1. science
2. chemistry

THE OPEN UNIVERSITY OF TANZANIA
FACULTY OF SCIENCE, TECHNOLOGY AND ENVIRONMENTAL
STUDIES
OEV 104: GENERAL CHEMISTRY
Prof. Ward J. MAVURA
Department of Chemistry
EGERTON UNIVERSITY
P.O. BOX 536, EGERTON
KENYA
DECLARATION
The Open University of Tanzania
P. O. Box 23409
Dar-es-Salaam
Tanzania
© The Open University of Tanzania, 2008
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system, or transmitted in any form or by any means, electronic, mechanical,
photocopying, recording or otherwise, without the prior permission of the copyright
owner.
TABLE OF CONTENTS
INTRODUCTION ....................................................... Error! Bookmark not defined.
Course Rationale ......................................................................................................... 1
Learning outcomes ...................................................................................................... 1
1.1. CHAPTER ONE – INORGANIC CHEMISTRY ............................................. 3
1.1.1.
Atomic Weights and Atomic Numbers ..................................................... 3
1.1.2.
Formula Weight ....................................................................................... 4
1.2. ELECTRON CONFIGURATION .................................................................... 5
1.3. QUANTUM NUMBERS AND ELECTRON CONFIGURATION .................. 8
1.3.1.
Quantum Numbers ................................................................................... 8
1.3.2.
Rules Governing the Allowed Combinations of Quantum Numbers.......... 9
1.3.3.
Shells and Sub shells of Orbitals............................................................. 10
1.3.4.
Possible Combinations of Quantum Numbers ......................................... 11
1.3.5.
The Relative Energies of Atomic Orbitals .............................................. 13
1.3.6.
The Aufbau Principle, Degenerate Orbitals, and Hund's Rule ................. 14
1.3.7.
Exceptions to Predicted Electron Configurations .................................... 18
1.3.8.
Electron Configurations and the Periodic Table ...................................... 19
1.4. IONIC AND COVALENT BONDS ............................................................... 20
1.4.1.
Ionic Bonds ............................................................................................ 21
1.4.2.
Some atoms with multiple valences. ....................................................... 22
1.4.3.
Some atoms with only one common valence: ......................................... 23
1.4.4.
Radicals or Polyatomic Ions ................................................................... 23
1.4.5.
Acids of Some Common Polyatomic Ions. ............................................. 24
1.4.6.
Writing Ionic Compound Formulas ........................................................ 25
1.4.7.
Binary Covalent Compounds .................................................................. 26
1.5. FORMAL CHARGE AND RESONANCE STRUCTURES........................... 29
1.5.1.
Deducing the not so easy structures ........................................................ 29
1.5.2.
Calculating the Formal Charge ............................................................... 29
1.5.3.
Resonance Structures ............................................................................. 30
1.6. THE VSEPR MODEL ................................................................................... 31
1.6.1.
Electron and Molecular Geometries ........................................................ 32
1.6.2.
Predicting Electron-Pair Geometry ......................................................... 32
1.6.3.
Using Electron Geometry to Figure Molecular Geometry ....................... 34
1.7. MOLARITY .................................................................................................. 39
1.8. ACIDS AND BASES..................................................................................... 44
1.8.1.
Properties of Acids ................................................................................. 46
1.8.2.
Properties of Bases ................................................................................. 47
1.8.3.
Strong Acids and Strong Bases ............................................................... 47
1.8.4.
The definition of pH and pOH ................................................................ 49
1.8.5.
Water Ion Product .................................................................................. 51
1.8.6.
Acid/Base Equilibrium ........................................................................... 52
1.8.7.
Bronsted-Lowry's acids and bases .......................................................... 53
1.9. BUFFER SOLUTION.................................................................................... 54
1.9.1.
Calculating pH of a buffer ...................................................................... 55
1.9.2.
Illustration of buffering effect: Sodium acetate/acetic acid...................... 56
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1.9.3.
Applications ........................................................................................... 56
2.1. CHAPTER TWO – PHYISICAL CHEMISTRY ............................................ 58
2.2. KINETIC MOLECULAR THEORY ............................................................. 60
2.2.1.
Postulates of Kinetic Theory................................................................... 60
2.2.2.
The Ideal Gas Law from Kinetic Theory ................................................ 61
2.3. HENRY’S LAW ............................................................................................ 62
2.4. RAOULT’S LAW.......................................................................................... 62
2.5. CHEMICAL THERMODYNAMICS............................................................. 65
2.5.1.
Internal Energy....................................................................................... 65
2.5.2.
Enthalpy and Enthalpy Change ............................................................... 66
2.5.3.
Entropy and Entropy Change .................................................................. 68
2.5.4.
The Second Law of Thermodynamics..................................................... 69
2.6. CHEMICAL KINETICS ................................................................................ 72
2.6.1.
Definition of a Reaction Rate ................................................................. 72
2.6.2.
Dependence of rate on concentration ...................................................... 74
2.6.3.
Reaction Order ....................................................................................... 75
2.6.4.
Determining the rate law ........................................................................ 75
2.6.5.
Change of concentration with time ......................................................... 75
2.6.6.
First Order Rate Law .............................................................................. 75
2.6.7.
Second Order Rate Law.......................................................................... 76
2.6.8.
Half-lives ............................................................................................... 77
2.7. HESS’S LAW OF HEAT SUMMATION ...................................................... 78
2.7.1.
Thermochemical Equations .................................................................... 79
2.7.2.
Heat of Formation .................................................................................. 80
3.1. CHAPTER THREE – ORGANIC CHEMISTRY ........................................... 82
3.1.1.
Alkanes .................................................................................................. 82
3.1.2.
Methane, the Simplest Hydrocarbon ....................................................... 82
3.1.3.
The alkane series .................................................................................... 83
3.1.4.
Nomenclature of Alkanes ....................................................................... 85
3.1.5.
Alkenes .................................................................................................. 88
3.1.6.
Alkynes .................................................................................................. 90
3.2. REACTIONS OF HYDROCARBONS .......................................................... 93
3.2.1.
Oxidation ............................................................................................... 93
3.2.2.
Substitution reactions of alkanes............................................................. 93
3.2.3.
Addition Reactions of Alkenes ............................................................... 94
3.2.4.
Addition Reactions of Alkynes ............................................................... 95
3.2.5.
Oxidation of Alkynes ............................................................................. 96
3.3. FUNCTIONAL GROUPS.............................................................................. 96
3.4. ELIMINATION REACTIONS .................................................................... 104
3.4.1.
E1 and E2 Reactions ............................................................................ 104
3.4.2.
Carbocations. ....................................................................................... 105
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PREFACE
Course Rationale
This course is designed to introduce the basic concepts of inorganic, organic and physical
chemistry upon which understanding of modern chemistry depends. These concepts
include atomic structure, ideal gas behavior and its deviation; covalent and ionic bonding;
the concept of reaction mechanism in the context of key reactions of organic and
inorganic chemistry, and the principles governing chemical processes in terms of
thermodynamic properties, kinetics and thermo-chemistry. The text also summarizes the
key functional groups in organic chemistry. The material is designed for students who
are not necessarily majoring in Chemistry but who require the fundamentals of this
subject in order to understand important processes in their scientific field of study, in this
particular case, aspects of environmental science studies. It is expected that the students
will use the knowledge from this course as a foundation upon which they can build
higher level knowledge as well as use it to scientifically interpret observations in their
own field of specialization.
Learning outcomes
After studying this course material, students should be able to:
 Account for deviations of the behaviour of real systems from an ideal model,
 Account for the main types of intermolecular forces found in liquids and
solutions,
 State and interpret the three laws of thermodynamics and solve simple problems
involving their application,
 Define the relationship between Gibbs free energy and chemical equilibrium,
 Recognise, exemplify, systematically name, and diagrammatically represent the
common functional groups of organic chemistry,
 Draw mechanisms for some of the fundamental reactions of organic chemistry,
 Predict
chemical reactivity
from knowledge of acid/base and
nucleophile/electrophile properties,
 Recognize and apply the four quantum numbers and their allowed values.
 Derive the shapes of molecules using the VSEPR method.
Mode of Assessment
Continuous Assessment
Written assignments
Timed tests
Final examination
15%
25%
60%
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Reading List
1) Brown, T and Leamay H (2001). Chemistry: The Central Science. Pentice-Hall,
Inc. New Jersey.
2) Petrucia, R., Harwood, G.H. and Herring G., (2005). General Chemistry: 7th
Edition. McMillan Publishing Company, New York.
3) Raph, S. B and Wayne, E.W. (1980). General Chemistry, 2nd Edition Houghton
Mifflin Company, USA.
4) Ebbing, D.D (1993). General Chemistry, 7th Edition. Houghton Mifflin Company,
USA.
5) Brady J.E. and Humiston, G.R. (1986). General Chemistry: Principles and
Structure 4th Edition. John Willey and Sons, USA.
2
CHAPTER ONE
INORGANIC CHEMISTRY
1.1.
ATOMS
All the matter around is made of atoms, and all atoms are made of only three types of
subatomic particle, protons, electrons, and neutrons. Furthermore, all protons are
exactly the same, all neutrons are exactly the same, and all electrons are exactly the same.
Protons and neutrons have almost exactly the same mass. Electrons have a mass that is
about 1/1835 the mass of a proton. Electrons have a unit negative charge. Protons each
have a positive charge. These charges are genuine electrical charges. Neutrons do not
have any charge.
The neutrons and protons are in the center of the atom in a nucleus. The electrons are
outside the nucleus in electron shells that are in different shapes at different distances
from the nucleus. The atom is mostly empty space. Ernest Rutherford shot subatomic
particles at a very thin piece of gold. Most of the particles went straight through the gold.
Almost all the mass of an atom is concentrated in the tiny nucleus. The mass of a proton
or neutron is 1.66 × 10 -24 grams or one AMU, atomic mass unit. The mass of an electron
is 9.05 × 10-28 grams. This number is a billionth of a billionth of a billionth of a gram. It
is not possible for anyone or any machine that uses light to actually see a proton using
visible light. The wavelength of light is too large to be able to detect anything that small.
1.1.1. Atomic Weights and Atomic Numbers
The integer that you find in each box of the Periodic Chart is the atomic number. The
atomic number is the number of protons in the nucleus of each atom. Another number
that you can often find in the box with the symbol of the element is not an integer. It is
oversimplifying only a little to say that this number is the number of protons plus the
average number of neutrons in that element. The number is called the atomic weight or
atomic mass.
3
The number of protons defines the type of element. If an atom has six protons, it is
carbon. If it has 92 protons, it is uranium. The number of neutrons in the nucleus of an
element can be different, though. Carbon 12 is the commonest type of carbon. Carbon 12
has six protons (naturally, otherwise it wouldn’t be carbon) and six neutrons. The mass of
the electrons is negligible. Carbon 12 has a mass of twelve. Carbon 13 has six protons
and seven neutrons. Carbon 14 has six protons and eight neutrons. Types of an element in
which every atom has the same number of protons and the same number of neutrons are
called isotopes.
1.1.2. Formula Weight
Other similar terms used: Molecular Weight or Formula Mass or Molar Mass. We can
consider matching up atoms on a mass-to-mass basis. Let’s take hydrogen chloride, HCl.
