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
Question
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Chapter 8.5, Problem 77SSC
Interpretation Introduction

Interpretation:

The Lewis structures of SF4 and SF6 molecules needs to be determined by interpreting the polarity.

Concept introduction:

The Lewis structure of an organic compound represents the bonding of atoms with lone pairs (if any). It indicates the bonds with atoms and also arrangement of atoms in molecule. The polarity of molecule can be determined by the arrangements of bonded atoms in the molecule.

Expert Solution & Answer
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Answer to Problem 77SSC

Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  1Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  2

Polar molecule

Non-polar molecule

Explanation of Solution

In the SF4 molecule, Satom contains 6 valence electrons and F has 7 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom S (valence electron in S) + Number of atom F (valence electron in F) = (1×6)  +  (4×7) = 34

Hence the Lewis structure must be:

Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  3

Hybridization = Number of sigma bonds + Number of lone pairs on bonded atomsHybridization = 4 + 1 = sp3d  

Since the standard geometry must be trigonal bipyramid yet the geometry changes to see-saw due to presence of one lone pair on S atom.

Since the polarity directions are towards F atom therefore overall it is a polar molecule.

In the SF6 molecule, S atom contains 6 valence electrons, and F has 7 valence electrons. The total number of valence electron can be determined in it as follows:

Number of valence electron = Number of atom S (valence electron in S) + Number of atom F (valence electron in F) = (1×6)+ (6×7) =  48

Hence the Lewis structure must be:

Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  4

Hybridization = Number of sigma bonds + Number of lone pairs on bonded atomsHybridization = 6 + 0 = sp3d2 = octahedral

Due to symmetrical structure of molecule it is non-polar molecule.

Conclusion

Thus,

Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  5Chemistry: Matter and Change, Chapter 8.5, Problem 77SSC , additional homework tip  6

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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