Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
Question
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Chapter U2, Problem SIII4RE

(a)

Interpretation Introduction

Interpretation: Comprehend the Lewis dot structure of HBr and label dipole in the molecule.

Concept Introduction: The chemical compounds can be classified as covalent compound and ionic compound. Ionic compounds have complete negative and positive charges on it whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to electronegativity difference between bonded atoms, partial charges induce on the bonded atoms. The partial charges effect the physical properties of the polar molecules.

(a)

Expert Solution
Check Mark

Answer to Problem SIII4RE

  Living By Chemistry: First Edition Textbook, Chapter U2, Problem SIII4RE , additional homework tip  1

Explanation of Solution

In the Lewis structure of HBr molecule, Br has 7 valence electrons and H has 1 electron.

Thus, total valence electrons in HBr molecule = 1 +7 = 8 valence electrons

In HBr molecule, Br is more electronegative element than H atom. This is because Br is a non-metal with 7 valence electrons. Therefore Br tends to accept electrons to get octet configuration. That makes it more electronegative than H atom. Due to this difference in electronegativity of bonded atoms, Br gets partial negative charge and H gets partial positive charge.

  Living By Chemistry: First Edition Textbook, Chapter U2, Problem SIII4RE , additional homework tip  2

(b)

Interpretation Introduction

Interpretation: Comprehend the attraction of HBr towards charged wand and formation of round drop on waxed paper.

Concept Introduction: The chemical compounds can be classified as covalent compound and ionic compound. Ionic compounds have complete negative and positive charges on it whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to electronegativity difference between bonded atoms, partial charges induce on the bonded atoms. The partial charges effect the physical properties of the polar molecules.

(b)

Expert Solution
Check Mark

Explanation of Solution

In HBr molecule, Br is more electronegative element than H atom. This is because Br is a non-metal with 7 valence electrons. Due to this difference in electronegativity of bonded atoms, Br gets partial negative charge and H gets partial positive charge.

  Living By Chemistry: First Edition Textbook, Chapter U2, Problem SIII4RE , additional homework tip  3

Because of the presence of partial charges on HBr molecule, it will attract towards charged wand.

Since it is a polar molecule therefore it will be soluble in polar solvents like water. Whereas it will be insoluble in non-polar solvents therefore it will make round drop on waxed paper as wax is a non-polar molecule.

(c)

Interpretation Introduction

Interpretation: Comprehend the type of bond exists between H and Br in HBr molecule.

Concept Introduction: The chemical compounds can be classified as covalent compound and ionic compound. Ionic compounds have complete negative and positive charges on it whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to electronegativity difference between bonded atoms, partial charges induce on the bonded atoms. The partial charges effect the physical properties of the polar molecules.

(c)

Expert Solution
Check Mark

Answer to Problem SIII4RE

  HBr molecule has polar covalent bond.

Explanation of Solution

In the Lewis structure of HBr molecule, Br has 7 valence electrons and H has 1 electron.

Thus, total valence electrons in HBr molecule = 1 +7 = 8 valence electrons

To get the octet configuration, Br share one of its valence electron with H to form covalent bond and H also shares its electron to get duplet configuration that leads to formation of polar covalent bond in molecule.

  Living By Chemistry: First Edition Textbook, Chapter U2, Problem SIII4RE , additional homework tip  4

Since it is a polar molecule therefore it will be soluble in polar solvents like water. Whereas it will be insoluble in non-polar solvents therefore it will make round drop on waxed paper as wax is a non-polar molecule.

(d)

Interpretation Introduction

Interpretation: Comprehend the smell of HBr molecule.

Concept Introduction: The chemical compounds can be classified as covalent compound and ionic compound. Ionic compounds have complete negative and positive charges on it whereas covalent compounds are formed by equal sharing of electrons between bonded atoms.

The polarity of molecule depends on the presence of electropositive and electronegative atoms present in the molecule.

Due to electronegativity difference between bonded atoms, partial charges induce on the bonded atoms. The partial charges effect the physical properties of the polar molecules.

