Chemistry
CHEM 1201
General
Chemistry 1
Professor J. B. Hopkins
127 Choppin Hall
225-578-3478
chhopk@lsu.edu
Fall 2019 v9.1
Table of Contents
TABLE OF CONTENTS
1
Introduction .
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1
Signi cant Figures in Multiplication and Division
Signi cant Figures in Addition and Subtraction
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Signi cant Figures
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Dimensional Analysis
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Valuable Tools for Careful Workers
Converting cubic units
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Density .
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Conversion Factors
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Atoms, Molecules, Ions
Atoms, Electrons, and Protons
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Atoms
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A Simplistic View of Atomic Structure
Atoms and Elements .
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Isotopes
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Mass Number
Ions
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The Periodic Table
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Groups
Periods
Main Group and Transition Metals.
Metals, non-Metals, and Metalloids
Properties of Metals
Properties of non-Metals
Properties of Metalloids
Information in the Periodic Table
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Molecules Elements and Compounds
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Molecules of Elements
Compounds
Molecular Compounds
Ionic Compounds
Empirical Formulas and Molecular Formulas
Polyatomic Ions
Predicting Formulas of Ionic Compounds
Naming Inorganic Compounds
Organic Molecules
Inorganic Molecules
Naming Molecular Compounds
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Naming Ionic Compounds
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Stoichiometry and Mole-Mass Calculations
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A chemical reaction is a rearrangement of atoms
Chemical Equations .
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Balancing Reactions
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Types of Reactions
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Formula Mass
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Working in amu
The 12 C Scale
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52
Percent Composition
Converting between Mass and Moles
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An Example from Everyday Life
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Limiting Reagent
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Theoretical Yields
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72
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Strategy for Limiting Reagent Problems
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86
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101
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Properties of Aqueous Solutions
Ionic Compounds in Water
Molecular Compounds in Water
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74
Acids and Bases
Strong, Weak, and non-Electrolytes
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76
Solubility Rules
79
Net Ionic Equations
80
Types of Solution Reactions
Homogeneous and Heterogeneous Solutions
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86
Concentration (Molarity)
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Experimental Preparation of a Solution
Calculations with Molarity
Solution Stoichiometry
Titrations
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Mixing Solutions
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Mixture Example Problem
100
How to Think about Word Problems
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Reading Skills for Word Problems
Electronic Structure of Atoms .
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Finding the concentrations
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Molar Mass
Aqueous Reactions .
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47
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Empirical Analysis
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Quantitative Calculations
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44
Patterns of Chemical Reactivity
The Mole
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43
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102
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104
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108
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111
Bohr’s Equation for the Lines in a Hydrogen Spectrum
Energy Levels of Hydrogen
116
Problems with Bohr’s Theory
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112
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119
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126
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130
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141
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144
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153
20th Century Physics
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Light as a Wave
Quantum Effects
Light as a Particle
Wave Particle Duality
Heisenberg Uncertainty Principle
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Spectrum of the Hydrogen Atom
The Bohr Atom
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Modern Quantum Theory
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1. Pauli Exclusion Principle
Electron Con gurations of H, He, Li, Be
2. Hund’s Rule
Writing Electron Con gurations
Short Hand Notation
Electron Con gurations of Transition Metals
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Periodic Table
Electron Shells
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127
The Aufbau Principle
Periodic Properties .
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121
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126
Electron Screening
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118
Summary
Orbital Shapes
Orbital Orientation in Space
Energy Levels and Shells
Lowest and Highest Energy Levels
Relationships between Quantum Numbers
Electron Spin
Nodes
Multi-Electron Atoms
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109
Modern View of the Atom
Electron Orbitals
Orbitals: An Analogy
Quantum Numbers
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106
106
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Electron Shells for He and Ar
Electron Screening
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142
Energy of an Electron
Effective Nuclear Charge
General Trends for Atomic Sizes
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Ion Sizes
Electron Con gurations of Ions
Electron Con guration of Transition Metal Ions
Ionization Energy
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Notes about Ionization Energies
Second Ionization Energies (I2 )
Comparing Ionization Energies
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144
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Comparing Ionization Energies for Na and Mg
Chemical Implications of Ionization Energy
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147
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Electron Af nity
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Comparison of Ionization Energy and Electron Af nity
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Metals, Nonmetals, Metalloids
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Metallic Character
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157
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158
Chemistry of Important Groups in the Periodic Table
Chemical Reactivity Alkali Metals
Hydride Ion
Alkaline Earths
Oxygen
Halogens
Noble Gases
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Introduction to Chemical
Bonding
Valence Electrons and Lewis Symbols
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165
Lewis Dot Symbol
Writing Lewis Dot Symbols
The Octet Rule and Chemical Bonding
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170
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171
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180
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Summary
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Charges on the atoms in a molecular compound
Resonance .
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190
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195
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Electronegativity
Ionic Bonding
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When do Ionic Bonds Form?
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Covalent Bonding
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Example 1, OF2
Example 2, HCN
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Exceptions to the Octet
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1. Odd Number of Total Valence Electrons
2. Less Electrons than an Octet
3. More Electrons than an Octet
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Bond Energy and Enthalpy
Potential Energy
What is Enthalpy?
Calculations Using Bond Dissociation Enthalpy
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181
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192
Bond Dissociation Enthalpies
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186
186
Resonance Hybrid
Resonance in Benzene
Benzene Resonance Hybrid
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178
Using Formal Charges to Choose Lewis Structures
Molecular Geometry
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173
Lewis Dot Structures of Molecules
Writing Lewis Structures .
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Formal Charge
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170
When does Covalent Bonding Occur?
Metallic Bonds
Polar Bonds
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200
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Electron Domains
VSEPR Model
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206
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246
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261
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269
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Why is CO2 Linear and NO2 Bent?
207
VSEPR MODEL
207
Molecular Geometry and Electron Domain Geometry
208
NO2 Resonance Hybrid
209
Predicting Geometries from Lewis Structures
Bond Angles
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Two Domains
Three Domains
Four Domains
Five Domains
Six Domains
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214
215
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Polar and NonPolar Molecules
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Polar Bonds
Polar Molecules
When do Dipoles Cancel?
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Covalent Bonding
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Molecular Orbitals
Hybrid Orbitals
sp2 Hybrid Orbitals in BH3
Determining the Type of Hybridization
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Multiple Bonding
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Boyles Law
Charles Law
Combined Gas Law
Ideal Gas Law
Avogadro’s Law of Combining Volumes
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Partial Pressure Equations
Mole Fractions
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Graham’s Law of Effusion
Diffusion
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267
269
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The van der Waals Equation
11 InterMolecular Forces
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262
263
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Effusion and Diffusion
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248
250
250
251
Gas Law Example Problems .
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Gas Density .
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Quantitative Gas Law Calculations
Partial Pressures
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235
237
241
Kinetic-Molecular Theory of Gases
Units of Pressure
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Gas Laws
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Real Gases
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227
230
232
Sigma Bonds
Pi bonds
Bonding in Acetylene
Molecular Orbitals
10 Gases
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219
220
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270
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Why are InterMolecular Bonds Important?
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Types of InterMolecular Forces
Ionic Forces
Ion-Dipole Forces
Dipole-Dipole
London Dispersion Forces
Hydrogen Bonding
Comparing Strengths of Intermolecular Forces
Summary
Examples
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Solids Liquids and Gases
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275
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Solids
Liquids
Gases
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Example
Melting
Boiling
Sublimation
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286
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288
289
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294
295
291
292
292
293
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Heating and Phase Transitions
What is boiling?
Index
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286
286
287
The Concept of Temperature
Phase Changes .
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Real Gases .
Phase Diagrams
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275
277
279
279
282
284
284
285
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297
299
301
PHYSICAL CONSTANTS .......................................................... 305
1 Introduction
Signi cant Figures
Numbers that are measured have a certain precision that is often
the result of the accuracy of the measuring device. Here are some
examples of how signi cant gures are used for experimental numbers. The best way to remember these is not by memorizing rules.
Instead, try to understand the idea of when a digit is signi cant and
when it is not.
1. Contains 4 signi cant gures. All nonzero digits are signi cant.
1278
2. Contains 1 signi cant gure. The zeros are not signi cant. Their
function is simply to indicate that the 1 is multiplied by 103 We
do not know their true value so they are not signi cant.
3. Contains 4 signi cant gures. The decimal point means that the
Note that these two numbers
both represent the same thing.
Both have 4 signi cant gures.
1278
1278.
zeros are known to have the exact value of zero so they are signi cant.
2 Chapter 1 Introduction
In nite sig gs
a. A Be atom has 4 protons.
4. Contains 5 signi cant gures. Zeros in between nonzero numbers
are signi cant.
The number 4 was obtained
10 001
by counting and we are absolutely sure the value is ex-
5. Contains 6 signi cant
gures. Zeros at the end of the number
and to the right of a decimal point are signi
the number was measured with a more accurate device than the
number in item 4.
10 0010
actly 4, not 4.13 or 3.9
b. A dozen contains 12 things.
This is an exact de nition.
c. A meter contains 100 cm. This
is an exact de nition.
d. A week has 7 days. This is
6. Contains 1 signi cant
gure. The zeros to the left of the rst
nonzeror digit are not signi cant. The function of the zerosz is
simply to scale the number by 10 3
an exact de nition.
e. Mary, Bob, and Jack submitted their homework to the teacher.
The number 3 is exact because it represents an exact
counting. We know the names
of everyone who submitted
homework.
f. About 25 birds were seen ying through the sky. This is
not an exact counting! The
7. The conversion factor 2.54 cm/inch is an exact number. It is a
de ned quantity and has an in nite number of signi cant gures.
8. A number in scienti c notation has as many signi cant gures as
there are in the pre x. For example, this number has 4 signi cant
gures. See also rule 5.
in the value so 2 sig gs is
appropriote for this value.
Here are more examples:
1000
1
103
1000
1 000
103 4 sig gs
0 001
1
10 3
1 sig g
0 0010
10
10 3
2 sig gs
1 sig g
Signi cant Figures in Multiplication and
Division
Identify the number in the calculation with the least number of
3
signi cant gures. Then round the answer to the same number of
signi cant gures.
Example 1.1
Calculate the area of a circle with a radius of 5.4
cm. The formula is
r2
area
3 14159 5 4 2
91 6088 cm2
Solution
The number in the calculation with the least number
of signi cant gures is 5.4. So the answer must be rounded to 2
signi cant gures.
Common Mistake: The number 5.4 has 2 signi cant gures.
Do Not round the answer to
the rst decimal place and report 91.6. Round to 2 signi cant gures and report 92.
area
92 cm2
Signi cant Figures in Addition and Subtraction
Round the answer to the least signi cant digit. That is not the
same thing as rounding to the least number of signi cant gures.
Example 1.2
Find the sum of 1.23, 100.1, 34.337, and 0.003
with the correct number of signi cant gures.
4
Chapter 1 Introduction
Solution
Least Signi cant Digit
Sig Figs Can
Change
In mulitiplication and division sig gs are soley determined
by the value with the least number of sig gs.
1 21 1 1
13
In addition and subtraction
the number of sig gs can change.
1 21
1
1 20
9
0 01
10
1 0 0 . 1 ?
?
1 . 2 3
?
3 4 . 3 3
7
0 . 0 0
3
1 3 5 . 6 7
0
Round to Least Signi cant Digit
The correct answer is 135 7. The number with the smallest number
of signi cant gures is 0.003. The column containing the least
Common Mistake: Students of- signi cant digit is identi ed by starting from the right most column
ten nd the number of signi cant gures in the answer by inspecting the numbers in the word of numbers and nding the rst column where all of the numbers
problem itself. That is not the
correct approach!! The num- are known. In the numbers above, the verticle columns of numbers
bers in the problem are 1.23, 100.1,
34.337, 0.003. Inspecting these containing 7,3 and 3,3,0 are not signi cant because some of the
numbers leads to the erroneous
conclusion that since 0.003 has
1 sig g the answer should be entries in the coulmn are missing. The rst column where all of the
rounded to 1 sig g. This is
wrong!! You must inspect the values are known is indicated with the arrows. Round the answer to
calculation to see how the numbers are used. In this case the the least signi cant decimal place.
numbers are being added together
and there are 4 signi cant gures in the answer.
Dimensional Analysis
A conversion factor such as
1 gal
3 785 L
is often used as a ratio. This ratio could be written in two ways
Dimensional Analysis 5
depending on how the conversion factor is needed in a particular
calculation.
1 gal
3 785 L
OR
3 785 L
1 gal
To convert 5 L to gallons the calculation below would be performed.
5L
1 gal
3 785 L
1 321 gal
Notice that the conversion factor was used in such a way that the
units cancelled to give the answer in the desired units of gallons. If
the conversion factor is used incorrectly, as below, the units in the
answer would be incorrect. In this incorrect calculation the units
come out to be L2 gal which is not the desired result.
5L
3 785 L
1 gal
L2 gal
WRONG!!
Valuable Tools for Careful Workers
1. Make it a habit to always use dimensions in your calculations.
2. Careful workers always ask themselves if the answer makes sense.
Example 1.3
Calculate the mass of gasoline in a full tank with
a volume of 15.0 gal. Use the conversion factor: 1 gal = 3.785 L
The density of gasoline is 0.703 g/mL
Solution
Units can be used to guide your calculation even when
you cannot see the entire calculation in your head. Start with the
given information and using some undetermined conversion factors
6
Chapter 1 Introduction
set your equation equal to the desired units of grams.
15 0 gal
??
??
??
??
grams
Inspect the equation above to determine how the conversion factors
should be used. Combine conversion factors to make the units of
gallons cancel and produce the units of grams.
3 785 L
1 gal
15 0 gal
1000 mL
1L
0 703 g
1 mL
grams
Notice that the units helped guide our thinking and allowed us to see
how the density should be used in the calculation. It is important
to ask yourself what you guess the answer will be before actually
putting the numbers into your calculator. Imagine picking up a gas
tank lled with 15 gallons of gas! How much would it weigh? I
guess its maybe 100 lbs or so. This would be about 50 kg or 50,000
g. Now enter the numbers into your calculator.
3 785 L
1 gal
15 0 gal
1000 mL
1L
0 703 g
1 mL
39 900 grams
Must know this!
The calculated answer compares reasonably well with our initial
guess.
However, if we had obtained a value 10 times larger or
smaller than our guess then we would have suspected something
was wrong!
Converting cubic units
To convert 15 cm3 min to units of m3 s we must convert cm3 to
Density 7
m3 and 1 min to 1 s Focus your attention on converting one unit
at a time.
Convert cm3 to m3
15 cm3
min
1m
100 cm
3
15
10 5 m3 min
Notice that to convert cm3 to m3 the conversion factor
must be cubed to produce this conversion factor
1m
100 cm
1m
100 cm
3
The
units and the number inside the parenthesis are all cubed.
Convert min to s
15
10 5 m3 min
1 min
60 s
25
10 7 m3 s
Density
Density is a physical property that measures the mass of an object
per unit volume. The density of gasoline is
density = 0 703 g/mL
Just by inspecting the units you should be able to guess that density
is a physical quantity that connects mass and volume.
Solid
styrofoam has a low density and lead has a high density. A brick
sized piece of solid lead has a mass of about 20 lbs due to the high
density of lead. A Styrofoam brick of the same volume would have
a mass about 400 times smaller than the lead brick. Density can
8
Chapter 1 Introduction
It is sometimes usefull to think be used as a conversion factor in these two ways.
of density as a conversion factor.
Density is a physical quantity that converts between mass
and volume.
density = 0 703 g/mL
0 703 g
1 mL
OR
1 mL
0 703 g
Conversion Factors
You must memorize these metric conversion factors.
You do
not have to memorize the conversion factors between metric and
between cm and inches.
All of the conversion factors below are
de ned quantities that have an in nite number of signi cant gures.
Conversion Factors 9
METRIC SYSTEM
1 Gm = 1 0
109 m
1 gigameter = 1 0
109 meters
M (mega) 1 Mm = 1 0
106 m
1 megameter = 1 0
106 meters
k (kilo)
1 km = 1 0
103 m
1 kilometer = 1 0
103 meters
d (deci)
1 dm = 1 0
10 1 m
1 decimeter = 1 0
10 1 meters
c (centi)
1 cm = 1 0
10 2 m
1 centimeter = 1 0
10 2 meters
m (milli)
1 mm = 1 0
10 3 m 1 millimeter = 1 0
10 3 meters
m=10
10 6 m
1 micrometer = 1 0
10 6 meters
n (nano)
1 nm = 1 0
10 9 m
1 nanometer = 1 0
10 9 meters
p (pico)
1 pm = 1 0
10 12 m 1 picometer = 1 0
10 12 meters
f (femto)
1 fm = 1 0
10 15 m
G (giga)
(micro) 1
1 femtometer = 1 0
10 15 meters
The table above contains the relationship
1 km = 1 0
103 m
From that relationship two conversion factors could be constructed.
1 km
1 0 103 m
OR
10
103 m
1 km
You must also know these volume relationships:
METRIC SYSTEM
1 cm3 = 1 0 mL
1 cubic centimeter = 1 0 mL
1000 cm3 = 1 0 L 1 megameter = 1 0
106 meters
2 Atoms, Molecules, Ions
Atoms, Electrons, and Protons
could be produced in an evacuated tube when a high potential voltage was applied to two electrodes. The beam of particles could be
de ected by both magnetic and electric elds applied to the outside
of the tube. The direction of de ection indicated that the particles
were negatively charged. This was the discovery of the electron by
J.J. Thomson. Since many materials could be used for the cathode, Thomson concluded that all of the tested materials contained
Cathode ray tubes were eventually modi ed for practical applications as CRT devices used
in televisions and, computer displays. A phosphor screen that
glows when struct by electrons
is used at the anode of the tube.
Electric
the electron beam to create patterns that we recognize as a TV
image. Red, green, and blue
phosphors are used to create colors.
electrons.
The metal electrodes in their natural state have no charge.
Af-
11
ter physicists learned that metals were contained negatively charged
electrons it was obvious that the same metals must also be composed of positive particles called protons.
Rutherford red a beam of alpha particles containing 2 positive protons and 2 neutrons at a very thin gold foil.
Most of the alpha
particles passed through the foil unde ected. However, some were
The plum pudding model was de
a competing theory for the structure of the atom in the early 20th
century. This model proposed
that the electrons and protons were
uniformily dispersed throughout
the volume of the atom like raisens
mixed throughout pudding. Rutherford’s experiment disproved this
model by proving the existence
of the nucleus.
ected at a very high scattering angle.
The high angle scattering could only result from the collision of
an alpha particle with an extremely high density of positive charge.
This showed that all of the positive protons in the gold atom were
concentrated in a very small volume that we now call the nucleus
of the atom. The vast majority of alpha particles are unde ected
and simply pass right through the gold foil. Rutherford interpreted
this result to mean that much of the atom is empty space allowing
the electrons to pass unde ected.
12
Chapter 2 Atoms, Molecules, Ions
The charge of an electron was measured by Robert Millikan.
A
beam of very small oil droplets was allowed to enter a chamber
where a beam of x-rays created negatively charged droplets. The
droplet was allowed to fall through the air where its terminal velocity was measured from the time required to pass two crosshairs.
Next an electric eld was applied to pull the droplet in the upward
direction.
Again the terminal velocity was measured. From this
information the charge on the droplet was computed.
Atoms 13
Atoms
14
Chapter 2 Atoms, Molecules, Ions
A Simplistic View of Atomic Structure
The picture of the atom presented here is simplistic and is only
intended to give you a rough idea of how to think about atoms. In
Chapter 6 the modern theory of the atom will be discussed in much
more detail.
1. Protons (+ charge) and neutrons (0 charge) are lumped together
in the nucleus.
2. Electrons (- charge) occupy the rest of the volume of the atom.
3. Most of the atomic mass is in the nucleus.
4. Protons and neutrons have approximately the same mass. An
electron has only 1/2000th the mass of a proton.
5. Most of the atom is empty space.
6. Atoms are incredibly small with dimensions on the order of 10
billionths of a meter. (1
10 10 m)
7. Atoms have an extremely tiny mass. A hydrogen atom has a mass
of 1 millionth of 1 billionth of 1 billionth of a gram!
8. Atoms are the building blocks for all materials in the universe.
Mass: most of the mass is in the nucleus of the atom. There are
7 nuclei and each nuclei has a mass of approximately 1 amu so the
mass will be about 7 amu.
Diameter:
The diameter of an H atom is 1.2 Å. The atom in
the cartoon has more protons, neutrons, nad electrons so it will be
somewhat larger. The point is that we can guess that the diameter
could not be smaller than 0.1 Å or larger than 100 Å.]Dimensions
Atoms 15
You must develop a common
sense about the orders of magnitude of numbers. Guess the
diameter and mass of the atom
in the cartoon on the last page.
Mass: most of the mass is
in the nucleus of the atom. There
are 7 nuclei and each nuclei has
a mass of approximately 1 amu
so the mass will be about 7 amu.
Diameter: The diameter of
an H atom is 1.2 Å. The atom
in the cartoon has more protons,
neutrons, nad electrons so it will
be somewhat larger. The point
is that we can guess that the diameter could not be smaller than
0.1 Å or larger than 100 Å.
and Units for the Atomic World
Angstrom 1 = 1
10 10 m
a hydrogen atom has a diameter of
amu
1 amu = 1 66
charge
1.2
10 24 g
1 proton
1 amu
1 neutron
1 amu
1 proton = 1 602
10 19 C
1
1 electron =
10 19 C
1 602
1
Example 2.4
Consider an atom with a radius of 2.7 Å. Find
the radius of this atom in meters.
Solution
Use the conversion factor in the table.
27
27
Example 2.5
10 10 m
1
10 10 m
1
What is the mass of a 5.2 g object in amu?
16
Chapter 2 Atoms, Molecules, Ions
Solution
Use the conversion factor in the table.
52g
3 13
1 amu
1 66 10 24 g
1024 amu
Atoms and Elements
Each of the atoms in the gure above is made up of the same elemental particles: protons, neutrons, and electrons. These subatomic
particles are the building blocks of all atoms.
I All atoms of the same element have the same number of protons.
All He atoms must have 2 protons.
Isotopes 17
I The number of protons determines the element.
Isotopes
Each element can have di erent types of atoms where the number of neutrons varies. These atoms have di erent masses and are
called isotopes.
I Each unique isotope has a di erent number of neutrons.
I All isotopes of the same element have the same number of protons.
All isotopes of He must have 2 protons.
I Naturally occurring elements often contain several isotopes. The
average mass of the element is calculated from the masses and
relative abundance of its isotopes.
avg mass =
Recall also that
abundance isotope mass
abundance = 1 0
2.1
18
Chapter 2 Atoms, Molecules, Ions
Mass Number
In order to identify a speci c isotope a notation is used where the
number of protons and neutrons are both speci ed.
I Remember that protons and neutrons each have a mass of about 1
amu. A He atom with a mass number of 4 (2 protons + 2 neutrons)
will have a mass of about 4 amu. The mass of electrons is so small
in comparison that we just ignore their contribution to the mass of
the atom.1
Example 2.6
How many protons, neutrons, and electrons does
the isotope 35
17 C have? Determine the approximate mass?
Solution
1. The atomic number is the number of protons so the number of
protons must be 17.
2. The mass number is the sum of protons + neutrons so 35
17
1 The mass of an atom can never be calculated by simply adding up the masses of the protons, neutrons, and
electrons. The reason is related to Einstein’s famous E
mc 2 equation. Some of the mass of the particles in the
nuclues is converted to the energy that holds the atom together. So the mass of the atom is always less than the sum of
the masses of the subatomic particles.
Isotopes 19
18 is the number of neutrons.
3. The atom has no charge so the electrons must be the same as
the number of protons or 17.
4. Since each proton and neutron has a mass of approximately 1
amu, the approximate mass of this isotope is 35 amu.
Example 2.7
Potassium has two naturally occurring isotopes,
39 K with mass 38.9637 amu and abundance 93.258 % and 41 K with
mass 40.9618 amu and abundance 6.730 %. What is the average
mass of naturally occurring potassium?
Strategy
This problem is like asking the following question.
Suppose there is a community of 100 workers where. 93 % make
$50,000 per year and 7 % make $100,00 per year. What is the
average salary? The average salary for this community is
% 1 salary1
% 2 salary2
avg salary
0 93 $50 000
0 07 $100 00
$53 500
The average salary is close to the salary of the rst group because
almost all of the workers (93 %) earn that salary.
Solution
For the isotope problem the average mass is calculated
from a similar equation. (Eqn. 2.1)
% 1 mass1
0 93258 38 9637 amu
% 2 mass2
0 06730 40 9618 amu
avg mass
39 0935 amu
20
Chapter 2 Atoms, Molecules, Ions
Ions
The charge of an atom is determined by the balance between the
number of electrons and protons.
Charge
0
Condition e
Name
p
e
anion
p
atom
p
e
cation
I Remember, a POSITIVE ion is a ca ion.
Example 2.8
Determine the symbol for the isotope that has 7
protons, 8 neutrons and 10 electrons.
Solution
The Periodic Table 21
1. According to the periodic table atomic number 7 is nitrogen.
2. The mass number is protons + neutrons = 7
8
15.
3. This atom has more electrons than protons so the charge is nega-
tive. The charge will be electrons - protons =
10
7
3
3
4. The symbol for this isotope is 15
7 N
The Periodic Table
All of the known elements are arranged by atomic number in the
periodic table. Much of the chemistry of an element can be determined by nding its location in the table.
Groups
Elements in the columns or Groups have similar chemical properties.
22
Chapter 2 Atoms, Molecules, Ions
Periods
Periods are elements arranged in the rows of the table.
Main Group and Transition Metals.
I Notice the location of the Main Group elements, transition metals, and inner transition metals.
Metals, non-Metals, and Metalloids
I Know the location of these types of elements. Note H is a non-
The Periodic Table 23
Metal.
Properties of Metals
1. Conduct electricity
2. Conduct heat
3. Shiny (high luster)
4. Bendable
5. Malleable (can pound into thin sheets)
6. Ductile (can pull into thin wires)
7. Like to lose electrons when forming ions
Properties of non-Metals
1. Brittle
2. Many are gases or powders
3.
4. Like to gain electrons when forming ions
24
Chapter 2 Atoms, Molecules, Ions
Properties of Metalloids
1. These elements have properties that are in-between the metals /
non-metals.
2. Partially conduct heat and electricity
3. Can be shiny but are usually brittle
4. Semiconductor electronics (computer chips) based on these ma-
terials.
Information in the Periodic Table
Common Mistake: Students
often confuse average mass with
mass number. Mass number is
always an interger because it represents the number of protons and
neutrons. The actual mass of
the isotope is not an interger. The
average mass is the average of
all isotopes.
The box for each element in the table contains the following useful
information.
The average mass is used because elements as they are found in
nature usually consist of many isotopes with di erent masses.
Example 2.9
Naturally occurring magnesium consists almost
entirely of 3 isotopes. Calculate the mass of 26 Mg using the given
The Periodic Table 25
isotopic information.
Isotope Abundance
Mass
24 Mg
78 70%
23 98504 amu
25 Mg
10 13%
24 98584 amu
26 Mg
11 17%
? amu
Solution
This is a good example of a probRemember that the average atomic mass of Mg is
lem where you have to think!
When the way to solve the prob- the number given in the periodic table. This number is the average
lem is not clear ask yourself “What
do I know about the relationship
between abundanaces and iso- of all isotopes and can be calculated from equation 2.1 above.
topic mass?” They give you the
average mass. “Could I use this
0 7870 23 98504
0 1013 24 98584
0 1117x
to nd the unknown mass?” Yes, 24 305 amu
the equation for average mass
will include the unknown isoThe 24 305 amu is the average atomic mass from the periodic table.
topic mass.