One hydrogen atom is attached to one chlorine atom, but they have different masses. A
hydrogen atom has a mass of 1.008 AMU and a chlorine atom has a mass of 35.453
AMU. Practically speaking, one AMU is far too small a mass for us to weigh in the lab.
We could weigh 1.008 grams of hydrogen and 35.453 grams of chlorine, and they would
match up exactly right. There would be the same number of hydrogen atoms as chlorine
atoms. They could join together to make HCl with no hydrogen or chlorine left over. If
we take one gram of a material for every AMU of mass in the atoms of just one of them,
we will have a mol (or mole) of that material. One mol of any material, therefore, has the
same number of particles of the material named, this number being Avogadro’s number,
6.022 ×1023.
The formula weight is the most general term that includes atomic weight and molecular
weight. In the case of the HCl, we can add the atomic weights of the elements in the
compound and get a molecular weight. The molecular weight of HCl is 36.461 g/mol, the
sum of the atomic weights of hydrogen and chlorine. The unit of molecular weight is
grams per mol. The way to calculate the molecular weight of any formula is to add up the
atomic weights of all the atoms in the formula.
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1.2.
ELECTRON CONFIGURATION
Protons have a positive charge and electrons have a negative charge. Free (unattached)
uncharged atoms have the same number of electrons as protons to be electrically neutral.
The protons are in the nucleus and do not change or vary except in some nuclear
reactions. The electrons are in discrete pathways or shells around the nucleus. There is a
ranking or hierarchy of the shells, usually with the shells further from the nucleus having
a higher energy. As we consider the electron configuration of atoms, we will be
describing the ground state position of the electrons. When electrons have higher energy,
they may move up away from the nucleus into higher energy shells. As we consider the
electron configuration, we will be describing the ground state positions of the electrons.
A hydrogen atom has only one proton and one electron. The electron of a hydrogen atom
travels around the proton nucleus in a shell of a spherical shape. The two electrons of
helium, element number two, are in the same spherical shape around the nucleus. The
first shell only has one subshell, and that subshell has only one orbital, or pathway for
electrons. Each orbital has a place for two electrons. The spherical shape of the lone
orbital in the first energy level has given it the name s orbital. Helium is the last element
in the first period. Being an inert element, it indicates that that shell is full. Shell number
one has only one s sub shell and all s sub shells have only one orbital. Each orbital only
has room for two electrons. So the first shell, called the K shell, has only two electrons.
Beginning with lithium, the electrons do not have room in the first shell or energy level.
Lithium has two electrons in the first shell and one electron in the next shell. The first
shell fills first and the others more or less in order as the element size increases up the
Periodic Chart, but the sequence is not immediately obvious. The second energy level has
room for eight electrons. The second energy level has not only an s orbital, but also a p
sub shell with three orbitals. The p sub shell can contain six electrons. The p sub shell has
a shape of three dumbbells at ninety degrees to each other, each dumbbell shape being
one orbital. With the s and p sub shells the second shell, the L shell can hold a total of
eight electrons. You can see this on the periodic chart. Lithium has one electron in the
outside shell, the L shell. Beryllium has two electrons in the outside shell. The s sub shell
5
fills first, so all other electrons adding to this shell go into the p sub shell. Boron has three
outside electrons, carbon has four, nitrogen has five, oxygen has six, and fluorine has
seven. Neon has a full shell of eight electrons in the outside shell, the L shell, meaning
the neon is an inert element, the end of the period.
Beginning again at sodium with one electron in the outside shell, the M shell fills its s
and p sub shells with eight electrons. Argon, element eighteen, has two electrons in the K
shell, eight in the L shell, and eight in the M shell. The fourth period begins again with
potassium and calcium, but there is a difference here. After the addition of the 4s
electrons and before the addition of the 4p electrons, the sequence goes back to the third
energy level to insert electrons in a d shell.
The shells or energy levels are numbered or lettered, beginning with K. So K is one, L is
two, M is three, N is four, O is five, P is six, and Q is seven. As the s shells can only have
two electrons and the p shells can only have six electrons, the d shells can have only ten
electrons and the f shells can have only fourteen electrons. The sequence of addition of
the electrons as the atomic number increases is as follows with the first number being the
shell number, the s, p, d, or f being the type of sub shell, and the last number being the
number of electrons in the sub shell.
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6d10 7p6
It is tempting to put an 8s2 at the end of the sequence, but we have no evidence of an R
shell. One way to know this sequence is to memorize it. There is a bit of a pattern in it.
The next way to know this sequence is to SEE IT ON THE PERIODIC CHART. As you
go from hydrogen down the chart, the Groups 1 and 2 represent the filling of an s sub
shell. The filling of a p sub shell is shown in Groups 3 through 8. The filling of a d sub
shell is represented by the transition elements (ten elements), and the filling of an f sub
shell is shown in the lanthanide and actinide series (fourteen elements).
Here is a copy of the periodic chart as you have usually seen it.
6
And here is the same chart re-arranged with the Lanthanides and Actinides in their right
place and Group I and II afterward. Both of these charts are color coded so that the
elements with the 2s sub shell on the outside (H and He) are turquoise. All other elements
with an s sub shell on the outside (Groups I and II) are outlined in blue. Lanthanides and
actinides are in grey. Other transition elements are in yellow, and all of the elements that
have a p sub shell as the last one on the outside are in salmon color.
7
You may be able to see it better with the sub shell areas labeled.
There are several other schemes to help you remember the sequence.
The shape of the s sub shells is spherical. The shape of the p sub shells is the shape of
three barbells at ninety degrees to each other. The shape of the d and f sub shells is very
complex.
1.3.
QUANTUM NUMBERS AND ELECTRON CONFIGURATION
1.3.1. Quantum Numbers
The Bohr model was a one-dimensional model that used one quantum number to describe
the distribution of electrons in the atom. The only information that was important was the
size of the orbit, which was described by the n quantum number. Later models allowed
the electron to occupy three-dimensional space. It therefore required three coordinates, or
three quantum numbers, to describe the orbitals in which electrons can be found. The
three coordinates are the principal (n), angular (l), and magnetic (m) quantum numbers.
These quantum numbers describe the size, shape, and orientation in space of the orbitals
on an atom.
8
The principal quantum number (n) describes the size of the orbital. Orbitals for which n =
2 are larger than those for which n = 1, for example. Because they have opposite
electrical charges, electrons are attracted to the nucleus of the atom. Energy must
therefore be absorbed to excite an electron from an orbital in which the electron is close
to the nucleus (n = 1) into an orbital in which it is further from the nucleus (n = 2). The
principal quantum number therefore indirectly describes the energy of an orbital.
The angular quantum number (l) describes the shape of the orbital. Orbitals have shapes
that are best described as spherical (l = 0), polar (l = 1), or cloverleaf (l = 2). They can
even take on more complex shapes as the value of the angular quantum number becomes
larger.
There is only one way in which a sphere (l = 0) can be oriented in space. Orbitals that
have polar (l = 1) or cloverleaf (l = 2) shapes, however, can point in different directions.
We therefore need a third quantum number, known as the magnetic quantum number (m),
to describe the orientation in space of a particular orbital. (It is called the magnetic
quantum number because the effect of different orientations of orbitals was first observed
in the presence of a magnetic field).
1.3.2. Rules Governing the Allowed Combinations of Quantum Numbers
The three quantum numbers (n, l, and m) that describe an orbital are integers: 0, 1, 2, 3,
and so on. The principal quantum number (n) cannot be zero. The allowed values of n are
therefore 1, 2, 3, 4, and so on. The angular quantum number (l) can be any integer
between 0 and (n – 1). If n = 3, for example, l can be either 0, 1, or 2. The magnetic
quantum number (m) can be any integer between -l and +l. If l = 2, m can be either -2, -1,
0, +1, or +2.
Practice Problem 1:
Describe the allowed combinations of the n, l, and m quantum numbers when n = 3.
9
1.3.3. Shells and Sub shells of Orbitals
Orbitals that have the same value of the principal quantum number form a shell. Orbitals
within a shell are divided into sub shells that have the same value of the angular quantum
number. Chemists describe the shell and sub shell in which an orbital belongs with a twocharacter code such as 2p or 4f. The first character indicates the shell (n = 2 or n = 4).
The second character identifies the sub shell. By convention, the following lowercase
letters are used to indicate different sub shells.
s: l = 0
p: l = 1
d: l = 2
f: l = 3
Although there is no pattern in the first four letters (s, p, d, f), the letters progress
alphabetically from that point (g, h, and so on). Some of the allowed combinations of the
n and l quantum numbers are shown in the figure below.
The third rule limiting allowed combinations of the n, l, and m quantum numbers has an
important consequence. It forces the number of sub shells in a shell to be equal to the
principal quantum number for the shell. The n = 3 shell, for example, contains three sub
shells: the 3s, 3p, and 3d orbitals.
10
1.3.4. Possible Combinations of Quantum Numbers
There is only one orbital in the n = 1 shell because there is only one way in which a
sphere can be oriented in space. The only allowed combination of quantum numbers for
which n = 1 is the following.
n
l
m
1
0
0
1s
There are four orbitals in the n = 2 shell.
n
l
m
2
0
0
2
1
-1
2
1
0
2
1
1
2s
2p
There is only one orbital in the 2s sub shell. But, there are three orbitals in the 2p sub
shell because there are three directions in which a p orbital can point. One of these orbital
is oriented along the X axis, another along the Y axis, and the third along the Z axis of a
coordinate system, as shown in the figure below. These orbitals are therefore known as
the 2px, 2py, and 2pz orbitals. There are nine orbitals in the n = 3 shell.
n
l
m
3
0
0
3
1
-1
3
1
0
3
1
1
11
3s
3p
3
2
-2
3
2
-1
3
2
0
3
2
1
3
2
2
3d
There is one orbital in the 3s sub shell and three orbitals in the 3p sub shell. The n = 3
shell, however, also includes 3d orbitals. The five different orientations of orbitals in the
3d sub shell are such that one of these orbitals lies in the XY plane of an XYZ coordinate
system and is called the 3dxy orbital. The 3dxz and 3dyz orbitals have the same shape, but
they lie between the axes of the coordinate system in the XZ and YZ planes. The fourth
orbital in this sub shell lies along the X and Y axes and is called the 3dx2-y2 orbital. Most
of the space occupied by the fifth orbital lies along the Z axis and this orbital is called the
3dz2 orbital. The number of orbitals in a shell is the square of the principal quantum
number: 12 = 1, 22 = 4, 32 = 9. There is one orbital in an s sub shell (l = 0), three orbitals
in a p sub shell (l = 1), and five orbitals in a d sub shell (l = 2). The number of orbitals in
a sub shell is therefore 2(l) + 1.
Before we can use these orbitals we need to know the number of electrons that can
occupy an orbital and how they can be distinguished from one another. Experimental
evidence suggests that an orbital can hold no more than two electrons.
To distinguish between the two electrons in an orbital, we need a fourth quantum number.
This is called the spin quantum number (s) because electrons behave as if they were
spinning in either a clockwise or counterclockwise fashion. One of the electrons in an
orbital is arbitrarily assigned an s quantum number of +1/2, the other is assigned an s
quantum number of -1/2. Thus, it takes three quantum numbers to define an orbital but
four quantum numbers to identify one of the electrons that can occupy the orbital. The
allowed combinations of n, l, and m quantum numbers for the first four shells are given in
the table below. For each of these orbitals, there are two allowed values of the spin
quantum number, s.