(d)

Expert Solution
Check Mark

Answer to Problem SIII4RE

  HBr molecules have pungent smell.

Explanation of Solution

In HBr molecule, Br is more electronegative element than H atom. This is because Br is a non-metal with 7 valence electrons. Due to this difference in electronegativity of bonded atoms, Br gets partial negative charge and H gets partial positive charge.

  Living By Chemistry: First Edition Textbook, Chapter U2, Problem SIII4RE , additional homework tip  5

Because of the presence of partial charges on HBr molecule, it is a polar molecule and soluble in water. A polar molecule can easily stick with receptor site in the nose and also soluble in mucous membrane of the nose therefore they have pungent smell.

Chapter U2 Solutions

Living By Chemistry: First Edition Textbook

Ch. U2.2 - Prob. 3ECh. U2.2 - Prob. 4ECh. U2.2 - Prob. 5ECh. U2.2 - Prob. 6ECh. U2.2 - Prob. 7ECh. U2.3 - Prob. 1TAICh. U2.3 - Prob. 1ECh. U2.3 - Prob. 2ECh. U2.3 - Prob. 3ECh. U2.3 - Prob. 4ECh. U2.3 - Prob. 5ECh. U2.4 - Prob. 1TAICh. U2.4 - Prob. 1ECh. U2.4 - Prob. 2ECh. U2.4 - Prob. 3ECh. U2.4 - Prob. 4ECh. U2.4 - Prob. 5ECh. U2.4 - Prob. 6ECh. U2.4 - Prob. 7ECh. U2.4 - Prob. 8ECh. U2.5 - Prob. 1TAICh. U2.5 - Prob. 1ECh. U2.5 - Prob. 2ECh. U2.5 - Prob. 3ECh. U2.5 - Prob. 4ECh. U2.5 - Prob. 5ECh. U2.5 - Prob. 6ECh. U2.5 - Prob. 7ECh. U2.5 - Prob. 8ECh. U2.5 - Prob. 9ECh. U2.6 - Prob. 1TAICh. U2.6 - Prob. 1ECh. U2.6 - Prob. 2ECh. U2.6 - Prob. 3ECh. U2.6 - Prob. 4ECh. U2.6 - Prob. 5ECh. U2.6 - Prob. 6ECh. U2.6 - Prob. 7ECh. U2.6 - Prob. 8ECh. U2.6 - Prob. 9ECh. U2.6 - Prob. 10ECh. U2.6 - Prob. 11ECh. U2.7 - Prob. 1TAICh. U2.7 - Prob. 1ECh. U2.7 - Prob. 2ECh. U2.7 - Prob. 3ECh. U2.7 - Prob. 4ECh. U2.7 - Prob. 5ECh. U2.8 - Prob. 1TAICh. U2.8 - Prob. 1ECh. U2.8 - Prob. 2ECh. U2.8 - Prob. 3ECh. U2.8 - Prob. 4ECh. U2.8 - Prob. 5ECh. U2.8 - Prob. 6ECh. U2.9 - Prob. 1TAICh. U2.9 - Prob. 1ECh. U2.9 - Prob. 2ECh. U2.9 - Prob. 3ECh. U2.9 - Prob. 4ECh. U2.9 - Prob. 5ECh. U2.9 - Prob. 6ECh. U2.9 - Prob. 7ECh. U2.10 - Prob. 1TAICh. U2.10 - Prob. 1ECh. U2.10 - Prob. 2ECh. U2.10 - Prob. 3ECh. U2.10 - Prob. 4ECh. U2.10 - Prob. 5ECh. U2.10 - Prob. 6ECh. U2.10 - Prob. 7ECh. U2.10 - Prob. 8ECh. U2.10 - Prob. 9ECh. U2.11 - Prob. 1TAICh. U2.11 - Prob. 1ECh. U2.11 - Prob. 2ECh. U2.11 - Prob. 3ECh. U2.11 - Prob. 4ECh. U2.11 - Prob. 5ECh. U2.11 - Prob. 6ECh. U2.11 - Prob. 7ECh. U2.12 - Prob. 1TAICh. U2.12 - Prob. 1ECh. U2.12 - Prob. 2ECh. U2.12 - Prob. 3ECh. U2.12 - Prob. 4ECh. U2.12 - Prob. 5ECh. U2.12 - Prob. 6ECh. U2.12 - Prob. 7ECh. U2.13 - Prob. 1TAICh. U2.13 - Prob. 1ECh. U2.13 - Prob. 2ECh. U2.13 - Prob. 3ECh. U2.13 - Prob. 4ECh. U2.13 - Prob. 5ECh. U2.14 - Prob. 1TAICh. U2.14 - Prob. 1ECh. U2.14 - Prob. 2ECh. U2.14 - Prob. 3ECh. U2.14 - Prob. 4ECh. U2.14 - Prob. 5ECh. U2.14 - Prob. 6ECh. U2.14 - Prob. 7ECh. U2.14 - Prob. 8ECh. U2.15 - Prob. 1TAICh. U2.15 - Prob. 1ECh. U2.15 - Prob. 2ECh. U2.15 - Prob. 4ECh. U2.15 - Prob. 5ECh. U2.15 - Prob. 6ECh. U2.16 - Prob. 1TAICh. U2.16 - Prob. 1ECh. U2.16 - Prob. 2ECh. U2.16 - Prob. 3ECh. U2.16 - Prob. 4ECh. U2.16 - Prob. 6ECh. U2.16 - Prob. 7ECh. U2.17 - Prob. 1TAICh. U2.17 - Prob. 1ECh. U2.17 - Prob. 2ECh. U2.17 - Prob. 3ECh. U2.17 - Prob. 4ECh. U2.17 - Prob. 5ECh. U2.17 - Prob. 6ECh. U2.17 - Prob. 7ECh. U2.17 - Prob. 8ECh. U2.17 - Prob. 9ECh. U2.18 - Prob. 1TAICh. U2.18 - Prob. 1ECh. U2.18 - Prob. 2ECh. U2.18 - Prob. 3ECh. U2.18 - Prob. 4ECh. U2.18 - Prob. 5ECh. U2.18 - Prob. 6ECh. U2.18 - Prob. 7ECh. U2.19 - Prob. 1TAICh. U2.19 - Prob. 1ECh. U2.19 - Prob. 2ECh. U2.19 - Prob. 3ECh. U2.19 - Prob. 4ECh. U2.19 - Prob. 5ECh. U2.19 - Prob. 6ECh. U2.19 - Prob. 7ECh. U2.19 - Prob. 8ECh. U2.19 - Prob. 9ECh. U2.20 - Prob. 1TAICh. U2.20 - Prob. 1ECh. U2.20 - Prob. 2ECh. U2.20 - Prob. 3ECh. U2.20 - Prob. 4ECh. U2.20 - Prob. 5ECh. U2.20 - Prob. 6ECh. U2.20 - Prob. 7ECh. U2.20 - Prob. 8ECh. U2.20 - Prob. 9ECh. U2.21 - Prob. 1TAICh. U2.21 - Prob. 1ECh. U2.21 - Prob. 2ECh. U2.21 - Prob. 3ECh. U2.21 - Prob. 4ECh. U2.21 - Prob. 6ECh. U2 - Prob. SI1RECh. U2 - Prob. SI2RECh. U2 - Prob. SI3RECh. U2 - Prob. SI4RECh. U2 - Prob. SI5RECh. U2 - Prob. SII1RECh. U2 - Prob. SII2RECh. U2 - Prob. SII3RECh. U2 - Prob. SII4RECh. U2 - Prob. SII5RECh. U2 - Prob. SIII1RECh. U2 - Prob. SIII2RECh. U2 - Prob. SIII3RECh. U2 - Prob. SIII4RECh. U2 - Prob. SIII5RECh. U2 - Prob. SIII6RECh. U2 - Prob. SIV1ECh. U2 - Prob. SIV2ECh. U2 - Prob. SIV3ECh. U2 - Prob. SIV4ECh. U2 - Prob. SIV5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RE
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