbon is 4 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(2+12(4))=0

The formal charge on each atom of resonating structure (c) of CON is calculated as,

Number of valence electrons in nitrogen is 5 .

Number of lone pair electrons in nitrogen is 2 .

Number of bond pair electrons in nitrogen is 6 .

To calculate the formal charge on nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=5(2+12(6))=0

Number of valence electrons in oxygen is 6 .

Number of lone pair electrons in oxygen is 2 .

Number of bond pair electrons in oxygen is 8 .

To calculate the formal charge on oxygen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=6(2+12(8))=0

Number of valence electrons in carbon is 4 .

Number of lone pair electrons in carbon is 4 .

Number of bond pair electrons in carbon is 2 .

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

Formalcharge=4(4+12(2))=1

According to formal charge calculations, resonating structure (a) and (c) least electronegative atom carbon possess 1 formal charge. In resonating structure (b), more electronegative atom nitrogen carries 1 formal charge. Hence resonating structure (b) is preffered for CON ion.

Conclusion

The preferred resonance form in the CNO ion is resonating structure (a).

The preferred resonance form in the NCO ion is resonating structure (b).

The preferred resonance form in the CON ion is resonating structure (b).

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!

Chapter 8 Solutions

Chemistry

Ch. 8.7 - Prob. 11PECh. 8.8 - Prob. 12PECh. 8 - Prob. 8.1VPCh. 8 - Prob. 8.2VPCh. 8 - Prob. 8.3VPCh. 8 - Prob. 8.4VPCh. 8 - Prob. 8.5VPCh. 8 - Prob. 8.6VPCh. 8 - Prob. 8.7VPCh. 8 - Prob. 8.8VPCh. 8 - Prob. 8.9VPCh. 8 - Prob. 8.10VPCh. 8 - Prob. 8.11VPCh. 8 - Prob. 8.12VPCh. 8 - Prob. 8.13VPCh. 8 - Prob. 8.14VPCh. 8 - Prob. 8.15VPCh. 8 - Prob. 8.16VPCh. 8 - Prob. 8.17VPCh. 8 - Prob. 8.18VPCh. 8 - Prob. 8.19QPCh. 8 - Prob. 8.20QPCh. 8 - Prob. 8.21QPCh. 8 - Prob. 8.22QPCh. 8 - Prob. 8.23QPCh. 8 - Prob. 8.24QPCh. 8 - Prob. 8.25QPCh. 8 - Prob. 8.26QPCh. 8 - Prob. 8.27QPCh. 8 - Prob. 8.28QPCh. 8 - Prob. 8.29QPCh. 8 - Prob. 8.30QPCh. 8 - Prob. 8.31QPCh. 8 - Prob. 8.32QPCh. 8 - Prob. 8.33QPCh. 8 - Prob. 8.34QPCh. 8 - Prob. 8.35QPCh. 8 - Prob. 8.36QPCh. 8 - Prob. 8.37QPCh. 8 - Prob. 8.38QPCh. 8 - Prob. 8.39QPCh. 8 - Prob. 8.40QPCh. 8 - Prob. 8.41QPCh. 8 - Prob. 8.42QPCh. 8 - Prob. 8.43QPCh. 8 - Prob. 8.44QPCh. 8 - Prob. 8.45QPCh. 8 - Prob. 8.46QPCh. 8 - Prob. 8.47QPCh. 8 - Prob. 8.48QPCh. 8 - Prob. 8.49QPCh. 8 - Prob. 8.50QPCh. 8 - Prob. 8.51QPCh. 8 - Prob. 8.52QPCh. 8 - Prob. 8.53QPCh. 8 - Prob. 8.54QPCh. 8 - Prob. 