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 141A
Interpretation Introduction

Interpretation:

The Lewis structure of ethylene glycol is to be drawn. The sigma and pi bonds in ethylene glycol are to be identified.

Concept introduction:

The strategy for drawing Lewis structure is mention below.

  • Calculate the number of valence electrons present in the molecule.
  • Calculate the electron pairs by diving number of valence electrons by 2.
  • Determine the bond pairs.
  • Determine the lone pairs.
  • Check whether the central atom satisfies octet rule.

Expert Solution & Answer
Check Mark

Answer to Problem 141A

The Lewis structure of ethylene glycol is drawn below.

Chemistry: Matter and Change, Chapter 8, Problem 141A , additional homework tip  1

All the bonds in ethylene glycol are sigma bond. Due to the miscibility of ethylene glycol in water, the freezing point of water decreases. Therefore, ethylene glycol acts as a antifreeze-coolant.

Explanation of Solution

The chemical formula of ethylene glycol is C2H6O2. The number of valence electron in ethylene glycol is the sum of valence electron in carbon, hydrogen, and oxygen atoms. The valence electron in carbon is four, the valence electron in hydrogen is one, and the valence electron in oxygen is six. Therefore, the valence electron in ethylene glycol is calculated as shown below.

No.ofvalenceelectron=2C+6H+2O=2×4e+6×1e+2×6e=26e

The number of electron pair is calculated as shown below.

electronpair=26e2=13epairs

Among thirteen electron pairs, nine electron pairs are involved in bonding and four electron pairs remain as lone pairs.

Therefore, the Lewis structure of ethylene glycol is drawn below.

Chemistry: Matter and Change, Chapter 8, Problem 141A , additional homework tip  2

It is known that in single bond only sigma bond is present. In double bond, one bond is sigma and one is pi bond. In triple bond, one bond is sigma and the remaining two bonds are pi bond. The lone pairs are non bonding electrons represented by two dots.

All the bonds present in ethylene glycol are single bond. Therefore, all the bonds in ethylene glycol are sigma bond.

Chemistry: Matter and Change, Chapter 8, Problem 141A , additional homework tip  3

Due to presence of hydroxyl group (OH) in ethylene glycol, it acts as a polar molecule. Therefore, it is soluble in water. Due to the miscibility of ethylene glycol in water, the freezing point of water decreases. Therefore, ethylene glycol acts as an antifreeze-coolant.

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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