Chemistry: Matter and Change
Chemistry: Matter and Change
1st Edition
ISBN: 9780078746376
Author: Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher: Glencoe/McGraw-Hill School Pub Co
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Chapter 8, Problem 117A

(a)

Interpretation Introduction

Interpretation:

The more polar bond between C-S and C-O needs to be determined by circling the negative end of the dipole.

Concept introduction:

The polar bond is formed when two atoms with electro negativity differences are bonded to each other. For pure covalent bonds, the electro negativity difference should be less than 0.4; for polar covalent bonds, this difference should be between 0.4 to 1.8 and for ionic bonds, the difference should be more than 1.8.

(a)

Expert Solution
Check Mark

Answer to Problem 117A

More polar bond is C-O.

Explanation of Solution

The given pair is C-S and C-O. The more polar bond can be determined by comparing the electro negativity between S and O atoms as C is common in both.

Comparing S and O atoms, both belong to the same group but O is placed above S in the periodic table. On moving down the group, due to an increase in the atomic radius, electro negativity decreases. Thus, S is less electronegative than O. Therefore, C-O bond is more polar than C-S bond.

The negative end that is O should be circled.

b.

Interpretation Introduction

Interpretation:

The more polar bond between C-F and C-N needs to be determined by circling the negative end of the dipole.

Concept introduction:

The polar bond is formed when two atoms with electro negativity difference are bonded to each other. For pure covalent bonds, the electro negativity difference should be less than 0.4, for polar covalent bonds, this difference should be between 0.4 to 1.8 and for ionic bonds, the difference should be more than 1.8.

b.

Expert Solution
Check Mark

Answer to Problem 117A

More polar bond is C-F.

Explanation of Solution

The given pair is C-F and C-N. The more polar bond can be determined by comparing the electro negativity between F and N atoms as C is common in both.

Comparing F and N atoms, F belongs to the 17th group and N belongs to the 15th group. On moving left to right in the periodic table, electro negativity increases due to an increase in nuclear charge. Thus, the electro negativity of F is more than N. Also, C-F bond is more polar than C-N bond.

The negative end that is F should be circled.

C.

Interpretation Introduction

Interpretation:

The more polar bond between P-H and P-Cl needs to be determined by circling the negative end of the dipole.

Concept introduction:

The polar bond is formed when two atoms with electro negativity differences are bonded to each other. For pure covalent bonds, the electro negativity difference should be less than 0.4, for polar covalent bonds, this difference should be between 0.4 to 1.8 and for ionic bonds, the difference should be more than 1.8.

C.

Expert Solution
Check Mark

Answer to Problem 117A

More polar bond is P-Cl.

Explanation of Solution

The given pair is P-H and P-Cl. The more polar bond can be determined by comparing the electro negativity between H and Cl atoms as P is common in both.

Comparing H and Cl atoms, Cl belongs to the 17th group and H belongs to the 1st group. On moving left to right in the periodic table, electro negativity increases due to an increase in nuclear charge. Thus, the electro negativity of Cl is more than H. Also, P-Cl bond is more polar than P-H bond.

The negative end that is Cl should be circled.