8.55QPCh. 8 - Prob. 8.56QPCh. 8 - Prob. 8.57QPCh. 8 - Prob. 8.58QPCh. 8 - Prob. 8.59QPCh. 8 - Prob. 8.60QPCh. 8 - Prob. 8.61QPCh. 8 - Prob. 8.62QPCh. 8 - Prob. 8.63QPCh. 8 - Prob. 8.64QPCh. 8 - Prob. 8.65QPCh. 8 - Prob. 8.66QPCh. 8 - Prob. 8.67QPCh. 8 - Prob. 8.68QPCh. 8 - Prob. 8.69QPCh. 8 - Prob. 8.70QPCh. 8 - Prob. 8.71QPCh. 8 - Prob. 8.72QPCh. 8 - Prob. 8.73QPCh. 8 - Prob. 8.74QPCh. 8 - Prob. 8.75QPCh. 8 - Prob. 8.76QPCh. 8 - Prob. 8.77QPCh. 8 - Prob. 8.78QPCh. 8 - Prob. 8.79QPCh. 8 - Prob. 8.80QPCh. 8 - Prob. 8.81QPCh. 8 - Prob. 8.82QPCh. 8 - Prob. 8.83QPCh. 8 - Prob. 8.84QPCh. 8 - Prob. 8.85QPCh. 8 - Prob. 8.86QPCh. 8 - Prob. 8.87QPCh. 8 - Prob. 8.88QPCh. 8 - Prob. 8.89QPCh. 8 - Prob. 8.90QPCh. 8 - Prob. 8.91QPCh. 8 - Prob. 8.92QPCh. 8 - Prob. 8.93QPCh. 8 - Prob. 8.94QPCh. 8 - Prob. 8.95QPCh. 8 - Prob. 8.96QPCh. 8 - Prob. 8.97QPCh. 8 - Prob. 8.98QPCh. 8 - Prob. 8.99QPCh. 8 - Prob. 8.100QPCh. 8 - Prob. 8.101QPCh. 8 - Prob. 8.102QPCh. 8 - Prob. 8.103QPCh. 8 - Prob. 8.104QPCh. 8 - Prob. 8.105QPCh. 8 - Prob. 8.106QPCh. 8 - Prob. 8.107QPCh. 8 - Prob. 8.108QPCh. 8 - Prob. 8.109QPCh. 8 - Prob. 8.110QPCh. 8 - Prob. 8.111QPCh. 8 - Prob. 8.112QPCh. 8 - Prob. 8.113QPCh. 8 - Prob. 8.114QPCh. 8 - Prob. 8.115QPCh. 8 - Prob. 8.116QPCh. 8 - Prob. 8.117QPCh. 8 - Prob. 8.118QPCh. 8 - Prob. 8.119QPCh. 8 - Prob. 8.120QPCh. 8 - Prob. 8.121QPCh. 8 - Prob. 8.122QPCh. 8 - Prob. 8.123QPCh. 8 - Prob. 8.124QPCh. 8 - Prob. 8.125QPCh. 8 - Prob. 8.126QPCh. 8 - Prob. 8.127QPCh. 8 - Prob. 8.128QPCh. 8 - Prob. 8.129QPCh. 8 - Prob. 8.130QPCh. 8 - Prob. 8.131QPCh. 8 - Prob. 8.132QPCh. 8 - Prob. 8.133QPCh. 8 - Prob. 8.134QPCh. 8 - Prob. 8.135QPCh. 8 - Prob. 8.136QPCh. 8 - Prob. 8.137QPCh. 8 - Prob. 8.138QPCh. 8 - Prob. 8.139APCh. 8 - Prob. 8.140APCh. 8 - Prob. 8.141APCh. 8 - Prob. 8.142APCh. 8 - Prob. 8.143APCh. 8 - Prob. 8.144APCh. 8 - Prob. 8.145APCh. 8 - Prob. 8.146APCh. 8 - Prob. 8.147APCh. 8 - Prob. 8.148APCh. 8 - Prob. 8.149APCh. 8 - Prob. 8.150APCh. 8 - Prob. 8.151APCh. 8 - Prob. 8.152APCh. 8 - Prob. 8.153APCh. 8 - Prob. 8.154APCh. 8 - Prob. 8.155APCh. 8 - Prob. 8.156APCh. 8 - Prob. 8.157APCh. 8 - Prob. 8.158APCh. 8 - Prob. 8.159APCh. 8 - Prob. 8.160APCh. 8 - Prob. 8.161APCh. 8 - Prob. 8.162APCh. 8 - Prob. 8.163APCh. 8 - Prob. 8.164APCh. 8 - Prob. 8.165APCh. 8 - Prob. 8.166APCh. 8 - Prob. 8.167APCh. 8 - Prob. 8.168APCh. 8 - Prob. 8.169APCh. 8 - Prob. 8.170APCh. 8 - Prob. 8.171APCh. 8 - Prob. 8.172APCh. 8 - Prob. 8.173APCh. 8 - Prob. 8.174APCh. 8 - Prob. 8.175APCh. 8 - Prob. 8.176AP
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
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