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1.3.5. The Relative Energies of Atomic Orbitals
Because of the force of attraction between objects of opposite charge, the most important
factor influencing the energy of an orbital is its size and therefore the value of the
principal quantum number, n. For an atom that contains only one electron, there is no
difference between the energies of the different sub shells within a shell. The 3s, 3p, and
3d orbitals, for example, have the same energy in a hydrogen atom. The Bohr model,
which specified the energies of orbits in terms of nothing more than the distance between
the electron and the nucleus, therefore works for this atom.
The hydrogen atom is unusual, however. As soon as an atom contains more than one
electron, the different sub shells no longer have the same energy. Within a given shell,
the s orbitals always have the lowest energy. The energy of the sub shells gradually
becomes larger as the value of the angular quantum number becomes larger.
Relative energies: s < p < d < f
As a result, two factors control the energy of an orbital for most atoms: the size of the
orbital and its shape.
13
1.3.6. The Aufbau Principle, Degenerate Orbitals, and Hund's Rule
The electron configuration of an atom describes the orbitals occupied by electrons on the
atom. The basis of this prediction is a rule known as the aufbau principle, which assumes
that electrons are added to an atom, one at a time, starting with the lowest energy orbital,
until all of the electrons have been placed in an appropriate orbital.
A hydrogen atom (Z = 1) has only one electron, which goes into the lowest energy
orbital, the 1s orbital. This is indicated by writing a superscript "1" after the symbol for
the orbital.
H (Z = 1): 1s1
The next element has two electrons and the second electron fills the 1s orbital because
there are only two possible values for the spin quantum number used to distinguish
between the electrons in an orbital.
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He (Z = 2): 1s2
The third electron goes into the next orbital in the energy diagram, the 2s orbital.
Li (Z = 3): 1s2 2s1
The fourth electron fills this orbital.
Be (Z = 4): 1s2 2s2
After the 1s and 2s orbitals have been filled, the next lowest energy orbitals are the three
2p orbitals. The fifth electron therefore goes into one of these orbitals.
B (Z = 5): 1s2 2s2 2p1
When the time comes to add a sixth electron, the electron configuration is obvious.
C (Z = 6): 1s2 2s2 2p2
However, there are three orbitals in the 2p sub shell. Does the second electron go into the
same orbital as the first, or does it go into one of the other orbitals in this sub shell?
To answer this, we need to understand the concept of degenerate orbitals. By definition,
orbitals are degenerate when they have the same energy. The energy of an orbital
depends on both its size and its shape because the electron spends more of its time further
from the nucleus of the atom as the orbital becomes larger or the shape becomes more
complex. In an isolated atom, however, the energy of an orbital doesn't depend on the
direction in which it points in space. Orbitals that differ only in their orientation in space,
such as the 2px, 2py, and 2pz orbitals, are therefore degenerate.
Electrons fill degenerate orbitals according to rules first stated by Friedrich Hund. Hund's
rules can be summarized as follows:
One electron is added to each of the degenerate orbitals in a sub shell before two
electrons are added to any orbital in the sub shell.
Electrons are added to a sub shell with the same value of the spin quantum number until
each orbital in the sub shell has at least one electron.
When the time comes to place two electrons into the 2p sub shell we put one electron into
each of two of these orbitals. (The choice between the 2px, 2py, and 2pz orbitals is purely
arbitrary.)
C (Z = 6): 1s2 2s2 2px1 2py1
15
The fact that both of the electrons in the 2p sub shell have the same spin quantum number
can be shown by representing an electron for which s = +1/2 with an
arrow pointing up and an electron for which s = -1/2 with an arrow pointing down.
The electrons in the 2p orbitals on carbon can therefore be represented as follows.
When we get to N (Z = 7), we have to put one electron into each of the three degenerate
2p orbitals.
N (Z = 7): 1s2 2s2 2p3
Because each orbital in this sub shell now contains one electron, the next electron added
to the sub shell must have the opposite spin quantum number, thereby filling one of the
2p orbitals.
O (Z = 8): 1s2 2s2 2p4
The ninth electron fills a second orbital in this sub shell.
F (Z = 9): 1s2 2s2 2p5
The tenth electron completes the 2p sub shell.
Ne (Z = 10): 1s2 2s2 2p6
There is something unusually stable about atoms, such as He and Ne that have electron
configurations with filled shells of orbitals. By convention, we therefore write
abbreviated electron configurations in terms of the number of electrons beyond the
previous element with a filled-shell electron configuration. Electron configurations of the
next two elements in the periodic table, for example, could be written as follows.
Na (Z = 11): [Ne] 3s1
Mg (Z = 12): [Ne] 3s2
16
Practice Problem 2:
Predict the electron configuration for a neutral tin atom (Sn, Z = 50).
The aufbau process can be used to predict the electron configuration for an element. The
actual configuration used by the element has to be determined experimentally. The
experimentally determined electron configurations for the elements in the first four rows
of the periodic table are given in the table in the following section.
The Electron Configurations of the Elements
(1st, 2nd, 3rd, and 4th Row Elements)
Atomic Number
Symbol
Electron Configuration
1
H
1s1
2
He
1s2 = [He]
3
Li
[He] 2s1
4
Be
[He] 2s2
5
B
[He] 2s2 2p1
6
C
[He] 2s2 2p2
7
N
[He] 2s2 2p3
8
O
[He] 2s2 2p4
9
F
[He] 2s2 2p5
10
Ne
[He] 2s2 2p6 = [Ne]
11
Na
[Ne] 3s1
12
Mg
[Ne] 3s2
13
Al
[Ne] 3s2 3p1
14
Si
[Ne] 3s2 3p2
15
P
[Ne] 3s2 3p3
16
S
[Ne] 3s2 3p4
17
17
Cl
[Ne] 3s2 3p5
18
Ar
[Ne] 3s2 3p6 = [Ar]
19
K
[Ar] 4s1
20
Ca
[Ar] 4s2
21
Sc
[Ar] 4s2 3d1
22
Ti
[Ar] 4s2 3d2
23
V
[Ar] 4s2 3d3
24
Cr
[Ar] 4s1 3d5
25
Mn
[Ar] 4s2 3d5
26
Fe
[Ar] 4s2 3d6
27
Co
[Ar] 4s2 3d7
28
Ni
[Ar] 4s2 3d8
29
Cu
[Ar] 4s1 3d10
30
Zn
[Ar] 4s2 3d10
31
Ga
[Ar] 4s2 3d10 4p1
32
Ge
[Ar] 4s2 3d10 4p2
33
As
[Ar] 4s2 3d10 4p3
34
Se
[Ar] 4s2 3d10 4p4
35
Br
[Ar] 4s2 3d10 4p5
36
Kr
[Ar] 4s2 3d10 4p6 = [Kr]
1.3.7. Exceptions to Predicted Electron Configurations
There are several patterns in the electron configurations listed in the table in the previous
section. One of the most striking is the remarkable level of agreement between these
configurations and the configurations we would predict. There are only two exceptions
among the first 40 elements: chromium and copper. Strict adherence to the rules of the
aufbau process would predict the following electron configurations for chromium and
copper.
18
predicted electron configurations: Cr (Z = 24): [Ar] 4s2 3d4
Cu (Z = 29): [Ar] 4s2 3d9
The experimentally determined electron configurations for these elements are slightly
different.
actual electron configurations: Cr (Z = 24): [Ar] 4s1 3d5
Cu (Z = 29): [Ar] 4s1 3d10
In each case, one electron has been transferred from the 4s orbital to a 3d orbital, even
though the 3d orbitals are supposed to be at a higher level than the 4s orbital. Once we
get beyond atomic number 40, the difference between the energies of adjacent orbitals is
small enough that it becomes much easier to transfer an electron from one orbital to
another. Most of the exceptions to the electron configuration predicted from the aufbau
diagram shown earlier therefore occur among elements with atomic numbers larger than
40. Although it is tempting to focus attention on the handful of elements that have
electron configurations that differ from those predicted with the aufbau diagram, the
amazing thing is that this simple diagram works for so many elements.
1.3.8. Electron Configurations and the Periodic Table
When electron configuration data are arranged so that we can compare elements in one of
the horizontal rows of the periodic table, we find that these rows typically correspond to
the filling of a shell of orbitals. The second row, for example, contains elements in which
the orbitals in the n = 2 shell are filled.
Li (Z = 3):
[He] 2s1
Be (Z = 4):
[He] 2s2
B (Z = 5):
[He] 2s2 2p1
C (Z = 6):
[He] 2s2 2p2
N (Z = 7):
[He] 2s2 2p3
19
O (Z = 8):
[He] 2s2 2p4
F (Z = 9):
[He] 2s2 2p5
Ne (Z = 10): [He] 2s2 2p6
There is an obvious pattern within the vertical columns, or groups, of the periodic table as
well. The elements in a group have similar configurations for their outermost electrons.
This relationship can be seen by looking at the electron configurations of elements in
columns on either side of the periodic table.
Group IA
Group VIIA
H
1s1
Li
[He] 2s1 F
[He] 2s2 2p5
Na
[Ne] 3s1 Cl
[Ne] 3s2 3p5
K
[Ar] 4s1 Br
[Ar] 4s2 3d10 4p5
Rb
[Kr] 5s1 I
[Kr] 5s2 4d10 5p5
Cs
[Xe] 6s1 At
[Xe] 6s2 4f14 5d10 6p5
The figure below shows the relationship between the periodic table and the orbitals being
filled during the aufbau process. The two columns on the left side of the periodic table
correspond to the filling of an s orbital. The next 10 columns include elements in which
the five orbitals in a d sub shell are filled. The six columns on the right represent the
filling of the three orbitals in a p sub shell. Finally, the 14 columns at the bottom of the
table correspond to the filling of the seven orbitals in an f subshell.
Practice Problem 3:
Predict the electron configuration for calcium (Z = 20) and zinc (Z = 30) from their
positions in the periodic table.
1.4.
IONIC AND COVALENT BONDS
A bond is an attachment among atoms. Atoms may be held together for any of several
reasons, but all bonds have to do with the electrons, particularly the outside electrons, of
atoms. There are bonds that occur due to sharing electrons. There are bonds that occur
due to a full electrical charge difference attraction. There are bonds that come about from
20
partial charges or the position or shape of electrons about an atom. But all bonds have to
do with electrons. Since chemistry is the study of elements, compounds, and how they
change, it might be said that chemistry is the study of electrons. If we study the changes
brought about by moving protons or neutrons, we would be studying nuclear physics. In
chemical reactions, the elements do not change from one element to another, but are only
rearranged in their attachments.
A compound is a group of atoms with an exact number and type of atoms in it arranged
in a specific way. Every bit of that material is exactly the same. Exactly the same
elements in exactly the same proportions are in every bit of the compound. Water is an
example of a compound. One oxygen atom and two hydrogen atoms make up water. Each
hydrogen atom is attached to an oxygen atom by a bond. Any other arrangement is not
water. If any other elements are attached, it is not water. H2O is the formula for that
compound. This formula indicates that there are two hydrogen atoms and one oxygen
atom in the compound. H2S is hydrogen sulfide. Hydrogen sulfide does not have the same
types of atoms as water. It is a different compound. H2O2 is the formula for hydrogen
peroxide. It might have the right elements in it to be water, but it does not have them in
the right proportion. It is still not water. The word formula is also used to mean the
smallest bit of any compound. A molecule is a single formula of a compound joined by
covalent bonds. The Law of Constant Proportions states that a given compound always
contains the same proportion by weight of the same elements.