Chapter 8 Solutions

Chemistry: Matter and Change

Ch. 8.1 - Prob. 11SSCCh. 8.1 - Prob. 12SSCCh. 8.1 - Prob. 13SSCCh. 8.2 - Prob. 14PPCh. 8.2 - Prob. 15PPCh. 8.2 - Prob. 16PPCh. 8.2 - Prob. 17PPCh. 8.2 - Prob. 18PPCh. 8.2 - Prob. 19PPCh. 8.2 - Prob. 20PPCh. 8.2 - Prob. 21PPCh. 8.2 - Prob. 22PPCh. 8.2 - Prob. 23PPCh. 8.2 - Prob. 24PPCh. 8.2 - Prob. 25PPCh. 8.2 - Prob. 26PPCh. 8.2 - Prob. 27PPCh. 8.2 - Prob. 28PPCh. 8.2 - Prob. 29PPCh. 8.2 - Prob. 30PPCh. 8.2 - Prob. 31SSCCh. 8.2 - Prob. 32SSCCh. 8.2 - Prob. 33SSCCh. 8.2 - Prob. 34SSCCh. 8.2 - Prob. 35SSCCh. 8.2 - Prob. 36SSCCh. 8.3 - Prob. 37PPCh. 8.3 - Prob. 38PPCh. 8.3 - Prob. 39PPCh. 8.3 - Prob. 40PPCh. 8.3 - Prob. 41PPCh. 8.3 - Prob. 42PPCh. 8.3 - Prob. 43PPCh. 8.3 - Prob. 44PPCh. 8.3 - Prob. 45PPCh. 8.3 - Prob. 46PPCh. 8.3 - Prob. 47PPCh. 8.3 - Prob. 48PPCh. 8.3 - Prob. 49PPCh. 8.3 - Prob. 50SSCCh. 8.3 - Prob. 51SSCCh. 8.3 - Prob. 52SSCCh. 8.3 - Prob. 53SSCCh. 8.3 - Prob. 54SSCCh. 8.3 - Prob. 55SSCCh. 8.4 - Prob. 56PPCh. 8.4 - Prob. 57PPCh. 8.4 - Prob. 58PPCh. 8.4 - Prob. 59PPCh. 8.4 - Prob. 60PPCh. 8.4 - Prob. 61SSCCh. 8.4 - Prob. 62SSCCh. 8.4 - Prob. 63SSCCh. 8.4 - Prob. 64SSCCh. 8.4 - Prob. 65SSCCh. 8.4 - Prob. 66SSCCh. 8.4 - Prob. 67SSCCh. 8.5 - Prob. 68SSCCh. 8.5 - Prob. 69SSCCh. 8.5 - Prob. 70SSCCh. 8.5 - Prob. 71SSCCh. 8.5 - Prob. 72SSCCh. 8.5 - Prob. 73SSCCh. 8.5 - Prob. 74SSCCh. 8.5 - Prob. 75SSCCh. 8.5 - Prob. 76SSCCh. 8.5 - Prob. 77SSCCh. 8 - Prob. 78ACh. 8 - Prob. 79ACh. 8 - Prob. 80ACh. 8 - Prob. 81ACh. 8 - Prob. 82ACh. 8 - Prob. 83ACh. 8 - Prob. 84ACh. 8 - Prob. 85ACh. 8 - Prob. 86ACh. 8 - Prob. 87ACh. 8 - Prob. 88ACh. 8 - Prob. 90ACh. 8 - Prob. 91ACh. 8 - Prob. 92ACh. 8 - Prob. 93ACh. 8 - Prob. 94ACh. 8 - Prob. 95ACh. 8 - Prob. 96ACh. 8 - Prob. 97ACh. 8 - Prob. 98ACh. 8 - Prob. 99ACh. 8 - Prob. 100ACh. 8 - Prob. 101ACh. 8 - Prob. 102ACh. 8 - Prob. 103ACh. 8 - Prob. 104ACh. 8 - Prob. 105ACh. 8 - Prob. 106ACh. 8 - Prob. 107ACh. 8 - Prob. 108ACh. 8 - Prob. 109ACh. 8 - Prob. 110ACh. 8 - Prob. 111ACh. 8 - Prob. 112ACh. 8 - Prob. 113ACh. 8 - Prob. 114ACh. 8 - Prob. 115ACh. 8 - Prob. 116ACh. 8 - Prob. 117ACh. 8 - Prob. 118ACh. 8 - Prob. 119ACh. 8 - Rank the bonds according to increasing polarity....Ch. 8 - Prob. 121ACh. 8 - Prob. 122ACh. 8 - Use Lewis structures to predict the molecular...Ch. 8 - Prob. 124ACh. 8 - Prob. 125ACh. 8 - Prob. 126ACh. 8 - Prob. 127ACh. 8 - Prob. 128ACh. 8 - Prob. 129ACh. 8 - Prob. 130ACh. 8 - Prob. 131ACh. 8 - Prob. 132ACh. 8 - Prob. 133ACh. 8 - Prob. 134ACh. 8 - Prob. 135ACh. 8 - Prob. 136ACh. 8 - Prob. 137ACh. 8 - Prob. 138ACh. 8 - Prob. 139ACh. 8 - Prob. 140ACh. 8 - Prob. 141ACh. 8 - Prob. 142ACh. 8 - Prob. 143ACh. 8 - Prob. 144ACh. 8 - Prob. 145ACh. 8 - Prob. 1STPCh. 8 - Prob. 2STPCh. 8 - Prob. 3STPCh. 8 - Prob. 4STPCh. 8 - Prob. 5STPCh. 8 - Prob. 6STPCh. 8 - Prob. 7STPCh. 8 - Prob. 8STPCh. 8 - Prob. 9STPCh. 8 - Prob. 10STPCh. 8 - Prob. 11STPCh. 8 - Prob. 12STPCh. 8 - Prob. 13STPCh. 8 - Prob. 14STPCh. 8 - Prob. 15STPCh. 8 - Prob. 16STPCh. 8 - Prob. 17STPCh. 8 - Prob. 18STPCh. 8 - Prob. 19STP

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