It is a known quantity that we must have to solve the problem. But
we must be clever enough to realize that and then go look it up in
the periodic table. Solving for x we nd
x
25 9419 amu
Naturally occuring boron consists of 2 isotopes. When you think
about the atoms in a sample of boron you should picture two different types of boron atoms. The mass listed in the periodic table
for boron is an average mass of the two isotopes in the sample.
26
Chapter 2 Atoms, Molecules, Ions
Molecules Elements and Compounds
Molecules of Elements
I A molecule is more than one atom bonded together. The atoms
in the molecule are chemically joined together.
I When all of the atoms are the same (# of protons) it is called an
element.
I When the number of molecules bonded together is very large we
no longer call it a molecule. Instead it is referred to as a solid or
crystal.
Molecules Elements and Compounds 27
I The subscript in O3 means there are 3 atoms in the molecule.
Compounds
1. When there is more than one type of atom in the molecule it is
called a compound.
2. All water molecules have the formula H2 O. The proportions of
the atoms in the molecules are always the same.2
3. When atoms are grouped together in di erent ways the chemical
properties change. H2 O2 is a di erent molecule from water with
di erent chemical and physical properties.
2
This is Daltons’s law of de nite composition.
28
Chapter 2 Atoms, Molecules, Ions
3
The properties of elements are di erent from those of the compound. Hydrogen and oxygen are colorless gases while water is a
liquid that acts as an excellant solvent for many materials.
Molecular Compounds
I Elements and compounds made from only non-metals are called
molecular compounds because they exist as individual moleMolecular compounds are formed
using covalent bonding. This will
be described in detail in Chapter 7.
I
cules.
non-metals can be combined in a wide number of proportions to
3
This is Dalton’s Law of Multiple Proportions.
Molecules Elements and Compounds 29
make di erent molecular compounds.
Atoms
Compounds
H,O
H2 O H2 O2
N,O
NO2 N2 O4 N2O5
C,H
CH4 C2 H4 C2H4
Ionic Compounds
I Compounds made from metal and non-metal atoms or polyatomic
ions are ionic.
I The atoms in the compound are charged.
I The proportions of the atoms in the compound are xed and are
determined by the charge.
I The chemical formula represents the empirical formula because
the actual molecule is a crystal that contains many formula units
combined together.
30
Chapter 2 Atoms, Molecules, Ions
Ionic Crystals
Ionic compounds are formed using ionic bonding which involves
charged atoms. This occurs when
the difference in electronegativity between atoms is very large.
This idea will be described in
detail in Chapter 7.
Table salt or NaC is made from sodium and chlorine atoms.
1. The picture below shows an electron jumping from a sodium atom
to a chlorine atom creating charged ions.
2. In a solid crystal of NaC the atoms are charged.
3. The charges create attractive forces. These forces are the glue
that holds the crystal together.
4. Since the number of atoms bonded together is very large it is
referred to as an NaC crystal. There is no individual NaC
molecule. The formula NaCl is the empirical formula which indicates the simplest ratio of atoms..
Molecules Elements and Compounds 31
Empirical Formulas and Molecular Formulas
I Molecular formulas describe the actual number of atoms in a
molecule.
I Empirical formulas describe the simplest whole number ratio of
atoms in a molecule.
Actual Molecule Molecular Formula Empirical Formula
N2 O4
N 2 O4
NO2
C 6 H6
C6 H6
CH
H2 O
H2 O
H2 O
H2 O2
H 2 O2
HO
C8 H18
C8 H18
C4 H9
32
Chapter 2 Atoms, Molecules, Ions
Notice that
Molecular Formula
Negative Polyatomic Ions
OH
NO3
NO2
SO23
SO24
HSO4
CO23
PO34
HCO3
C O
C O2
C O3
C O4
CN
hydroxide
nitrate
nitrite
sul te
sulfate
hydrogen sulfate
carbonate
phosphate
hydrogen carbonate
hypochlorite
chlorite
chlorate
perchlorate
cyanide
Positive Polyatomic Ions
NH4
Hg22
ammonium
mercury (I)
n Empirical Formula
where n is some integer that multiplies the subscripts. For example
C6 H6
6 CH
where the integer 6 multiplies the subscripts of the atoms in the
empirical formula CH.
Polyatomic Ions
Certain groups of atoms prefer to stay bonded together in a charged
form. They are called polyatomic ions. Hg
is the mercuride
atomic ion and Hg22 is the mercury (I) polyatomic ion. (More than
1 atom in the ion.) Remember also that ionic compounds can be
made from all nonmetal atoms by using polyatomic ions in the compound.
I You must memorize the names, formulas, and charges of the
polyatomic ions in the table. These are the most common polyatomic ions we will use in Chem 1201.
Molecules Elements and Compounds 33
Predicting Formulas of Ionic Compounds
Additional Polyatomic Ions
CH3CO2
HPO24
H2 PO4
MnO4
CrO24
Cr2 O27
S2 O23
O22
acetate
hydrogen phosphate
dihydrogen phosphate
permanganate
chromate
dichromate
thiosulfate
peroxide
O2
superoxide
Remember that metal atoms combined with non-metal atoms make
ionic compounds. The atoms in the molecule are charged but all
of the positive and negative charges are balanced so that the net
charge is zero. Molecules made with polyatomic ions are also ionic.
The empirical formula of table salt is NaC . The
is cancelled by the
1 charge of C
1 charge of Na
so that the net charge of the
empirical formula is zero.
One way to determine the empirical formula of an ionic molecule
is to write both ions and their charges. Then cross the charges as
in the examples here. The charge on the Na becomes the subscript
of the sulfur atom and the charge on the sulfur atom becomes the
subscript of the sodium atom.
This gives the empirical formula
Na2 S with an overall charge of zero.
Here are some more examples.
34
Chapter 2 Atoms, Molecules, Ions
Determining Ionic Charges
Certain ions always have the same charge when they form ionic
molecules. The tendency of an ion to have a certain number of
electrons arises from the desire of the atom to have a lled shell of
electrons. We will learn more about this in Chapter 6.
I Memorize the charges on the ions in this periodic table.
Naming Inorganic Compounds 35
Hydrogen is an exception. H For example, Mg is in column 2A and will have a
2 charge when
can be -1 when it forms hydrides
with metals like NaH and +1 when it forms an ion. O almost always has a
2 charge as an ion.
it combines with nonmetals like
in HC
Natural Forms of the Elements
You should know the common natural forms of these elements.
Naming Inorganic Compounds
Organic Molecules
Molecules made up of primarily C, H, O, N, and S atoms. What
Organic Molecules
Oil, gasoline, alcohol, plastics, biomolecules
distinguishes organic molecules is the presence of C atoms often
bonded together in a chain of atoms. These molecules are extremely
important. All of the biological processes that life is based on involve
organic molecules. The extensive chemistry of organic molecules is
the subject of another course.
36
Chapter 2 Atoms, Molecules, Ions
Inorganic Molecules
Inorganic molecules are everything other than organic molecules.
Naming Molecular Compounds
Recall that molecular compounds result from the combination
# Atoms
Pre x
1
2
3
4
5
6
7
8
mono
di
tri
tetra
penta
hexa
hepta
octa
of 2 or more non-metal atoms. They can be combined in more
than one way so the name must indicate how many of each type of
atom are in the molecule. Here is the procedure for determining the
name. Pre xes that indicate the number of atoms in the molecule
are listed in the table at left.
Note that when the rst element has only 1 atom the mono pre x
is dropped. The correct name for NO2 is nitrogen dioxide.
Naming Inorganic Compounds 37
Examples of Naming Molecular Compounds
N2 O4 For the N atom the pre x must be 2 or di. The rst part of
name is dinitrogen.
For the O the pre x must be 4 or tetra. The second part of the
Molecule
Name
N 2 O4
dinitrogen tetroxide
N2 O
dinitrogen monoxide
NO2
nitrogen dioxide
SF6
sulfur hexa uoride
CO2
carobon dioxide
C 2O
dichlorine monoxide
Naming Ionic Compounds
Recall that ionic compounds result from the combination of
metal ions with non-metal ions. Alternatively, one of the ions
could be a polyatomic ion. The proportion of atoms in the molecule is xed and is determined by the charges on the ions. No pre x
is needed in the name to specify the number of atoms. For many
transition metal atoms the metal can have more than one possible
charge. For this course, the most important exceptions to this rule
are scandium, nickel, silver and zinc which can only be Ni2
Ag
38
Chapter 2 Atoms, Molecules, Ions
and Zn2 . Formulas that include transition metal atoms other than
Ag or Zn must also include the charge. Here is the procedure for
With ionic compounds the mono,
di, tri pre xes are not used because the proportions of atoms
in the compounds are determined
by the charges. For example, a
compound with Magnesium and
Bromide ions would combine to
give the empirical formula
determining the name.
Mg2 and Br
gives
MgBr2
The ions always combine to make
a compound with an overall charge
of zero. The only possible combination would be MgBr2
Silver ion is always Ag
Zinc ion is always Zn2
Nickel ion is always Ni 2
Naming Inorganic Compounds 39
Examples of Naming Ionic Formulas
Formula
Name
MgC 2
magnesium chloride
Na2 S
sodium sul de
FeBr3
iron (III) bromide
AgI
silver iodide
NH4 C
ammonium chloride
Na2 SO4
sodium sulfate
Finding the Charge on a Transition Metal
Common Mistakes: Students
often determine the charge on
the metal ion by “uncrossing”
the subscripts in the formula. Look-Recall that we learned above how to
ing at the formula
charges were known. For Fe3 and C
FeC 3
nd the ionic formula when the
the charges are crossed and
we immediately see that the Fe
the formula must be FeC 3 To work backwards and nd the charge
must be 3 and C must be 1
just from uncrossing the subscripts
to the charges on the ion. This on the metal when the formula is known set up a little algebraic
works almost all of the time. However, in some cases like MgO equation. The charges on the ions must sum to the charge on
uncrossing the subscripts to nd
the charges incorrectly predicts
the neutral compound which equals zero. The charge on the C is
that Mg = 1. This is wrong!!
The formula method however, is
known. (C is in a column of the periodic table where the charge
never wrong.
Mg
2
Mg
0
on the ion is always
2
1)
40
Chapter 2 Atoms, Molecules, Ions
Finding the compound from the name
Consider the compound name trinitrogen diphosphide. The use of
tri and di in the name below indicate this is a molecular compound
made up of 2 nonmetals. Note the ide ending always indicates an
element so you immediately know that phosphide = P.
Consider the compound name iron (II) phosphate. The name does
not use mono, di, tri pre xes so this must be an ionic compound.
The second name does not end in ide suggesting that phosphate
is a polyatomic ion.
Looking up phosphate in the table of poly-
atomic ions we nd that phosphate = PO 34
Use the charges on
Naming Inorganic Compounds 41
the ions to nd out how many of each ion is in the neutral compound. Crossing the charges gives Fe3 PO4 2
3 Stoichiometry and Mole-Mass
Calculations
A chemical reaction is a rearrangement of atoms
States of matter
solid
s
liquid
aqueous
aq
gas
g
1. Atoms are conserved in a chemical reaction.
2. Mass is conserved in a chemical reaction.
43
Chemical Equations
Chemical reactions are written as in this example. Reactants are on
the left and products on the right. The states of matter are important in chemistry and are given in parenthesis for each ingredient.
1. A chemical reaction is a rearrangement of atoms.
2. When atoms are combined in di erent ways or in di erent proChemical properties only become evident during a chemcial
reaction when a substance changes
chemical identity.
Physical properties are measured quantities such as mass,
density, boiling temperature, and
color.
portions the chemical properties change.
and H2 O2
The molecules H2 O
have di erent chemical properties because the arrangement of
atoms is not the same.
3. All molecules of the same compound have the same proportions
of atoms. For example, all water molecules have the formula
H2 O. In every water molecule the proportion of atoms is
2 H atoms to 1 O atom
An equivalent statement called The law of de nite proportions
states that a chemical compound always contains the same proportions of elements by mass. All samples of water have
2 1 01 g
2 1 01 g +16 0 g
16 g
2 1 01 g +16 0 g
100
11 2% by mass H
100
88 8% by mass O
4. The number in front of the ingredient indicates how many mole-
cules of that type are used in the reaction. These numbers are
called the Stoichiometry of the reaction.
5. Atoms are neither created nor destroyed in chemical reactions.
44
Chapter 3 Stoichiometry and Mole-Mass Calculations
Mass is conserved.4
2H2 g + O2 g
2 H2 O
4 H atoms
4 H atoms
2 O atoms
2 O atoms
Balancing Reactions
Sometimes reactions are written without the stoichiometry and these
reactions are not balanced.
CH4 + O2
CO2 + H2 O
4 H atoms
2 H atoms
1 C atom
1 C atom
2 O atoms
3 O atoms
I Atoms are neither created nor destroyed. The atoms are not balanced in the equation because the stoichiometry is missing.
Exercise 3.10
Balance the reaction
CH4 + O2
4
This is one of Daltons Laws for chemical reactions.
CO2 + H2O
45
Strategy
1. Balance all atoms that appear in only one place on both sides of
the equation. Start with C or H but not O.
a. The ratio of CH4 to CO2 must be 1:1 in order to balance the
C atom.
b. The ratio of CH4 to H2 O must be 1:2 to balance the H atoms.
CH4 + O2
CO2 + 2H2 O
2. Now that all the atoms found in one place (C,H) are balanced,
the remaining atoms (O) can be balanced. Note: do not change
the numbers in front of the CH4 CO2 H2 O because the C,H are
already balanced.
CH4 + 2O2
CO2 + 2H2 O
The reaction is now balanced.
Note:
When balancing reactions adjust the stoichiometry
only
CH4 + 2O2
CO2 + 2H2 O
Do not adjust the subscripts which determine the number of atoms
in each molecule.
Exercise 3.11
Balance the reaction
C2 H2 + O2
CO2 + H2 O
46
Chapter 3 Stoichiometry and Mole-Mass Calculations
Solution
1. Again, C,H are balanced
rst because they appear in only one
place on each side of the reaction.
C2 H2 + O2
2CO2 + H2O
2. O is balanced last because it appears in two places on the right
side of the reaction.
C 2 H2
O2
2CO2 + H2 O
2 O atoms
5 O atoms
Somehow we need to have 5 O atoms on the left. But on the left
oxygen is in the form of O2 So 5/2 O2 are required.
5
C2 H2 + O2
2
2CO2 + H2 O
3. The reaction is now balanced but the fraction 5/2 is generally not
preferred, especially on exams! So every ingredient is multiplied
by 2 to clear the fraction.
2C2 H2 + 5O2
4CO2 + 2H2 O
Patterns of Chemical Reactivity
1. Chemical reactivity is similar for elements of the same group.
2. Products are similar for every member of this group (except
hydrogen)..
Patterns of Chemical Reactivity 47
Note: H is a non-metal whereas
the other elements in the group
are metals. H is an exception
and does not react like the others in the group.
The engine in cars and lawn mowers is called an internal combustion engine because it relies on
the combustion of hydocarbons
(gasoline) as a sourse of energy.
The gases created in the reaction are used to drive the pistons
which in turn propels the car.
2Na s + 2H2 O
2NaOH aq + H2 g
2K s + 2H2 O
2KOH aq + H2 g
2Rb s + 2H2 O
2RbOH aq + H2 g
Types of Reactions
Combustion (Burning)
Cx H y +
O2
xCO2 +
y
H2 O
2
Combustion can occur for many reactants, however, hydrocarbons
like Cx H y and C x H y Oz are the most common. For simple hydrocarbons the reaction always looks like this.
Combination Reactions
A+B
2Mg s + O2 g
C
2MgO s
48
Chapter 3 Stoichiometry and Mole-Mass Calculations
Decomposition Reactions
AB
C
CaCO3 s
CaO s + CO2 g
I You should be able to recognize these three type of reactions. You
should also be able to
gure out the reaction for yourself. For
example, Write the decomposition reaction for CaCO3 s .
The Mole
In everyday life we count things like eggs with simple integers 1 2 3 4
For larger quantities the counting might be done in dozens or gross.
To covert between counting schemes a conversion factor is used.
12 eggs = 1 dozen
How many dozen are in 6 eggs?
6 eggs
1 doz
12 eggs
0 5 doz
How many eggs are in 2 doz?
2 doz
12 eggs
1 doz
24 eggs
For atoms and molecules we use a similar method, but instead of
counting by dozens we use a much larger group of things called a
The Mole 49
mole.
NA
6 022
1023
things / mole
Consider two samples of CO2
1 doz molecules contains
1 mole molecules contains
12 CO2 molecules
6 02
1023 molecules
12 C atoms
6 02
1023 C atoms
2
3
12
24 O atoms
12
36 atoms
2
3
6 02
1023 O atoms
6 02
1023 atoms
Molar Mass
The molar mass is the mass of one mole of molecules. To nd
the molar mass add up all of the atomic masses of each element.
Remember that the atomic masses
given in the periodic table are
average quantities that take into
account all of the naturally occuring isotopes of each element.
Exercise 3.12
Solution
Find the molar mass of C12 H22 O11
The equation to calculate the molecular mass (mass
50
Chapter 3 Stoichiometry and Mole-Mass Calculations
of 1 molecule) is
Mass =
12 atoms C mass 1 C atom
22 atoms H mass 1 H atom
11 atoms O mass 1 O atom
Looking up the atomic masses in the periodic table
Mass =
12 12 01 amu
22 1 008 amu
11 15 9994 amu
342 0 amu/molecule
To calculate the molar mass (mass of 1 mole of molecules) the same
formula is used except the units of mass from the periodic table are
now g/mol.
Mass =
12 12 01 g/mol
22 1 008 g/mol
11 15 9994 g/mol
342 0 g/mol
Formula Mass
Formula mass is the same idea as molar mass but is used for
empirical formulas. Recall that table salt, or NaC is not a single
molecule but exists instead as a crystal containing many atoms. The
The Mole 51
mass of 1 mole of the formula NaC does not represent the mass of
a mole of actual molecules. (There are no single NaC molecules.)
Instead the mass represents the mass of 1 empirical formula unit
that is written as NaC
Exercise 3.13
Solution
COMMON MISTAKE:
Students often forget to account
for the 2 in the subscript of the
formula. Mg(NO3 2 has 1 Mg
atom, 2 N atoms, and 6 O atoms
because there are 2, NO3 ions
in each formula unit.
What is the formula mass of Mg(NO3 2 ?
The problem is worked in exactly the same way as
for molar mass.
Mass =
1 Mg atom mass 1 mol Mg atom
2 [ 1 N atom mass 1 mol N atom
3 O atoms mass 1 mol O atom ]
24 31 g/mol
2 1 14 01 g/mol
3 15 994 g/mol
148 29 g/mol
Working in amu
Recall that the masses in the periodic table also represent the mass
of 1 atom in amu instead of a mole of atoms. The formula mass and
molar mass can also be obtained in amu by using atomic masses in
amu. The result will have the units amu/molecule. In other words,
52
Chapter 3 Stoichiometry and Mole-Mass Calculations
the formula mass of Mg(NO3 2 is also 148 29 amu/molecule.
The 12C Scale
Atomic masses found in the periodic table are measured against a
standard using the isotope 12
6 C. Atomic mass is de ned
12.0000 g
12.0000 amu
1 mole 12
6 C atoms
1 12
6 C atom
Both the unit of gram and the amu are de ned in this way.
Percent Composition
The percent composition by mass is calculated in the same way any
percent is calculated. The percent by mass of C in the molecule
C12 H22 O11 is
%C
Mass of all C
Total Formula Mass of C12 H22 O11
12 Mass 1 carbon
Formula Mass
12 12 01 g/mol
342 0 g/mol
100
100
42 1%
Recall that the formula mass 342.0 g/mol was calculated above.
The Mole 53
Converting between Mass and Moles
conversion factors in the calculations.
Grams to Moles
Exercise 3.14
How many moles are in a 15.0 g sample of
CO2 ?
Solution
Molar mass is the conversion factor. Molar mass = 12.01 g/mol +
Note: putting the units into the
formula helps you to know when
to multiply by formula mass and
when to divide. Using units in
your calculations is a mature way
to work.
2(16 g/mol) = 44.01 g/mol
15 g
Moles to Grams
1 mole
44 01 g
0 34 mol
54
Chapter 3 Stoichiometry and Mole-Mass Calculations
Exercise 3.15
How many grams CO2 are required to have 0 5
mol?
Solution
Molar mass is the conversion factor.
0 5 mol
44 01 g
1 mol
22 g
Moles to Particles
Exercise 3.16
How many O atoms are in 0.7 mol sample of
CO2 ?
Solution
must rst nd the number of CO2 molecules.
0 7 mol
6 02
1023 CO2 molecules
1 mole
4 21 1023 CO2 molecules
The Mole 55
To nd the number of O atoms in the sample multiply the number
of CO2 molecules by 2 since there are 2 O atoms in each CO2
molecule.
4 21 1023 CO2 molecules
2 O atoms
CO2 molecule
8 43 1023 O atoms
Particles to Moles
Looking at bikes in a bikerack you count 8 tires. How
many bikes are present?
Exercise 3.17 A sample of CO2 contains 8 43 1023 CO2 molecules. How many moles of CO2 does the sample contain?
Solution
8 tires
1 bike
2 tires
4 bikes
moles
A sample of CO2 has 8 O atoms.
How many CO2 molecules are
present?
8 43
1023 CO2 molecules
6 02
1 mole
1023 CO2 molecules
1 40 moles
1 CO2 molecule
2 O atoms
4 CO2 molecules
8 O atoms
Exercise 3.18
A sample of CO2 contains 8 43
What is the mass of the sample?
Solution
Ask yourself the question:
1023 O atoms.
56
Chapter 3 Stoichiometry and Mole-Mass Calculations
1. The key point to solving this problem is to realize that: To
nd
the mass of the sample we need to nd the moles of CO2 and
then multiply by the molar mass of CO2 The sample is described
by the problem as the number of O atoms. To convert that to
CO2 molecules we must take into account that each molecule
contains 2 O atoms. So the sample contains half as many CO2
molecules.
2. The second key point to understand is that to convert from mole-
mass =
8 43
6 02
1023 O atoms
1 CO2 molecule
2 O atoms
1 mole
1023 CO2 molecules
44 01 g/mol
30 8 g
Empirical Analysis
The formula of a compound can be deduced by chemical analysis.
When the percent composition by mass of the atoms is measured
Empirical Analysis 57
experimentally the formula can be calculated. Remember that one
the same ratio of atoms. The ratio of atoms in the molecule must
Analogy: An alien asks a human what a bike consits of. The
human, trying to fool the alien
says, “A dozen bikes contains
1 doz frames and 2 doz wheels
and all bikes are the same.” The
clever alien quickly deduces that
each bike must be frame1 wheel2
In other words the ratio of atoms
in a single molecule is the same
as the ratio of moles of atoms in
a large sample of molecules.
be the same as the ratio of moles in a large sample.
Example 3.19
Experimental analysis nds that the percent
composition of Hg in Hg x C y is 73.9%. determine the empirical
formula.
Strategy
1. Assume 100 g of material.
100 g = 73.9 g Hg
0 739
100 g = 26.1 g C
a. 0 739
b. 1
2. Convert these masses to moles.
3. Ratio x y in a Hgx C y molecule is the same as the mole ratio
Hg/C in the 100 g sample.
4. Convert to whole numbers by dividing both x and y by the smaller
value.
Solution
73 9 g
1 mol Hg
200 g Hg
0 368 mol Hg
26 1 g
1 mol C
35 5 g C
0 735 mol C
The molecule must be Hg0 368 C 0 735 To convert to whole numbers
58
Chapter 3 Stoichiometry and Mole-Mass Calculations
divide both values by the smallest.
Hg 0 368 C 0 735
0 368
0 368
OR
Hg1 C 2
Remember that this is an empirical formula. All we know is that the
ratio of Hg to C is 1:2. We do not know what the true molecule
looks like. It could be Hg5 C 10 Since this molecule is ionic the
actual crystal will have many Hg atoms and C atoms. But the
ratio will be 1:2. Recall the crystal structure of NaC
For the interior atoms in this crystal each Na is surrounded by 4 C
atoms (8 atoms in 3 dimensions) and each C is surrounded by the
same number of Na atoms. The empirical formula is the simplest
ratio of atoms or NaC
Empirical Analysis 59
Exercise 3.20
A compound is found to have the following com-
position by mass: 62.1% C, 5.21% H, 12.1% N, 20.7% O. Determine
the empirical formula.
Solution
Assume 100 g and calculate the moles of each atom
in the 100 g sample.
1.
2.
3.
4.
62 1 g
12 011 g/mol
5 1703 mol C
5 21 g
1 008 g/mol
5 1687 mol H
12 1 g
14 007 g/mol
0 8638 mol N
20 7 g
16 00 g/mol
1 2939 mol O
5. The molecule must have the atom ratios
C5 1703 H5 1687 N 8639 O1 2939
6. To convert to whole numbers divide each number by the smallest
or 0.8639. This gives the formula
C5 98 H5 98 N1 O1 50
These numbers can be rounded to
C6 H6 N1 O1 50
The number1.50 has 3 signi cant gures so it should not be
rounded to the whole number 2. Normally, for problems like this
in Chem 1201 it should not be necessary to round by more than
01
7. To convert 1.50 to a whole number
nd the smallest possible
whole number, n that allows the product, n 1 5 to be rounded
60
Chapter 3 Stoichiometry and Mole-Mass Calculations
Suppose empirical analysis
produced this empirical formula.
to a whole number. In this case the smallest whole number is
n
2 because 2 1 5
3 Multiply every subscript in the
formula by n 2
C6 H6 N1 O1 50 2
gi es
C12 H12 N2 O3
Pb1 PO4 1 33
To convert the subscript 1.33 to
a whole number multiply both
subscripts by n 3 This gives
Pb3 PO4 3 99
The subscript 3 99 is within
0 1 of 4 so it can be rounded
up to 4.
Pb3 PO4 4
Molecular Formula from Empirical Formula
The actual molecular formula can be found from the empirical formula by comparing the mass of the empirical formula to the mass
of the molecular formula.
Molecular Formula
An Analogy: A train consists
of many train cars that each contain one box with 4 wheels. So
we could write the empirical formula as
10 tons
1 car =
Box1Wheel4
Empirical Formula
C2 H2
2 CH
26 g/mol
2 13 g/mol
The table shows an actual molecular formula and the simplest whole
ratio of atoms described by the empirical formula. Since the molar
mass (given) is twice the mass of the empirical formula the actual
Suppose we knew that each car
was 10 tons and the total mass molecular formula can be found by multiplying the subscripts in the
of the train was 100 tons. That
tells us there are actually 100 tons/10
tons = 10 cars in the train. So empirical formula by 2.
the formula for the whole train
would be
Another way to describe this idea is
100 tons
1 train =
Box10 Wheel40
That fromula is equivalent to a
molecular formula that describes
the actual train whereas the empirical formula expains the simplest whole number ratio.
actual molar mass
empirical molar mass
n
The actual molecular formula is found by multiplying the subscripts
Quantitative Calculations 61
in the empirical formula by the number n
n
empirical formula
molecular formula
For the example above with the empirical formula CH and actual
molar mass given as 26 g/mol
26 g/mol
13 g/mol
2
CH
C2 H2
and
2
Quantitative Calculations
An Example from Everyday Life
How many cups of hot chocolate can be made from 6 tsp of cocoa
according to the following recipe?
2 tsp cocoa + 2 tsp sugar + 3/4 cup milk
1 glass H.C.
All of the relationships we need are in the recipe.
6 tsp cocoa
1 glass H.C.
2 tsp cocoa
3 glass H.C.