1.4.1. Ionic Bonds
Some atoms, such as metals tend to lose electrons to make the outside ring or rings of
electrons more stable and other atoms tend to gain electrons to complete the outside ring.
An ion is a charged particle. Electrons are negative. The negative charge of the electrons
can be offset by the positive charge of the protons, but the number of protons does not
change in a chemical reaction. When an atom loses electrons, it becomes a positive ion
because the number of protons exceeds the number of electrons. Non-metal ions and most
of the polyatomic ions have a negative charge. The non-metal ions tend to gain electrons
to fill out the outer shell. When the number of electrons exceeds the number of protons,
21
the ion is negative. The attraction between a positive ion and a negative ion is an ionic
bond. Any positive ion will bond with any negative ion. They are not fussy. An ionic
compound is a group of atoms attached by an ionic bond that is a major unifying portion
of the compound. A positive ion, whether it is a single atom or a group of atoms all with
the same charge, is called a cation. A negative ion is called an anion. The name of an
ionic compound is the name of the positive ion (cation) first and the negative (anion) ion
second.
The valence of an atom is the likely charge it will take on as an ion. The names of the
ions of metal elements with only one valence, such as the Group 1 or Group 2 elements
are the same as the name of the element. The names of the ions of nonmetal elements
(anions) develop an -ide on the end of the name of the element. For instance, fluorine ion
is fluoride, oxygen ion is oxide, and iodine ion is iodide. There are a number of elements,
usually transition elements that having more than one valence, that has a name for each
ion, for instance ferric ion is an iron ion with a positive three charge. Ferrous ion is an
iron ion with a charge of plus two. There are a number of common groups of atoms that
have a charge for the whole group. Such a group is called a polyatomic ion or radical.
1.4.2. Some atoms with multiple valences.
Note: There are two common names for the ions. You should know both the stock
system and the old system names.
STOCK
OLD
STOCK
OLD
SYSTEM
SYSTEM
SYSTEM
SYSTEM
Fe2+
iron II
ferrous
Fe3+
iron III
ferric
Cu+
copper I
cuprous
Cu2+
copper II
cupric
Au+
gold I
aurous
Au3+
gold III
auric
Sn2+
tin II
stannous
Sn4+
tin IV
stannic
Pb2+
lead II
plumbous
Pb4+
lead IV
plumbic
Hg+
mercury I
mercurous
Hg2+
mercury II
mercuric
ION
ION
22
Cr2+
chromium II chromous
Cr3+
chromium III chromic
Mn2+
manganese II manganous
Mn3+
manganeseIII manganic
The ions by the Stock system are pronounced, copper one, copper two, etc. Notice that
the two most likely ions of an atom that has multiple valences have suffixes in the old
system to identify them. The smallest of the two charges gets the -ous suffix, and the
largest of the two charges has the -ic suffix.
1.4.3. Some atoms with only one common valence:
ALL GROUP 1 ELEMENTS ARE +1
ALL GROUP 2 ELEMENTS ARE +2
ALL GROUP 7 (HALOGEN) ELEMENTS ARE -1 WHEN IONIC
Oxygen and sulfur (GROUP 6) are -2 when ionic
Hydrogen is usually +1
Al3+, Zn2+, and Ag+
1.4.4. Radicals or Polyatomic Ions
The following radicals or polyatomic ions are groups of atoms of more than one kind of
element attached by covalent bonds. They do not often come apart in ionic reactions. The
charge on the radical is for the whole group of atoms as a unit. These are common
radicals you should learn WITH THEIR CHARGE AND NAME.
(NH4)+ AMMONIUM - Do not confuse with NH3, AMMONIA GAS)
(NO3)- NITRATE (Do not confuse with NITRIDE (N3-) or NITRITE)
(NO2)- NITRITE (Do not confuse with (N3-) or NITRATE)
(C2H3O2)- ACETATE (NOTE - This is not the only way this may be written.)
(ClO3)- CHLORATE (Do not confuse with CHLORIDE (Cl-) or CHLORITE)
(ClO2)- CHLORITE (Do not confuse with CHLORIDE (Cl-) or CHLORATE)
(SO3)2- SULFITE (Do not confuse with (S2-) or SULFATE)
(SO4)2- SULFATE (Do not confuse with SULFIDE (S2-) or SULFITE)
(HSO3)- BISULFITE (or HYDROGEN SULFITE)
(PO4)3- PHOSPHATE (Do not confuse with P3-, PHOSPHIDE)
23
(HCO3)- BICARBONATE (or HYDROGEN CARBONATE)
(CO3)2- CARBONATE
(HPO4)2- HYDROGEN PHOSPHATE
(H2PO4)- DIHYDROGEN PHOSPHATE
(OH)- HYDROXIDE
(CrO4)2- CHROMATE
(Cr2O7)2-DICHROMATE
(BO3)3- BORATE
(AsO4)3- ARSENATE
(C2O4)2- OXALATE
(ClO4)- PERCHLORATE
(CN)- CYANIDE
(MnO4)- PERMANGANATE
1.4.5. Acids of Some Common Polyatomic Ions.
These are written here with the parentheses around the polyatomic ions to show their
origin. Usually these compounds are written without the parentheses, such as HNO3 or
H2SO4. Note that the polyatomic ions with a single negative charge only have one
hydrogen. Polyatomic ions with two negative charges have two hydrogens.
H(OH) WATER
H(NO3) NITRIC ACID
H(NO2) NITROUS ACID .
H(C2H3O2) ACETIC ACID
H2(CO3) CARBONIC ACID
H2(SO3) SULFUROUS ACID
H2(SO4) SULFURIC ACID
H3(PO4) PHOSPHORIC ACID
H2(CrO4) CHROMIC ACID
H3(BO3) BORIC ACID
H2(C2O4) OXALIC ACID
24
1.4.6. Writing Ionic Compound Formulas
In the lists above, the radicals and compounds have a small number after and below an
element if there is more than one of that type of that atom. For instance, ammonium ions
have one nitrogen atom and four hydrogen atoms in them. Sulfuric acid has two
hydrogens, one sulfur, and four oxygens.
Knowing the ions is the best way to identify ionic compounds and to predict how
materials would join. People who do not know of the ammonium ion and the nitrate ion
would have a difficult time seeing that NH4NO3 is ammonium nitrate. Let’s consider
what happens in an ionic bond using electron configuration, the octet rule, and some
creative visualization. A sodium atom has eleven electrons around it. The first shell has
two electrons in an s sub shell. The second shell is also full with eight electrons in an s
and a p sub shell. The outer shell has one lonely electron, as do the other elements in
Group 1. This outside electron can be detached from the sodium atom, leaving a sodium
ion with a single positive charge and an electron. A chlorine atom has seventeen
electrons. Two are in the first shell, eight are in the second shell, and seven are in the
outside shell. The outside shell is lacking one electron to make a full shell, as are all the
elements of Group 7. When the chlorine atom collects another electron, the atom
becomes a negative ion. The positive sodium ion missing an electron is attracted to the
negative chloride ion with an extra electron. The symbol for a single unattached electron
is e-.
Cl2 + Na → Cl + e- + Na+ → Cl- + Na+ → Na+Cl- → NaCl
Any compound should have a net zero charge. The single positive charge of the sodium
ion cancels the single negative charge of the chloride ion. The same idea would be for an
ionic compound made of ions of plus and minus two or plus and minus three, such as
magnesium sulfate or aluminum phosphate
Mg2+ + (SO4)2- → Mg2+(SO4)2- → Mg(SO4) or MgSO4
Al3+ + (PO4)3- → Al3+(PO4)3- → Al(PO4) or AlPO4
25
But what happens if the amount of charge does not match? Aluminum bromide has a
cation that is triple positive and an anion that is single negative. The compound must be
written with one aluminum and three bromide ions. AlBr3. Calcium phosphate has a
double positive cation and a triple negative anion. If you like to think of it this way, the
number of the charges must be switched to the other ion. Ca3(PO4)2. Note that there must
be two phosphates in each calcium phosphate, so the parentheses must be included in the
formula to indicate that. Each calcium phosphate formula (Ionic compounds do not make
molecules.) has three calcium atoms, two phosphate atoms, and eight oxygen atoms.
There are a small number of ionic compounds that do not fit into the system for one
reason or other. A good example of this is magnetite, an ore of iron, Fe3O4. The
calculated charge on each iron atom would be +8/3, not a likely actual charge. The
deviance from the system in the case of magnetite could be accounted for by a mixture of
the common ferric and ferrous ions.
1.4.7. Binary Covalent Compounds
The word binary means that there are two types of atom in a compound. Covalent
compounds are groups of atoms joined by covalent bonds. Binary covalent compounds
are some of the very smallest compounds attached by covalent bonds. A covalent bond
is the result of the sharing of a pair of electrons between two atoms. The chlorine
molecule is a good example of the bond, even if it has only one type of atom. Chlorine
gas, Cl2, has two chlorine atoms, each of which has seven electrons in the outside ring.
Each atom contributes an electron to an electron pair that make the covalent bond. Each
atom shares the pair of electrons. In the case of chlorine gas, the two elements in the bond
have exactly the same pull on the electron pair, so the electrons are exactly evenly shared.
The covalent bond can be represented by a pair of dots between the atoms, Cl:Cl, or a
line between them, Cl-Cl. Sharing the pair of electrons makes each chlorine atom feel as
if it has a completed outer shell of eight electrons. The covalent bond is much harder to
break than an ionic bond. The ionic bonds of soluble ionic compounds come apart in
water, but covalent bonds do not usually come apart in water. Covalent bonds make real
molecules, groups of atoms that are genuinely attached to each other. Binary covalent
26
compounds have two types of atom in them, usually non-metal atoms. Covalent bonds
can come in double (sharing of two pairs of electrons) and triple (three pairs of electrons)
bonds.
FORMULA
COMMON NAME
SYSTEM NAME
N2O
nitrous oxide
dinitrogen monoxide
NO
nitric oxide
nitrogen monoxide
N2O3
nitrous anhydride
dinitrogen trioxide
NO2
nitrogen dioxide
nitrogen dioxide
N2O4
nitrogen tetroxide
dinitrogen tetroxide
N2O5
nitric anhydride
dinitrogen pentoxide
NO3
nitrogen trioxide
nitrogen trioxide
With the compounds of nitrogen and oxygen to use as examples, we see that there are
often more ways for any two elements to combine with each other by covalent bonds than
by ionic bonds. Many of the frequently seen compounds already have names that have
been in use for a long time. These names, called common names, may or may not have
anything to do with the makeup of the material, but more of the common names of
covalent compounds are used than of the ionic compounds.
*GP
number
*GP
number
*GP
number
*GP
number
mono-
one
di-
two
tri-
three
tetra-
four
penta-
five
hexa-
six
hepta-
seven
octa-
eight
nona-
nine
deca-
ten
undeca-
eleven
dodeca-
twelve
The system names include numbers that indicate how many of each type of atom are in a
covalent molecule. The Greek Prefixes (GP’s above in the chart) are used to indicate the
number. It would be wise of you to know the GP’s. In saying or writing the name of a
binary covalent the GP of the first element is said, then the name of the first element is
27
said, then the GP of the second element is said, and the name of the second element is
said, usually with the ending -ide on it. The only notable exception for the rule is if the
first mentioned element only has one atom in the molecule, in which case the monoprefix is omitted. CO is carbon monoxide. CO2 is carbon dioxide. In both cases, there is
only one carbon in the molecule, and the mono- prefix is not mentioned. For oxygen, the
last vowel of the GP is omitted, as in the oxides of nitrogen in the above table.
