EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
EBK ORGANIC CHEMISTRY: PRINCIPLES AND M
2nd Edition
ISBN: 9780393630817
Author: KARTY
Publisher: W.W.NORTON+CO. (CC)
bartleby

Concept explainers

Question
Book Icon
Chapter 1, Problem 1.24P
Interpretation Introduction

(a)

Interpretation:

For the given species, the complete Lewis structure is to be completed by adding multiple bonds and/or lone pairs.

Concept introduction:

In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms to bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have a maximum of four bonds; Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bonds in case of halogen is one; while there will be three lone pair of electrons on halide atoms.

Expert Solution
Check Mark

Answer to Problem 1.24P

The complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  1

Explanation of Solution

The given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  2

The formula for the species above is C2H5NO. The total number of valence electrons for the structure is 24. The arrangement of atoms is given which attributes to the distribution of 18 valence electrons. Thus, there remain 6 electrons which need to be distributed.

The carbon atom on the left side has four bonds, thus, its octet is complete. The carbon atom in the middle has four bonds, hence, its octet is also complete. The nitrogen atom has three bonds, thus, its octet is not complete. There should be one lone pair of electron on nitrogen. The double bonded oxygen atom has got two bonds. Thus, in order to complete its octet, it should possess two lone pair of electrons. Thus, all the 24 electrons have been distributed. Thus, the complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  3

Conclusion

The complete Lewis structure for the given species including multiple bonds and lone pairs is shown in Figure 1 above.

Interpretation Introduction

(b)

Interpretation:

For given species, the complete Lewis structure is to be completed by adding multiple bonds and/or lone pairs.

Concept introduction:

In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. For a charged species, each negative charge increases the number of valence electrons by one while each positive charge decreases the number of valence electrons by one. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms to bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have maximum of four bonds. Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bond in case of halogen is one, while there will be three lone pair of electrons on halide atoms.

Expert Solution
Check Mark

Answer to Problem 1.24P

The complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  4

Explanation of Solution

The given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  5

The formula for the species above is C2H6N. The total number of valence electrons for the structure is 19. One negative charge adds to the total valence electron, count to make it 20. The arrangement of atoms is given which attributes to distribution of 16 valence electrons. Thus, there remain 4 electrons which need to be distributed.

The carbon atom on the left as well as on the right has four bonds, thus, their octets are complete. The nitrogen atom has two bonds and a negative formal charge. This suggests that the remaining four electrons should be present on the nitrogen atom so as to complete its octet and have a negative formal charge. Thus, the complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  6

Conclusion

The complete Lewis structure for the given species including multiple bonds and lone pairs is shown in Figure 2 above.

Interpretation Introduction

(c)

Interpretation:

For given species, the complete Lewis structure is to be completed by adding multiple bonds and/or lone pairs.

Concept introduction:

In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. For a charged species, each negative charge increase the number of valence electrons by one while each positive charge decrease the number of valence electrons by one. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms to bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have maximum of four bonds. Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bond in case of halogen is one, while there will be three lone pair of electrons on halide atoms.

Expert Solution
Check Mark

Answer to Problem 1.24P

The complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  7

Explanation of Solution

The given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  8

The formula for the species above is CH6N+. The total number of valence electrons for the structure is 15. One positive charge decreases the total valence electron count to make it 14. The arrangement of atoms is given which attributes to distribution of 14 valence electrons. Thus, no electron remains, and all the electrons have been distributed. Thus, the given structure is a complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  9

Conclusion

The complete Lewis structure for the given species including multiple bonds and lone pairs is shown in Figure 2 above.

Interpretation Introduction

(d)

Interpretation:

For given species, the complete Lewis structure is to be completed by adding multiple bonds and/or lone pairs.

Concept introduction:

In order to draw a Lewis structure for a molecule, start by counting the total number of valence electrons in a molecule. The number of valence electrons by each atom is the same as its group number. For a charged species, each negative charge increase the number of valence electrons by one while each positive charge decrease the number of valence electrons by one. For the given skeleton of the molecule, distribute the remaining electrons as lone pairs. In doing so, start with the outer atoms and work inwards. Try to achieve an octet on each atom other than hydrogen. If there is an atom with less than an octet, increase the atom’s share of electrons by converting lone pairs from neighboring atoms into bonding pairs thereby creating double or triple bonds. For an uncharged atom, carbon atoms will have maximum of four bonds. Nitrogen will have three bonds and one lone pair, while oxygen will have two bonds and two lone pairs. Hydrogen always contributes to one bond. The number of bond in case of halogen is one, while there will be three lone pair of electrons on halide atoms.