I Notice that to use the recipe the amount of cocoa must be known
in tsp! If the value was given in grams the mass would rst have to
62
Chapter 3 Stoichiometry and Mole-Mass Calculations
be converted to tsp.
Chemical calculations work the same way. The moles of one ingredient are related to moles of another ingredient by the stoichiometry
given in the chemical recipe.
Suppose we wish to calculate the moles of B that will be produced
from the reaction of y moles of A. The calculation would be:
Quantitative Calculations 63
Example 3.21
Calculate the moles of O3 that can be produced
from 4 moles of O2 in the reaction
3O2
Strategy
2O3
Since the question gives the moles of O2 and asks
about moles of O3 this relationship can be found directly using the
chemical recipe.
Solution
4 moles O2
Example 3.22
2 moles O3
3 moles O2
2 67 mol O3
Calculate the moles of O3 that can be produced
from 16 g of O2 in the reaction
64
Chapter 3 Stoichiometry and Mole-Mass Calculations
Strategy
To nd the relationship between the moles of O2 and
moles of O3 the chemical recipe must be used. Since the chemical
recipe only works in units of moles we must rst convert 16 g of O2
to moles.
Stoichiometry
The stoichiometry in the balanced chemical equation could
be used to represent moles or molecules. Under certain conditions,
as we will learn in Chapter 10,
it can also be used to represent
other quantities such as the relationships between the volumes
of gases.
Solution
16 g
1 mol O2
32 g/mol
2 mol O3
3 mol O2
0 33 mol O3
Exercise 3.23
1. How many moles of CO 2 can be produced when 0.3 mol C6 H12 O6
undergo reaction?
Quantitative Calculations 65
2. How many grams of C 6 H12 O6 are needed to make 2.00 g of C2 H5 OH?
C6 H12 O6
Solution Part 1
2C2 H5 OH + 2CO2
Moles of CO2 can be compared to moles of
C6 H12 O6 directly from chemical recipe.
0 3 mol C6 H12 O6
Solution Part 2
2 mol CO2
1 mol C6 H12 O6
0 60 mol CO2
Grams of C6 H12 O6 cannot be compared
directly to g of C2H5 OH using the chemical recipe. The chemical recipe only works in moles. The rst step is to convert grams
C2 H5 OH to moles then use the recipe to nd the moles C6 H12 O6
required. Once the moles are known grams can be found from the
formula mass.
66
Chapter 3 Stoichiometry and Mole-Mass Calculations
Common Mistakes: Pay attention to the units so the conversion factors are used correctly.
Is the mole conversion factor 21
or 12 ? Putting the units will help
you decide. Using the units in
the calculation as shown here forces
the correct conversion factor 12
to be used.
g C6 H12 O6
2.00 g
1 mol C2 H5 OH
46 07 g
1 mol C6 H12 O6
2 mol C2 H5 OH
180 16 g
mol C6 H12 O6
3 91 g
Limiting Reagent
I There will be a limiting reagent problem on every exam including
the nal!
Limiting Reagent 67
In the picture above it is clearly seen that 6 wheels + 2 frames
can only make 2 bikes. The frames are called the limiting reagent
because they are completely used up and frames limit how much
product can be made. The wheels are the reagent in excess because
there are 2 wheels left over when the reaction is complete.
Strategy for Limiting Reagent Problems
1. Calculate how much product is produced assuming reactant 1 is
the limiting reagent and is all used up in the reaction.
6 tires
1 bike
2 tires
3 bikes
2. Calculate how much product is produced assuming reactant 2 is
the limiting reagent and is all used up in the reaction.
2 frames
1 bike
1 frame
2 bikes
3. The limiting reagent is the reactant that produced the smaller
amount of product. In this case the smaller product in step 1
and 2 is 2 bikes. So the actual bikes produced is 2 and frames is
the limiting reagent.
4. To
nd out how much of the other reactant is left over work
backwards from the actual product produced to nd how much
of that reactant was required. Then use the equation Excess =
Initial - Required.
2 bikes
2 tires
1 bike
4 tires required
Excess
Initial
Required
6
2 tires left over
4
68
Chapter 3 Stoichiometry and Mole-Mass Calculations
Exercise 3.24
How much CaO (s) is expected from the reaction
of 4.35 g of Ca and 2.8 g O 2 ? How much excess reactant is left over?
2Ca s + O2 g
Strategy
2CaO s
Since the amounts of two reactants are given we
know this is a limiting reagent problem.
Solution
Common Mistakes: Students
1. Calculate the amount of product expected assuming Ca is the
often make the mistake of simply choosing the reactant with
limiting reagent and completely reacts:
the smaller amount of grams as
the limiting reagent. This strat1 mol Ca
2 mol CaO
egy does not work. The limiting
4 35 g
reagent also depends on the sto40 08 g
2 mol Ca
ichiometry and the molar mass
of the reactant. It is very hard to
56 08 g
simply guess which reactant is
limiting. That is why the calcu1 mol CaO
lations are done for both reactants as in the model calculation
here. This is the only proper
6 1 g CaO
way to determine the limiting reagent.
2. Calculate the amount of product expected assuming O2 is the
Theoretical Yields 69
limiting reagent and completely reacts:
1 mol O2
32 00 g
28g
2 mol CaO
1 mol O2
56 08 g
1 mol CaO
9 8 g CaO
3. The smaller amount of product calculated in steps 1 and 2 will
be the actual amount that will be obtained. So 6 1 g CaO is the
predicted amount of product and Ca is the limiting reagent.
4. The reactant in excess must be O2 since it produces the larger
amount of product in step 2. To nd the amount of O2 used up
in the reaction calculate how much oxygen is required to produce
the actual 6 1 g of CaO.
61g
1 mol CaO
56 08 g
1 mol O2
2 mol CaO
32 00 g
1 mol O2
1 7 g O2 required
The excess oxygen left over after reaction can be calculated
Excess
Theoretical Yields
Initial
Required
28g 17g 11g
70
Chapter 3 Stoichiometry and Mole-Mass Calculations
Sometimes chemical reactions do not produce the expected amount
of product. Side reactions can occur where reactants are converted
to other products by way of a completely di erent reaction. Also in
a real experimental situation there is almost always some product
left behind in the container because it is not entirely scraped from
the walls. The Experimental Yield is de ned
Actual Yield
100
Theoretical Yield
1. Actual Yield is determined by the actual experiment. It is usually
the given quantity. In the cartoon above the actual yield is 1 bike.
%Yield =
2. Theoretical yield is what the limiting reagent calculation predicts
we should get based on the reactants available in the problem. In
the cartoon 2 bikes were theoretically expected assuming tires and
frames go together to make normal bikes and nothing unexpected
Theoretical Yields 71
happens. (i.e. we did not expect to get a defective bike.)
1 bike
2 bikes
%Yield =
100
50% Yield
Exercise 3.25
1.4 mol of H2 and 0.8 mol of O2 were reacted
to produce water. The actual amount of water obtained was 0.5 mol.
What was the yield?
2H2 + O2
2H2 O
Solution
1. First the theoretical yield must be calculated for this limiting
reagent problem.
a. Product expected if H2 is the limiting reagent:
1 4 mol H2
2 mol H2 O
2 mol H2
1 4 mol H2 O
b. Product expected if O2 is the limiting reagent:
0 8 mol O2
2 mol H2 O
1 mol H2
1 6 mol H2 O
c. The smaller amount of product is 1.4 mol H2 O so this is the
actual amount expected and H2 is the limiting reagent.
2. Calculate the Yield
%Yield
0 5 mol
1 4 mol
35 7%
100
4 Aqueous Reactions
Properties of Aqueous Solutions
When a small amount of chemical material is dissolved in a pure liquid, the SOLVENT is the majority material and the SOLUTE the
minority species. The entire mixture is called the SOLUTION.
Ionic Compounds in Water
When ionic compounds are dissolved in water the ions separate and
become surrounded by water molecules. Here are some typical ways
73
in which ionic salts dissolve in water.
NaC
s
Na
aq
C
aq
MgBr2 s
Mg2
aq
2Br
aq
K2 S s
2K
aq
S2
aq
1. In water, the solid K2 S crystal comes apart into ions. The ions
separate from each other.
2. Each ion (cation and anion) is surrounded by a sphere of water
molecules called a hydration sphere.
Water, H2 O is a polar molecule. The oxygen atom has a
partial negative charge and the
hydrogen atoms have partial positive charges. Polar molecules
will be discussed in Chapter 9.
3. Ions carry charge, so their movement through the solution allows
the solution to conduct electricity. Since all of the K2 S breaks
apart into ions that carry charge, K2 S is called a strong electrolyte.
74
Chapter 4 Aqueous Reactions
4. The symbols S2
aq refers to separate sul de ions surrounded
by a sphere of water molecules.
5. The symbol K2 S aq
implies that the K
and S2
ions have
separated.
Molecular Compounds in Water
Common Mistake
Molecular compounds made up of nonmetal elements do not break
Alcohol molecules like CH3 OH
contain mostly C and H atoms
up into separate ions when they dissolve in water.
with an OH group. This is not
the OH polyatomic ion!!! Alcohol molecules are not ionic molecules that exhibit charges.
Here are some examples of molecular compounds that do not make
ions when they dissolve in water.
sugar
C12 H22 O11
methanol
CH3 OH
ethyl alcohol CH3 CH2 OH (see footnote5 )
Recall that molecules that are molecular compounds do not contain
charged ions. When these compounds dissolve in water the com5
Ethyl alcohol is a very weak acid so there is a tiny amount of dissociation.
CH3 CH2 OH aq
H
aq
CH3 CH2 O
aq
75
pound remains intact in an uncharged form.
No ions are formed
when the compound dissolves.
Acids and Bases
Some acids and bases are molecular molecular compounds that do
Strong Acids
Name
HC
HBr
HI
HC O3
HC O4
HNO3
H2 SO4
Weak Acids
hydrochloric
hydrobromic
hydroiodic
chloric
perchloric
nitric
sulfuric
Name
make ions in water. In general, acids donate H
accept H
aq
HC
HCHO2
HC7 H5 O2
aq and bases
aq
H
aq
C
aq
Formic
Benzoic
Strong Bases
LiOH
NaOH
KOH
RbOH
CsOH
Ca(OH)2
Sr(OH)2
NH3 aq
H2 O aq
NH4 aq
OH
aq
Weak Bases
NH3
An acid or base not listed in
as a strong acid or base would
be considered a weak acid. or
base.
I Note: NH3 is a molecular compound that is an exception. It partially makes ions in solution because it is a weak base.
Strong and Weak Acids/Bases
Most acids are molecular compounds that create ions when in water.
76
Chapter 4 Aqueous Reactions
1. Strong Acids and Bases completely dissociate into ions.
HBr aq
H
aq
Br
aq
2. Weak Acids and Bases only partially dissociate into ions. Under
typical conditions, many weak acids have less than 1% dissociation.
Common strong acids are listed in the table in the margin. To identify the strength of an unknown acid such as formic acid, HCHO2
look up the acid in the table of strong acids. HCHO2
pear in the table of strong acids so you should assume that it is a
weak acid.
HCHO2 aq
H
aq
CHO2 aq
Strong, Weak, and non-Electrolytes
Materials are classi ed as strong, weak, or nonelectrolyte depending
on the electrical conductance of a solution of the compound in
water. The electrical current is carried by the charged ions in the
solution. In general, the more ions a solution contains the higher
the electrical conductance.
77
non-Electrolytes
These are typically molecular compounds whose solution does
not conduct electricity well. The compound dissolves in water but
does not create ions necessary to carry the charge. Examples
are sugar, methanol, and ethyl alcohol.
Common Mistake: Students often believe that materials have
to be insoluble in order to be a
non-Electrolyte. That is not true.
Non-Electrolytes like sugar are
highly soluble but they do not
produce ions in water.
Strong Electrolyte
Ionic compounds that fully dissolve and completely ionize
are strong electrolytes because they produce ions that carry charge.
Examples are salts like NaC
strong acids like HC
and strong
bases like NaOH.
Weak Electrolytes
Ionic materials that are highly soluble but only partially ionize are
78
Chapter 4 Aqueous Reactions
weak electrolytes. Weak acids like HCHO2 HF and weak bases
like NH3 are examples of weak electrolytes. Only a small percent actually dissociate in
ions making a solution with a low concentration of ions that is poorly
conducting. Again, the material can be highly soluble. However,
only a small percent ionize.
Example 4.26
Determine whether the following materials will
be strong, weak, or non-electrolytes when dissolved in water. KC
HC
HF, KOH, CO2
1. KC is an ionic salt. It is not a weak acid or base. It completely
ionizes so it is a strong electrolyte.
2. HC is a strong acid according to the table on p. 75
3. HF is an acid but it is not in the strong acid table on p. 75. This
means HF is a weak acid that only partially ionizes. This is a
weak electrolyte.
4. KOH is a strong base. It is listed in the table on p. 75. Strong
bases fully ionize so this is a strong electrolyte.
79
5. CO2 is a molecular compound and does not ionize when dissolved
in water. This is a nonelectrolyte.
Strategy
The way to do this problem is to determine if ions are
made when the material dissolves. If so, is the concentration large
or small. Then refer to the diagram above. Most of the time the
following owchart will work.
Solubility Rules
In order for salts to ionize in water energy must be used to pull the
ions in the crystal apart. Some energy is released when the ions
become solvated or surrounded by water molecules. For now it
is enough to know that if the net energy required to separate the
atoms is too large the process will not occur and the material is not
soluble. (The second law of thermodynamics is required to explain
Example of low solubility where
very little of the solid dissolves.
this result and will be discussed in Chapter 19.) Here are the rules
that govern which ionic materials are soluble in water.
80
Chapter 4 Aqueous Reactions
Soluble
Li Na
K Rb
Exceptions
None
NH4 NO3 C O4
None
acetate = C2 H3 O2
None
C
SO24
Br I
Insoluble
OH
Exceptions
Li Na
K Rb
,NH4
Ca2 Sr2 Ba2
O2
Li Na
Ag Pb2 Cu Hg22
K Rb
,NH4
Sr2 Ba2
Ca2 Sr2 Ba2 Pb2
CO23 PO34
Li Na
K Rb
NH4
Hg22
S2
Li Na
K Rb
, NH4
Ca2 Sr2 Ba2
I You must memorize this table. Please note how the rules are related
to each other.
Exercise 4.27
Which of these salts are soluble and which are
insoluble in water: NaC
AgBr, Pb(NO3 2 K2 SO4 CaS?
Solution
1. The
rst two rules take care of most situations. NaC K2 SO4
are soluble because all compounds with alkali metals are soluble.
2. NO3 is always soluble so Pb(NO3 )2 is soluble.
3. S2
is usually insoluble but Ca2 is one of the exceptions so CaS
is soluble. (Most compounds with alkali metal or alkaline earths
are soluble.)
4. Br is usually soluble but Ag is an exception so AgBr is insoluble.
Net Ionic Equations
AgNO3 and NaC are both soluble in water. However, when they
are mixed, the insoluble compound AgC is formed. Since AgC is
81
insoluble, it forms a solid material called a precipitate that falls
out of the solution.
Here is the overall reaction:
AgNO3 aq
NaC
aq
AgC
s
NaNO3 aq
The reaction equation that describes the actual chemical changes
that occur in this reaction is
Ag
aq
C
aq
AgC s
A reaction written in this way is called a Net Ionic Equation. It
shows only the chemical species that change in the reaction. Notice that Na
aq and NO3 aq are not included in the Net Ionic
Reaction. They are Spectator Ions and do not change chemically
in any way during the reaction.
I
spectator ions.
82
Chapter 4 Aqueous Reactions
Writing Net Ionic Equations
1. Start with a balanced overall reaction
MgBr2 aq
Na2 CO3 aq
MgCO3 s
2NaBr aq
2. Strong electrolytes in the aq
form are broken up into the
separate ions. The s
g forms cannot be broken up into
ions. Weak electrolytes such as HF aq and non electrolytes
like CH3 OH aq would not be broken up into ions.
Mg2
aq
2Br
MgCO3 s
aq
2Na
2Na
aq
aq
2Br
CO23
aq
aq
3. Cross out species that are in the same form on both sides of
the equation. These are the spectator ions. The remaining
83
equation is the Net Ionic Equation.
Mg2
aq
CO23
The spectator ions are Na
Example 4.28
aq
MgCO3 s
aq and Br
aq
Write the net ionic equation for the reaction
Na2 S aq
ZnSO4 aq
?
Strategy
1. To nd the possible products exchange the ions. Take the positive
ion from the Na2 S, with the negative ion from ZnSO4 . This gives
Na2 SO4
2. Then take the positive ion from the ZnSO4 with the negative
ion from the Na2 S. This gives ZnS.
3. Now apply the solubility rules from p. 80.
a. Na2 SO4 is soluble and will exist in the aq form.
b. ZnS is insoluble and will exist as a solid.
Solution
1. The overall balanced reaction is
Na2 S aq
ZnSO4 aq
Na2 SO4 aq
ZnS s
84
Chapter 4 Aqueous Reactions
2. Break up the aq forms into separate ions.
2Na
aq
S2
2Na
aq
SO24
aq
Zn2
aq
SO24
aq
ZnS s
aq
3. Cross out ions in the same form on both sides of the equation.
This results in the net ionic equation:
Zn2
aq
S2
aq
ZnS s
4. The spectator ions are the chemically unchanged species: Na
and SO24
aq
aq
Types of Solution Reactions
Exchange Reactions (Metathesis Reactions)
To Find Products
1. Exchange the ions
2. Cross charges to de-
termine the formula
Exchange reactions involve the exchange of positive ions from one
compound with negative ions from another compound.
example below, the Cu2
In the
ion from CuC 2 combines with the S2
ion from K2 S to make the product CuS s
85
Common Mistake
Notice in the balanced exchange reaction below the subscripts in
the reactant compounds are not the same as the subscripts in the
product compounds. The subscripts depend on the charges of the
ions combined to make the compound. To nd the subscripts in the
product compounds you must consider the charges on both ions in
the product compound. In the gure below the method of crossing
the charges (p. 33) is used to nd the subscripts.
Acid Base Neutralization Reaction
HC aq
NaOH aq
H2 O aq
Acid
Base
Salt
NaC
aq
water
86
Chapter 4 Aqueous Reactions
The net ionic reaction shows that the net chemical change is the
Common Mistake
neutralization of the acid H and the base OH to make water.
CH3OH and C2H5 OH are alcohols. They are not bases!!!
H
aq
OH
aq
H2 O l
Homogeneous and Heterogeneous Solutions
1. Homogeneous solutions have the same composition through-
out the mixture. A salt dissolved in pure water is an example of
a homogeneous solution.
2. The composition of heterogeneous solutions varies from point
to point throughout the mixture. River water is an example of a
heterogeneous solution. There is sand, bug parts, and so on at
di erent points throughout the mixture. Air is an example of a
heterogeneous mixture since certain regions have higher concentrations of perfume, smoke, etc.
Concentration (Molarity)
Solutions are di erent from solid samples in the respect that in order
to deliver a certain quantity of solute the volume and concentration of the solution must be known. The most common unit of
concentration in chemistry is Molarity (M) de ned
M=
moles solute (mol)
mol
=
total volume (L)
L
Experimental Preparation of a Solution
Concentration (Molarity) 87
The solution is prepared by adding the desired number of moles
of solute to a volumetric ask and then diluting with water until
the total volume is exactly the desired quantity. Due to adhesive
forces (Chapter 11) the water is attracted to the walls of the glass
forming a negative meniscus (a downward curve) that can be very
accurately aligned with a calibrated mark on the container wall.
Calculations with Molarity
Finding the Concentration
88
Chapter 4 Aqueous Reactions
Exercise 4.29
Common Mistakes: Students often forget to
covert to liters and wrongly divide by 500 instead of 0.500 L.water
is added to 15 g of NaC so that the total volume is 500 mL, what is
the concentration of NaC ?
Solution
Molarity is de ned
M=
moles solute (mol)
total volume (L)
Common Mistakes: Students The total volume is given so the moles of NaC must be calculated.
have a very hard time remembering the stoichiometry when
15 g /58.44 g/mol
ionic compounds dissolve. ReM=
0 500 L
member for a molecule like CuC 3 aq
the C aq conc. is 3 the Cu aq
conc.
0 513 M
Exercise 4.30
If water is added to 7.5 g of K2 S so that the total
volume is 250 mL, what is the concentration of K and S2 ?
Concentration (Molarity) 89
Solution
1. When K2 S dissolves here is the process that occurs.
K2 S aq
2K
S2
aq
aq
The stoichiometry is important because the concentration of K
is twice that of S2 aq
M=
aq
moles solute (mol)
total volume (L)
The total volume is given so the moles of K2S must be calculated.
M=
7 5 g /110.27 g/mol
0 250 L
0 272 M K2 S
2. To
nd the concentration of K aq we look at the chemical
formula, K2 S, to see how many K aq ions are in each formula
unit. There are 2 K ions in each formula unit.
0 272 M K2 S
2 mol K
1 mol K2 S
3. The same thing is also done to
0 544 M K
nd the S2
aq
aq concentration.
0 272 M K2 S
1 mol S2
1 mol K2 S
0 272 M S2
Species
Conversion
Concentration
Factor
K2 S aq
K
aq
S2
aq
1
2 mol K
1 mol K2 S
1 mol S2
1 mol K2 S
0 272 M
0 544 M
0 272 M
aq
90
Chapter 4 Aqueous Reactions
Liters to Moles
The conversion factor used to convert from liters to moles or vice
versa is the Molarity.
M mol/L
Exercise 4.31
V L
mol
How many moles of HNO 3 are contained in 2.00
L of 0.200 M HNO3 ?
Solution
M mol/L
V L
0 200 mol/L 2 00 L
mol
0 400 mol HNO3
Moles to Liters
Exercise 4.32
2.0 mol HNO3 ?
What volume of 0.3 M HNO3 is needed to supply
Concentration (Molarity) 91
Solution
M mol/L
V L
mol
0 3 mol/L V
V
2 0 mol
2 0 mol
0 3 mol/L
67L
Dilutions
The basic idea of a dilution problem is that water is added to a
solution and the volume changes but the moles of solute remain the
same. Sometimes it is not always easy to gure out that an exercise
is a dilution problem. You have to convert the words of the problem
92
Chapter 4 Aqueous Reactions
to a picture in your head describing what is really taking place in
order to be able to understand that what is being described is a
dilution problem.
Exercise 4.33
How much 1.0 M CuSO4 solution is required to
make 250 mL of a 0.1 M CuSO 4 solution?
Solution
This is a dilution problem in disguise. If you under-
stand the process described by the words of the exercise you will see
this picture in your head and realize this is a dilution problem. Notice that after water is added to the container, the concentration
decreases from 1.0 M to 0.1 M.
To solve for the unknown volume use the equation
Note that this equation only works in a dilution problem where
Solution Stoichiometry 93
no reaction takes place.
10M V
V
0 1 M 0 250 L
0 025 L
Solution Stoichiometry
In Chapter 3 we performed calculations using the stoichiometry of a
reaction to convert from mass of reactant to mass of product. The
diagram illustrates how to nd the mass of product B produced from
a certain mass of reactant A. To convert from moles of A to moles
of B requires the balanced chemical recipe. Before the recipe can
be used the mass of A must be converted to moles of A.
The calculation in solution is very similar except that CONCEN-
94
Chapter 4 Aqueous Reactions
TRATION
MOLE relationships are involved using Molarity and
Volume. Here is a diagram illustrating how to nd the concentration
of product B produced from a certain volume and concentration of
reactant A.
Example 4.34
MgO is a common antacid used to relieve stom-
ach acid by neutralization. Let us imagine that stomach acid is HC
and is present with a concentration of 0.15 M. What volume of this
acid can be neutralized with a 250 mg tablet of MgO?
MgO s
Strategy
2HC
aq
MgC 2 aq
H2 O
Solution Stoichiometry 95
1. Mass MgO s can be converted to moles using the formula mass.
2. Use the chemical equation to convert moles MgO to moles HC
aq
neutralized.
3. Use the Volume Concentration = Moles to
nd the unknown
volume of 0.15 M HCl
Solution
0 250 g
1 mole MgO
40 31 g
Now nd the volume of HC
2 mol HC
1 mol MgO
0 0124 mol HC
96
Chapter 4 Aqueous Reactions
V L
M mol/L
mol HC
V
0 0124 mol HC
0 15 M
V
0 083 L or 83 mL
Titrations
Titrations are special solution reactions that are used to nd out
information about an unknown solution.
1. Suppose we have a solution of base NaOH but the concentration
is unknown.
TITRATION
ANALOGY
2. A known concentration of the acid HC is added to neutralize the
base.
Have you ever participated
in a contest to guess the numHC aq
NaOH aq
H2 O aq
NaC aq
ber of jelly beans in large jar?
Suppose we gured it out this
way. We add 10 insects to the
3. From the reaction stoichiometry
jar and each one can eat 1 jelly
bean. Then we ask the question
1 mol base
“Are there any jelly beans left?”
Moles acid added
Moles unknown base
1 mol acid
If there are we add 1 more insect and so on until all the jelly
beans are gone. We know how
4. The point where enough moles of acid have been added to react
many insects we added to eat all
with all of the NaOH is called the equivalence point.
of the jelly beans so we know
how many jelly beans were in
the jar. That’s how a titration
5. Note: Suppose the chemical equation was
works. In a titration the color
change of an indictor tells us when
2Acid + Base
Neutralization
all of the reactant is gone.
Then this equation would be used.
Moles acid added
1 mol base
2 mol acid
Moles unknown base
Actually, there are no special fromulas that are needed to solve these
problems. The same solution equations you already know will work
Titrations 97
By now you should be an expert
at converting between mL and
L. You should be able to do the
conversion in your head without writing it down or referring
to a calculator.
just ne.
Example 4.35
35 mL of an unknown concentration of an
HBr solution is titrated with 0.1 M Ca OH 2 and 73.7 mL is required to reach the equivalence point. What is the unknown acid
concentration?
Strategy
1. Use Volume Concentration = Moles to
nd moles of Ca OH 2
2. Use the chemical equation to convert moles Ca OH 2 to moles
HBr aq neutralized.
3. Use Volume Concentration = Moles to
nd the unknown con-
centration of HBr
Solution
At the equivalence point just enough Ca OH 2 has
98
Chapter 4 Aqueous Reactions
been added to react with all of the HBr.
The moles of HBr are
determined from the moles of Ca OH 2 used to react with all of the
HBr.
2 HBr aq
Ca OH 2
moles of Ca OH 2
aq
2 H2 O aq
2 mol HBr
1 mol Ca OH 2
MBase VBase
0 1 M 0 0737 L
2 mol acid
1 mol base
2 mol acid
1 mol base
Ca Br 2
aq
mol HBr
mol HBr
0 01474 moles HBr
Once the unknown moles of HBr have been determined use this
equation to nd the unknown concentration.
vol
conc
MAcid 0 035 L
MAcid
moles HBr
0 01474 moles HBr
0 421 M
Mixing Solutions
Suppose we have a box that contains a volume of 1000 L of water. Into this box we put 1 bicycle and 2 tricycles. The bikes come
apart in the water into frames and wheels. What is the concentration of wheels and frames in the box?