Common Names of Binary Covalent Compounds
H2O water
NH3 ammonia
N2H4 hydrazine
CH4 methane
C2H2 acetylene
In an attempt to simplify, some books may seem to suggest that covalent and ionic bonds
are two separate and completely different types of attachment. A covalent bond is a
shared pair of electrons. The bond between the two atoms of any diatomic gas, such as
chlorine gas, Cl2, is certainly equally shared. The two chlorine atoms have exactly the
same pull on the pair of electrons, so the bond must be exactly equally shared. In cesium
fluoride the cesium atom certainly donates an electron and the fluoride atom certainly
craves an electron. Both the cesium ion and the fluoride ion can exist independently of
the other. The bond between a cesium and a fluoride ion is clearly ionic.
The amount of pull on an atom has on a shared pair of electrons, called electronegativity,
is what determines the type of bond between atoms. Considering the Periodic Chart
without the inert gases, electronegativity is greatest in the upper right of the Periodic
Chart and lowest at the bottom left. The bond in francium fluoride should be the most
ionic. Some texts refer to a bond that is between covalent and ionic called a polar
covalent bond. There is a range of bond between purely ionic and purely covalent that
depends upon the electronegativity of the atoms around that bond. If there is a large
difference in electronegativity, the bond has more ionic character. If the electronegativity
of the atoms is more similar, the bond has more covalent character.
28
1.5.
FORMAL CHARGE AND RESONANCE STRUCTURES
1.5.1. Deducing the not so easy structures
By now, you should be fairly comfortable with the notion of creating Lewis dot diagrams
for most basic structures. But what happens when there is more than one way to draw a
dot structure in which all of the electrons are accounted for and all of the octets filled, as
in the following example?
OR
Both of the structures above should be satisfactory by the rules laid out for determining
Lewis dot structures. But in fact, one of the structures is actually more correct than the
other. To determine which it is, we will use the Formal Charge calculation.
1.5.2. Calculating the Formal Charge
The formal charge of an atom is the charge, which an atom would have if all atoms had
the same electronegativity. The formal charge of an atom is equal to the number of
valence electrons in an original atom (i.e. 4 in carbon) minus the number of electrons
assigned to the atom in the Lewis structure. The sum of the formal charges of all the
atoms in the molecule give the overall charge of the molecule. Let's see how that works
with the CO2 example shown above. We will use the following formula to determine the
formal charge of the atoms in the molecule and the overall charge of the molecule in
order to determine which one is more stable and therefore the more preferred structure.
Formal Charge Formula
Formal
charge
=
# of valence
electrons in the
neutral
unbonded atom
29
-
(all unshared electrons +1/2 of
all shared electrons
Using this formula, we see that the CO2 structures, labeled A & B, respectively, would
have the following formal charges on their respective atoms:
A
B
Structure A
Structure B
O (left) = 6 - (4 + 1/2(4))= 0
O (left) = 6 - (2 + 1/2(6)) = +1
O (right) = 6 - (4 + 1/2(4)) = 0
O (right) = 6 - (6 + 1/2(2)) = -1
C = 4 - (0 + 1/2(8)) = 0
C = 4 - (0 + 1/2(8)) = 0
TOTAL CHARGE = 0
TOTAL CHARGE = 0
Due to the fact that CO2 is a neutral molecule, either of the structures above could work
as a Lewis dot structure. In these cases, however, the most stable structure will be the
one that:
1. Has atoms which bear the smallest formal charge and,
2. Has any negative charges residing on the more electronegative atom.
Since the structure on the left above bears no formal charges, it is the more stable
conformation for CO2.
The formal charge system allows us a way to choose between structures when they have
more than one way that they can satisfactorily be drawn, but have one structure that is
more stable than the other.
1.5.3. Resonance Structures
Now we'll look at another situation which arises sometimes, that is a structure that can
not be described accurately by one Lewis structure. These structures are said to be
30
resonance structures. The resonance theory was developed by Linus Pauling in the
1930s and basically says that many molecules and ions are best described by a
combination of Lewis structures rather than by one single structure. These molecules are
considered to be a composite of their resonance structures.
Individual Lewis structures are known as contributing structures, and the real molecule
that is described by resonance structures is known as a resonance hybrid. Contributing
structures are linked by a double headed arrow, and combine to give the hybrid structure.
This is more easily understood with a picture.
In the picture above, NO2- is a hybrid structure which is shown as a combination of two
resonance structures. Neither of the structures is a true representation of what the
molecule actually looks like, but since we can't draw the actual structure, we use the two
contributing structures to symbolize the hybrid structure.
1.6.
THE VSEPR MODEL
In order to understand and study the properties of molecules, one must first be able to
recognize and understand the orientation of atoms within a molecule. The way atoms
within a molecule are arranged in three dimensions determines the structure of the
molecule. This can in turn define other properties of the molecule such as its polarity.
The valence-shell electron pair repulsion (VSEPR) model gives us a way of creating the
correct 3-D model of a molecule by helping us determine the correct placement of atoms
and nonbonding electrons in the molecule based on the repulsions of electrons in the
molecule. The most stable conformation for a molecule is the one which has the electron
pairs as far away from each other as possible.
31
As you know by now, atoms in a molecule that are bonded to each other share a pair of
valence electrons. Some molecules also have atoms with nonbonding electron pairs.
Electron pairs repel each other, and they want to take up a position in space that will
minimize their interactions with other electron pairs, bonded or nonbonded. The goal of
the VSEPR model is to arrange the electron pairs around the central atom so that
there is the least amount of repulsions among them. This occurs when the electron
pairs are as far away from each other as possible.
1.6.1. Electron and Molecular Geometries
First, let's review bonding and nonbonding electrons. In water (H2O) for example, there
are bonds between the two hydrogens and oxygen which consist of bonding electrons. In
addition to these, oxygen has two pairs of nonbonding electrons. This is displayed in the
following:
The picture drawn above is an example of a Lewis dot diagram. The arrangement of
atoms within the molecule is known as the electron-pair geometry. We can use the
electron-pair geometry to help us determine molecular geometry, the position of the
atoms of a molecule in space. As mentioned before, the whole goal of the VSEPR model
is to correctly determine the position of the atoms in a molecule based on electron
repulsion. We can do this using some simple rules and some basic tables.
1.6.2. Predicting Electron-Pair Geometry
There are a few simple steps to follow when predicting the electron-pair geometry of a
molecule:
32
 First, determine the correct Lewis dot structure of the molecule, including figuring
out the most stable conformation using formal charge calculations.
 Second, count the total number of electron pairs around the central atom, bonded and
nonbonded. Use this number and Table 1 below to determine the arrangement of the
electron pairs that minimizes electron-pair repulsions. When you come upon double
or triple bonds in a molecule, these are counted as one bonding pair, not two or three.
Table 1
Number of Electron Pairs
Picture of Electron Pair
Electron Pair Geometry
Arrangement
2
Linear
3
Trigonal Planar
4
Tetrahedral
5
Trigonal bypyramidal
6
Octahedral
Let's look at an example of how this process works:
The organic compound formaldehyde has the molecular formula
CH2O. The correct Lewis dot diagram of formaldehyde is:
The structure's central atom (C) has two single bonded electron pairs and one double
bonded set of electron pairs to it. Remember, however, that a double or triple bond is
33
considered the same as a single bond when determining the geometry of a molecule.
Thus, carbon has three bonded sets of electrons and no nonbonded pairs of electrons.
Using the table above, we see that this corresponds to a trigonal planar electron
geometry
1.6.3. Using Electron Geometry to Figure Molecular Geometry
Once you have figured the correct electron geometry for a molecule, you can then
proceed to figure its molecular geometry. First, the total number of bonded electron pairs
and nonbonded electron pairs must be counted separately. Once this is done you can use
the following tables to predict the correct molecular geometry. Table 2 can be used for
molecules with four or less pairs of electrons; this covers most molecules which follow
the octet rule. As we will see later, there are cases where the molecule does not follow
the octet rule and thus it can exist in other conformations.
Table 2
Nonbonding
Total Electron Electron Pair Bonding Pairs
Molecular
Pairs
of
Pairs
Geometry
of Electrons
Geometry
Electrons
2
Linear
2
0
Linear
3
Trigonal
planar
3
0
Trigonal planar
2
1
Bent
34
4
Tetrahedral
4
0
Tetrahedral
3
1
Trigonal
pyramidal
2
2
Bent
When a central atom of a molecule comes from the third period of the periodic table or
beyond, that atom is capable of having more than four pairs of electrons around it. In
other words, the atom does not follow the octet rule. In these cases, the molecular
geometry is a little more difficult to figure out, but the following generalities and
information from Table 3 should help.
For molecules with a central atom with five electron pairs, the most stable electron-pair
geometry it can exist in is the trigonal bypyramidal geometry. A picture of this, with
bond angles included, is shown below.
35
For molecules with a central atom with six electron pairs, the most stable electron-pair
geometry it can exist is an octrahedral geometry. A picture of this, with bond angles
included, is shown below.
Depending upon the number of bonding and nonbonding pairs around the central atom,
molecules with more than four electron pairs around the central atom will reside in the
following molecular geometries.
Table 3
Total Electron Electron-Pair
Pairs
Geometry
5 pairs
Trigonalbypyramidal
Bonding
Nonbonding
Pairs
of Pairs
of Molecular Geometry
Electrons
Electrons
5
0
Trigonal-bypyramidal
36
4
1
Seesaw
3
2
T-shaped
2
3
Linear
6 pairs
Octahedral
6
0
Octahedral
5
1
Square pyramidal
37
4
2
Square planar
Molecular geometry is vital in order to understand the polarity of molecules. Also, the
geometry of molecules is crucial to understanding reactions in organic, inorganic and
biochemistry. Many reactions proceed the way they do because of the nature of the
geometry of a molecule. Many times reactions occur at certain places in a molecule
based on the positioning of the molecules. Some sites on the molecule are more open to
reaction than other sites. In order to deduce which sites are open for reaction you must
first know the correct orientation of the atoms in the molecule. The VSEPR model gives
you the tools to be able to do this.
38
1.7.
MOLARITY
Molarity is simply a measure of the "strength" of a solution. A solution that we would
call "strong" would have a higher molarity than one that we would call "weak".
A solution is made up of two parts. The solute is what gets mixed into the solution, like
powdered drink mix. The solvent is that which does the dissolving, like water.
When you are doing any type of quantitative analysis in the lab, you want to be as
accurate as possible.
# of moles of solute
Molarity = ---------------------Liters of solution
The unit for molarity is M and is read as "molar". (i.e. 3 M = three molar)
Molarity problems vary quite a bit. Pay careful attention to the wording of the problem,
and focus on what you are given and what the problem is asking for. Start with the
original molarity formula shown above, but be prepared to modify it when the need
arises.
