Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
Question
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Chapter 12, Problem 53E
Interpretation Introduction

(a)

Interpretation:

The labeling of each atom in the polar covalent bond CH is to be done using delta notation (δ+ and δ).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Expert Solution
Check Mark

Answer to Problem 53E

The polar CH bond using delta notation (δ+ and δ).is shown below.

δδ+CH

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure 12.9 the electronegativity value of C is 2.5 and that of H is 2.1. As carbon is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on carbon and a corresponding positive charge on hydrogen. The polar CH bond using delta notation is shown below.

δδ+CH

Conclusion

The polar CH bond using delta notation (δ+ and δ) is shown below.

δδ+CH

Interpretation Introduction

(b)

Interpretation:

The labeling of each atom in the polar covalent bond SeO is to be done using delta notation (δ+ and δ).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Expert Solution
Check Mark

Answer to Problem 53E

The polar SeO bond using delta notation is shown below.

δδ+OSe

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure 12.9 the electronegativity value of Se is 2.4 and that of O is 3.5. As oxygen is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on oxygen and a corresponding positive charge on selenium. The polar SeO bond using delta notation is shown below.

δδ+OSe

Conclusion

The polar SeO bond using delta notation is shown below.

δδ+OSe

Interpretation Introduction

(c)

Interpretation:

The labeling of each atom in the polar covalent bond PI is to be done using delta notation (δ+ and δ).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Expert Solution
Check Mark

Answer to Problem 53E

The polar PI bond using delta notation is shown below.

δδ+IP

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure 12.9 the electronegativity value of P is 2.1 and that of I is 2.5. As iodine is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on iodine and a corresponding positive charge on phosphorous. The polar PI bond using delta notation is shown below.

δδ+IP

Conclusion

The polar PI bond using delta notation is shown below.

δδ+IP

Interpretation Introduction

(d)

Interpretation:

The labeling of each atom in the polar covalent bond HBr is to be done using delta notation (δ+ and δ).

Concept introduction:

Electronegativity is defined as the tendency of an atom to attract electrons towards it. Polarized bonds are a result of the electronegativity difference between bonding atoms. The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires partial positive charge while showing charge distribution.

Expert Solution
Check Mark

Answer to Problem 53E

The polar HBr bond using delta notation is shown below.

δδ+BrH

Explanation of Solution

The electronegativity of an element increases along the period from left to right and decreases down in the group. From the figure 12.9 the electronegativity value of Br is 2.8 and that of H is 2.1. As bromine is the more electronegative element, the bonding electron pair is shifted towards it and a slightly negative charge is generated on bromine and a corresponding positive charge on hydrogen. The polar HBr bond using delta notation is shown below.

δδ+BrH

Conclusion

The polar HBr bond using delta notation is shown below.

δδ+BrH

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Chapter 12 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 12 - Prob. 11CECh. 12 - Prob. 12CECh. 12 - Prob. 1KTCh. 12 - Prob. 2KTCh. 12 - Prob. 3KTCh. 12 - Prob. 4KTCh. 12 - Prob. 5KTCh. 12 - Prob. 6KTCh. 12 - Prob. 7KTCh. 12 - Prob. 8KTCh. 12 - Prob. 9KTCh. 12 - Prob. 10KTCh. 12 - Prob. 11KTCh. 12 - Prob. 12KTCh. 12 - Prob. 13KTCh. 12 - Prob. 14KTCh. 12 - Prob. 15KTCh. 12 - Prob. 16KTCh. 12 - Prob. 17KTCh. 12 - Prob. 18KTCh. 12 - Prob. 19KTCh. 12 - Prob. 20KTCh. 12 - Prob. 21KTCh. 12 - Prob. 22KTCh. 12 - Prob. 23KTCh. 12 - Prob. 24KTCh. 12 - Prob. 25KTCh. 12 - Prob. 26KTCh. 12 - Prob. 27KTCh. 12 - Prob. 28KTCh. 12 - Prob. 29KTCh. 12 - Prob. 1ECh. 12 - Prob. 2ECh. 12 - Prob. 3ECh. 12 - Prob. 4ECh. 12 - Prob. 5ECh. 12 - Prob. 6ECh. 12 - Prob. 7ECh. 12 - Prob. 8ECh. 12 - Prob. 9ECh. 12 - Prob. 10ECh. 12 - Prob. 11ECh. 12 - Prob. 12ECh. 12 - Prob. 13ECh. 12 - Prob. 14ECh. 12 - Prob. 15ECh. 12 - Prob. 16ECh. 12 - Prob. 17ECh. 12 - Prob. 18ECh. 12 - Prob. 19ECh. 12 - Prob. 20ECh. 12 - Prob. 21ECh. 12 - Prob. 22ECh. 12 - Prob. 23ECh. 12 - Prob. 24ECh. 12 - Prob. 25ECh. 12 - Prob. 26ECh. 12 - Prob. 27ECh. 12 - Prob. 28ECh. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Prob. 34ECh. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - Prob. 37ECh. 12 - Prob. 38ECh. 12 - Prob. 39ECh. 12 - Prob. 40ECh. 12 - Prob. 41ECh. 12 - Prob. 42ECh. 12 - Prob. 43ECh. 12 - Prob. 44ECh. 12 - Prob. 45ECh. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Prob. 50ECh. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - Prob. 54ECh. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - Prob. 58ECh. 12 - Prob. 59ECh. 12 - Prob. 60ECh. 12 - Prob. 61ECh. 12 - Prob. 62ECh. 12 - Prob. 63ECh. 12 - Prob. 64ECh. 12 - Prob. 65ECh. 12 - Prob. 66ECh. 12 - Prob. 67ECh. 12 - Prob. 68ECh. 12 - Prob. 69ECh. 12 - Prob. 70ECh. 12 - Prob. 71ECh. 12 - Prob. 72ECh. 12 - Prob. 73ECh. 12 - Prob. 74ECh. 12 - Prob. 75ECh. 12 - Prob. 76ECh. 12 - Prob. 77ECh. 12 - Prob. 78ECh. 12 - Prob. 79ECh. 12 - Prob. 80ECh. 12 - Prob. 81ECh. 12 - Prob. 82ECh. 12 - Prob. 83ECh. 12 - Prob. 84ECh. 12 - Prob. 85ECh. 12 - Prob. 86ECh. 12 - Prob. 87ECh. 12 - Prob. 88ECh. 12 - Prob. 89ECh. 12 - Prob. 90ECh. 12 - Prob. 91ECh. 12 - Prob. 92ECh. 12 - Prob. 93ECh. 12 - Prob. 94ECh. 12 - Prob. 95ECh. 12 - Prob. 96ECh. 12 - Prob. 97ECh. 12 - Prob. 98ECh. 12 - Prob. 99ECh. 12 - Prob. 100ECh. 12 - Prob. 1STCh. 12 - Prob. 2STCh. 12 - Prob. 3STCh. 12 - Prob. 4STCh. 12 - Prob. 5STCh. 12 - Prob. 6STCh. 12 - Prob. 7STCh. 12 - Prob. 8STCh. 12 - Prob. 9STCh. 12 - Prob. 10STCh. 12 - Prob. 11STCh. 12 - Prob. 12STCh. 12 - Prob. 13STCh. 12 - Prob. 14STCh. 12 - Prob. 15STCh. 12 - Prob. 16STCh. 12 - Prob. 17STCh. 12 - Prob. 18ST
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