Mixing Solutions 99
Finding the concentrations
1. To nd the number of wheels Add the wheels from both sources.
We have to consider the fact that bicyles have 2 wheels per bike
2. To
nd the concentration add up all wheels and divide by the
total volume. To do that we calculate
conc wheels =
bikes*2 + tricycles*3
1000 L
8 wheels
1000 L
3. The calculation is slightly di erent for frames because each bike
has only 1 frame.
conc frames =
bikes + tricycles
1000 L
3 frames
1000 L
100
Chapter 4 Aqueous Reactions
Mixture Example Problem
Suppose we have SOLUTION 1: 100 mL of 0.15 M NaC added to
SOLUTION 2: 250 mL of 0.2 M MgC 2 Using the analogy above,
conc C
=
To nd the total moles of C
total moles of C
total volume
we have to add the moles from
both (1) NaC and (2) MgC 2 Remember, moles = molarity
volume .
conc C
conc C
mol C
from NaC
mol C from MgC 2
total volume
0 1 L 0 15 mol/L
2 0 250 L 0 2 mol/L
0 1 L + 0 250 L
0 329 M
How to Think about Word Problems 101
Caution: Remember that to nd moles of C
tiply moles MgC 2
from MgC 2 mul-
2 moles C
1 mole MgC 2
to nd wheels/bike or wheels/tricycle!! That is where the 2 comes
from in the numerator of the last equation.
Caution: Remember to work in liters!
The equation for the concentration of Na is a little di erent. You
have to think! For Na all of the ions comes from the NaC only
But the volume changes when the two solutions are added.
conc Na =
total moles of Na
total volume
0 1 L 0 15 mol/L
0 1 L + 0 250 L
0 043 M
How to Think about Word Problems
The strategy for solving word problems involves making connections
between the facts given in the problem and chemical principles that
connect these facts together. Students often make the mistake of
simply memorizing procedures for solving problems. This is a very
bad approach! On exams, I am going to give you problems you have
never seen before and expect you to gure out how to solve them.
My intention is to nd out if you really understand the concepts or
if you have you simply memorized how to do certain calculations.
102
Chapter 4 Aqueous Reactions
Reading Skills for Word Problems
The way you read a word problem and sort the information is crucial
to guring out the solution. Here is a basic strategy.
1. You must rst understand the chemical principles explained in the
notes.
2. Every piece of information in the problem can be used to do
something. Make a list of those connections as you read the
problem.
a. For example, when you read 15 g of MgO you should think
with that I can calculate moles using the formula mass
15 g
formula mass
b. When you read What volume of 0.2 M KBr you should think
the connection between volume and molarity is
moles =
Molarity
Volume = moles
I have Molarity, I want to nd Volume so I know I am looking
for a way to nd moles of KBr
c. When you read 15 molecules of CO2 ” you should think with
that information I can
number 6 02 1023
15 molecules
moles =
6 02 1023
3. Sometimes it is useful to remember the de nitions of terms.
a. When you read what is the molar mass of . . . think molar
mass is g/mol so as I read the rest of the problem I know what
I am looking for. I need g of sample and moles of sample.
b. When you read what is the molarity of . . . think molarity is
moles/L so as I read the rest of the problem I know what I am
looking for. I need a way to nd moles of sample and volume
of sample.
How to Think about Word Problems 103
Examples
An unknown compound Mg3 X2 with mass 15 g contains 0.11121
moles of molecules. What is the unknown element?
A solution contains 0.115 M C
ion. How many grams of CuC 3
must have been added to 0.250 L to prepare this solution?
6 Electronic Structure of Atoms
20th Century Physics
Light as a Wave
I Light is an electromagnetic wave. The electric eld oscillates in one
plane and the magnetic eld oscillates in a perpendicular plane. The
x axis in the above picture could be either distance at one instant
in time or time, t at one xed position.
The equation that relates frequency, wavelength, and speed of travel
105
for a wave is
c
1.
is the wavelength (m)
2.
is the frequency (1/s)
3. c is the speed. For light c
30
108 m/s
The Young double slit experiment demonstrates the wave nature of
light. Light is passed through 2 slits. An interference pattern is
observed where the crests of the waves add at certain points. At
other points crests and valleys of the wave cancel resulting in a dark
section of light intensity.
QUANTUM
ANALOGY
1. A baseball can be thrown
with any velocity and
energy.
2. Most blenders are ex-
amples of quantized
systems. There is a
button for o , chop ,
liquify . The blade
spins with those energies and nothing in
between.
3. Quantum e ects are
usually not observed
in the macroscopic
world because the dif- Quantum Effects
ferences in steps are
too small to be obMax Planck explained light being emitted from heated objects.
served.
106
Chapter 6 Electronic Structure of Atoms
Planck said the atoms in solid objects are like springs that can hold
energy. Planck proposed a radical new behavior of atoms where
only certain quantized energies are possible. When the energy is released from the solid it is emitted in the form of light.
Light as a Particle
Albert Einstein, in his explanation of the photoelectric e ect was
the rst to prove that light can also behave like a particle.
E
1. h
6 63
10 34 J
2. The symbol,
h
PHOTON
is the frequency of light. Units are s 1
Wave Particle Duality
deBroglie Particle Wavelength
deBroglie proposed that particles have wave properties with a wave-
107
length related to the momentum of the particle.
1.
h
m
is the wavelength of the particle (m)
s)
2. h
3. m is the mass of the particle (kg)
4.
is the particle velocity (m/s)
Davisson Germer Experiment
Davisson and Germer proved that particles have wave properties
by observing an interference pattern when particles (electrons) are
scattered from a Cu solid.
108
Chapter 6 Electronic Structure of Atoms
Heisenberg Uncertainty Principle
Heisenberg proposed that certain quantities could not be measured
simultaneously.
x means the uncertainty in position.
x p
h
4
E t
h
4
1. Position, x, and momentum, p cannot be precisely known at the
same time.
2. Time, t, and energy, E cannot be simultaneously measured.
Our universe is constructed in such a way that measuring position
changes momentum and vice versa. The more accurately x is
measured the more uncertain p becomes. Note: this principle is
the result of the laws of physics for our universe. A clever device
will never be invented that gets around the limitations imposed by
the Heisenberg Uncertainty Principle.
Uncertainty Principle Analogy
Suppose we wish to measure
1. The taste of a lemon.
2. The quality of a kiss.
Heisenberg Uncertainty Principle 109
As the mind concentrates on the lemon, the mouth gives up on
the kiss and vice versa. Measuring one quantity changes the other
because they are related to each other.
Spectrum of the Hydrogen Atom
The energy storage ability of individual atoms can be investigated
by this simple experiment.
H + electrical energy
H
H + light
Analyzing the light, just as in black body radiation, tells us about
the energy states of the atom.
E2
Light energy
Electrical energy
E1
The resulting spectrum consists of lines indicating only certain energy states are possible. The energy states of the atom are quantized. The Rydberg formula predicts the energies of the lines in the
spectrum. Rydberg guessed at the result empirically rather than de-
110
Chapter 6 Electronic Structure of Atoms
riving a formula on the basis of a physical understanding of how the
atom behaved.
1
109680
1
n 21
1
n22
In this formula the terms represent
frequency of light in cm 1
wavelength of light
109680
R H in units of cm 1
n1
quantum number of upper energy level
n2
quantum number of lower energy level
It is important to note that Johannes Rydberg, who proposed this
equation, only understood the n values as something to make the
equation work. He did not understand the signi cance in terms of
quantum numbers describing energy levels in the atom.
No energy
in between
is possible
light observed at the checkmarks
The Bohr Atom 111
The Bohr Atom
Niels Bohr proposed the rst modern model of the hydrogen atom.
1. This model is now known not to be completely correct.
CENTRIFUGAL
FORCE?
The centrifugal force felt by
the electron as it moves on a circular orbit is the same outward
force you would feel driving a
car around a circle.
2. Electrostatic
attraction balances the centrifugal
force.
3. This theory broke with classical physics by suggesting only certain
orbits are possible because the electron orbit must contain an integer number of electron wavelengths. (Remember that particles
have wave properties so a moving electron is a wave.)
112
Chapter 6 Electronic Structure of Atoms
4. Resulting equation for the allowed energies of electrons has only
certain energies. The energies are quantized.
E
RH
n2
a. In this equation n is an integer called the quantum number.
2 18 10 18 J. The Rydberg constant, R H is speci c
for hydrogen.
b. R H
5.
a. Exactly predicts the observed light coming out of an excited
hydrogen atom.
b. Is the beginning of a new physics to describe atomic properties.
Bohr’s Equation for the Lines in a Hydrogen
Spectrum
The Bohr Atom 113
c
energy.
RH
n2
When an excited hydrogen atom changes from a higher energy orbit
E
to a lower orbit energy is emitted in the form of light. The equation
for the energy change of the atom when a transition (jump) is made
from initial level, i to the nal level f is
E atom
Ef
RH
n 2f
for absorption
Ei
RH
n 2i
for emission
The energy change in the hydrogen atom is positive for absorption
114
Chapter 6 Electronic Structure of Atoms
and negative for emission.
Now we consider the equations for the light emitted from or absorbed by a hydrogen atom when a transition is made between two
quantum levels. The energy, frequency, and wavelength of the pho-
EMISSION
ton of light are always positive. The values are forced to be positive
numbers by putting the upper (bigger) quantum number rst and
the lower (smaller) quantum number second.
E phot on =Eu -E =
In emission the electron drops
to a lower energy orbit and the
excess energy is emitted as a photon of light.
The notation u and
R H RH
- 2
n 2u
n
stand for upper (larger) and lower (smaller)
quantum number. The equations for the frequency and wavelength
of the light are obtained by using these equations:
h for a photon of light
E
and
c
for a light wave
The result is
E phot on
h
E photon
h
RH
h
1
n 2u
hc
E photon
1
n2
1
E phot on
hc
1
RH
hc
1
n 2u
1
n2
The Bohr Atom 115
Exercise 6.36
What is the wavelength of light emitted from a
hydrogen atom when a transition is made from the n
3 to n
1
level?
Solution
1
Use the equation
RH
hc
1
n 2u
6 63
2 18 10 18 J
10 34 J s 3 0 108 m/s
1
1
9 7424
1
n2
1
32
1
12
106 m 1
Rearranging to nd the wavelength
1
9 7424 106 m 1
102 6
10 9 m
102 6 nm
Exercise 6.37
What is the frequency of light absorbed by a
hydrogen atom when a transition is made from the n
level?
Solution
ABSORPTION
Use the equation
1 to n
3
116
Chapter 6 Electronic Structure of Atoms
RH
h
1
1
n 2big
n 2small
2 18 10 18 J
6 63 10 34 J s
2 92
1
32
1
12
1015 s 1
Energy Levels of Hydrogen
1. The diagram shows the energies of the hydrogen atom.
2. The highest energy level when the proton and electron are sepa-
rated by an in nite distance has an energy of zero and n
3. As the electron gets closer to the nucleus the energy levels become
Modern Quantum Theory 117
more negative.
1 is the lowest energy, smallest orbit called the
ground state.
4. The level n
5. Higher energy levels, n
1 are called excited states.
Problems with Bohr’s Theory
Many additional properties of atoms could not be explained by
modern quantum theory.
1. 4 quantum numbers are needed to describe each electron where
Bohr theory only had one.
2. It became clear that electrons did not move on precise xed orbits.
3. The shapes of the regions in space occupied by electrons could
4. The arrangement of electrons in the energy levels could not be
Modern Quantum Theory
Modern quantum theory used a new approach to describe the physics
of atoms.
1. Electrons are described using a mathematical function
treats the electron as a wave.
that
2. A wave equation (The Schrodinger equation) is used to solve for
the physical properties of the electron.
3. No assumptions are made like electrons move in orbits.
118
Chapter 6 Electronic Structure of Atoms
Modern View of the Atom
1. It is impossible to know the motion of electrons in atoms.
2. Each electron is described by a mathematical wavefunction,
3. The wavefunction has no physical signi cance. However,
2 is
the probability of nding the electron in a certain region of space.
4. The energies are quantized.
5. Four quantum numbers n
m m s describe each electron.
Electron Orbitals
We cannot know precisely how the electron moves. Instead, we
think of the regions in space where the probability for nding the
electron is high.
I An orbital is a map in space of the high probability regions for
The shape of the orbital is
found by plotting the probability ( 2) as a function of coordinates x,y,z or more commonly
in spherical coordinates r,
nding an electron.
Orbitals: An Analogy
Suppose there is a hive of yellow bees that only works at night.
We cannot watch the bees in action or know how an individual bee
moves. But as they work they leave behind a trace of yellow color.
In the morning when the sun comes up we can get a picture of
where they have been. There are yellow dots everywhere but some
regions where they have been very active have thick yellow color.
That picture is a map of probability very much like an orbital. Now
Quantum Numbers 119
in the case of electrons their path is completely unknown to us. It is
important to realize that the physics of electrons in atoms does not
allow us to know their movement. No clever device will ever make
this possible.
I The probabilities of nding an electron in an orbital are known.
I The movement of electrons is not known.
Quantum Numbers
Summary
Quantum Number
n
n
m
ms
Principle
1 2 3
0 1
1
2
Name
Determines
orbital size, energy, shell #
1
Azimuthal orbital shape, angular momentum
1
Magnetic
orientation of orbital in space
Spin
electron spin
120
Chapter 6 Electronic Structure of Atoms
Orbital Shapes
The angular momentum quantum number
determines the shape
of the orbital. The angular momentum quantum number characterizes the angular momentum of the orbital.
Orbital Notation
I You must know everything in this gure.
Orbital Orientation in Space
The m quantum number determines the orientation of the orbital
in space. There is no simple connection between m and the Cartesian x y z axis. The magnetic quantum number m describes the
projection of the angular momentum on an axis.
Quantum Numbers 121
DEGENERACY
The three p orbitals have the
same energy. They are energy
degenerate. However, in the presence of a magnetic eld the energies split because each orientation interacts differently with
the magnetic eld.
Energy Levels and Shells
For a hydrogen atom the energies of the levels depend on the principle quantum number, n only. This picture shows the relationship
between the energy levels and quantum numbers in the n
1 2 3
shells of the hydrogen atom.
1. As n gets larger the size of the orbital increases.
2. As n gets larger the orbital energy increases.
3. As n gets larger there are more possible suborbitals (shapes and
orientations described by
1. Each shell has n subshells.
0 to
n 1
2. Each subshell has 2
1
suborbitals.
3. Each shell has 2n2 orbitals.
and m ).
4. The idea of a shell (same n) is only for convenience of grouping
orbitals together. It does not mean there is one egg shell inside
of another shell. All of these orbitals are intermixed in space.
5. A subshell means the same n but di erent
contains the 3s, 3 p, 3d subshells.
The n
3 shell
122
Chapter 6 Electronic Structure of Atoms
Lowest and Highest Energy Levels
The picture below is an analogy for the energy levels of the hydrogen
atom.
Quantum Numbers 123
Imagine one man near the top of a series of steps and another near
the bottom. We choose to de ne the top step with an energy of
zero so the bottom step will have a negative value of energy. To
climb out of the hole they must each do work against the force of
gravity. The man near the top has less work to do to climb out of
the hole because he is standing closer to the top on a higher energy
step. The man at the bottom has more work to do as he is standing
on a lower energy step.
Relationships between Quantum Numbers
Looking at the Figure above you can see certain relationships between quantum numbers. The 2nd shell is described by n
shell only has an s and p orbital. Apparently, when n
and
1 but
Notice also when
2. This
2
0
2
1 the type of orbital is a p orbital. When
124
Chapter 6 Electronic Structure of Atoms
1, m can be m
m is the projection of on an
axis. When
0 the only projection is 0. When
1 there
are 3 projections, m
1 0 1
This is the reason m depends
on
1 0 1 But m
2 2
The following table summarizes these relationships.
Quantum Number Range
n
n
1 2 3
0 1
m
n
m
1
OR m
ms
1
0
1
2
1
2
ms
I You must know how these quantum numbers depend on each other.
Exercise 6.38
A certain electron has a quantum number m
2 What are the possible values for n
Solution
Since
ms ?
depends on m and
depends on n we
work upwards (in the table) to determine the relationships. When
m
2, must be at least 2 (When
When
3 m
3
for are
2 3
In other words,
than or equal to 2. If
2
2 m
2
1 0 1 2
1 0 1 2 3) So the possible values
could be any number greater
2 n must be at least
1
3 So the
possible values of n are 3 4
The only possible values of m s are
Exercise 6.39
1
1
2 or 2
Which of these sets of quantum numbers is not
Quantum Numbers 125
allowed?
A. n
2
0 m
0 ms
B. n
0
0 m
0 ms
C. n
2
1 m
2 ms
D. n
2
1 m
B. Not allowed because n
0 and n
1
2
1
2
1
2
0 ms
0
Solution
A. Allowed
C. Not allowed because m can be at most
be 1 if
1
D. Not allowed because
0. Also ms 0
So at most m could
cannot be negative. Smallest possible
is
Electron Spin
The property of electron spin is a relativistic e ect that requires Ein-
Nevertheless, we imagine the electron spinning like a top creating
an angular momentum accompanied by a small magnetic eld. The
spin is quantized with two possibile values. Normally, we refer to
the two possibilities as spin up
and spin down
Electrons in a
magnetic eld can be separated into these two types.
ms
1 2
ms
1 2
126
Chapter 6 Electronic Structure of Atoms
Nodes
Orbitals have some regions in space where the probability of nding
an electron is zero. The point on the curve below where the probability goes to zero occurs when
2
0 This point where
2
0
is called a node.
Multi-Electron Atoms
In single electron atoms like H, He
Li2 all of the orbitals in the
same shell (Shell = all orbitals with the same value of principle
quantum number, n) have the same energy.
We say that the
orbitals with the same energy are energy degenerate.
The gure
below shows that in hydrogen the 3s, 3p, and 3d orbitals all have
the same energy.
Multi-Electron Atoms 127
In multi-electron atoms such as He, Li, Be. . . the energies of orbitals
within the same shell have di erent energies. The energies are no
longer degenerate. The gure below shows that the 3s, 3p, and
3d orbitals have di erent energies in a multielectron atom.
Electron Screening
This plot shows the probability of nding an electron,
distances from the nucleus.
2 at various
128
Chapter 6 Electronic Structure of Atoms
WHY SCREENING
AFFECTS ENERGY
Electron screening is the reason why the 2s and 2p orbitals do not
have the same energy in multi-electron atoms. An electron in the 1s
or 2s orbital is much closer to the nucleus than a 2p electron. When
electrons exist in the 1s orbital, a 2p electron does not feel as much
of the nuclear charge. The 1s electron “screens” or blocks the
Recall that in the Bohr atom
the
attraction of the
electron for the nucleus was a
major factor in determining the
energy of the electron. In multielectron atoms screening can reduce the nuclear charge felt by
an electron. This reduces the
energy of that electron.
nucleus o to the 2p electron. As a result of the presence of the 1s
electron, the 2p electron decreases in energy. (The orbital moves
upwards towards 0 energy.) Notice that the screening e ect is larger
for the 2p electron than the 2s electron. The reason is the 2s orbital
penetrates closer to the nucleus. The 1s electron cannot get in the
way of the 2s electron as much as it can for the 2p electron.
Multi-Electron Atoms 129
The gure below compares the energies of the orbitals in hydrogen
Screening Increases
Within the same shell screeing increases.
s
p
d
f
So a 4f orbital experiences
more screening than a 4s orbital.
As a result the 4f orbital is higher
in energy than a 4s orbital in a
multielecrton atom.
and lithium. There are three key points to recognize in this gure.
1. As protons are added to the nucleus all of the orbitals in the atom
attraction has increased.
2. When electrons are added to the nucleus the orbitals move to
higher (less negative) energies as a result of increased electronelectron repulsions.
3. Orbitals with more electron screening have the most movement
towards higher energies. The gure below shows that the n 3
shell in lithium experiences the largest increase in energy. Electron screening increases 3s
3p
3d. So the 3d orbital in
lithium will have the highest energy and the 3s the lowest energy.
130
Chapter 6 Electronic Structure of Atoms
The Aufbau Principle
The Aufbau principle explains how electrons arrange themselves into
the available atomic orbitals. This arrangement of electrons is important because it accounts for all of the chemical and physical
properties of the elements. There are two key principles that determine the arrangement of electrons.
1. Pauli Exclusion Principle
Each electron must have a unique set of quantum numbers. This is
the Pauli Exclusion Principle.
How many electrons can be located in a 1s orbital?
A 1s orbital has n
1
1
2 and 2
1,
0 m
0 The possibilities for ms are
According to the Pauli Exclusion Principle each electron
The Aufbau Principle 131
must have a unique set of the 4 quantum numbers. The only way
electrons can have a unique set of quantum numbers in a 1s orbital
is:
1. n
1,
0 m
0 ms
1
2
2. n
1,
0 m
0 ms
1
2
No additional electrons can be placed in a 1s orbital with a set of
quantum numbers di erent from these two! So only 2 electrons can
be placed in a 1s orbital. This same idea applies to ALL orbitals.
I The consequence of the Pauli Exclusion Principle is that at most
2 electrons can be placed in an orbital and the spins must be in
opposite directions.
I Three electrons cannot be placed in the same orbital because it violates the Pauli exclusion principle.
132
Chapter 6 Electronic Structure of Atoms
Electron Con gurations of H, He, Li, Be
Electron con gurations are the way in which the electrons are arranged
in the available atomic orbitals.
1. The orbitals are arranged in energy with the lowest energy 1s
orbital at the bottom.
2. For each element take the number of available electrons and put
a maximum of 2 electrons in each orbital with opposite spins.
3. Start at the 1s orbital and go up in energy to the next highest
orbital until all of the electrons have been arranged in the appropriate orbitals. This is the lowest energy or ground state electron
con guration.
4. Other electron con gurations with higher energies are possible.
A con guration that is higher in energy than the ground state is
called an excited state.
The Aufbau Principle 133
Paired Spins
2. Hund’s Rule
UnPaired Spins When lling electrons into p d, and f orbitals there are several
Paramagnetic
choices for how the electrons can be arranged.
In Li the electrons in the 1s
orbital are paired with one spin
up and the other spin down. The
electron in the 2s is unpaired.
Atoms with unpaired spins have
magnetic moments and are called
paramagnetic.
Diamagnetic
In He all of the electrons are
paired resulting in almost no magnetic moment. When all of the
electrons have paired spins the
atom is called diamagnetic.
Hund’s rule says (for degenerate p d f orbitals) ll electrons
one per orbital with the same spin until all of the orbitals are
half lled. If there are more electrons then begin to pair them 2 per
134
Chapter 6 Electronic Structure of Atoms
orbital with opposite spins. This minimizes the repulsions between
electrons which creates the lowest energy electron con guration.
I
guration will have
the smallest possible number of paired electrons
the maximum number of unpaired electrons
con guration
per orbital and
per orbital So the
is lower in energy than
.
Writing Electron Con gurations
Electron con gurations are a notation for showing the arrangement
of electrons in all of the orbitals. The electron con guration for Li
is 1s 2 2s 1 The orbitals are written in increasing energy.
The Aufbau Principle 135
In order to write electron con gurations you must memorize the
order of the orbitals in increasing energy. The easiest way to do
that is to use the periodic table as a guide.
Example 6.40 Write the electron con guration for Ge.
Solution
Find Ge in the periodic table. Follow the atomic
136
Chapter 6 Electronic Structure of Atoms
numbers of the elements from H to Ge to nd the order of the orbitals in energy.
The orbitals for Ge in increasing energy are
Orbital
suborbitals
e
s
p
d
f
1
3
5
7
2
6
10
14
1s
2s
2p
4s
3d
4p
3s
3p
To write the electron con guration ll in the number of electrons in
each orbital until all of the electrons for Ge are used up.
1s 2 2s 2 2 p6 3s 2 3 p 6 4s 2 3d 10 4 p 2
Note that the 4p orbital is not lled. Ge is the second element that
uses a 4 p orbital so there will be 2 electrons in the 4p. As a check
count up all of the electrons and make sure the sum is the same as
the atomic number.
2
2
6
2
6
2
10
2
32
Knowing how to look at the periodic table to identify electron congurations is a very powerful way of quickly identifying the electronic
The Aufbau Principle 137
properties of an element. Note how fast these questions can be answered by just looking at the periodic table.
1. How many unpaired electrons does Ge have? All of the electrons
must be paired except the ones in the last un lled orbital. Acthe answer is 2.
2. Which element has 5 electrons in the 5p orbital? I
3. Which of these elements has a lled 4d orbital Ga or In? In
Short Hand Notation
The electron for Ge is
A short hand notation is sometimes used. The last lled noble gas
electron con guration is written followed by the rest of the remaining
orbitals. For Ge, Ar is the last lled noble gas with the con guration
[Ar]
1s 2 2s 2 2 p 6 3s 2 3 p 6
So the electron con guration for Ge in short hand notation is
[Ar] 4s 2 3d 10 4 p 2
The inner core electrons are represented by [Ar] The outer electrons 4s 2 3d 10 4 p 2 are called the valence electrons. The valence
138
Chapter 6 Electronic Structure of Atoms
electrons are involved in chemical bonding. Although in this course
most of the molecules will not use electrons in atomic d orbitals
when making bonds.
Valence electrons are the outer (highest energy) electrons involved in chemical bonding. For Ge these would be the 4s 2 3d 10 4 p2
electrons.
Example 6.41
What is the short hand electron con guration
of Rb? What orbital is the valence elctron in?
Solution
The electron con guration of Rb is [Kr]5s 1 The
valence electron is the electron in the highest energy outer most
orbital. In this case that is the 5s orbital.
Electron Con gurations of Transition Metals
The transition metals have several exceptions where the expected
electron con guration is not observed. The s and d orbitals are
very close in energy. As electrons are lled into the d orbitals the
energies of the orbital shifts from electron-electron repulsions. This
creates unexpected electron con gurations.
The Aufbau Principle 139
I The easiest way to remember where the exceptions occur is to remember that the d orbital likes to be half lled and completely lled.
When possible an electron is removed from the s orbital to make
the d orbital half or completely lled.
Element Expected
Sc
[Ar]4s 2 3d 1
Ti
[Ar]4s 2 3d 2
V
[Ar]4s 2 3d 3
Cr
[Ar]4s 2 3d 4
Mn
[Ar]4s 2 3d 5
Fe
[Ar]4s 2 3d 6
Co
[Ar]4s 2 3d 7
Ni
[Ar]4s 2 3d 8
Cu
[Ar]4s 2 3d 9
Zn
[Ar]4s 2 3d 10
Exception
[Ar]4s 1 3d 5
[Ar]4s 1 3d 10
Note that Mo follows the same behavior as Cr and Ag follows the
same behavior as Cu.
There are many more exceptions in the d
140
Chapter 6 Electronic Structure of Atoms
and f block.
7 Periodic Properties
Periodic Table
1. Mendeleev / Meyer developed 1st periodic table .
2. Used increasing atomic mass.
3. Elements with similar chemical properties were aligned in columns.
4. Some elements were unknown at the time. Spaces were left in
the table predicting their discovery.
5. Henry Moseley discovered a way to measure the number of pro-
tons in an atom. He made the connection between element and
number of protons. Moseley modi ed the table by arranging elements by increasing atomic number instead of mass.
142
Chapter 7 Periodic Properties
Electron Shells
1.
nucleus is bigger
pulls the 1s orbital closer (+/- attraction).
2. Ne is bigger than He for two reasons.
2
n 1, bigger n
b. screening of 2s 2 p electrons by 1s electrons
shell of Ne even bigger.
a. n
larger orbital.
makes n 2
Electron Shells for He and Ar
1s
Ar
2s,2p
3s,3p
He
0.0
0.5
1.0
1.5
Distance From Nucleus (Angstroms)
2.0
Electron Screening 143
1.
2. Can identify certain regions of space where electrons are mostly
from one particular shell.
3. Note how close the 1s orbital is to the nuclues in Ar compared to
He. This is a result of the much larger nucleus in Ar which pulls
the inner shell 1s orbital closer.
Electron Screening
Energy of an Electron
The energy of each electron depends on (1) electron
attraction and (2) electron
electron repulsion. Both of these
quantities depend on what orbital the electron is in.
Coulomb Law for
Electrons
1. Energy increases with
nuclear charge.
