Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 8.4, Problem 29SP

(a)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in H and Br .

Concept Interpretation:

Molecules also contain positive and negative ends; these are known as polar molecules. If not, they are referred as non-polar.

(a)

Expert Solution
Check Mark

Answer to Problem 29SP

The moderately polar covalent compound

Explanation of Solution

  Br-Br is a highly covalent bond whereas HBr is a slightly polar covalent bond.

  ΔEN= 2.8-2.1 = 0.7 for H and Br.

Therefore, a moderately polar covalent connection between H and Br would exist.

(b)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in Cl and F .

Concept Interpretation:

When two atoms exchange one or more pairs of electrons, a covalent connection is formed. The two atomic nuclei are concurrently drawing these electrons to them.

(b)

Expert Solution
Check Mark

Answer to Problem 29SP

  Cl-F is a moderately polar covalent bond.

Explanation of Solution

In general, nonpolar covalent bonds have an electronegativity difference of 0-0.5 , covalent bonds have an Δ EN(Electronegativity difference) of 0.5-2.0 , and ionic bonds have an EN difference of 2.0 or higher.

Atoms are held together in the form of compounds by chemical bonds, which are strong forces of attraction. Using the electronegativity difference (EN) between the atoms, it is possible to determine the type of bond.

  CL and F :- ΔEN=4-3=1

Therefore, this bond is moderately polar covalent.

(c)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in K and CL .

Concept Interpretation:

Ions with opposing charges electrostatically attract one another to form ionic bonds, sometimes referred to as electrovalent bonds, in chemical molecules.

(c)

Expert Solution
Check Mark

Answer to Problem 29SP

  K and CL is an Ionic bond.

Explanation of Solution

In general, nonpolar covalent bonds have an electronegativity difference of 0-0.5 , covalent bonds have an Δ EN(Electronegativity difference) of 0.5-2.0 , and ionic bonds have an EN difference of 2.0 or higher.

  K and Cl :- ΔEN=3.0-0.8=2.2

Hence this bond is Ionic.

(d)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in Li and O .

Concept Interpretation:

Ions with opposing charges electrostatically attract one another to form ionic bonds, sometimes referred to as electrovalent bonds, in chemical molecules.

(d)

Expert Solution
Check Mark

Answer to Problem 29SP

  Li and O is an ionic bond.

Explanation of Solution

In general, nonpolar covalent bonds have an electronegativity difference of 0-0.5 , covalent bonds have an Δ EN(Electronegativity difference) of 0.5-2.0 , and ionic bonds have an EN difference of 2.0 or higher.

  Li and O :- ΔEN=3.5-1.0=2.5

Therefore, this bond is ionic.

(e)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in C and O .

Concept Interpretation:

A polar covalent bond is a bond created by exchanging electrons between two atoms with different electronegativities.

(e)

Expert Solution
Check Mark

Answer to Problem 29SP

  C and O is a moderately polar covalent bond.

Explanation of Solution

In general, nonpolar covalent bonds have an electronegativity difference of 0-0.5 , covalent bonds have an Δ EN(Electronegativity difference) of 0.5-2.0 , and ionic bonds have an EN difference of 2.0 or higher.

  C and O :- ΔEN=3.5-2.5=1.0

Therefore, C and O is a moderately polar covalent bond.

(f)

Interpretation Introduction

Interpretation:

Identify the bonds between atoms of each pair of the element as a nonpolar covalent, moderately polar covenant, very covalent, or ionic in Br and Br .

Concept Interpretation:

When two atoms exchange one or more pairs of electrons, a covalent connection is created.

(f)

Expert Solution
Check Mark

Answer to Problem 29SP

  Br and Br is a highly covalent bond.

Explanation of Solution

In general, nonpolar covalent bonds have an electronegativity difference of 0-0.5 , covalent bonds have an Δ EN(Electronegativity difference) of 0.5-2.0 , and ionic bonds have an EN difference of 2.0 or higher.

  Br and Br :- ΔEN=2.8-2.8=0

Hence Br and Br is a highly covalent bond.

Chapter 8 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 8.2 - Prob. 11LCCh. 8.2 - Prob. 12LCCh. 8.2 - Prob. 13LCCh. 8.2 - Prob. 14LCCh. 8.2 - Prob. 15LCCh. 8.2 - Prob. 16LCCh. 8.2 - Prob. 17LCCh. 8.2 - Prob. 18LCCh. 8.2 - Prob. 19LCCh. 8.2 - Prob. 20LCCh. 8.3 - Prob. 21LCCh. 8.3 - Prob. 22LCCh. 8.3 - Prob. 23LCCh. 8.3 - Prob. 24LCCh. 8.3 - Prob. 25LCCh. 8.3 - Prob. 26LCCh. 8.3 - Prob. 27LCCh. 8.3 - Prob. 28LCCh. 8.4 - Prob. 29SPCh. 8.4 - Prob. 30SPCh. 8.4 - Prob. 31LCCh. 8.4 - Prob. 32LCCh. 8.4 - Prob. 33LCCh. 8.4 - Prob. 34LCCh. 8.4 - Prob. 35LCCh. 8.4 - Prob. 36LCCh. 8.4 - Prob. 37LCCh. 8.4 - Prob. 38LCCh. 8 - Prob. 39ACh. 8 - Prob. 40ACh. 8 - Prob. 41ACh. 8 - Prob. 42ACh. 8 - Prob. 43ACh. 8 - Prob. 44ACh. 8 - Prob. 45ACh. 8 - Prob. 46ACh. 8 - Prob. 47ACh. 8 - Prob. 48ACh. 8 - Prob. 49ACh. 8 - Prob. 50ACh. 8 - Prob. 51ACh. 8 - Prob. 52ACh. 8 - Prob. 53ACh. 8 - Prob. 54ACh. 8 - Prob. 55ACh. 8 - Prob. 56ACh. 8 - Prob. 57ACh. 8 - Prob. 58ACh. 8 - Prob. 59ACh. 8 - Prob. 60ACh. 8 - Prob. 61ACh. 8 - Prob. 62ACh. 8 - Prob. 63ACh. 8 - Prob. 64ACh. 8 - Prob. 65ACh. 8 - Prob. 66ACh. 8 - Prob. 67ACh. 8 - Prob. 68ACh. 8 - Prob. 69ACh. 8 - Prob. 70ACh. 8 - Prob. 71ACh. 8 - Prob. 72ACh. 8 - Prob. 73ACh. 8 - Prob. 74ACh. 8 - Prob. 75ACh. 8 - Prob. 76ACh. 8 - Prob. 77ACh. 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. 89ACh. 8 - Prob. 90ACh. 8 - Prob. 91ACh. 8 - Prob. 92ACh. 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. 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. 14STP
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