Expert Solution
Check Mark

Answer to Problem 1.24P

The complete Lewis structure for the given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  10

Explanation of Solution

The given species is:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  11

The formula for the species above is C2H5+. The total number of valence electrons for the structure is 13. One positive charge decreases the total valence electron, count to make it 12. The arrangement of atoms is given which attributes to the distribution of 12 valence electrons. Thus, no electron remains, and all the electrons have been distributed. Thus, the given structure is a complete Lewis structure for the given species:

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M, Chapter 1, Problem 1.24P , additional homework tip  12

Conclusion

The complete Lewis structure for the given species including multiple bonds and lone pairs is shown in Figure 4 above.

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!
Students have asked these similar questions
Compute the formal charge (FC) on each atom in the following structures.(a) Methane (CH4)
16. Which of the following structures is the CORRECT resonance structure of the following. molecule: (A) (B) (C) (D) CH3- CH₂ CH3 -H CH₂CH3 CH₂CH-CH₂ CH3 CH3
Compute the formal charge (FC) on each atom in the following structures. (b) The hydronium ion, H3O+

Chapter 1 Solutions

EBK ORGANIC CHEMISTRY: PRINCIPLES AND M

Ch. 1 - Prob. 1.11PCh. 1 - Prob. 1.12PCh. 1 - Prob. 1.13PCh. 1 - Prob. 1.14PCh. 1 - Prob. 1.15PCh. 1 - Prob. 1.16PCh. 1 - Prob. 1.17PCh. 1 - Prob. 1.18PCh. 1 - Prob. 1.19PCh. 1 - Prob. 1.20PCh. 1 - Prob. 1.21PCh. 1 - Prob. 1.22PCh. 1 - Prob. 1.23PCh. 1 - Prob. 1.24PCh. 1 - Prob. 1.25PCh. 1 - Prob. 1.26PCh. 1 - Prob. 1.27PCh. 1 - Prob. 1.28PCh. 1 - Prob. 1.29PCh. 1 - Prob. 1.30PCh. 1 - Prob. 1.31PCh. 1 - Prob. 1.32PCh. 1 - Prob. 1.33PCh. 1 - Prob. 1.34PCh. 1 - Prob. 1.35PCh. 1 - Prob. 1.36PCh. 1 - Prob. 1.37PCh. 1 - Prob. 1.38PCh. 1 - Prob. 1.39PCh. 1 - Prob. 1.40PCh. 1 - Prob. 1.41PCh. 1 - Prob. 1.42PCh. 1 - Prob. 1.43PCh. 1 - Prob. 1.44PCh. 1 - Prob. 1.45PCh. 1 - Prob. 1.46PCh. 1 - Prob. 1.47PCh. 1 - Prob. 1.48PCh. 1 - Prob. 1.49PCh. 1 - Prob. 1.50PCh. 1 - Prob. 1.51PCh. 1 - Prob. 1.52PCh. 1 - Prob. 1.53PCh. 1 - Prob. 1.54PCh. 1 - Prob. 1.55PCh. 1 - Prob. 1.56PCh. 1 - Prob. 1.57PCh. 1 - Prob. 1.58PCh. 1 - Prob. 1.59PCh. 1 - Prob. 1.60PCh. 1 - Prob. 1.61PCh. 1 - Prob. 1.62PCh. 1 - Prob. 1.63PCh. 1 - Prob. 1.64PCh. 1 - Prob. 1.65PCh. 1 - Prob. 1.66PCh. 1 - Prob. 1.67PCh. 1 - Prob. 1.68PCh. 1 - Prob. 1.69PCh. 1 - Prob. 1.70PCh. 1 - Prob. 1.71PCh. 1 - Prob. 1.72PCh. 1 - Prob. 1.73PCh. 1 - Prob. 1.74PCh. 1 - Prob. 1.75PCh. 1 - Prob. 1.76PCh. 1 - Prob. 1.77PCh. 1 - Prob. 1.78PCh. 1 - Prob. 1.79PCh. 1 - Prob. 1.80PCh. 1 - Prob. 1.81PCh. 1 - Prob. 1.82PCh. 1 - Prob. 1.1YTCh. 1 - Prob. 1.2YTCh. 1 - Prob. 1.3YTCh. 1 - Prob. 1.4YTCh. 1 - Prob. 1.5YTCh. 1 - Prob. 1.6YTCh. 1 - Prob. 1.7YTCh. 1 - Prob. 1.8YTCh. 1 - Prob. 1.9YTCh. 1 - Prob. 1.10YTCh. 1 - Prob. 1.11YTCh. 1 - Prob. 1.12YTCh. 1 - Prob. 1.13YTCh. 1 - Prob. 1.14YTCh. 1 - Prob. 1.15YTCh. 1 - Prob. 1.16YTCh. 1 - Prob. 1.17YT
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
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
Text book image
Organic Chemistry: A Guided Inquiry
Chemistry
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Cengage Learning
Text book image
Pushing Electrons
Chemistry
ISBN:9781133951889
Author:Weeks, Daniel P.
Publisher:Cengage Learning