2. Energy decreases with
distance from nucleus.
nucleus
144
Chapter 7 Periodic Properties
Effective Nuclear Charge
The e ective nuclear charge, Z e f f is the NET (average) electron
nucleus attraction. Z e f f characterizes the average nuclear charge
an electron feels including the average e ects of all electron
electron repulsions.
1. Electron screening does not change too much within the same
subshell (s p d, or f orbital).
2. Electron screening is more important for an orbital whose shape
is further away from the nucleus. Screening is more important for
f
d
p s
3. As the orbital gets larger screening becomes more important.
Screening increases with principle quantum number n
General Trends for Atomic Sizes
1. Recall from chapter 6 that orbital size increases with principle
quantum number, n Principle quantum number n increases down
the column so the orbital size increases down the column. As a
result the atomic radius increases going down the column.
2. In the periods (rows) the e ective nuclear charge experienced by
General Trends for Atomic Sizes 145
the valence electrons increases towards the right as the nucleus
grows. When the e ective nuclear charge increases the orbitals
get pulled closer to the nucleus and atom size decreases.
Ion Sizes
As the number of electrons goes up in an atom electron
electron
repulsions increases and the atom size increases. Said another way,
the e ective nuclear charge decreases as the number of electrons
goes up causing atomic size to increase.
I Cations are smaller than the neutral atom. Anions are larger than
146
Chapter 7 Periodic Properties
the neutral atom.
Electron Con gurations of Ions
To nd the electron con gurations of cations remove the outer
most (highest energy) electrons.
Atom Electron Con guration
Na
1s 2 2s 2 2 p 6 3s 1
Na
1s 2 2s 2 2 p 6
Ne
1s 2 2s 2 2 p 6
I Isoelectronic means two di erent elements with the same electron
con guration. Na is isoelectronic with Ne.
Atom Electron Con guration
A
1s 2 2s 2 2 p 6 3s 2 3 p 1
A 3
1s 2 2s 2 2 p 6
Ne
1s 2 2s 2 2 p 6
To nd the electron con guration of anions add an electron into
the next highest available orbital.
Atom Electron Con guration
O
1s 2 2s 2 2 p 4
O2
1s 2 2s 2 2 p 6
Ne
1s 2 2s 2 2 p 6
I Certain ions take on a speci c charge when they make ionic molecules. They do this to achieve a full shell or noble gas electron
General Trends for Atomic Sizes 147
con guration. This is the reason why O is always O2
and Na is
always Na when they form ionic molecules.
Electron Con guration of Transition Metal Ions
Review of Main Group Ions
With main group elements the ion electron con gurations seem
reasonable. Cation con gurations are made by just removing the
outer most electrons and anion con gurations are made by adding
additional electrons.
Transition Metal Ions
1. The highest energy electron that is removed is from the last lled
s orbital.
2. The d orbital electrons are removed last.
Atom Electron Con guration
Fe
[Ar] 4s 2 3d 6
Fe
[Ar] 4s 1 3d 6
Fe2
[Ar] 4s 0 3d 6
Fe3
[Ar] 4s 0 3d 5
These electron con gurations may seem unexpected. The reason
has to do with the very close energies of the s and d orbitals.
The electron con guration of Mn and Fe is not the same. Although they both have the same number of electrons, Fe has 1
more proton. This makes enough di erence in the energies of the
orbitals to make the lowest energy electron con guration of Fe
[Ar] 4s 1 3d 6 and the electron con guration of Mn [Ar] 4s 2 3d 5
148
Chapter 7 Periodic Properties
Ionization Energy
Ionization energy measures the energy required to remove the highest energy (outer most) electron in an atom.
Li (g)
Li (g) + e
1st Ionization, I1
Li (g)
Li2 (g) + e
2nd Ionization, I2
In the process of ionization
energy is added to move the highest energy electron to the n
level where it can separate
from the atom.
1. Going down a column of the periodic table electron is further
away from nucleus
easier to ionize
drops going down the periodic table.
Ionization Energy
2. Electrons get closer to nucleus going across periodic table towards
right (radius decreases from left to right)
atom is harder to
ionize
Ionization Energy increases going across towards right.
Ionization Energy 149
Notes about Ionization Energies
1. Noble gases have the highest ionization energy.
2. Alkali metals have lowest ionization energy.
3. He
Ne
Ar because atomic radius is increasing from He to Ar
4. Ne
F
O
Li because atomic radius increases from Ne to Li
5. There are two important kinks in the in the Figure above at the
points labeled 1 and 2 . At point 1, the electron being ionized
from the B atom is removed from a higher energy 2p orbital.
This results in a lower ionization energy for B.
150
Chapter 7 Periodic Properties
6. The kink in the trend, labeled 2 occurs where electron spins pair
in the 2p orbital of the O atom.
Pairing the spins raises the energy of the electron due to increased electron-electron repulsions. Less energy is therefor
required to remove the electron when ionization occurs. This
is the reason why the ionization energy of O is less than the
ionization energy of N.
Example 7.42
Which atom Na or Ne has the lowest ionization
energy and which the highest ionization energy?
Second Ionization Energies (I2 ) 151
Solution
Looking at the periodic table we see that Na is much larger than
Ne. So we expect the outer electron in Na to be more easily ionized
(lower energy) than an electron in Ne.
Second Ionization Energies (I2)
Na
Na
Na
Na2
I1 (First Ionization)
e
e
I2 (Second Ionization)
To understand the trends in second ionization energy we must remember that the starting point for second ionization is the ion with
one electron already removed (Na
For example, to think about
how the second ionization energy for Na compares with Ne you must
rst consider the electron arrangement for Na and Ne .
Example 7.43
Which atom Na or Ne has the lowest second
ionization energy and which the highest?
152
Chapter 7 Periodic Properties
Solution
For second ionization we must compare Na and Ne
These are the starting points for the second ionization process.
Na
Na2
e
Ne
Ne2
e
COMMON MISTAKE:
Many students believe that to compare rst ionization energies you
rst remove one electron. Then
they compare the +1 cations. This
is wrong! The starting point for
rst ionization is the neutral atom.
Compare the neutral atoms.
X
X
e
Many students believe that
to compare second ionization energies you rst remove two electrons. Then compare the 2 cations.
This is wrong! The starting point
for second ionization is the 1
cation. Compare the elements
isoelectronic with the 1 cations.
X
X2
e
Notice that Na (11 protons + 10 electrons) has same number of
electrons as Ne. So Na
behaves electronically like Ne. Also Ne
(10 protons, 9 electrons) is electronically like F.
1. Na has the highest I2 (because Na behaves like Ne).
2. Ne has a lower I2 (because Ne
Example 7.44
behaves like F).
Determine which of these elements has the high-
est rst ionization enegy and which has the highest second ionization
energy. Kr, Se, Sr, Br
Comparing Ionization Energies 153
Solution
1. For rst ionization compare the trends of the neutral atoms. For
elements in the same row the noble gas is the highest.
Se
Br
Kr
The elements in the next row have lower ionization so the overall
trend in rst ionization energy is
Sr
Se
Br
Kr
2. For second ionization energy the starting point of the process
is the 1 cation. We must compare the ionization energies
of Kr 1 , Se 1 , Sr 1 , Br 1 The easiest way to compare these
ions is to determine what neutral elements they are isoelectronic
with. The neutral atom isoelectronic with the ion is in parenthesis. Kr 1 Br , Se 1 As , Sr 1 Rb , Br 1 Se Comparing these
neutral atoms the trend in ionization energy is
Rb
As
Se
Br
Rb has the lowest ionization energy in the list and it is isoelectronic with Sr 1 , so Sr will have the lowest second ionization. Br
has the highest ionization energy in the list and it is isoelectronic
with Kr 1 so Kr has the highest second ionization energy. The
overall trend in second ionization energy is
Sr
Se
Br
Kr
Comparing Ionization Energies
The diagram below shows why the ionization energy of helium is
higher than the ionization energy of hydrogen.
154
Chapter 7 Periodic Properties
1. All of the energy levels in He are lower in energy than the cor-
responding levels in hydrogen. The reason why is simple! As
more protons are added to the nucleus the positive charge of the
nucleus increases. As a result, the attractive force between the
electrons and the nucleus increases. The electrons in He are more
tightly held to the nucleus than the electron in hydrogen. The
orbital energies have larger negative energies in He. For lithium
with 3 protons the orbital energies are even more negative than
in He as a result of the increased attractive force resulting from
the three protons in the lithium nucleus.
2. The length of the arrows in the gure above represent the magni-
tude of the ionization energy. For lithium the ionization energy is
smaller compared to helium. In lithium the electron being ionized
comes from a higher energy 2s orbital. In helium the electron
being ionized comes from a lower energy 1s orbital.
Comparing Ionization Energies for Na and Mg
In the diagram below, the length of the arrows is proportional to
Comparing Ionization Energies 155
the ionization energy.
1. For Na, rst ionization removes a 3s electron. The second ioniza-
tion energy of sodium increases dramatically because the second
electron comes from the lower energy 2p orbital.
2. For Mg, the
rst two electrons are removed from the 3s orbital.
The energy required to remove the third electron (third ionization
energy) increases dramatically because the third electron comes
from the much lower energy 2p orbital.
This diagram illustrates the general trend that ionization energy always increases dramatically when the electron being removed comes
from a lled shell.
For Mg, the third electron is removed from a
lled n = 2 shell and this results in a very high ionization energy.
156
Chapter 7 Periodic Properties
Chemical Implications of Ionization Energy
The magnitude of the ionization energies is responsible for determining the typical charges observed on atomic ions. The plot below
shows a comparison of the ionization energies for Na and Mg.
1. Na typically exists as a +1 cation because the energy required
to convert Na to Na ( rst ionization energy) is relatively small.
Since Na is isoelectronic with Ne, the energy rquired to make
Na2 (second ionization energy) is much higher so the +2 and
+3 cations are rarely observed.
2. Mg typically exists as a +2 cation because the energy required
to convert Mg to Mg2 (second ionization energy) is relatively
small. Since Mg2 is isoelectronic with Ne, the energy rquired to
make Mg3 (third ionization energy) is much higher so the +3
and +4 cations are rarely observed.
Electron Af nity
Electron A nity is the energy required to gain an electron (usually
Electron Af nity 157
negative). Electron a nity measures the attraction of an electron
to an atom.
C has the highest electron a nity.
C (g) + e
C
(g) Electron A nity
Comparison of Ionization Energy and Electron
Af nity
Note the di erence between electron a nity and ionization energy.
1. Ionization Energy is the energy required to pull o an electron
(usually positive).
2. Electron Af nity is the energy required to attach an electron
(usually negative).
158
Chapter 7 Periodic Properties
C (g)
C
C (g) + e
(g) + e
C
Ionization Energy
(g) Electron A nity
Metals, Nonmetals, Metalloids
1. Review your notes p. 15-16 and know these concepts.
2. Metals are usually easily oxidized (lose electrons is oxidation)
because ionization energy is low.
3. Nonmetals tend to (means usually, not always) gain electrons
(reduction). Ionization energy is high.
Metallic Character
As elements exhibit more pronounced properties of metals (like lower
ionization energy) they are said to have more metallic character. As
in the periodic table towards the element Fr which has the highest
metallic character.
Chemistry of Important Groups in the Periodic Table 159
Chemistry of Important Groups in the Periodic Table
I Elements in the same Groups of the periodic table have similar
electron con gurations. This is the reason why the chemistry
of elements in the same group is similar.
I Chemistry is most a ected by the electrons in the outer orbitals.
160
Chapter 7 Periodic Properties
These electrons are called valence electrons.
Group
Electron Con guration
Alkali Metals
ns 1
Alkaline Earths ns 2
3A
ns 2 np 1
4A
ns 2 np 2
5A
ns 2 np 3
6A
ns 2 np 4
Halogens
ns 2 np 5
Noble Gases
ns 2 np 6
Electron con gurations are common for elements in the same
column of the periodic table.
Chemical Reactivity Alkali Metals
Alkali metals like to react to form
1 cations that have a lled shell
or noble gas electron con guration.
The reason why
1 cations are formed is because the energy re-
quired to make the
1 cation ( rst ionization energy) is small for
alkali metals. Whereas the energy required to make
2 cations
(second ionization energy) is very large for alkali metals. For this
Metals React by
Losing Electrons.
Lower ionization
energies increase
reactivity.
reason alkali metals like to be +1 cations. Heavier alkali metals react more vigorously because the ionization energy decreases down
the column of the periodic table. Lower ionization energies make
it easier to form the +1 cation.
Chemistry of Important Groups in the Periodic Table 161
2M (s) + H2
2MH
2M (s) + S
M2 S
2M (s) + Cl2
2MCl
M is any alkali metal (Na K . . . )
Reactions with Water
2M (s) + 2H2 O
2MOH aq + H2 g
Compounds with Oxygen
Sodium Peroxide
Na2 O2 contains the O2 2
peroxide ion. Find the unknown
charge on the O2 using the known
charge on the sodium ion.
2
1
x
x
0
2
Potassium
Superoxide
Li2 O
O2
Lithium Oxide
Na2 O2 Sodium Peroxide
KO2
Potassium Superoxide (O2
I Because ionization (M
bigger
(O2 )2
M
Only Na, K, Rb, Cs
Only K, Rb, Cs
e ) gets easier as atoms gets
reactivity of alkali metals increases as you go down the
KO2 contains the O2 superoxide ion. Find the unknown
column in the table.
charge on the O2 using the known
charge on the potassium ion.
1
x
x
0
1
Hydride Ion
I Note: Hydrogen has a much higher ionization energy than the rest
of the alkali metals. For this reason H does not act like a metal and
162
Chapter 7 Periodic Properties
lose electrons.
Hydrogen makes negative ions called hydrides when it reacts with
metals.
NaH sodium hydride
LiH
lithium hydride
KH
potassium hydride
Alkaline Earths
Alkaline earth metals like to form
2 cations that have a lled shell
or noble gas electron con guration.
Metals React by
Losing Electrons
The reason why
2 cations are formed is because the energy re-
quired (the second ionization energy) is relatively low. The third
ionization energy is very high so alkaline earth elements tend to be
2 cations.
Heavier alkaline earth metals react more vigorously
because the ionization energy decreases down the column of the periodic table. Lower ionization energies make it easier to form the
+2 cation
Chemistry of Important Groups in the Periodic Table 163
Ca (s) + H2
CaH2
Ca (s) + S
CaS
Ca (s) + Cl2
CaCl2
Mg (s) + O2
2MgO
Reactions with Water
Be + H2 O
No Reaction
Mg + H2 O
No Reaction
Mg + H2 O steam
MgO s + H2 g
Ca s + 2H2 O
Ca(OH)2 aq + H2 g
Ba s + 2H2 O
Ba(OH)2 aq + H2 g
Oxygen
Oxygen likes to have a charge of
gas electron con guration.
2 to achieve a lled shell noble
164
Chapter 7 Periodic Properties
O2 oxide ion
O2 2 peroxide ion
O2
superoxide ion
Halogens
NonMetals React by
Gaining Electrons
Like to form
1 anions to achieve a lled noble gas electron con-
guration.
2Na
F2
2NaF
I Reactivity increases going up the periodic table because nonmetals
gain electrons and the attraction for electrons gets stronger as the
NonMetals React by
Gaining Electrons
atom gets smaller.
Noble Gases
Unreactive because the electron con gurations have lled shells.
Larger noble gases like Kr Xe form stable compounds. For example,
XeF4 is a stable crystal.
Noble Gases React
Using Dispersion
Effects (Chapter 11)
8 Introduction to Chemical
Bonding
Valence Electrons and Lewis Symbols
Inner core electrons are pulled so close to the nucleus that they do
not participate much in the chemistry. The outer shell or valence
electrons are responsible for most of chemical behavior. The reason why chemistry is similar in the same group (column) of the
periodic table is that all of the elements in the same column have
the same number of valence electrons.
I Elements in the columns of the periodic table have the same number of valence electrons. These electrons greatly in uence the
chemistry.
I Elements in the columns of the periodic table have di erent core
electrons. These electrons have only a small e ect on the chemistry.
166
Chapter 8 Introduction to ChemicalBonding
Cl
He
0
1
2
Distance From Nucleus (Angstroms)
Lewis Dot Symbol
Be has the electron con guration
The valence electrons are written as dots around the element symbol. The dots help us to visualize the electrons that in uence the
167
chemistry.
Writing Lewis Dot Symbols
The alkali metals have 1 valence electron. The noble gases on
the right side of the periodic table are where the shell lls with 8
electrons. To nd the number of valence electrons for any element
just count from the left. Nitrogen is 5 elements from the left and
has 5 valence electrons. Ignore d electrons when counting valence
electrons for Lewis structures.
I Elements in the columns of the periodic table have the same number of valence electrons. The elements di er only in the core
168
Chapter 8 Introduction to ChemicalBonding
electrons which have only a minor e ect on the chemistry.
Exercise 8.45
Write the Lewis Dot symbol for the atoms As, I
and Kr.
Solution
1. Ignore the d electrons and just count from the left to
nd the
number of valence electrons.
2. The element, As is in the same column as N and will have the
same number of valence electrons as N (5).
3. Iodine is in the same column as F and will have the same number
of valence electrons (7).
The Octet Rule and Chemical Bonding
I The driving force behind chemical bonding is the desire for elements
to achieve a full shell of 8 electrons. (We are only counting s and
p electrons.)
Electronegativity
I Electronegativity is a measure of the pull on an electron from an
Electronegativity 169
element IN A CHEMICAL BOND.
Fluorine has the highest electronegativity because it has a small radius so the nucleus is more accessible. As the atom size grows going
down the column in the Halogen family the atom has more electron
shells. The nucleus becomes more shielded with core electrons and
electronegativity drops. Going across the row towards the right effective nuclear charge increases. The desire for electrons increases
from left to right.
170
Chapter 8 Introduction to ChemicalBonding
Ionic Bonding
I Ionic bonding results from
electrostatic forces. Electrons
are transferred to and from the atoms making them ionic. The
oppositely charged ions stick together to make ionic compounds.
1. Sodium gives up 1 electron to achieve a
lled shell electron con-
guration of [Ne].
2. Chlorine accepts 1 electron to achieve a lled shell electron con-
guration of [Ar].
3. The ions are held together in an ionic bond that results from the
electrostatic force.
When do Ionic Bonds Form?
When the difference between the electronegativity of the two atoms
is large ionic bonds tend to form. If the di erence in electronegativity
is
2 0 the bond is probably ionic. (See also covalent bonding.)
This usually means the atoms are metals reacting with nonmetals.
Covalent Bonding 171
The di erence in electronegativity,
E N
40
E N for NaF is
09
31
A di erence greater than 2.0 is an ionic bond.
Covalent Bonding
I Covalent Bonding results from a sharing of electrons.
Covalent bonding results from the overlap of half lled atomic orbitals. It is not easy to explain why this sharing of electrons results
in a bond. The bond is a purely quantum mechanical e ect aris-
172
Chapter 8 Introduction to ChemicalBonding
ing from the physics of small particles. These e ects are sometimes
di cult to explain in the macroscopic physics that we understand
from daily experience. Note that the H atoms in the gure are both
exactly the same. The electronegativity is the same. Electrons are
shared rather than being transferred from one atom to another.
When does Covalent Bonding Occur?
Covalent bonding occurs when the difference in elecrotonegativity
between the two atoms in the bond is
2 0. This typically happens
when bonds are formed between two nonmetal atoms.
The di erence in electronegativity,
E N
35
E N for NO2 is
30
05
The type of bond that occurs with a small electronegativity di erence of 0.5 is classi ed as polar covalent.
Covalent Bonding 173
Metallic Bonds
Bonding between metal atoms is called metallic bonding. Orbitals
from many metal atoms overlap which allow electrons to migrate
from atom to atom. This gives rise to the high conductivity observed
for metals.
Polar Bonds
When the electronegativity of two atoms in a bond is di erent, the
electrons will be more strongly pulled towards the atom with the
higher electronegativity.
DIPOLE
We say the HF bond is polar because the shape of the molecular
orbital is skewed towards the atom with the higher electronegativity
Two point charges make a dipole.
(F in this case). This creates a partial negative charge on the F
and a partial postive charge on the H atom. The partial charges are
indicated by the
symbol.
174
Chapter 8 Introduction to ChemicalBonding
As the di erence in electronegativity ( E N) becomes larger the
bond becomes more polar. As a general rule, when the di erence in
electronegativity (see footnote6 ) is:
1. E N
0 and
0 4 the bond is nonpolar covalent with an
equal sharing of electrons.
2.
EN
0 4 and
2 0 the bond is polar covalent.
3.
EN
2 0 the bond is ionic.
Bond
E N Type of Bond
HC
0.9
polar covalent
Br2
0.0
nonpolar covalent
CO
1.0
polar covalent
CH
0.4
nonpolar covalent
NaH
1.1
polar covalent
NaC
2.1
ionic
For HC the di erence in electronegativity, E N can be calculated by subtracting the electronegativity values listed for the
element H and C in the table on page ??.
EN
30
21
09
The C atom has the higher electronegativity compared to hydrogen. This tells us that the C atom pulls more strongly on
the electrons in the bond giving the C atom a partial negative
charge and the H atom a partial positive charge.
EN
09
indicates that the bond is polar covalent.
6 Bonding is very complex. Many factors like ionization energy, electronegativity, electron af nity, and overlap
of molecular orbitals must be considered to fully understand the nature of a bond. By itself, the difference in
electronegativity is only a rough guide. For example, the difference in electronegativity between the atoms in HF is 1.9.
From that you might conclude that the bond is nearly ionic. The bond is very polar, however, the ionization potential of
the hydrogen atom is far too high for the H atom to lose the electron completely and become H HF is a polar covalent
bond. Sometimes it is useful to think of the bond as partially ionic and partially covalent. The bonding in HC is about
83% covalent and 17% ionic. Even in NaC the electron is never completely transferred to the C atom. NaC is about
80% ionic.
Lewis Dot Structures of Molecules 175
In the CO molecule the di erence in electronegativity is
EN
35
25
10
The O atom has the higher electronegativity compared to carbon.
This tells us that the O atom pulls more strongly on the electrons
in the bond giving the O atom a partial negative charge and the
C atom a partial positive charge.
EN
1 0 indicates that the
bond is polar covalent. The CO bond is slightly more polar than
the bond in HC
Lewis Dot Structures of Molecules
Lewis structures of molecules show how the valence electrons are
involved in the chemistry of the molecule. These structures are
useful for indicating:
1. The arrangement of the atoms.
2. Electrons in shared covalent bonds
3. Electrons in nonbonding molecular orbitals.
a. Nonbonding electrons do not contribute to bonding.
b. Nonbonding electrons in uence the chemistry and geometry
as we will see in later chapters.
4. Lewis structures do not indicate the geometry and angles between
176
Chapter 8 Introduction to ChemicalBonding
atoms.
Writing Lewis Structures
1. Sum up the valence electrons from each atom in the molecule.
a. If the atom is charged add 1 electron for every negative charge.
b. Subtract one electron for every positive charge.
Molecule S
O
Charge Total Valence e
SO2
6
26
0
6
26
0
18
SO24
6
46
2
6
46
2
32
SO2
6
26
1
6
26
1
17
A
1 charge means 1 e is missing so subtract 1 e
A
2 charge means 2 extra e exist so add 2 e
2. Write symbols for the atoms to show the arrangement of the
atoms in space.
a. The central atom is usually the atom with the lowest elec-
tronegativity.
b. Higher electronegativity atoms like to be on the outside be-
Writing Lewis Structures 177
make as many bonds.)
c. Draw lines to represent the bonds between atoms.
Molecule Atom Arrangement
OF2
F
O
F
H
NH3
H
N
H
O
O
SO24
S
O
O
3. Place the valence electrons on the OUTSIDE atoms attached to
the central atom. Each atom should have an octet (8) of electrons. When counting electrons each bond counts as 2 electrons.
These electrons are shared between both atoms in the bond. For
example, the bond between F and O in OF2 counts as 2 electrons
on the F and two electrons on the O.
a. The H atom has a maximum of 2 electrons, not 8.
4. After making octets on all of the outside atoms, ALL left over
valence electrons go on the central atom. Sometimes the octet
will have more than 8 electrons. This will be explained later.
5. Check the central atom to see if it has an octet. If not, remove
nonbonding electrons from the outside atoms to make multiple
bonds.
a. The atoms that most often make multiple bonds are C, N, O
and the atoms beneath them in the table.
b. H, F never make double bonds.
Example 1, OF2
1. Total valence electrons: F has 7, O has 6. Total valence electrons
=72
6
20
178
Chapter 8 Introduction to ChemicalBonding
2. The atom arrangement for this molecule is
O
F
F
a. Note: Usually, the central atom is the atom with the lowest
electronegativity. In this case O has the lowest electronegativity.
b. Also, F cannot make a two bonds. We conclude O is the
central atom.
3. Use the 20 available valence electrons to make octets on the
outside atoms rst. Note that 4 electrons have already been used
for the two bonds so we only have 16 left.
O
F
F
4. Place the remaining electrons on the central atom. We have used
16 valence electrons. 4 electrons are in bonds and 12 electrons
are nonbonding. We have 20 16 4 electrons left. These go
on the central atom.
O
F
F
5. The central atom already has an octet so we do not have to go
further and worry about multiple bonds.
Example 2, HCN
1. Total valence electrons: H has 1, C has 4, N has 5. Total valence
electrons =1
4
5
10
Writing Lewis Structures 179
2. The atom arrangement for this molecule is
a. H has the lowest electronegativity but can never be the central
atom since H can only make 1 bond.
b. C has a lower electronegativity than N so C is the central atom.
3. Place valence electrons on the outside atoms to make octets. (H
already has its maximum of 2 electrons.)
H
C
N
4. The remaining valence electrons go on the central atom. We
have used 10 electrons so there are no valence electrons left for
the central atom.
5. The central atom does not have an octet. Nonbonding electrons
on the N must be converted to multiple bonds. This way the
electrons can be shared with the C atom giving the C atom an
octet.
MULTIPLE BONDS
a. Stronger than single
bonds.
b. Shorter length than
single bonds.
H
C
N
H
C
N
The central carbon atom and the nitrogen now both have an
octet. The electrons in the bonds are shared between both atoms.
180
Chapter 8 Introduction to ChemicalBonding
Formal Charge
Formal charge is an electron bookkeeping method. It is used to
check if the arrangement of electrons around a particular atom in
a Lewis structure makes sense. Formal charge does not give the
actual charge on the atom. Each atom in a Lewis structure has a
formal charge that is calculated from this formula.
For any speci c atom in the Lewis Structure
Formal
Charge = # of valence electrons in the isolated atom
# of nonbonding electrons on that atom
1 electron for each bond to that atom
Formal Charges of OF2
F
O
F
An isolated O atom has 6 valence electrons. In this OF2 molecule
there are 4 nonbonding electrons on the O atom and 2 bonds to the
O atom. The formal charge of the O atom is 6
4
2
0
An isolated F atom has 7 valence electrons. In this OF2 molecule
there are 6 nonbonding electrons on the F atom and 1 bond to the
Formal Charge 181
F atom. The formal charge of the F atom is 7
6
1
0
I The sum of the formal charges on all of the atoms in the molecule
must be equal to the charge on the molecule. This can be used as
a check that the formal charges were calculated correctly.
Formal Charge Example 2
A student wrote this improper Lewis Structure for OCS. The gure
shows the formal charge calculations.
F.C. = 6 - 4 - 2 = 0
O
C
S
F.C. = 6 - 6 - 1 = -1
F.C. = 4 - 0 - 3 = +1
Check: The sum of the formal charges must be the charge on the
molecule. 0
1
1
0 and the charge on the molecule is, in fact,
0.
Using Formal Charges to Choose Lewis
Structures
Sometimes more than one possible Lewis structure can be drawn.
Formal charges are used to decide which structure is more likely to
be the best Lewis structure.