I. Basic molarity problems where the molarity is the unknown.
Example 1. What is the molarity of a 5.00 liter solution that was made with 10.0 moles
of KBr ?
Solution: We can use the original formula. Note that in this particular example, where
the number of moles of solute is given, the identity of the solute (KBr) has nothing to do
with solving the problem.
# of moles of solute
Molarity = ---------------------Liters of solution
Given: # of moles of solute = 10.0 moles. Liters of solution = 5.00 liters
10.0 moles of KBr
Molarity = -------------------------- = 2.00 M
5.00 Liters of solution
39
Answer = 2.00 M
Example 2. A 250 ml solution is made with 0.50 moles of NaCl. What is the Molarity of
the solution?
Solution: In this case we are given ml, while the formula calls for L. We must change
the ml to Liters as shown below:
1 liter
250 ml x
--------
= 0.25 liters
1000 ml
A 250 ml (0.25 L) solution is made with 0.50 moles of NaCl.
Molarity?
Now, we solve the problem as we solved example 1.
# of moles of solute
Molarity = ---------------------Liters of solution
Given: Number of moles of solute = 0.50 moles of NaCl. Liters of solution = 0.25 L of
solution
0.50 moles of NaCl
Molarity = --------------------- = 2.0 M solution
0.25 L
Answer = 2.0 M solution of NaCl
II. Basic molarity problems where volume is the unknown.
This is similar to when we studied density, we have a formula with three possible
unknowns. When the molarity of the solution and the number of moles of solute are
given, but the volume is unknown, we must adjust our original formula to isolate the
unknown variable. Observe:
# of moles of solute
Molarity = ---------------------Liters of solution
40
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
Molarity x Liters of solution = # of moles of solute
----------
--------------------
Molarity
Molarity
# of moles of solute
Liters of solution = -------------------Molarity
Example 1. What would be the volume of a 2.00 M solution made with 6.00 moles of
LiF?
Solution:
# of moles of solute
Liters of solution = -------------------Molarity
Given: # of moles of solute = 6.00 moles. Molarity = 2.00 M (moles/L)
6.00 moles
Liters of solution = ----------2.00 moles/L
Answer = 3.00 L of solution
Now, you must also be prepared for the fact that the number of moles is not always given
to you. Sometimes you will be given the mass of the solute and you will need to
determine the number of moles by dividing the mass given by the Molar mass of the
solute. In these cases, use the formula below.
41
mass given
# of moles = ----------------Molar mass
Example 2. What is the volume of 3.0 M solution of NaCl made with 526g of solute?
Solution:
First find the molar mass of NaCl.
Na = 23.0 g x 1 ion per formula unit = 23.0 g
Cl = 35.5 g x 1 ion per formula unit = 35.5 g
---------58.5 g
Now find out how many moles of NaCl you have:
mass of sample
# of moles = ----------------Molar mass
Given: mass of sample = 526. Molar mass = 58.5 g
526 g
# of moles of NaCl = -----------58.5 g
Answer: # of moles of NaCl = 8.99 moles
Finally, go back to your molarity formula to solve the problem:
# of moles of solute
Liters of solution = -------------------Molarity
Given: # of moles of solute = 8.99 moles. Molarity of the solution = 3.0 M (moles/L)
42
8.99 moles
# of Liters of solution = ------------3.0 moles/L
Final Answer = 3.0 L
III. Basic molarity problems where the number of moles is the unknown.
Of course, the total number of moles used in the creation of a solution might be
unknown to you. However, given the molarity and the volume of the solution, you can
determine the number of moles of solute. Observe:
# of moles of solute
Molarity = ---------------------Liters of solution
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
# of moles of solute = Molarity x Liters of solution.
Example 1. How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3 of
a 5.0M solution?
Notice that the volume is given in cm3. Since there are 1000 cm3 in 1 liter, 500 cm3 must
be equal to 0.500 liters. Make that change right in the problem.
Example 1. How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3
(0.500 L) of a 5.0M solution?
Now you are ready to solve.
Solution:
# of moles of solute = Molarity x Liters of solution.
Given: Molarity = 5.0 M (moles/L). Volume = 0.500 L
# of moles of CaCl2 = 5.0 moles/L x 0.500 moles
Answer = 2.5 moles of CaCl2
43
Notice that the identity of the solute does not work into the math of the problem.
However, if the wording was different, it would. Observe example # 2.
Example 2. How many grams of CaCl2 would be used in the making of 5.00 x 102 cm3 of
a 5.0M solution?
In this case, what they are looking for is different. You could start to solve this problem
the same way you did example 1, but the end would require you to change the number of
moles of CaCl2 to the mass of CaCl2. You would use the formula below.
mass of sample
# of moles = ----------------Molar mass
mass of sample
# of moles x Molar mass = ----------------- x Molar mass
Molar mass
mass of sample = # moles of solute x Molar mass
Given: # of moles of solute = 2.5 moles (from our answer to example 1.)
Molar mass of solute (CaCl2) = 111 g/mole (from the periodic table)
Mass of CaCl2 = 2.5 moles x 111 g/mole
Answer: Mass of CaCl2 = 280 g (when rounded correctly).
1.8.
ACIDS AND BASES
By the 1884 definition of Arrhenius, an acid is a material that can release a proton or
hydrogen ion (H+). Hydrogen chloride in water solution ionizes and becomes hydrogen
ions and chloride ions. If that is the case, a base, or alkali, is a material that can donate a
hydroxide ion (OH-). Sodium hydroxide in water solution becomes sodium ions and
hydroxide ions. By the definition of both Lowry and Brønsted working independently in
1923, an acid is a material that donates a proton and a base is a material that can accept a
proton. We are going to use the Arrhenius definitions most of the time. The LowryBrønsted definition is broader, including some ideas that might not initially seem to be
44
acid and base types of interaction. Every ion dissociation that involves a hydrogen or
hydroxide ion could be considered an acid- base reaction. Just as with the Arrhenius
definition, all the familiar materials we call acids are also acids in the Lowry- Brønsted
model. Lewis idea of acids and bases is broader than the Lowry- Brønsted model. The
Lewis definitions are: Acids are elecron pair acceptors and bases are electron pair donors.
We can consider the same idea in the Lowry- Brønsted fashion. Each ionizable pair has a
proton donor and a proton acceptor. Acids are paired with bases. One can accept a proton
and the other can donate a proton. Each acid has a proton available (an ionizable
hydrogen) and another part, called the conjugate base. When the acid ionizes, the
hydrogen ion is the acid and the rest of the original acid is the conjugate base. Nitric acid,
HNO3, dissociates (splits) into a hydrogen ion and a nitrate ion. The hydrogen almost
immediately joins to a water molecule to make a hydronium ion. The nitrate ion is the
conjugate base of the hydrogen ion. In the second part of the reaction, water is a base
(because it can accept a proton) and the hydronium ion is its conjugate base.
HNO3
+
H2O
BASE
NO3CONJUGATE
BASE
+
H3O+
CONJUGATE
ACID
In a way, there is no such thing as a hydrogen ion or proton without anything else. They
just do not exist naked like that in water solution. Remember that water is a very polar
material. There is a strong partial negative charge on the side of the oxygen atom and a
strong partial negative charge on the hydrogen side. Any loose hydrogen ion, having a
positive charge, would quickly find itself near one of the oxygens of a water molecule. At
close range from the charge attraction, the hydrogen ion would find a pair (its choice of
two pairs) of unshared electrons around the oxygen that would be capable of filling its
outer shell. Each hydrogen ion unites with a water molecule to produce a hydronium ion,
H3O+, the real species that acts as acid. The hydroxide ion in solution does not combine
with a water molecule in any similar fashion. As we write reactions of acids and bases, it
45
is usually most convenient to ignore the hydronium ion in favor of writing just a
hydrogen ion.
1.8.1. Properties of Acids
For the properties of acids and bases, we will use the Arrhenius definitions.
 Acids release a hydrogen ion into water (aqueous) solution.
 Neutralize bases in a neutralization reaction.
 An acid and a base combine to make a salt and water. A salt is any ionic compound
that could be made with the anion of an acid and the cation of a base. The hydrogen
ion of the acid and the hydroxide ion of the base unite to form water.
 Corrode active metals. Even gold, the least active metal is attacked by an acid. When
an acid reacts with a metal, it produces a compound with the cation of the metal and
the anion of the acid and hydrogen gas.
 Turns blue litmus to red. Litmus is one of a large number of organic compounds that
change colors when a solution changes acidity at a particular point. Litmus is the
oldest known pH indicator. It is red in acid and blue in base. Litmus does not change
color exactly at the neutral point between acid and base, but very close to it. Litmus is
often impregnated onto paper to make 'litmus paper.'
 Taste sour. Stomach acid is hydrochloric acid. Although tasting stomach acid is not
pleasant, it has the sour taste of acid. Acetic acid is the acid ingredient in vinegar.
Citrus fruits such as lemons, grapefruit, oranges, and limes have citric acid in the
juice. Sour milk, sour cream, yogurt, and cottage cheese have lactic acid from the
fermentation of the sugar lactose.
46
1.8.2. Properties of Bases
 Release a hydroxide ion into water solution. (Or, in the Lowry or Brønsted model,
cause a hydroxide ion to be released into water solution by accepting a hydrogen ion
in water.)
 Neutralize acids in a neutralization reaction. The word reaction is: Acid plus base
makes water plus a salt. Symbolically, where 'Y' is the anion of acid 'HY,' and 'X' is
the cation of base 'XOH,' and 'XY' is the salt in the product, the reaction is:
HY + XOH
HOH + XY
 Denature protein. This accounts for the "slippery" feeling on hands when exposed to
base. Strong bases that dissolve in water well, such as sodium or potassium lye are
very dangerous because a great amount of the structural material of human beings is
made of protein. Serious damage to flesh can be avoided by careful use of strong
bases.
 Red litmus to blue. This is not to say that litmus is the only acid- base indicator, but
that it is likely the oldest one.
 Taste bitter. There are very few food materials that are alkaline, but those that are
taste bitter.
1.8.3. Strong Acids and Strong Bases
The common acids that are almost
HNO3
HCl
-
HBr
hydrochloric
-
HClO4
hundred
nitric
-
H2SO4
one
-
percent
ionized
are
acid
acid
sulfuric
acid
perchloric
acid
hydrobromic
acid
HI - hydroiodic acid
The acids on this short list are called strong acids, because the amount of acid quality of
a solution depends upon the concentration of ionized hydrogens. You are not likely to
47
see much HBr or HI in the lab because they are expensive. You are not likely to see
perchloric acid because it can explode if not treated carefully. Other acids are
incompletely ionized, existing mostly as the unionized form. Incompletely ionized acids
are called weak acids, because there is a smaller concentration of ionized hydrogens
available in the solution. Do not confuse this terminology with the concentration of acids.
The differences in concentration of the entire acid will be termed dilute or concentrated.
Muriatic acid is the name given to an industrial grade of hydrochloric acid that is often
used in the finishing of concrete.
In the list of strong acids, sulfuric acid is the only one that is diprotic, because it has two
ionizable hydrogens per formula (or two mols of ionizable hydrogen per mol of acid).
(Sulfuric acid ionizes in two steps. The first time a hydrogen ion splits off of the sulfuric
acid, it acts like a strong acid. The second time a hydrogen splits away from the sulfate
ion, it acts like a weak acid.) The other acids in the list are monoprotic, having only one
ionizable proton per formula. Phosphoric acid, H3PO4, is a weak acid. Phosphoric acid
has three hydrogen ions available to ionize and lose as a proton, and so phosphoric acid is
triprotic. We call any acid with two or more ionizable hydrogens polyprotic.