1. Lower formal charges are better than higher formal charges. Ideally, all atoms should have a zero formal charge.
182
Chapter 8 Introduction to ChemicalBonding
2. Negative formal charges should be on the more electronegative
atoms in the structure. In other words, a structure with a positive
formal charge on the most electronegative atom would be unlikely.
Example of Using Formal Charges to Decide Structure
Here are two perfectly valid Lewis structures for CO2 Every atom
has an octet. Both structures could have resulted by following the
rules given above. Which is more correct?
F.C. = 6 - 4 - 2 = 0
A
O
C
O
F.C. = 6 - 4 - 2 = 0
F.C. = 4 - 4 - 0 = 0
F.C. = 6 - 6 - 1 = -1
B
O
C
O
F.C. = 6 - 2 - 3 = +1
F.C. = 4 - 4 - 0 = 0
1. Structure A has ideal formal charges of zero on each atom.
2. Structure B has two problems.
a. The formal charges are higher than structure A.
b. The oxygen with the triple bond has a +1 formal charge. So
the atom with the highest electronegativity has a +1 formal
charge. This is not a good situation.
Understanding the Meaning of Formal Charges
Atoms combine together to make molecules.
Each atom shares
Formal Charge 183
electrons to make covalent bonds. The number of bonds made by
a particular atom is determined by how many electrons the atom
requires to reach a lled valence shell. The table below shows the
number of bonds these elements typically make to achieve an octet.
Number of Bonds
Keep these numbers in mind
when drawing Lewis structures.
HCN will look like this
because H will make 1 bond, C
typically makes 4 bonds, and N
typically makes 3 bonds to achieve
a lled valence shell.
What does formal charge tell us? Consider the Lewis structure
of NH3 .
Nitrogen has 5 valence electrons and needs 3 more electrons to make
an octet. By making 3 covalent bonds nitrogen is able to achieve an
octet by sharing 3 electrons with the hydrogen atoms. Looking at
the electrons arranged around nitrogen notice that 5 came from the
5 valence electrons initially on nitrogen and 3 from the hydrogens.
184
Chapter 8 Introduction to ChemicalBonding
Since no electrons were actually transferred completely to nitrogen
its formal charge is zero.
Now look at the Lewis structure of CO.
According to the periodic table above, oxygen typically makes 2
Formal Charge of
1
bonds to achieve an octet. But in CO the oxygen atom makes 3
A formal charge of 1 on C bonds. How is this possible? We could imagine that O transfers
means the arrangement of electrons on C uses the 4 valence
electrons from C plus 1 extra elec- one electron to C. This makes the O have a +1 charge and C have
tron from O.
a
Formal Charge of
1
A formal charge of 1 on O
means the arrangement of electrons on O only uses 5 of the 6
valence electrons from O.
1 charge.
But now C
and O
are both isoelectronic with
N. The C and O atoms can now each make 3 bonds to achieve an
octet. The formal charge is
1 on carbon and +1 on oxygen.
Notice that the carbon atom in the CO Lewis structure has an
octet of 8 electrons.
The triple bond contains 2 electrons from
each bond for a total of 6 electrons that are being shared between
the carbon and oxygen atoms. But if we ask how the 8 electrons
on C determine the charge on the carbon atom it must concluded
Exceptions to the Octet 185
that only the two non bonding electrons and one electron from each
electrons. Since an isolated C atom only has 4 valence electrons it
must be concluded that the formal charge on C is
1. The gure
above shows that the oxygen atom contributes 5 electrons to the
8 electrons arranged around the oxygen atom.
Since an isolated
O atom has 6 valence electrons but only uses 5 in this electron
arrangement in CO it must be concluded that the formal charge on
O is
1.
Exceptions to the Octet
1. Odd Number of Total Valence Electrons
RADICALS
Highly reactive radicals are
responsible for the destruction
of ozone in the upper atmosphere.
When the total valence electrons is an odd number there is no way
to satisfy each atom with an octet of electrons. These molecules
are called radicals. They are very reactive species and some are
responsible for cell damage in the human body. Here are some
examples or radicals.
Molecule Total Valence Electrons
NO
5
6
11
OH
6
1
7
C O2
4
5
9
186
Chapter 8 Introduction to ChemicalBonding
Using Formal Charges in Lewis Structures of Radicals
With an odd number of valence electrons one of the atoms in the
Lewis structure will not have an octet and will be missing an electron. But which atom will be missing the electron?
A
F.C = 6 - 3 - 2 = +1
B
N O
F.C = 6 - 4 - 2 = 0
N O
F.C = 5 - 5 = 0
F.C = 5 - 4 - 2 = -1
The formal charges for this radical indicate that structure B will be
more likely since it has formal charges of zero for both atoms.
2. Less Electrons than an Octet
The atoms H, He, Li, Be, B will have less than an octet of electrons.
B cannot have an octet in BF3
Here is an example molecule.
Do not be tempted to use nonbonding electrons on F to make
a double bond. F is too electronegative to make a double bond.
F B
F
F
3. More Electrons than an Octet
The elements in the third row and below in the periodic table can
have more than 8 electrons in a Lewis structure. These elements use
Exceptions to the Octet 187
their d orbitals for making covalent bonds and storing nonbonding
electrons. This is called expanded valence.
Cl
Cl
P
Cl
Cl
Cl
These are the elements that can have an expanded valence. Note
that expanded valence occurs more often with electronegative atoms
O,F,C attached to one of the atoms below as a central atom..
188
Chapter 8 Introduction to ChemicalBonding
Summary
I Multiple bonds result from molecules with too few valence electrons.
I More than an octet results from too many valence electrons.
Charges on the atoms in a molecular compound
There are several ways to determine which atoms in a molecule, if
any, have charges.
1. Investigate the formal charges.
2. Compare the electronegativities of atoms in each bond. If there
is a di erence in electronegativity of more than 0.4 the higher
electronegativity atom will have a partial negative charge. (The
higher electronegativity atom has a stronger pull on the electrons.)
Exercise 8.46
Determine which atoms are charged or have
Summary 189
partial charges in these molecules.
Solution
In the molecule on the left, NH2 CH3 all of the atoms
have a formal charge of zero.
The atom with the highest elec-
tronegativity is nitrogen with E N = 3.0 (See table on p.169.). H
has E N = 2.1, and C has E N = 2.5 so all of the bonds to nitrogen will have
EN
04
These are polar bonds with partial
charges. The higher E N nitrogen atom will have the partial negative charge and all atoms attached to nitrogen will have partial
positive charges.
EN
30
21
09
EN
30
25
05
For the molecule on the right, NH3 CH3 the nitrogen atom has a
formal charge of +1.
All of the other atoms have a zero formal
charge. The molecule is really [NH3 CH3 ] with the positive charge
located on the nitrogen atom.
190
Chapter 8 Introduction to ChemicalBonding
Resonance
The Lewis structure at the top of this gure has a central oxygen
atom with only 6 electrons. To give that oxygen atom an octet of
8 electrons a double bond must be made.
Notice that there are
two equivalent ways to accomplish this. The double bond could be
made on the left or on the right and both structures are equivalent..
A double headed arrow.is used
to indicate resonance.
These are called resonance structures.
1. The arrangement of electrons is similar in both structures except
that electrons are shifted to di erent atoms.
2. The only di erence between resonance structures is the position
of the electrons. The atoms cannot be moved.
3. The condition of resonance is indicated by the double headed
Resonance 191
arrow shown inside the highlight box.
Resonance Hybrid
I The signi cance of resonance is that the actual molecule will not
look like any of the individual resonance structures. Instead, the
actual molecule will be an average of the resonance structures and
is called a resonance hybrid.
O
O
O
The dashed line represents a 2 electron bond that spans all three
oxygen atoms. The actual bonds in ozone are both identical and
Resonance in Benzene
Benzene has the molecular formula C6 H6 The carbon atoms form
a ring structure with alternating single and double bonds.
192
Chapter 8 Introduction to ChemicalBonding
H
H
H
H
H
H
H
H
H
H
H
H
These are resonance structures. The benzene ring is unusually stable
because of the existence of resonance.
Benzene Resonance Hybrid
The actual bonds in the ring of benzene are all the same. The bonds
H
H
H
H
H
H
H
H
H
H
H
H
The molecular structure of benzene is often drawn with a circle in the
ring to represent 6 electrons that form a bond which spans the entire
ring. Another way to imagine resonance in benzene is to picture
the molecule hopping back and forth between the two resonance
not an accurate picture it does provide a way to see why the bonds
Resonance 193
Exercise 8.47
Draw Lewis structures for SO3 Determine if
the molecule has resonance. Guess what the resonance hybrid looks
like.
Solution
This molecule has 24 total valence electrons. Af-
ter drawing the arrangement of the atoms and putting the valence
electrons on the outside atoms we obtain this structure.
SO3 RESONANCE
HYBRID
O
The actual SO3 molecule is
an average of all of the resonance
structures and includes a small
contribution from these two resonance structures.
S
O
O
There are no valence electrons left so nonbonding electrons must
be converted to a double bond in order to give the central sulfur an
octet. Notice that there are three possible places where the double
bond can be positioned. This is what leads to resonance.
O
O
O
S
O
O
S
Resonance Hybrid
O
O
O
S
O
The resonance hybrid will be a mixture
of these three resonance structures. All of the bonds will be the
194
Chapter 8 Introduction to ChemicalBonding
O
O
S
O
Additional Resonance Structures
Since sulfur is allowed
to exhibit an expanded valence (have more than 8 electrons) the
following resonance structures are also possible.
O
O
S
O
O
O
S
O
According to the rules for drawing Lewis structures these are perfectly valid structures. A quantum mechanical calculation must be
done in order to gure out how important a particular resonance
structure will be. This is for professionals! The two structures
shown above have lower overall formal charges. However, a general
rule is that resonance structures that require the central atom to
Bond Dissociation Enthalpies 195
violate the octet are of lesser importance.
Bond Dissociation Enthalpies
One way to think about the chemical bond is to picture the overlap
of atomic orbitals creating a covalent bond.
The molecular orbitals in the molecule represent the probability of
nding electrons in the molecule. Of course the molecular orbitals
are somewhat di erent from the atomic orbitals. The molecular orbitals formed from the overlap of atomic orbitals look something
like this.
We get a more accurate picture of the bond when we remember
that the bond is very polar because of the large electronegativity of
uorine. The electrons in the bond are pulled more closely to the F
atom resulting in a polar covalent bond.
196
Chapter 8 Introduction to ChemicalBonding
Bond Energy and Enthalpy
The covalent bond represents the forces holding the two atoms toAttractive forces between two
magnets. To separate the magnets energy must be ADDED.
gether in a molecule. Imagine two magnets stuck together. To pull
them apart we must put in energy to overcome the magnetic force
holding the magnets together. The same is true for the covalent
bond in a H–F molecule. To pull the H and F atoms apart energy
must be added and this is called the bond energy.
Bond Dissociation Enthalpies 197
This gure shows what happens when a H and F atom that are
separated approach each other and form a bond. As the distance
between atoms gets smaller a covalent bond begins to form and the
energy decreases (becomes more negative.) The minimum of the
curve shows the bond dissociation energy is
Wait a minute, the graph shows
567 kJ/mol for H–F.
567 kJ/mol. What is wrong? The
graph is showing the energy stored in the bond when the bond is
formed. Bond dissociation energy means the energy added to an
existing H–F molecule to break the bond. Notice that the units
are given as kJ/mole which is the energy to break 6 022
1023
bonds. When this energy is added in the form of heat (at constant
pressure) it is called bond dissociation enthalpy.7
7 Note for the advanced student:
E
H
ork As long as we are just talking about the energy used to
break the bond and ignore the work done then E
H So in this discussion energy and enthalpy can be used
interchangably.
198
Chapter 8 Introduction to ChemicalBonding
I Bond Dissociation Enthalpy is the heat required to break 1
mole of bonds at constant pressure.
Potential Energy
Sometimes students are confused about energy because they only
understand kinetic energy which is energy in motion. They are
confused because they ask Where is the energy in a chemical bond,
Energy in Motion is Kinetic
Energy
I Energy in chemical bonds is stored energy called potential energy. When the fuel in a rocket ship burns the energy stored in
the chemical bonds is released and used to propel the rocket ship.
What is Enthalpy?
Enthalpy is just one way to talk about energy. Enthalpy is a very
useful concept for chemical reactions at constant pressure. Un-
Bond Dissociation Enthalpies 199
der these conditions,
enthalpy
heat of reaction
For example, when water is heated to 100 C at constant pressure
of 1 atm additional heat can then be added to make the water boil.
The extra heat added to make the water boil is used to overcome
the intermolecular forces in water. The molecules in the liquid can
then escape each other and expand to ll the volume as a gas. The
energy supplied in the form of heat to convert 1 mole of liquid
water to 1 mole of gas is called the molar enthalpy of vaporization. This is very similar to the bond dissociation enthalpy discussed
above. Except that in the case of boiling water the enthalpy is the
heat required for breaking intermolecular bonds. Where as Bond
Dissociation Enthalpy is energy in the form of heat required to break
1 mole of intramolecular bonds.
200
Chapter 8 Introduction to ChemicalBonding
Calculations Using Bond Dissociation Enthalpy
Consider this reaction
H2 g
C 2 g
2HC
g
We can calculate the enthalpy for this reaction from a table of Bond
Dissociation Enthalpies.
Bond Dissociation Enthalpy
H-H
C
DC
D H-C
436 kJ/mol
C
242 kJ/mol
431 kJ/mol
Before doing the calculation it helps to draw molecular structures
so we can see the bonds being broken and formed.
The heat of the reaction is the NET CHANGE in enthalpy in the
reaction. In other words, the Enthalpy of reaction is the enthalpy
(heat) required to break all of the bonds on the reactants side minus
the enthalpy (heat) released when bonds are formed on the prod-
bond and one C
bond and we form two
bonds. So the
Bond Dissociation Enthalpies 201
NET CHANGE in enthalpy is given by this equation:
enthalpy of rxn
D
D C
2D
Look up the bond dissociation enthalpies in the table and insert the
values into the equation.
enthalpy of reaction
D
enthalpy of reaction
Example 8.48
D C
436
242
2 431
2D
184 0 kJ/mol
Calculate the Enthalpy of reaction using bond
dissociation enthalpies.
Solution
First write out the Lewis Structures so the bonds can
be clearly seen.
202
Chapter 8 Introduction to ChemicalBonding
Next apply the equation
Looking at the bonds in the molecular formula we can see that this
equation becomes
enthalpy of reaction
4D
D C=C
D C
D
2D
4D
C
Look up the bond dissociation enthalpies in the table and substitute
them into the equation. Here is the table.
Bond
Bond Dissociation Enthalpy
413 kJ/mol
C=C
D C=C
614 kJ/mol
348 kJ/mol
C
C
D C
D
C
242 kJ/mol
328 kJ/mol
Bond Dissociation Enthalpies 203
Making the substitution of table enthalpies into the equation gives
enthalpy of reaction
4D
D C=C
D C
D
2D
4D
4 413
4 413
614
348
C
242
2 328
148 kJ/mol
Notice that the C=C double bond is much stronger than the C–C
single bond.
Example 8.49
Calculate the bond dissociation energy for the
O–O bond in O3 using this information. D O=O
and the enthalpy of reaction is
495 kJ/mol
142 5 kJ/mol for the reaction given
below.
3
2
O
O
O
O
O
Remember that O3 has resonance so the O–O bond in the product
is in between a double and single bond. So we expect to nd a
bond dissociation enthalpy for O3 that is higher than a O–O single
bond.
Solution
At rst, it might not be obvious how to approach this
problem. So let us begin by writing an equation for the enthalpy of
reaction using the bond dissociation energies. Then maybe a light
bulb will go on inside our heads and we will be able to see how to
204
Chapter 8 Introduction to ChemicalBonding
nd the missing D
in O3
3
reaction enthalpy = D O=O
2
2D O-O
Looking at the given information we see that we know some of the
information for this equation.
reaction enthalpy
=
142 5
3
D O=O
2D O-O
2
3
495 2D O-O
2
So we can rearrange the equation and solve for the missing D
in O3
3
2 495
D O-O
Resonance in ozone
142 5
2
300 kJ/mol
Here is a comparison of bond dissoci-
ation enthalpies for oxygen bonds:
Bond
Dissociation Enthalpy
O-O
3
O=O
146 kJ/mol
D O-O in O3
D O=O
300 kJ/mol
495 kJ/mol
Type of Bond
Single
In Between
Double
O3 does indeed have bonds that are in betweena double and single
bond due to resonance. The two Lewis structures of ozone look like
this.
Bond Dissociation Enthalpies 205
O
O
O
O
O
O
The actual molecule is a mixture of these two Lewis structures
called the resonance hybrid. Each bond is identical and is approximately 1.5 bonds because of the second bond shared in two
places. Our calculation in the example above shows that indeed
this is the case!
O
O
O
9 Molecular Geometry
Electron Domains
Electron domains are regions of space occupied by electrons. In
reality the electrons are located in molecular orbitals that can be
bonding or nonbonding.
3
1
2
The central nitrogen atom has three electron domains.
1. The multiple bond counts as 1 domain because both bonds occupy the same region of space.
2. The single bond is domain #2.
3. The nonbonding electron pair (sometimes called lone pair) is
domain #3.
VSEPR Model
207
Why is CO2 Linear and NO2 Bent?
Linear
Bent
1. CO2 has 2 electron domains about the central C atom.
2. NO2 has 3 electron domains about the central N atom.
3. Electron domains repel each other.
C
N
4. Two domains avoid each other by assuming a linear geometry.
This is why CO2 is linear.
5. Three domains avoid each other by assuming a planar geometry
with 120 angles between domains. This is why NO2 is bent.
VSEPR MODEL
I The Valence Shell Electron Pair Repulsion (VSEPR) model predicts that the optimum geometry of a molecule is established by
208
Chapter 9 Molecular Geometry
minimizing the repulsions of the electron domains on the central
atom. NO2 assumes a bent geometry to minimize the repulsions
between the lone pair electrons on the nitrogen and the 2 bonding
electron domains.
Molecular Geometry and Electron Domain
Geometry
imagine
nonbonding
electrons are
invisible
1. Electron domain geometry describes the geometry of all of the
electron domains about the central atom. The left diagram shows
all of the electrons. There are 3 domains and the geometry about
the central atom is trigonal planar.
2. The molecular geometry describes the geometry of the atoms
only. The diagram on the right shows the atoms. The nonbonding electrons help determine this geometry but are invisible in
the picture on the right. To determine the molecular geometry
pretend the nonbonding electrons are invisible. Use your thumb
to cover up the nonbonding electrons on the central atom and
determine the geometry of the atoms.
209
NO2 Resonance Hybrid
The actual NO2 molecule does not contain one N-O single bond
and one N=O double bond as is shown in this Lewis structure.
There are two N-O single bonds The real molecule is a mixture of these two resonance structures.
in NO2 Because of resonance
there is one more bond that is
shared in two places. Each N-O
bond is approximately 1.5 bonds
in strength.
As shown in the model below, both bonds in the actual molecule are
identical. They have the same bond length and the same bond dissociation enthalpy. Each bond is about 1.5 bonds in strength. The
geometry is bent because of the repulsions of the lone pair electrons
as predicted by the VESPR model.
210
Chapter 9 Molecular Geometry
Predicting Geometries from Lewis Structures
To nd the geometry of a molecule predicted from VESPER theory
1. Draw the Lewis structure.
2. Count the electron domains around the central atom in the Lewis
structure.
3. Identify the number of bonding and nonbonding domains around
the central atom.
4. Look up the geometry in the table below.
5. These geometries are derived by
nding the best way for the
electron domains to avoid each other to minimize electron
electron repulsions.
Total # of Bonding Nonbonding
Molecular
Bond
Domains Domains
Geometry
Angle
Domains
2
2
0
Linear
180
3
3
0
Trigonal Planar
120
3
2
1
Bent
120
4
4
0
Tetrahedral
109 5
4
3
1
Trigonal Pyramidal
109 5
4
2
2
Bent
109 5
5
5
0
Trigonal Bipyramidal
120 90
5
4
1
Seesaw
120
5
3
2
T-Shaped
5
2
3
Linear
180
6
6
0
Octahedral
90
6
5
1
Square Pyramidal
90
6
4
2
Square Planar
90
90
90
Predicting Geometries from Lewis Structures 211
212
Chapter 9 Molecular Geometry
Bond Angles 213
Bond Angles
Two Domains
1. Two domains avoid each other by being 180 apart.
Three Domains
1. Three domains avoid each other by being 120 apart.
2. All three bond angles are equivalent.
I Lone pairs occupy more space and repel the bonding domains more.
This reduces the bond angle.
I Multiple bonds also occupy more space than bonding domains.
Four Domains
1. Four domains avoid each other by arranging the domains at the
corners of a tetrahedron. The bond angles are 109.5
214
Chapter 9 Molecular Geometry
2. When lone pairs are present their larger size increases the repul-
sions and reduces the bond angle.
3. Two lone pairs reduce the bond angles even more. See Example
9.52 on p. 217.
Five Domains
1. Five domains are arranged at the corners of a trigonal bipyramid
SEESAW
GEOMETRY
geometry to minimize electron repulsions.
2. Three of the ve domains are arranged in a plane with 120 angles
and have equatorial bonds.
3. Two domains are arranged with 90 degree angles to the plane
and have axial bonds.
When the structure is tipped
on its side you can see the seesaw.
4. Nonbonding (lone pair electrons) always go in the plane at the
equatorial positions. The 120 equatorial angles minimize the
repulsions of the lone pairs better than in the axial position.
Bond Angles 215
Six Domains
1. Six domains are arranged in an octahedral geometry.
2. All of the positions are equivalent with 90 bond angles.
3. Since all positions are equivalent a lone pair domain can go in
any position.
Example 9.50
Determine the (1) molecular geometry and
(2) electron domain geometry of SO3 and PH3 Determine (3) the
approximate bond angles.
216
Chapter 9 Molecular Geometry
Solution
The Lewis structures for these two molecules are:
H
O
O
S
O
H
P
H
1. The Lewis structure of SO3 has 3 total electron domains about
the central S atom.
a. The electron domain geometry is trigonal planar.
b. There are 3 bonding domains and no nonbonding domains.
(Double bonds count as 1 domain, remember?) According to
the table on p. 210 the molecular geometry is trigonal planar.
Remember that this molecule will have 3 equivalent resonance
structures. In the resonance hybrid all of the bonds are the
same. (About 1.5 bonds)
c. The bond angles are all 120 because all three domains of the
trigonal plane geometry are equivalent.
2. The Lewis structure of PH3 has 4 total electron domains about
the central P atom.
a. The electron domain geometry is tetrahedral.
b. There are 3 bonding domains and 1 nonbonding domain around
the P atom. According to the table on p. 210 the molecular
geometry is trigonal pyramidal.
c. The bond angles of the tetrahedron are 109.5 The presence
of the lone pair reduces the angle slightly because of the increased repulsions. The angle is 109 5 because a lone pair
Bond Angles 217
is present.
Example 9.51
Determine the geometry of IF5
Solution
1. There are 42 valence electrons.
2. After making octets on the outer F atoms there are 2 electrons
left over.
3. These must go on the central I atom even though I is already
violating the octet rule. Iodine is one of those exception elements
that is allowed to have an extended valence (more than 8 electrons).
MEMORIZING
GEOMETRIES
F
Most of the geometries are
easy to remember if you pay attention to the structure and its
name. Can you see the square
pyramid in the IF5 structure?
F
F
F
F
4. According to the Table on p. 210, 5 bonding domains and 1 lone
pair has a square pyramidal molecular geometry.
Example 9.52
Rank the following molecules in order of in-
creasing bond angle: CH4 NH3 H2 O
218
Chapter 9 Molecular Geometry
Solution
Draw the Lewis structure, determine the molecular
geometry, and then consider the e ects of the lone pair electrons
H
H
C
H
on the bond angles. Refer also to the Table on p. 210.
1. CH4 has a tetrahedral molecular geometry. The bond angles are
109.5
2. NH3 has a tetrahedral electron domain geometry with one lone
H
pair. The lone pair makes the bond angles
109 5
3. H2 O also has a tetrahedral electron domain geometry. Since it
H
has 2 lone pairs the bond angles will be even less than NH3
H
N
H
The bond angles rank in the order from small to big:
H
O
H
H2 O
NH3
CH4
Polar and NonPolar Molecules
Polar Bonds
In Chapter 7 we discussed the concept of polar bonds.
Bond ( EN) Type of Bond
HC
0.9
polar covalent
Br2
0.0
nonpolar covalent
CO
1.0
polar covalent
CH
0.4
nonpolar covalent
Electronegativity measures the pull on electrons in a bond. When
the di erence in electronegativity is large electrons are pulled closer
to the more electronegative atom giving the atom a partial negative
Polar and NonPolar Molecules 219
charge. In general, when the atoms in the bond are di erent the
electronegativity is usually di erent and the bond is polar. More
precisely, if the di erence in electronegativity is
0 4 we say the
bond is polar. See table of electronegativities on p.169
Polar Molecules
DIPOLE ANALOGY
The bonds in CO2 are polar because the electronegativity of O is
higher than C. However, since the two dipoles are oriented in opposite directions the dipoles cancel. We call this a nonpolar molecule.
If the dogs have equal strength
and pull in opposite directions
their efforts cancel. But if you
have a big strong dog on one
side and a little wimpy dog on
the other side, the big dog wins!
220
Chapter 9 Molecular Geometry
The molecule OCS is polar because the dipoles do not cancel. The
dipole oriented towards O is much larger than the one oriented towards S because the electronegativity of O is much higher than S.
I To determine if a molecule is polar you must rst know the molecular
geometry.
When do Dipoles Cancel?
Dipoles cancel when all domains are the same and the electron
domain geometry is one of these basic types.
1. Linear
2. Trigonal Planar
3. Tetrahedral
4. Trigonal Bypyramidal
5. Octahedral
The next two pages show all of the cases that give rise to polar
and nonpolar molecules. For the more complex geometries it is not
always easy to see how the dipoles cancel . In high school physics you
Polar and NonPolar Molecules 221
learn about vectors. When the dipoles in a trigonal plane geometry
are written as vector components it is easier to see why they cancel
when all 3 domains are the same. This diagram is for BH3 which
has a trigonal planar molecular geometry.
A DIPOLE MADE
INTO
x,y VECTOR
COMPONENTS
For water the dipoles do not cancel and this results in a net dipole.
Water has a tetrahedral electron domain geometry. The dipoles cancel when all 4 domains are the same. For water 2 of the domains
are nonbonding. As you can see below there is a net dipole because
the individual dipoles do not cancel.
222
Chapter 9 Molecular Geometry
Polar and NonPolar Molecules 223
224
Chapter 9 Molecular Geometry
Polar and NonPolar Molecules 225
Example 9.53
Determine which of these molecules is polar
and which is nonpolar. PH3 BH3 CH4 SF4
Solution
First you must draw the Lewis structure and determine
the geometry about the central atom. Next consult the Tables
on the next two pages to determine if that geometry is polar or
nonpolar.
1. PH3 has a trigonal pyramidal molecular geometry and a tetrahedral electron domain geometry. One domain is di erent (the lone
pair) so the molecule is polar.
H
H
H
P
2. BH3 has a trigonal planar molecular geometry and a trigonal pla-
COMMON MISTAKE:
nar electron domain geometry. All of the domains are the same
so the molecule is nonpolar.
H
Note that you cannot simply look
at the molecule and decide if it
is polar or nonpolar. You must
determine the geometry and apply the rules discussed here.
B
H
H
a. PH3 is polar and BH3
is nonpolar.
b. CH4 is nonpolar and
3. CH4 has a tetrahedral molecular geometry and a tetrahedral elecSF4 is polar.
tron domain geometry. All of the domains are the same so the
molecule is nonpolar.
H
H
C
H
H
226
Chapter 9 Molecular Geometry
4. SF4 has a seesaw molecular geometry and a trigonal bipyramid
electron domain geometry. One of the domains in the equatorial
plane is di erent (the lone pair) so the molecule is polar.
F
S
F
F
F
Covalent Bonding
1. Covalent bonds are formed when electrons are shared in a bond.
2. Atomic orbitals must overlap in space in order for the electrons
to be shared between atoms.
3. New orbitals called molecular orbitals are formed.
Molecular
orbitals show the probability of nding electrons around the nuclei
in the molecule. Molecular orbitals can be thought of as a mixture
of atomic orbitals.
4. The shared electrons occupy the molecular orbitals.
5. The bond in the picture is made from the overlap of 2 atomic 1s
orbitals.
6. Bonds can also be made from the overlap of s and p orbitals.
Covalent Bonding 227
Molecular Orbitals
Covalent bonds in molecules are made from two half lled atomic
orbitals. These two atomic orbitals mix together to form a new orbital called a molecular orbital.
Hybrid Orbitals
H Be H
228
Chapter 9 Molecular Geometry
How are the bonds in BeH2 made? The Be-H bond cannot be explained as an overlap of atomic orbitals on H with atomic orbitals
on Be. A covalent bond requires the overlap of 2 orbitals that each
hold one electron. Be does not have any atomic orbitals with one
electron.
In addition, the 2s orbital on Be could not be used to
make 2 bonds with 1800 angles as in BeH2 .
To explain the bonding in BeH2 we allow the 2s and 2p atomic
orbitals on Be to mix together to form new orbitals. We call this
orbital mixing process hybridization. The two new orbitals are called
sp hybrids because they are a mixture of one 2s and one 2p atomic
orbital.
Hybridization
Covalent Bonding 229
1. Hybridization is a process where atomic orbitals are mixed to-
gether to make new hybrid orbitals.
2. For BeH2 the 2s and 2 p atomic orbitals are mixed together to
form two new molecular orbitals called sp hybrid orbitals.
3. The gure above shows the electron con gurations for the inital
atomic and the resulting hybrid orbitals.
4. After mixing, the 2s and 2p orbitals no longer exist as 2s and 2p
atomic orbitals.
5. Notice that the hybrid orbitals have two half
bitals that can be used to make two bonds.
6. Notice also energy levels
lled sp hybrid or-
230
Chapter 9 Molecular Geometry
This gure shows the two sp hybrid orbitals on the Be atom. Each
Be-H bond is made from the overlap of one sp hybrid on Be and one
1s orbital on H. The Be-H bond has 2 shared electrons. The 1s
atomic and sp hybrid orbital each contribute one unpaired electron
to the bond.
sp2 Hybrid Orbitals in BH3
The molecular orbitals used in making trigonal planar molecules are
constructed from one s and 2 separate p orbitals. They are called
s
p
p = sp 2 hybrid orbitals.
Covalent Bonding 231
This picture shows the atomic energy levels of the central B atom
before and after hybridization in the molecule BH3 .
The two 2p
and one atomic 2s orbitals are mixed together and become three
sp2 hybrid orbitals.
Each one of the sp 2 hybrid orbitals has one
unpaired electron so each hybrid orbital can be used to make one
bond.
The three H-B bonds are made from the overlap of an sp2 hybrid on
B and a 1s atomic orbital on H. The 1s orbital and sp2 hybrid orbital
each contribute one electron to the bond. The bond results from
the overlap of two orbitals that each contain one electron.
The
spins of the two electrons must be in opposite directions.
Note
that the bond angles are 1200 as expected for a central atom with
three domains.
232
Chapter 9 Molecular Geometry
Determining the Type of Hybridization
1. Write the Lewis structure
2. Determine the electron domain geometry
3. Look up the hybridization in this table
Electron Domain Geometry
Hybrid Orbitals Atomic Orbitals Used
2 domains (Linear)
sp
s
p
3 domains (Trigonal planar)
sp 2
s
p
p
4 domains (Tetrahedral)
sp 3
s
p
p
p
5 domains (Trigonal bipyramidal) sp 3 d
s
p
p
p
d
sp 3 d 2
s
p
p
p
d
6 domains (Octahedral)
2
I Note the use of ELECTRON DOMAIN Geometry not MOLECULAR Geometry in this table
Example 9.54
in BH3 and PH3 ?
What is the hybridization about the central atom
H
Covalent Bonding 233
B
H
Solution
1. BH3 has a trigonal planar electron domain geometry.
H
a. The entry in the table indicates that the hybridization is sp 2
b. Alternatively, since 3 domains are required 3 molecular orbitals
will be needed.
H
s
p
sp 2
p
3 hybrid orbitals
required atomic orbitals
H
P
H
The 3 domains on the central atom come from 3 hybrid orbitals. sp 2 hybrid orbitals come from s p p atomic orbitals.
Boron has a 1s 2 2s 2 2 p1 electron con guration. Valence electrons are used to construct the hybrid orbitals. So the sp 2
hybrid orbitals on boron come from one 2s and two 2p atomic
orbitals.
2. PH3 has a tetrahedral electron domain geometry.
(Note that
the molecular geometry is trigonal pyramidal. But we only care
about the domain geometry because we must determine all of the
orbitals used for both bonding and nonbonding electrons.)
a. The entry in the table indicates that the hybridization is sp 3
b. Alternatively, since 4 domains are required (3 bonding plus 1
nonbonding) the central atom requires 4 hybrid orbitals.
s
p
p
p
sp 3
4 hybrid orbitals
required atomic orbitals
COMMON MISTAKE:
There is no such thing as an sp4
or s 2 p hybrid orbital. There is
only one s atomic orbital and three
p atomic orbitals that can be mixed.
To make an sp 4 would require
four p orbitals. An sp3 mixes
all of the available s and p orbitals producing 4 hybrid orbitals.
To make 5 or 6 hybrid orbitals
requires mixing in a d orbital.
This would make sp 3 d or sp3 d 2
The 4 domains on the central atom come from 4 hybrid orbitals. sp 3 hybrid orbitals come from s p p p atomic
orbitals. Phosphorous has a 1s 2 2s 2 2 p 63s 2 3 p3 electron conguration. Valence electrons are used to construct the hybrid
orbitals. In the case of phosphorous the valence electrons are
the 3s 2 3 p 3 electrons. So the sp3 hybrid orbitals on Phosphorous come from one 3s and three 3p atomic orbitals. Three
hybrid orbitals make covalent bonds and one hybrid orbital
holds the nonbonding pair of electrons.
I You cannot know the hybridization of the molecule with-
234
Chapter 9 Molecular Geometry
out rst nding the electron domain geometry.
Multiple Bonding
Sigma Bonds
Covalent bonds result from quantum mechanical e ects that are difcult to conceptualize. Nevertheless, chemists use simple models
to obtain an approximate understanding of chemical bonding. Consider the ethylene molecule.
Earlier in this chapter we learned that each carbon atom has 3 electron domains with sp2 hybridization and 120 bond angles.
The
result of the overlap between a 1s atomic orbital on H with an sp2
sp2 hybrid orbitals.
Multiple Bonding 235
All of the covalent bonds in the picture above are called sigma bonds
because electron density (highest probability) is located along the
axis connecting the two nuclei in the bond.
Pi bonds
There is a second bond between the two carbon atoms in C2 H4 .
How does this bond come about? Remember that in making the 3
sp2 hybrid orbitals an electron is promoted from the 2s to the 2p.
The hybrid orbitals are formed by mixing together one 2s and two
2p atomic orbitals. There is one half lled atomic 2p orbital that
236
Chapter 9 Molecular Geometry
was not used in hybridization.
The left over atomic 2p orbital on each C atom is used to make a
covalent bond. Notice how the orbitals overlap in the gure below.
Because of the orientation of the 2p orbitals there is no electron
This type of bond is called a pi bond.
This gure shows all of the bonds in C 2 H4
bond. In almost all
cases multiple bonds are formed from sigma bond and additional pi
Multiple Bonding 237
bonds.
Bonding in Acetylene
Acetylene, C2 H2 has one carbon-carbon triple bond.
The central C atoms each have 2 domains and use sp hybridization.
The sigma bonds are shown in this gure. Notice that each bond
has the largest electron density along the axis connecting the nuclei
involved in the bond. This is the hallmark of a sigma bond.
238
Chapter 9 Molecular Geometry
The atomic orbitals used in sp-Hybridization are shown below. There
are two left over atomic p-oritals remaining on each C atom.
The two unused p-orbitals are used to make two pi bonds. Again,
each p-orbital has one unpaired electron so each pi bond represents
two shared electrons.
Multiple Bonding 239
Example 9.55
Determine how many sigma and pi bonds are
used by the CH2 O molecule.
Solution
bond.
First draw the Lewis structure.
240
Chapter 9 Molecular Geometry
The second bond between the C and O is a
bond formed from
the overlap of a 2p on carbon with a 2p atomic orbital on O. The
electron con guration for oxygen is shown below. There are 4 nonbonding electrons on the O atom. The nonbonding electrons are
located in the 2s and single lled 2p orbital. One of the remain-
Remember that only valence
electrons in half lled orbitals
are used to make covalent bonds.
ing half lled 2p orbitals is used to make the
bond with C. The
other oxygen 2p orbital is used to make the
bond with C.
Multiple Bonding 241
for the sigma bonds.
The
bond has no electron density along
Molecular Orbitals
Another way to think about bonding is to consider the mixing of
atomic orbitals on di erent atoms resulting in molecular orbitals.
The wavefunction of atomic orbitals has both negative and positive
signs.
When atomic orbitals combine there can be regions of space where
the wavefunction is enhanced (becomes more negative or positive)
and regions where the wave function will cancel.
242
Chapter 9 Molecular Geometry
Molecular orbitals that have an increase of electron density between
the atoms in the bond are called bonding molecular orbitals. Molecular orbitals that have a reduced electron density between the nuclei
are called antibonding molecular orbitals.
10 Gases
Most of the volume occupied by a gas is empty space.
In solids and liquids the atoms
are in close contact with very
little free space.
Kinetic-Molecular Theory of Gases
1. Gases contain a large number of molecules in continuous random
motion.
2. Most of the volume is empty space.
a. As a result, gases are very compressible.
3. The pressure of the gas comes from the force exerted by the
particle when it hit the container wall.
a. The force arises from the change in momentum (mv) that oc-
curs when the particle hits the wall and changes direction.
244
Chapter 10 Gases
4. TIn a monoatomic gas, the Kelvin temperature is a measure of
the average energy content.
He Maxwell Boltzmann Distribution
300 K
600 K
0
1000
2000
3000
Velocity (m/s)
4000
5000
245
a. There is a distribution of gas speeds called the Maxwell Boltz-
mann distribution.
b. As the gas is heated the distribution broadens and the peak
shifts to higher average speeds and therefor higher average
energies.
c. Hotter gases have higher average speeds.
d. At the same temperature, all gases have the same average kinetic energy. As a result, heavier gases have lower average
speeds.
Maxwell Boltzmann Distribution
O2
He
0
1000
2000
3000
4000
Speed (m/s)
5. There are forces between molecules that a ect pressure.
a. These forces are small at low pressure and are ignored in the
simple form of the kinetic theory.
246
Chapter 10 Gases
Units of Pressure
There are several di erent units for pressure that are commonly
used. Here are some conversion factors.
Unit
Unit name Conversion
atmosphere atm
1 atm = 14 7 lb/sq. in.
mm Hg
torr
1 atm = 760 torr
pascal
Pa
1 atm = 101 325 Pa
bar
bar
1 bar = 105 Pa
Gas Laws
Boyles Law
These gas laws were rst developed from experimental observations
before the Kinetic Theory of gases was understood. Today we under-
law says:
1. T = constant, moles
2. i
initial conditions,
pi Vi
pf Vf
constant
f
nal conditions
Boyles law predicts that when pressure is lowered by 1/2 volume
increases by 2 times. When pressure doubles volume decreases by
1/2 at constant temperature.
Units of Pressure 247
250
200
p
1/V
10
15
150
100
50
0
0
5
20
25
30
Volume (L)
As volume increases the gas becomes less dense. (Fewer molecules
per volume.) Collision frequency decreases and pressure drops.
248
Chapter 10 Gases
Charles Law
At constant pressure Charles Law says,
1. p
Vf
Vi
Ti
Tf
constant
constant, moles
2. T in kelvins
3. i
initial conditions,
f
nal conditions
Charles law says that volume is proportional to the kelvin temperature. When T doubles, V doubles at constant pressure. For ideal
gases the molecules themselves have a zero volume. As a result,
when T = 0 K, ideal gases have a zero volume. For real gases, the
molecules themselves have a nite volume. At T = 0 K the volume
of a real gas corresponds to the actual volume of the molecules in
a liquid or solid state.
Units of Pressure 249
When temperature increases the average speed of the gas increases.
He Maxwell Boltzmann Distribution
300 K
600 K
0
1000
2000
3000
4000
5000
Velocity (m/s)
At the higher speeds the average force of the collisions increases so
250
Chapter 10 Gases
pressure increases.
Combined Gas Law
the moles of gas do not change,
CONVERSION TO
KELVINS
T K
C 273 15
pi Vi
pf Vf
Ti
Tf
1. moles constant
2. T in Kelvins
3. i
initial conditions,
f
nal conditions
Ideal Gas Law
The ideal gas law can be derived from the Kinetic Molecular theory.
The two major assumptions that make this equation work only under
ideal conditions are:
1. The volumes of the gase molecules themselves are negligible.
2. The forces between molecules are negligible.
Neither of these statements are true. At high pressures and/or low
temperatures the ideal gas law exhibits strong deviations from the
observed behavior of many gases.
pV
1. n
n RT
moles
2. T in Kelvins
3. R
0 08206 atm L/(mol K)
Units of Pressure 251
4. p in atm
5. V in L
Avogadro’s Law of Combining Volumes
I
T p have
the same number of molecules.
2 L H2
2 L H2 O
2 L H2
2 L H2 O at constant T,P
We can construct a conversion factor to convert from liters of H2 to
liters of water produced by using the stoichiometry in the balanced
chemical equation. This works because moles is proportional to gas
The same method will work for pressure at constant temperature
252
Chapter 10 Gases
and volume.
2 atm H2
2 atm H2 O
2 atm H2
2 atm H2 O at constant T,V
Notice in the constant pressure and temperature reaction in the
above gure that:
1. 10 molecules of H2 occupy 2 L under these conditions of T p
2. The volume of O2 has half that volume (1 L) because there are
only 5 molecules.
3. The products occupy 2 L because Avogadros Laws says 10 mole-
cules of any kind must occupy 2 L under these conditions of
T p.
Common Mistake:You might have thought that the products would occupy 3L because adding the volumes of reactants we get 1L + 2L = 3L. If
we had simply added gases together without reaction at the same temperature
and pressure then the volume would have been 3L. But for chemical reactions
chemical reaction. (3 moles of reactants make 2 moles of product.) . AvoT P) have the same number of
molecules. The original volume of H2 and nal volume of H2 O must be the
same because the number of molecules of H2 reactant and H2 O product are
the same. We can understand why the volume of products is not 2 L + 1 L
if we remember that collisions of molecules with the walls create the pressure.
For the products, only H2
2 and
O2 molecules mixed in with the products to contribute to the pressure. Do not
add the volumes of reactants to nd the volume of products.
Gas Law Example Problems 253
Gas Law Example Problems
T is always in Kelvins, K = 0 C + 273.15
Keywords
Equation
T n constant
pi Vi
p f Vf
Vi
Vf
Ti
Tf
p f Vf
pi Vi
Ti
Tf
pV n RT
p n constant
p V T change
moles
Name
Charles Law
Combined
Ideal
WHEN YOU SEE
p,V,T THINK
I just need to use
n
pV RT
Example 10.56
(1) A gas occupies a volume of 7.50 L at
0.988 atm at 28 C What pressure will the gas exert when the volume
is reduced to 4.98 L at 28 C? (2) How many moles does this gas
contain?
1. Solution
C" means temperature is constant. This
pi Vi
0 988 atm 7 50 L
pf
p f Vf
p f 4 98 L
1 49 L
254
Chapter 10 Gases
2. Solution
pV
n
To nd moles the ideal gas equation is required.
n RT
pV
RT
1 49 atm 4 98 L
0 082 atm L/ mol K 273 15
0 30 mol
Example 10.57
28 K
At a certain temperature and pressure 0.70 L
of N2 reacts with excess H2 What volume of NH3 is produced under
COMMON MISTAKE!!!
Watch your units!!!!! these conditions?
1. T must be kelvins.
2. p must be atm
3. V must be L
Strategy
The actual pressure and temperature is not given so
the ideal gas law cannot be used to
law can be used to compare volumes of each gas.
Solution
1. We know the volume of N2 = 0.7 L
2. The chemical equation must be
N2 g
3H2 g
2NH3 g
3. According to the equation there will be twice as many NH3 g
molecules as there are N2 molecules that we started with. Avopressure have the same number of molecules. So the volume of
NH3 g will be 2 0 7 L
1 4 L.
Gas Law Example Problems 255
N2 g
3H2 g
2NH3 g
07L+3 07L
2 07L
In the same way that we can convert moles of N2 to moles of
NH3 using the balanced chemical equation we can also construct
a conversion factor from the balanced chemical equation to convert
liters of N2 to liters of NH3 produced.
We can do this because
law of combining volumes.)
0 7 L H2
2 L NH3 g
1 L N2
Example 10.58
1 4 L NH3 g at constant T,P
A sample of 2.0 L of He at 4.5 atm and 25 C
is heated to 200 C. The volume expands to 2.5 L. What is the nal
pressure?
Strategy
Since volume, pressure, and temperature all change
the combined gas equation is the best choice.
pi Vi
Ti
Solution
pf Vf
Tf
The unknown is p f Remember that T must be in
Kelvins!!
4 5 atm 2 0 L
25 273 15 K
pf 2 5 L
200 273 15 K
256
Chapter 10 Gases
Solving for p f we nd
pf
5 71 atm
Note: The ideal gas equation could have also been used to solve this
problem but the solution would have required slightly more work.
Example 10.59
What would happen to the pressure of a gas if
the volume were doubled at constant temperature?
Solution 1
When temperature is constant, Boyles Law applies.
pi Vi
If the volume is doubled V f
pi Vi
pf Vf
2Vi so
p f 2Vi
pi
pf
pf
pi
2
2Vi
Vi
The equation says the nal pressure must be 1/2 the initial pressure.
Solution 2
Another way to solve the problem is just to pic-
increases pressure decreases. Pressure and volume are inversely re250
lated.
200
p
1/V
10
15
150
100
50
Example 10.60
0
0
5
Volume (L)
20
25
30
A sample of 15 g of He occupies a volume of
Gas Density 257
2.0 L at 25 C. What is the pressure of this gas?
Strategy
1.
grams
OK I can nd moles using the formula mass.
2. Whenever you have moles of a gas you should be thinking about
the ideal gas equation.
pV
n RT
3. With moles, volume, and temperature the pressure can be calcu-
lated.
COMMON
MISTAKES:
T must be in Kelvins!!!
Solution
p
p
p
n RT
V
15 g/4 0 g/mol 0 082 atm L/ mol K
20L
45 8 atm
25
273 15 K
Gas Density
Density = mass/volume. We can use the gas laws to nd an equation that works for gases.
Density
mass/volume
nM
V
M is the molar mass of the gas. Now substitute the ideal gas
258
Chapter 10 Gases
equation Vn
p
RT
p
M
RT
This equation is useful because it gives a way to nd the molar mass
Density =
of a gas by just measuring the density, temperature and pressure.
Example 10.61
What is the molar mass of a gas that has a gas
density of 1.366 g/L at a pressure of 1.2 atm at 300K?
Solution
Density
1 366 g/L
M
p
RT
M
1 2 atm
0 082 atm L/ mol K 300 K
M
28 00 g/mol
Quantitative Gas Law Calculations
The keywords p V T in any problem should immediately make you
pV
n RT
Look at
the owchart to see how grams of reactant is converted to pressure
and volume of a product using the chemical recipie.
Quantitative Gas Law Calculations 259
Example 10.62
If 10 g of solid sodium is added to excess water
what volume of hydrogen will be collected at 1 atm and 300 K?
Solution
1. Use the strategy in the
owchart to nd moles of hydrogen just
like in Chapter 3.
10 g Na
22 99 g/mol
1 mol H2
2 mol Na
0 217 mol H2 g
2. Use the ideal gas equation to nd the pressure this gas will exert
260
Chapter 10 Gases
under the given conditions.
V
n RT
p
0 217 mol H2 g
5 3 L of H2 g
0 08206 atm L/mol K
1 atm
300 K
Partial Pressures 261
Partial Pressures
In a sample containing a mixture of gases it is often useful to consider what would happen if each individual gas was the only gas in
the container. The mixture in the gure above contains gas A and
gas B.
1. The partial pressure of gas A is the pressure gas A would exert
if it was alone in the container.
2. The partial pressure of gas B is the pressure gas B would exert
if it was alone in the container.
3. The total pressure is the sum of the partial pressures.
Ptotal
PGas A
PGas B
All of the gas laws, Boyles, Charles, ideal, and combined can be
262
Chapter 10 Gases
applied to one component of a gas mixture or the entire mixture.
With ideal gases pressure depends only on the total moles of gas
present and not what kind of gas is present. So 15 moles of He will
have the same pressure as a mixture of 10 moles of Ar and 5 moles
of nitrogen. (Volume and Temperature are constant.)
How does this principle work? Recall that pressure comes from the
collisions of gas particles against the walls of the container.
In a mixture of gases, we can imagine that the total pressure comes
from the pressure exerted by the collisions from only gas A plus the
pressure that results from the collisions of only Gas B. The partial
pressure of Gas A is the pressure that results from the collisions of
only Gas A with the walls of the container.
Partial Pressure Equations
1. Total pressure is the sum of the partial pressures.
ptotal
pA
pB
pC
2. Partial pressures can be calculated from the ideal gas law.
pA
n A RT
V
3. Total pressure can be calculated from the sum of the moles of
Partial Pressures 263
each gas present.
n total
ptotal
nA nB
n total RT
V
nC
Mole Fractions
Mole fractions, X are the percent by moles of a gas in a mixture.
The mole fraction of gas A in a mixture is
nA
XA
n A n B nC
1. Partial pressure of gas A is the mole percent of gas A in the
mixture times total pressure.
pA
X A ptotal
2. The sum of the mole fractions must equal 1. That is the same
thing as saying the sum of the percent compositions of each component in the mixture must be 1.
XA
Example 10.63
XB
XC
1
8 g of He and 14 g of N 2 gas are loaded into a
10 L container at 300 K. What are the partial pressures of each gas?
What is the total pressure?
Strategy
Two gases are discussed so this is a gas mixture and
partial pressures can be used. Find the pressure of each gas as if it
were the only gas in the container. These are the partial pressures.
The total pressure is the sum of the partial pressures.
264
Chapter 10 Gases
Solution
1. The partial pressure of He is
n He RT
V
Plugging in the numbers for He we nd
pHe
8 g/ 4 00 g/mol
pHe
pHe
0 082 atm L/ mol K
10 0 L
300 K
4 92 atm
2. The partial pressure of N2 is
n N2 RT
V
Plugging in the numbers for He we nd
pN2
pN 2
pN 2
14 g/ 28 02 g/mol
0 082 atm L/ mol K
10 0 L
300 K
1 23 atm
3. The total pressure is the sum of the partial pressures
ptotal
pHe
4 92
pN 2
1 23
6 15 atm
Note: The total pressure could have also been found using p
n total RT
V
Example 10.64
8 g of He and 14 g of N2 gas are loaded
into a container where the total pressure is 6.15 atm. What are the
partial pressures of each gas? What is the mole fraction of each
gas?
Solution
Effusion and Diffusion 265
1. The mole fraction of He is
X He
moles He
moles He + Moles N2
8 0 g 4 00 g/mol
80g
4 00 g/mol
14 g
28 02 g/mol
0 80
2. The mole fraction of N2 is
X He
moles N2
moles He + Moles N2
14 0 g 28 02 g/mol
80g
4 00 g/mol
14 g
28 02 g/mol
0 20
3. Check that the sum of the mole fractions add to 1. 0 8 0 2
10
4. The partial pressures are
pHe
X He ptotal
0 8 6 15 atm) = 4 92 atm
pN 2
X N2 ptotal
0 2 6 15 atm) = 1 23 atm
Effusion and Diffusion
The most probable speed of a gas molecule is the peak of the Boltzmann distribution discussed in the beginning of the Chapter.
266
Chapter 10 Gases
Maxwell Boltzmann Distribution
O2
He
0
1000
2000
3000
4000
Speed (m/s)
most
probable
speed
u
1. The average kinetic energy of a gas is determined by the equation
Ek
1 2
mu
2
2. The term u is speed from which the average kinetic energy is
calculated. However, u is not the average speed, nor is it the
most probable speed. The derivation of u nds the equation
1000
2000
3RT
M
We call u the root mean square or rms speed.
u
1. M is the molar mass of the gas molecules in kg.
COMMON MISTAKES
These equations use the MKS
system of measurements.
1. R 8 314 J/(mol K)
2. M is in kg
3. T is Kelvins
4. u in m/s
2. R is the MKS gas constant 8.314 J/(mol K)
1. Notice that the heavier molecules have slower rms speeds. Re-
member that all gases at the same temperature have the same
1
2
average kinetic energy. Since Ek
2 mu it makes sense that u
Effusion and Diffusion 267
should depend on
1
M
Graham’s Law of Effusion
E usion is the process where gas molecules leak through a small
During WWII the isotope 235 U hole into a vacuum.
was required to construct the atomic
bomb. To separate the low abundance 235U from 238 U effusion
of UF6 gas was used. A large
factory was built for this purpose.
Although the rates of effusion
are only slightly different (because the masses are so similar)
the project was eventually successful.
The rate of e usion depends directly on the rms speeds. The faster
the speeds the higher the rate of e
the rates of e usion for two gasses compare with each other. r1 is
the rate of e usion of gas 1 and r2 is the rate for gas 2.
r1
r2
M2
M1
usion rate directly de-
pends on the rms speed. Since the rms speed is u
ratio of the e usion rates is
r1
r2
3RT M1
3RT M2
M2
M1
3RT
M
the
268
Chapter 10 Gases
I Remember that the lighter molecule is faster and has the higher rate
of e usion.
Example 10.65
What is the rms speed of N2 at 300 K? What
is the rate of effusion of H2 compared to N2 ?
Solution
1. The rms speed is given by the equation
HOW DO YOU GET
m/s??
J/(mol K)
K
kg/mol
u
J/kg
3RT
M
For nitrogen the rms speed is (Remember to use MKS units!!.)
J/kg =
m2 s2
kg m2 s2 kg
m/s
u
8 314 J/mol K
300 K
28 02 10 3 kg/mol
516 8 m/s
3
2. To compare the rates of e usion use Grahams law of e usion.
r1
r2
M2
M1
r1
r2
28 02
2 02
37
Be careful now. Is the rate of e usion of H2 3.7 times faster than
N2 or is it the other way around? In the equation we used the
notation 1 for hydrogen and 2 for nitrogen. Or
r H2
r N2 3 7
Hydrogen is faster than nitrogen. Make sense? Yes, the lighter
molecule has the faster speed.
Real Gases 269
Diffusion
Di usion is the process where molecules migrate from one point to
another through a gas. When you smell perfume or food cooking
the molecules that reach your nose have diffused through the air
from their source.