Likewise, there is a short list of strong bases, ones that completely ionize into hydroxide
ions and a conjugate acid. All of the bases of Group I and Group II metals except for
beryllium are strong bases. Lithium, rubidium and cesium hydroxides are not often used
in the lab because they are expensive. The bases of Group II metals, magnesium, calcium,
barium, and strontium are strong, but all of these bases have somewhat limited solubility.
Magnesium hydroxide has a particularly small solubility. Potassium and sodium
hydroxides both have the common name of lye. Soda lye (NaOH) and potash lye (KOH)
are common names to distinguish the two compounds.
LiOH - lithium hydroxide
NaOH - sodium hydroxide
KOH - potassium hydroxide
RbOH - rubidium hydroxide
CsOH - cesium hydroxide
48
Mg(OH)2 - magnesium hydroxide
Ca(OH)2
- calcium hydroxide
Sr(OH)2
- strontium hydroxide
Ba(OH)2
- barium hydroxide
The bases of Group I metals are all monobasic. The bases of Group II metals are all
dibasic. Aluminum hydroxide is tribasic. Any material with two or more ionizable
hydroxyl groups would be called polybasic. Most of the alkaline organic compounds (and
some inorganic materials) have an amino group (-NH2) rather than an ionizable hydroxyl
group. The amino group attracts a proton (hydrogen ion) to become (-NH3)+. By the
Lowry- Brønsted definition, an amino group definitely acts as a base, and the effect of
removing hydrogen ions from water molecules is the same as adding hydroxide ions to
the solution. Memorize the strong acids and strong bases. Other acids or bases are weak.
1.8.4. The definition of pH and pOH
Every water solution contains H+ ions. Their concentration is one of the most important
parameters describing solution properties. Concentrations of H+ can change in a very
wide range, it can be 10 M as well as 10-15 M. Such numbers are inconvenient to use so
to simplify things the pH scale was introduced. The pH is the negative of logarithm base
10 of [H+]:
It is much easier to use pH definition and to say "pH of the solution is 4.1" than to use
concentrations - as in "H+ concentration is 0.000079M". Idea of using letter p to denote
concentrations and numbers that can vary by several orders of magnitude was widely
accepted and is used not only in pH definition, but for example also for displaying
dissociation constants values in tables. Not only H+ ions are present in every water
solution. Also OH- ions are always present, and their concentration can change in the
same very wide range. Thus it is also convenient to use similar definition to describe
[OH-]
49
Note: In real solutions not concentrations, but ion activities should be used for
calculations. Especially pH definition uses not minus logarithm of concentration, but
minus logarithm of activity. In diluted solutions activity is for all practical purposes
identical to concentration, but when the concentration increases, so does the activity. As
a rule of thumb if the concentration of charged ions present in the solution is below
0.001 M you don't have to be concerned about activities and you can use classic pH
definition.
The pH scale
The concentration of H+ ions impacts most of the chemical reactions. Depending on their
concentration hydrogen peroxide can behave as oxidizing or reducing agent. Pepsine one of the enzymes used for digestion - works best in strongly acidic conditions and is
inactive in neutral solutions. There are flowers which are either pink or blue, depending
on the acidity of the soil they grow in. Even tea changes its color when you add a slice of
lemon. Acidity of the solution is so important, that it was convenient to create a special
pH scale for its measurements. This pH scale uses pH definition. Concentration of H+ is
usually confined to 1-10-14M range. Thus pH scale contains values falling between 0 and
14. In some rare cases you may see pH lower than 0 or higher than 14, when the
concentration of H+ takes some extreme values.
On the pH scale pure water has pH 7, although you will probably never see water pure
enough for such pH. Air always contains small amounts of carbon dioxide which
dissolves in water making it slightly acidic - with pH of about 5.7. All values on the pH
scale lower than 7 denote solutions that are acidic - the lower the pH, the more acidic
solution. On the contrary solutions with pH above 7 are basic - the higher the pH the
more basic solution is. There are two things worth of remembering about pH scale. First,
as pH scale is logarithmic, 1 unit pH change means tenfold change in the H+ ion
concentration. Second, while only solution with pH=7.00 is strictly neutral, all solutions
with pH in the range 4-10 have real concentration of H+ and OH- lower than 10-4M which can be easily disturbed with small additions of acid and base.
50
The pH scale as described above is called sometimes "concentration pH scale" as
opposed to the "thermodynamic pH scale". Main difference between the two scales is that
in thermodynamic pH scale we are interested not in H+ concentration, but in H+ activity.
In fact what we measure in the solution - for example using pH electrodes - is just
activity, not the concentration. Thus it is thermodynamic pH scale that describes real
solutions, not the concentration one.
1.8.5. Water Ion Product
Not only acids and bases dissociate, water dissociates too:
And the equilibrium of this reaction is described by the equation
However, in practice we use simplified version of this equation, called water ionization
constant (sometimes called also water ion product)
……………..1
The reason for writing the equation in this format is that as long as we are talking about
diluted solutions we may safely assume water concentration is constant, and error
introduced into our calculations will be rarely higher then precision of dissociation
constants used in calculations. K given by the equation 1 has value 1.8 × 10-16. We know
there is 1000g of water per liter and it has molar mass of 18g, thus water concentration
[H2O] = 55.56M. If we assume water concentration doesn't change we can rewrite
equation 1 as
………..2
This new Kw constant will have value of 1.8 × 10-16 × 55.56 = 10-14. That's why Kw is
sometimes listed as 10-14 (pKw=14) and sometimes as 1.8 × 10-16 (pKw=15.7).
It is commonly assumed that neutral solution pH is 7.00. In most cases that's a very good
approximation, but - as the water dissociation constant varies with temperature - it is true
only for 25 °C. Neutral solution is defined as the one in which concentration of H+ is
51
identical with concentration of OH- and its pH may vary from 7.47 at 0 °C
to 6.14 at 100 °C.
1.8.6. Acid/Base Equilibrium
Most solutions containing different ions are in state of equilibrium - all concentrations are
constant and not changing in time. This equilibrium is dynamic which means we have
forward and reverse reactions going in the solution at the same rate, so that
concentrations are not changing. We are going to concentrate our attention on the
methods of finding the equilibrium in the solution being mixture of water, acid and base.
Finding such equilibrium is always a part of pH calculation - even if we are often
concentrated only on pH calculation, once we know pH we are usually able to calculate
concentrations of all other ions. In fact we could as well concentrate our efforts on any
other ion, but pH scale is very universal and knowing pH we can often tell a lot about the
solution.
Let's assume we have acid HA, base BOH, and their analytical concentrations (i.e.
concentrations off all forms present in solution) are respectively Ca and Cb.
For the acid dissociation reaction
Equilibrium is described by the acid dissociation constant defined as
For the base dissociation reaction
Base dissociation constant is defined as
It is commonly believed that strong acids and strong bases are fully dissociated. For most
practical purposes this is true, but they have their dissociation constants as well and in
some cases effects of partial dissociation can be observed.
Finally we will often need water ionization constant:
52
……………………………………………1
pH and pOH defined in the above section can be used not only as measure of the ions
concentration, but also in calculations. Let's take logarithm of both sides of equation (1):
……………………..3
Changing the signs and using the p notation gives:
……………………………………………4
Equation 4 is often used when we need to calculate pH but it is much easier to calculate
pOH and vice versa.
1.8.7. Bronsted-Lowry's acids and bases
As all reactions we are interested in take place in water and water dissociates itself into
H+ and OH- ions, classic definition of acid as a substance that dissociates producing H+
ions becomes a little bit problematic. Consider the solution of salt of weak base BOH.
Such solution contains B+ ions that are between products of BOH dissociation:
with equilibrium described by the already mentioned in the previous section base
dissociation constant:
……………………5
For the equilibrium BOH molecules are needed. As there are already OH - ions from the
water dissociation present in the solution, they will react with B+. This will lower OHconcentration, forcing water to dissociate further. Final solution in equilibrium will
contain some BOH molecules and some excess of H+ - so it will be acidic, even if we
haven't add any acid! Seems that B+ is an acid - even if it doesn't dissociate to give H+
ions.
To overcome this inconsistency Bronsted and Lowry proposed independently in 1923
new definitions of acid and base: acid is a substance that can donate the proton and base
is a substance than can accept the proton. The most important outcome of this definition
53
is the fact that every acid loosing its proton becomes a Bronsted-Lowry base (as it has a
free "slot" for the proton) and that every base when protonated becomes a BronstedLowry acid (it has a proton that is can release). These pairs of acid and base are called
conjugated. In other words every acid loosing proton becomes its conjugated base, and
every protonated base becomes its conjugated acid.
This approach has some interesting implications. Let's take a reaction of conjugated base
A- with water:
Its equilibrium constant is
………………6
(water concentration is assumed to be constant). Multiplying this equation by the
equation for acid dissociation constant we get
…….7
[A-] and [HA] cancel out leaving
…………….8
or
……………………..9
The most important lesson is that to describe acid/base properties of substance in the
solution (not necessarily water solution) we can use either Ka or Kb value. Sometimes it is
more convenient to use Kb for calculations, but whenever selection of constant doesn't
matter we will concentrate our efforts around Ka.
1.9.
BUFFER SOLUTION
Buffer solutions are solutions which resist change in hydrogen ion and the hydroxide ion
concentration (and consequently pH) upon addition of small amounts of acid or base, or
upon dilution. Buffer solutions consist of a weak acid and its conjugate base (more
54
common) or a weak base and its conjugate acid (less common). The resistive action is the
result of the equilibrium between the weak acid (HA) and its conjugate base (A−):
HA(aq) + H2O(l) → H3O+(aq) + A−(aq)
Any alkali added to the solution is consumed by hydrogen ions. These ions are mostly
regenerated as the equilibrium moves to the right and some of the acid dissociates into
hydrogen ions and the conjugate base. If a strong acid is added, the conjugate base is
protonated, and the pH is almost entirely restored. This is an example of Le Chatelier's
principle and the common ion effect. This contrasts with solutions of strong acids or
strong bases, where any additional strong acid or base can greatly change the pH.
When writing about buffer systems they can be represented as salt of conjugate base/acid,
or base/salt of conjugate acid. It should be noted that here buffer solutions are presented
in terms of the Brønsted-Lowry notion of acids and bases, as opposed to the Lewis acidbase theory. Omitted here are buffer solutions prepared with solvents other than water.
1.9.1. Calculating pH of a buffer
The equilibrium above has the following acid dissociation constant:
Simple manipulation with logarithms gives the Henderson-Hasselbalch equation, which
describes pH in terms of pKa:
In this equation
1. [A−] is the concentration of the conjugate base. This may be considered as
coming completely from the salt, since the acid supplies relatively few anions
compared to the salt.
2. [HA] is the concentration of the acid. This may be considered as coming
completely from the acid, since the salt supplies relatively few complete acid
molecules (A − may extract H + from water to become HA) compared to the
added acid.
55
Maximum buffering capacity is found when pH = pKa, and buffer range is considered to
be at a pH = pKa ± 1.