Perfume
molecule
collision with
air molecules
The illustration shows a perfume molecule on a random walk through
the air. Collisions with the air molecules change the direction and
speed of the motion. The average distance the molecule travels
between collisions is called the mean free path. It makes sense that
at higher pressures when the density of the gas is higher the mean
free path will be shorter.
Real Gases
The two weakest assumptions used when the ideal gas equation was
270
Chapter 10 Gases
derived are:
1. The volume of the gases are negligible.
2. The forces between molecules are negligible.
Neither of these statements is true. Real gases are very complex.
Forces between molecules (Chapter 11) become very important at
low temperatures. At high pressures where the density of gas molecules is large the volume of the molecules cannot be ignored. In
general, larger molecules, have larger deviations from ideal behavior. There are many equations that handle real gases. The degree
of success depends on the severity of the conditions.
The van der Waals Equation
The van der Waals equation is a very useful real gas equation. It is
easy to understand how this equation comes about.
Real Gases 271
p
n RT
V nb
n 2a
V2
1. a and b are di erent for each molecule.
2. b describes the volume of a mole of molecules.
3. a is a term that describes the e ect on pressure
resulting from forces (usually attractive) between
molecules.
Comparing the van der Waals equation to the ideal gas
equation.
p
n RT
V
p
n RT
V nb
Ideal gas equation
n 2a
V2
van der Waals
1. Attractive forces between molecules reduce the force of the col-
lisions. This e ect reduces the actual pressure of the real gas.
2
The Vn 2a term corrects the pressure of real gases for the e ects
of attractive forces.
2. The V
nb term represents the actual volume available to molecules. V is the container volume and nb is the total volume of
the molecules. The b term is the volume of 1 mole of molecules
272
Chapter 10 Gases
so nb is the total volume of the molecules.
Consider an experiment where a gas is compressed by continuously lowering the volume.
As the volume decreases pressure in-
creases. A plot is then made at each volume to show the actual
ratio P V n RT for real gas behavior as predicted by the van der
Waals equation.
As long as the gas behaves as an ideal gas the
ratio, P V n RT is always 1.
When the behavior of the gas is
dominated by attractive forces the real gas pressure will be lower
than ideal gas pressure and the ratio P V n RT is
P V n RT
1.
1 repulsive forces dominate the gas behavior.
When
Real Gases 273
The molecules in real gases
begin to repel each other when
the average distance between molecules becomes small. This will
occur at small volumes and high
pressures. Repulsive forces will
cause the real gas pressure to be
higher than calculated with the
ideal gas equation.
11 InterMolecular Forces
I IntraMolecular Bonds are strong covalent bonds responsible for
holding the atoms of a molecule together. These bonds are the normal chemical bonds that we have discussed throughout the course.
(Intramolecular means within the molecule.)
I InterMolecular bonds are weak bonds between molecules.
Types of InterMolecular Forces 275
Why are InterMolecular Bonds Important?
1. Intermolecular forces determine whether a compound is a solid,
liquid, or gas.
2. Melting and Boiling points are determined by intermolecular forces.
3. Solubility is partially determined by intermolecular forces.
4. Other physical properties such as surface tension and viscosity are
determined by intermolecular forces.
Types of InterMolecular Forces
Ionic Forces
Ionic crystals are held together by electromagnetic attractive forces.
The energy required to separate one mole of molecules (formula
units) from the crystal is called the lattice energy.
The strength of the ionic force (electromagnetic force) is described
276
Chapter 11 InterMolecular Forces
Q 1 Q 2 and the distance, d between the ions. As shown in the gure below d is the sum of the ionic radii of the two ions in the crystal
lattice. k is a term we will not calculate in this course. k depends
on the structure of the crystal and some physical constants.
ing points of ionic crystals. The larger the ionic force the larger the
lattice energy and the higher the temperature required to melt the
A larger ionic force, F increases the lattice energy. The
stronger the ions are held together
the more energy is required to
melt the crystal and pull the ions
apart. Therefore a larger force
causes a higher melting temperature to separate the ions.
salt.
In the following table you can see that the melting points decrease
from NaF to NaBr. The reason for this is the ionic radii increases
F
C
Br . The largest force for these three salts occurs in
Types of InterMolecular Forces 277
You should be able to predict these trends in ionic radii
and melting points!!!
NaF since F has the smallest radii of all of the halogen ions.
Salt
mp ( 0 C )
ionic radii ( )
NaF
993
F = 1.19
NaC
801
C
NaBr
747
Br = 1.82
= 1.67
Ions with higher charges usually have much stronger ionic forces and
much higher melting points. Compare the melting points for the
following salts.
As the charges on the ions increase the melting
point increases dramatically.
Salt
mp ( 0 C )
NaF
993
Na
F
CaF2
1423
Ca2
F
CaO
2500
Ca2
O2
Ion-Dipole Forces
charge on ions
278
Chapter 11 InterMolecular Forces
1. Ion-dipole forces require an ion and a polar molecule.
2. The polar molecule contains a dipole. (dipole means a
and a
pole
pole.)
3. The force arises from the
electrostatic attraction. This
is a coulomb force.
4. Ion-dipole forces occur most commonly when ions are dissolved
in polar solvents.
The example in the gure above is sodium ion dissolved in water.
This illustration describes what we mean when we write Na
aq
Water is polar and has a dipole. The O atoms have a slight negative
charge and the hydrogen atoms a slight positive charge. Notice in
the gure that all of the negatively charge O atoms are oriented
towards the positively charged ion.
Types of InterMolecular Forces 279
Dipole-Dipole
Two dipoles can produce an attractive dipole-dipole force when they
align so that the + part of one molecule attracts the
part of
another.
Dipoles in the liquid become
oriented by strong attractive forces.
London Dispersion Forces
London dispersion forces are also called van der Waals forces. All
atoms and molecules have this type of force. A charge or dipole is
not required.
280
Chapter 11 InterMolecular Forces
1. Temporary uctuations in time of electron orbitals create instan-
taneous dipoles.
2. In one instant they exist in the next they are gone.
3. A dipole on one atom induces (creates) a dipole on an adjacent
atom.
4. The two dipoles are oriented in opposite directions. This creates
a temporary attractive intermolecular force. (Dipole-dipole force)
5. When averaged over time all of these random uctuations create
a net attractive force.
6. The net force is called a London Dispersion Force.
Trends for Dispersion Forces
1. The larger the atom or molecule the larger the uctuations of the
orbital.
Types of InterMolecular Forces 281
a. The ability of an electron orbital to distort is called polariz-
ability.
b. Larger atoms have a larger polarizability.
2. London forces are larger for bigger atoms and molecules due to
their larger polarizability.
GRAPHITE
Solid graphite is made up of
large planes of 6 membered rings.
London Dispersion forces between
the rings allow the planes to slip
past each other. This is the reason why graphite acts as a powdered lubricant.
3. London forces explain the trends in chemical reactivity of the
noble gases (Chapter 7).
a. Only the larger noble gases react. Weak bonds are made using
London dispersion forces. For the lighter noble gases, the
small polarizability makes the London forces too weak for bond
formation in most cases.
4. London forces become the dominant force in very large molecules.
282
Chapter 11 InterMolecular Forces
Hydrogen Bonding
1. Hydrogen bonds can be surprisingly strong.
2. Hydrogen bonds are formed between:
a. A positively charged hydrogen atom and
b. The nonbonding electrons of a very electronegative atom.
Importance of Hydrogen Bonding
1. Hydrogen bonds are responsible for making water a liquid under
normal conditions. Without hydrogen bonds water would be a
gas and life on earth would be very limited or impossible.
2. Hydrogen bonds make water expand when it freezes, whereas
most materials compress on freezing.
3. Hydrogen bonds are used for the zipping and unzipping functions
that occur between the helixes of DNA. The biochemical functions
of DNA depend on hydrogen bonding.
Types of InterMolecular Forces 283
Which molecules can form hydrogen bonds?
I Hydrogen bonds require a highly polar negatively charged atom to
bond with the H atom.
I The negative atom must have a pair of nonbonding electrons.
I Hydrogen bonds can only exist with N-H, O-H, of F-H bonds. Other
combinations like C-H, S-H, etc. do not make hydrogen bonds.
1. Note: In the gure below of methanol CH3 OH, the C-H hydrogen cannot form hydrogen bonds because that bond is not highly
polar. The H is not positive enough because the C and H electronegativities are similar.
2. However, the O-H bond is highly polar and can form hydrogen
bonds.
284
Chapter 11 InterMolecular Forces
Comparing Strengths of Intermolecular Forces
Summary
Intermolecular
Requirements
Examples
Force
London
All molecules have London forces.
Larger molecules
larger London forces N2 Xe, CH4
Dipole-Dipole
Polar molecule
HC
PH3 OF2
Hydrogen bond
H-F, H-N, or O-H bonds
HF, H2 O
Ion-Dipole
1. Ion (Na
NH4 NO3 )
2. Polar molecule usually the solvent
NH4 in H2 O
Types of InterMolecular Forces 285
Examples
Molecule
Characteristics
F2
nonpolar molecule
London only
CO2
nonpolar molecule
London only
H2 S
polar molecule
Dipole-Dipole, London
NH3
1. polar molecule,
Hydrogen bond,
2. N high EN, has nonbonding e
Dipole-Dipole, London
ion, polar liquid
ion-dipole
K
NH3
Forces Present
Example 11.66
Rank the strengths of the intermolecular
forces for these molecules: F2 CO2 NH3 H2 S, K
dissolved in
NH3
Solution
1. All of these molecules are similar in size so London dispersion
e ects will be similar.
2. The weakest forces are London dispersion forces. F2 and CO2
only have London forces because they are both nonpolar. (See
table above.) CO2 is a little larger than F2 so the London forces
will be larger for CO2
3. H2 S and NH3 both have dipole-dipole forces but NH3 can also
hydrogen bond. So NH3 will have larger intermolecular forces.
can have ion-dipole forces so it has the largest forces
4. K
NH3
of all.
F2
CO2
H2 S
NH3
K
NH3
286
Chapter 11 InterMolecular Forces
Example 11.67
Rank the strengths of the intermolecular forces
for these molecules: F2 C 2 Br2 I2
Solution
Since these molecules have vastly di erent sizes we
must consider the fact that London forces for the larger molecules
may swamp the other types of forces. However, these molecules are
all nonpolar so London forces are the only forces that exist for these
molecules. London forces increase with size.
F2
C 2
Br2
I2
Note: I2 is so large that the London forces have become very strong.
In fact, the London forces are now stronger than the forces between
NH3 which has dipole-dipole, and hydrogen bonding forces.
Solids Liquids and Gases
Solids
1. Intermolecular bonds are strong.
2. Atoms are in xed positions. Molecules do not ow.
3. Density is high.
4. Solids are not very compressible.
Liquids
1. Intermolecular bonds are weaker than forces in solids.
2. Molecules are in not in
a. Molecules ow.
xed positions.
Solids Liquids and Gases 287
b. Liquids adopt the shape of the container.
3. Density is high.
4. Liquids are not very compressible.
Gases
1. Intermolecular bonds are very weak or nonexistent.
2. Molecules are not in xed positions. Molecules
ow.
3. Molecules expand to ll the volume.
4. Density is usually low under most conditions. (low pressure)
5. Gases are very compressible because they are mostly empty space.
288
Chapter 11 InterMolecular Forces
The Concept of Temperature
In Chapter 10 we learned that gas molecules are in constant motion
and there is a distribution of energies or speeds.
He Maxwell Boltzmann Distribution
300 K
600 K
0
1000
2000
3000
4000
5000
Velocity (m/s)
1. Temperature is a measure of the average energies of the atoms
or molecules.
2. Energy can be stored in a variety of ways as the molecule trans-
lates, rotates, and vibratates.
I The higher the temperature the larger the average energy that is
stored in the substance.
Phase Changes 289
Phase Changes
What’s Boiling???
Analogy: The force of gravity
holds the space shuttle on the
surface of the earth. If the rockets get the ship moving fast enough
it can escape the force of gravity and move into outer space.
Molecules in the liquid phase get
their energy from the surroundings. When the temperature is
increased by heating the molecules move faster. If the liquid
is heated enough—vaporization
occurs.
1. Phase changes from solid to liquid or liquid to gas occur when
the temperature of the substance is increased enough so that the
molecules have enough energy to overcome the intermolecular
forces.
a. When materials melt the energy in the molecules is just enough
to overcome the forces so that the material can
.
b. When materials boil the temperature is high enough so that
the molecules have enough energy to escape each other and
expand to ll the container.
290
Chapter 11 InterMolecular Forces
WHY WE SWEAT
2. Stronger intermolecular forces require more energy for the molecules to overcome the forces.
a. Stronger forces imply higher melting points.
b. Stronger forces also imply higher boiling points so that the
substances can have the needed energy to overcome the stronger
intermolecular forces.
When we sweat water molecules evaporate. This requires
energy. The water molecules must
pull energy from the human body
to overcome the intermolecular
forces in the liquid in order to
escape into the gas phase. Evaporating sweat cools off the body.
Why is it that people wipe off
their sweat with a towell? That’s
like turning off the body’s air
conditiong system!!!
Phase Changes 291
Example
At room temperature H2 O is a liquid and CO2 is a gas. The molar
mass of water is 18 g/mol. Carbon dioxide is heavier with a molar
mass of 44 g/mol. So why is the heavier CO2 a gas under the same
conditions where water is a liquid?
292
Chapter 11 InterMolecular Forces
1. Water has much stronger intermolecular forces.
a. Water has hydrogen bonding, dipole-dipole, and London forces.
b. CO2 is nonpolar and only has London forces.
2. CO2 has enough energy for the molecules to overcome the inter-
molecular forces and exist as a gas. Under the same conditions,
water molecules are still held tightly together as a liquid. Water molecules do not yet have enough energy to overcome the
stronger hydrogen bonding forces. However, if water is given
more energy by heating to 100 C at 1 atm then the conversion
to gas will occur.
Melting
I A process where the temperature has reached the point where the
energy of the molecules can overcome the intermolecular forces and
the molecules can ow.
Boiling
I A process where the temperature has reached the point where the
energy of the molecules can overcome the intermolecular forces and
Phase Changes 293
the molecules can escape each other and ll the volume.
Sublimation
I A process where the temperature has reached the point where the
energy of the molecules can overcome the intermolecular forces in
the solid and can escape directly to the gas phase.
Example 11.68
Which of these molecules has the highest
melting point? F2 CO2 NH3 H2 S
Solution
The molecules with the strongest forces will require
more energy to melt and will have the highest melting point. We
have already determined the ranking of forces in the Example Prob-
molecules have stronger forces. The melting points will rank in the
same order as the forces.
F2
Example 11.69
CO2
H2 S
NH3
Rank the order of the boiling points for these
molecules: F2 C 2 Br2 I2
Solution
The strategy is the same as in the last example.
The molecules with the strongest forces will require more energy to
escape into the gas phase and will have the highest boiling point.
294
Chapter 11 InterMolecular Forces
The forces and the boiling points will rank in the same order. The
molecules in the list are all nonpolar and only have London dispersion
forces. The heavier molecules have larger forces. The boiling points
will be ranked in this order.
F2
C 2
Br2
I2
Real Gases
In Chapter 10 there is a discussion of real gas behavior. The existence of attractive and repulsive intermolecular forces was shown to
a ect the real pressure of a gas. The plot below shows the real gas
pressure of several gases as the volume of the container is reduced.
The ratio P V n RT is plotted on the y axis and the actual measured pressure on the x axis. As long as the gas behaves as an
ideal gas the ratio, P V n RT is always 1. Attractive intermolecular forces are responsible for making the ratio P V n RT
The largest deviations observed when P V n RT
CO2
10
1 0 occur for
The reason for this is CO2 has the largest London forces
and therefore the largest attractive forces for this set of molecules.
The London forces for H2 are so weak that they never
causeP V n RT
1 0 under these conditions.
Phase Diagrams 295
Phase Diagrams
1. Phase diagrams show the relationships between the solid, liquid
and gas phases.
2. The lines on the diagram show the points where two phases co-
exist in equilibrium.
3. The regions marked solid, liquid, and gas represent the conditions
where those phases are stable.
4. At the triple point solid, liquid, and gas all coexist in equilibrium.
5. The normal melting point is the temperature where the solid and
liquid phases are in equilibrium at 1 atm pressure.
6. The normal boiling point is the temperature where the liquid and
gas phases are in equilibrium at 1 atm pressure.
7. The critical point is the temperature beyond which it is not pos-
sible to liquify the substance by compression alone.
8. The negative slope of the solid-lquid phase diagram is unusual.
The negative slope comes from the fact that water expands when
it freezes. (The change in volume on melting is negative.)
296
Chapter 11 InterMolecular Forces
Phase Diagrams 297
Heating and Phase Transitions
Suppose you take a chunk of ice at -25 C and 1 atm pressure, throw
it in a pot and place the pot on a stove burner. The curve below
shows what will happen.
298
Chapter 11 InterMolecular Forces
1. As energy is added to the ice its temperature increases because
the average energy content in the motions of the molecules
increases.
2. When the temperature of the ice reaches 0 C its temperature
Remember that temperature
is related to the average energy
content in the motions of the molecules.
stops changing when addition heat is added. Why? 0 C is
the normal melting point of ice. When additional heat energy
is added at this temperature the average energy content in the
motions of the molecules does not change. Instead, the added
energy is being used to cause the ice
liquid phase transition. In other words, at the melting point the added energy
is used to break intermolecular bonds rather than increase
the energy of the molecular motions so temperature remains constant.
3. Once all of the ice has melted additional heat increases the aver-
age energy of the motions of the water molecules so temperature
increases when heat energy is added.
4. At the normal boiling point of 100 C the water
gas phase
transition occurs. For a phase transition to occur heat energy
must be added to overcome intermolecular forces. The
temperature does not change when this heat is added because
the energy is not being used to increase the average energy of
Phase Diagrams 299
the molecular motions.
What is boiling?
When a liquid is placed in a closed container some evaporation
occurs.
This produces a gas whose partial pressure is called the
vapor pressure.
The total pressure inside the container comes
from the sum of the partial pressures of all gases inside the container.
When the vapor pressure equals the total pressure boiling will occur.
At 50.00 C water has a vapor pressure of 92.5 torr.
If the total
pressure inside the container is lowered to 92.5 torr the liquid water
will begin to boil. The vapor pressure at 50.00 C can be found by
locating the temperature on the x axis in the phase diagram below.
Draw a verticle line from the x axis at 50.00 C upwards until it meets
the gas-liquid equilibrium line. Next draw a horizontal line towards
the y axis and read the vapor pressure on the y axis.
300
Chapter 11 InterMolecular Forces
The phase diagram above shows that at 100.00 C the vapor pressure
of water is 760.0 torr. Remember that 760 torr = 1 atm. Under
normal conditions when Ptotal
because P ap
1 atm water boils at 100.00 C
Ptotal under those conditions
Phase Diagrams 301
Index
acid-base reactions, 75
acids
table of, 75
alkali metals
reactivity, 160
alkaline earths
reactivity, 162
amu, 15, 51
Angstrom, 15
anion, 20
aqueous, 72
atom, 14
atomic mass unit, 15
atomic orbitals, 226
atomic radius, 144
atomic size, 144
aufbau principle, 130
Avogadros Law, 251
axial equatorial, 214
balancing, 44
bases
table of, 75
benzene, 191
Bohr Atom, 111
boiling, 292, 299
boiling point
normal, 295
Boltzmann, 249
Boltzmann distribution, 245, 288
bond
pi, 235
sigma, 235
bond angle, 217
bond angles, 213
bond dissociation enthalpy, 195
bonding, 165
Boyles Law, 246
cation, 20
charge, 15
charges
nding, 39
charges on ions, 34
Charles Law, 248
combination reaction, 47
combustion reactions, 47
compound, 27
condensation, 296
Coulombs Law, 276
covalent bonding, 171, 226
critical point, 295
Dalton, 44, 57
Davisson and Germer, 107
decomposition reactions, 48
degeneracy, 121
degenerate, 127
density
gas, 257
deposition, 296
diamagnetic, 133
diffusion, 269
dimensional analysis, 4
dipole, 173
dipole-dipole, 279
dipoles, 219
dispersion, 280
effective nuclear charge, 144
effusion, 265, 268
Grahams Law, 267
Einstein, 106
electrolytes, 76
electromagnetic wave, 104
electron af nity, 156
electron con gurations, 130, 134
exceptions, 139
ions, 146
shorthand, 139
transition metal ions, 146, 147
transition metals, 138
electron domain, 206
electron domain geometry, 208
electron electron repulsions, 144
electron screening, 127, 143
electron spin, 125
electronegativity, 168, 218
electrons, 14
element, 16, 26
emperical formula, 30
empirical formulas, 56
energy levels
picture, 122
enthalpy, 198
equivalence point, 96
exchange reactions, 84
excited state, 132
expanded valence, 187
experimental yield, 70
formal charge, 180, 181
meaning of, 182
formula mass, 50
free radical, 185
freezing, 296
gases, 243, 287
302
Chapter 11 InterMolecular Forces
Grahams Law, 267, 268
ground state, 132
group, 21
groups, 159
halogens, 164
Heisenberg Uncertainty Principle, 108
heterogeneous, 86
homogeneous, 86
Hund’s rule, 133
hybrid orbitals, 227, 230
hybridization, 228, 232
table, 232
hydride ion, 162
hydrogen
energy levels, 116
hydrogen bonding, 282, 283
ideal gas law, 250
inner transition metals, 22
inorganic compounds
naming, 35
intermolecular forces, 274
strengths, 284
intermolecular forces, examples of, 285
intermolecular forces, table of, 285
ion charges, 34
ion sizes, 145
ion-dipole, 277
ionic, 174
ionic bonding, 170
ionic compound, 29
ionic compounds, 33
in water, 72
ionization energy, 148
ions, 20
electron con gurations, 146
isoelectronic, 146
isotope
abundance, 17
isotopes, 17
isotopic mass, 17
kinetic energy, 266
kinetic theory of gases, 243
Lattice Energy, 275
Lewis Dot, 166
Lewis structures
multiple bonds, 179
octet rule exceptions, 185
procedure for writing, 176
light, 104
limiting reagent, 66
liquids, 286
London dispersion, 279
London dispersion,
trends, 280
lone pair electrons, 206, 214
main group, 22
mass number, 18
mean free path, 269
melting, 289, 292, 296
melting point
normal, 295
Mendeleev, 141
metallic bonds, 173
metallic character, 158
metalliods
properties, 24
metalloids, 158
metals, 158
non-metals, metalloids, 22
properties, 23
metathesis reaction, 84
molar mass, 49
molarity, 86
mole, 48
mole fraction, 263, 264
molecular compound, 28
molecular compounds
in water, 74
naming, 36
molecular formula, 60
molecular geometry, 206, 208
molecular geometry, table of, 210, 211, 213
molecular orbital, 195, 219
molecular orbitals, 226, 227
molecule, 26
most probable speed, 266
multi-electron atoms, 126
multiple bonds, 177, 179
naming
molecular compounds, 36
organic molecules, 35
naming
ionic compounds, 37
net ionic equations, 80
neutrons, 14
noble gases, 164
nodes, 126
non-metals
properties, 23
nonbonding electrons, 175, 206
nonelectrolyte, 77
nonmetals, 158
nonpolar molecule, 219
nonpolar molecules, 225
nonpolar molecules, table, 223, 225
octet rule, 168
violations, 186
octet rule
violations, 186
octet rule exceptions, 185
orbital
shape, 120
Phase Diagrams 303
orbitals, 118
oxide, 161
oxygen, 163
ozone, 203
paired spins, 133
paramagnetic, 133
partial charge, 173
partial pressure, 262–264
partial pressures, 261
particles
wave properties, 107
Pauli exclusion principle, 130
percent composition, 52
period, 22
periodic table, 21, 24, 141
peroxide, 161
phase changes, 289
phase diagrams, 295
phase transitions
concept, 291
photons, 106
pi bond, 235
Planck, 106
polar, 173
polar bond
ionic, 174
polar covalent, 174
polar bonds, 173, 218
polar molecule, 278
polar molecules, 219, 225
polar molecules, table, 223, 225
polar, nonpolar bond
nonpolar covalent, 174
polyatomic ions, 32
table, 32
potential energy, 198
precipitate, 81
pressure
units, 246
probability, 119
protons, 14
quantum effects, 106
quantum number, 121
m , 120
m s , 125
quantum numbers, 118
l, 120
n, 119
principle, 119
table, 124
quantum theory, 117
radicals, 185
reactions
acid-base, 75
combination, 47
combustion, 47
decomposition, 48
exchange, 84
reactivity of metals, 160
real gases, 269, 270
resonance, 190, 193
resonance hybrid, 191–193, 205, 209
rms speed, 266, 268
Rydberg constant, 112
salt, 78
second ionization energies, 151
shell, 121
shells, 142
short hand electron con gurations, 137
sigma bond, 235
signi cant gures, 1
addition and subtraction, 3
multiplication and division, 2
solids, 286
solubility rules
ionic compounds, 79
table, 79
solute, 72
solution, 72
solvent, 72
spectator ions, 81
spins
paired, 133
strong electrolyte, 73, 77
sublimation, 293, 296
subshell, 121
superoxide, 161
temperature
concept of, 288
theoretical yield, 69
titrations, 96
transition metals, 22
triple point, 295
uncertainty principle, 108
valence electrons, 165, 167, 177
van der Waals
forces, 279
van der Waals equation, 270
vapor pressure, 299
vaporization, 296
VSEPR model, 206, 207
wavefunction, 118
wavelength of particle, 107
weak acids, 78
weak bases, 78
weak electrolyte, 78
yields, 69
Young double slit, 105
A
Physical Constants
Useful Conversion Factors and Relationships
Length
Mass
Volume
SI unit: meter (m)
SI unit: kilogram (kg)
SI unit: (m3
1 kilometer = 1000 m
1 kilogram = 1000 g
1 liter (L) = 0.001 m3
1 kilometer = 0.62137 mi
1 gram = 1000 mg
= 1000 cm3
1 meter = 1.0936 yd
1 pound = 453.59237 g
= 1.056710 qt
1 meter = 100 cm
1 pound = 16 oz
1 centimeter = 0.39370 in
1 ton = 2000 lbs
1 gallon = 4.00 qt
1 centimeter = 10 mm
1 kilogram = 2.2046 lbs
1 gallon = 3.7854 L
1 amu = 1 6605
= 1 dm3
10 24 g
1 cm3 = 1 mL
1 in3 = 16.4 cm3
1 inch = 2.54 cm (exactly)
1 Å = 10 10 m
1 mile = 5280 ft
1 mile = 1.6093 km
Temperature
SI unit: kelvin (K)
0K=
273 15 C
=
459 67 F
K = C + 273 15
C = 59 ( F - 32 )
F = 95 C + 32
Pressure
SI unit: pascal (Pa)
1 pascal = 1 N/m2
Energy
SI unit: joule (J)
= 1 kg / m s2
1 joule = 1 kg m2 / s2
1 atmosphere = 101.33 kPa
= 0.23901 cal
= 760 mm Hg
= 760 torr
= 14.70 lb / in2
= 1.01325 bar
1 bar = 105 Pa (exactly)
=1C
1V
1 calorie = 4.184 J
306
Appendix A Physical Constants
Physical Constants
NA
6 02214
1023
1 amu = 1 66054
10 24 g
Atomic mass unit
me
9 10939
10 28 g
electron mass
mp
1 67262
10 24 g
proton mass
mn
1 67493
10 24 g
neutron mass
1 60218
10 19 C
electron charge
h = 6 62608
10 34 J
c
2 99792
108 m/s
R
8 31451 J/mol K
gas constant in J
R
0 0820578 atm L/mol K
gas constant in atm L
F
96 485 3 C/mol
Faraday constant
RH
2 18
e
10 18 J
speed of light
Rydberg constant for H atom