1.9.2. Illustration of buffering effect: Sodium acetate/acetic acid
The acid dissociation constant for acetic acid-sodium acetate is given by the equation:
Since this equilibrium only involves a weak acid and base, it can be assumed that
ionization of the acetic acid and hydrolysis of the acetate ions are negligible. In a buffer
consisting of equal amounts of acetic acid and sodium acetate, the equilibrium equation
simplifies to
Ka = [H +],
and the pH of the buffer as is equal to the pKa.
To determine the effect of addition of a strong acid such as HCl, the following
mathematics would provide the new pH. Since HCl is a strong acid, it is completely
ionized in solution. This increases the concentration of H+ in solution, which then
neutralizes the acetate by the following equation.
The consumed hydrogen ions change the effective number of moles of acetic acid and
acetate ions:
After accounting for volume change to determine concentrations, the new pH could be
calculated from the Henderson-Hasselbalch equation. Any neutralization will result in a
small change in pH, since it is on a logarithmic scale...
1.9.3. Applications
Their resistance to changes in pH makes buffer solutions very useful for chemical
manufacturing and essential for many biochemical processes. The ideal buffer for a
particular pH has a pKa equal to the pH desired, since a solution of this buffer would
56
contain equal amounts of acid and base and be in the middle of the range of buffering
capacity.
Buffer solutions are necessary to keep the correct pH for enzymes in many organisms to
work. Many enzymes work only under very precise conditions; if the pH strays too far
out of the margin, the enzymes slow or stop working and can denature, thus permanently
disabling its catalytic activity. A buffer of carbonic acid (H2CO3) and bicarbonate
(HCO3−) is present in blood plasma, to maintain a pH between 7.35 and 7.45.
Industrially, buffer solutions are used in fermentation processes and in setting the correct
conditions for dyes used in coloring fabrics. They are also used in chemical analysis and
calibration of pH meters.
57
CHAPTER TWO
PHYSICAL CHEMISTRY
2.1.
REAL GASES and IDEAL GASES
Experiments have shown that the ideal gas law describes the behavior of a real gas quite
well at moderate pressures and low temperatures but not so well at high pressures and
low temperatures. You can see why this is so from the postulates of kinetic theory, from
which the ideal gas law can be derived. According to postulate 1, the volume of space
actually occupied by molecules is small compared with the total volume that they occupy
as a result of their motion through space. Further according to postulate 3, the molecules
are sufficiently far apart on the average, so the intermolecular forces can be disregarded.
Both postulates are satisfied by a real gas when its density is low. The molecules are far
enough apart to justify treating them as point particles with negligible intermolecular
forces (such forces diminish as the molecules become further separated). But postulates 1
and 3 apply less accurately as the density of a gas increases. As the molecules become
more densely packed, the space occupied by molecules is no longer negligible compared
with the total volume of gas. Moreover, the intermolecular forces become stronger as the
molecules come closer together.
When the ideal gas law is not sufficiently accurate for our purposes, we must use another
of the many such equations available. One alternative is the van der Waals equation,
which is an equation that relates P, T, V and n for non ideal gases (but is not valid for
very high pressures).

n 2a 
 P  2 V  nb   nRT
V 

here a and b are constants that must be determined for each kind of gas.
The equation above can be obtained from the ideal gas law, PV = nRT, by replacing P by
P + n2a/V2 and V by V – nb. To explain the change made to the volume term, note that
for the ideal gas law you find that the molecules have negligible volume. Thus, any
particular molecule can move through out the entire volume V of the container. But if the
58
molecules do occupy a small but finite amount of space, the volume through which any
particular molecule can move is reduced. Thus you subtract nb from V. the constant b is
essentially the volume occupied by a mole of molecules. To explain the change made in
the pressure term, you need to consider the effect of intermolecular forces on the
pressure. Such forces attract molecules to one another. A molecule about to collide with
the wall is attracted by other molecules, and this reduces its impact with the wall.
Therefore, the actual pressure is less than that predicted by the ideal gas law.
You can obtain this pressure correction by noting that the total force of attraction on any
molecule about to hit the wall is proportional to the concentration of molecules, n/V. if
this concentration is doubled, the total force on any molecule about to hit the wall is
doubled. However, the number of molecules about to hit the wall per unit wall area is
also proportional to concentration, n/V, so that the force per unit wall area, or pressure, is
reduced by a factor proportional to n2/V2. You can write this correction factor as an2/V2,
where a is a proportionality constant.
If you write the ideal gas equation corrected for the volume of molecules, you obtain for
the pressure
P
nRT
V  nb
Now if the pressure is reduced by the correction term an2/V2 to account for intermolecular
forces, you obtain
nRT
an 2
P

V  nb V 2
You can rearrange this to give the form of the van der Waals equation that we wrote
earlier.
59
2.2.
KINETIC MOLECULAR THEORY
According to this theory a gas consists of molecules in constant random motion. The
word kinetic describes something in motion. Thus, kinetic energy, Ek, is the energy
associated with the motion of an object of mass m. From physics
Ek 
1
2
m  speed 
2
2.2.1. Postulates of Kinetic Theory
Physical theories are often given in terms of postulates. These are the basic statements
from which all conclusions or predictions of a theory are deduced. The kinetic theory of
an ideal gas (a gas that follows the ideal gas law PV = nRT) is based on five postulates.
POSTULATE 1.
Gases are composed of molecules size is negligible compared with
the average distance between them. Most of the volume occupied by a gas is empty
space. This means that you can usually ignore the volume occupied by the molecules.
POSTULATE 2.
Molecules move randomly in straight lines in all directions and at
various speeds. This means that properties of a gas that depend on the motion of the
molecules, such as pressure, will be the same in all directions.
POSTULATE 3.
The forces of attraction or repulsion between two molecules
(intermolecular forces) in a gas are very weak or negligible, except when they collide.
This means that a molecule will continue moving in a straight line with undiminished
speed until it collides with another gas molecule or with the walls of the container.
POSTULATE 4.
When molecules collide with one another, the collisions are elastic. In
an elastic collision, the kinetic energy remains constant; no kinetic energy is lost. To
understand the difference between elastic and an inelastic collision, compare the collision
of two hard steel spheres with the collision of two masses of putty. The collision of steel
spheres is nearly elastic (that is, the spheres bounce off each other and continue moving),
but that of putty is not. Postulate 4 says that unless the kinetic energy of molecules is
removed from a gas- for example, as heat- the molecules will forever move with the same
average kinetic energy per molecule.
POSTULATE 5.
The average kinetic energy of a molecule is proportional to the
absolute temperature. This postulate establishes what we mean by temperature from a
60
molecular point of view: the higher the temperature, the greater the molecular kinetic
energy.
2.2.2. The Ideal Gas Law from Kinetic Theory
One of the most important features of kinetic theory is its explanation of the ideal gas
law. To show how you can get the ideal gas law from kinetic theory, we will first find an
expression for the pressure of a gas. According to kinetic theory, the pressure of a gas, P,
will be proportional to the frequency of molecular collisions with a surface and to the
average force exerted by a molecule in collision.
P  frequency of collision x average force
The average force exerted by a molecule during a collision depends on its mass m and its
average speed u – that is, on its average momentum mu. In other words, the greater the
mass of the molecules and the faster it is moving, the greater the force exerted during
collision. The frequency of collisions is also proportional to the average speed u, because
the faster a molecule is moving; the more often it strikes the container walls. Finally, the
frequency of collisions is proportional to the number of molecules N in the gas volume.
Putting these factors together gives
1


p   u   N   mu
 V

Bringing the volume to the left side you get
PV  Nmu2
Because the average kinetic energy of a molecule of mass m and the average speed u is
½mu2, PV is proportional to the average kinetic energy of a molecule. Moreover, the
average kinetic energy is proportional to the absolute temperature (postulate 5). Noting
that the number of molecules, N, is proportional to the moles of molecules, n, you have
PV  nT
You can write this as an equation by inserting a constant of proportionality, R, which you
can identify as the molar gas constant
PV = n RT
61
2.3.
HENRY’S LAW
The effect of pressure on the solubility of a gas in a liquid can be predicted quantitatively.
According to Henry’s law, the solubility of a gas is directly proportional to the partial
pressure of the gas above the solution. Expressed mathematically, the law is
S  kH P
Where S is the solubility of the gas (expressed as mass of solute per unit volume of
solvent), kH is Henry’s law constant for the gas for a particular liquid at a given
temperature, and P is the partial pressure of the gas.
2.4.
RAOULT’S LAW
The partial pressure of solvent, PA, over a solution equals the partial pressure of the pure
solvent, PA times the mole fraction of solvent, XA, in the solution.
PA 
P

A
XA
If the solute is nonvolatile, PA is the total vapour pressure of the solution. Because of the
mole fraction of solvent in a solution is always less than 1, the vapour pressure of the
solution of a nonvolatile solute is less than that for the pure solvent; the vapor pressure is
lowered. In general, Raoult’s law may hold for all mole fractions. You can obtain an
explicit expression for the vapor pressure lowering of a solvent in a solution assuming
Raoult’s law holds and that the solute is a nonvolatile nonelectrolyte. The vapor-pressure
lowering, ∆P is

P  P A  PA
Substituting Raoult’s law gives



P  P A  P A X A  P A 1  X A 
But the sum of the mole fractions of the components of a solution must be equal to 1; that
is, XA + XB = 1. So XB = 1 - XA. Therefore,

P  P A X B
62
From this equation you can see that the vapor-pressure lowering is a colligative propertyone that does not depend on the concentration, but on the nature, of the solute.
Actually, very few mixtures really obey Raoult’s law very closely over wide ranges of
compositions. Benzene and carbon tetrachloride, a pair of substances that do form such
mixtures, are said to yield ideal solutions. Mixtures that deviate from Raoult’s law are
called non-ideal. When the vapor pressure of a mixture is greater than the predicted, it is
said to exhibit a positive deviation from Raoult’s law; conversely, when a solution gives
a lower vapor pressure than we would expect from Raoult’s law, it is said to show a
negative deviation.
The origin of non-ideal behavior lies in the relative strengths of the interactions between
molecules of the solute and solvent. When the attractive forces between the solute and the
solvent molecules are weaker than those between solute molecules or between solvent
molecules, neither the solute nor solvent particles are held as tightly in the solution as
they are in the pure substances. The escaping tendency of each is therefore greater in the
solution than in the solute or solvent alone. As a result, the partial pressures of both of
them over the solution are greater than predicted by Raoult’s law, and the solution
exhibits a large vapour pressure than expected.
Just the opposite effect is produced when the solute-solvent interactions are stronger than
the solute-solute or solvent-solvent interactions. Each substance, in the presence of the
other, is held more tightly than in the pure materials, and their partial pressures over a
solution are therefore less than Raoult’s law would predict. The result is that such a
solution exhibits a negative deviation from ideality.
Since, in a solution that shows positive deviations from ideal behavior, the forces of
attraction between solute and solvent are weaker than those between both solute
molecules and solvent molecules, the formation of these solutions occurs with the
absorption of energy. Conversely, of course, mixtures that exhibit negative deviations
from Raoult’s law are formed with evolution of heat.
63
Summary of solution properties
Temperature
Relative
change when
Deviation
Attractive
solution is
from Raoult’s
forces
∆Hsoln
formed
law
Example
A-A, B-B = A-B
Zero
None
None (ideal
Benzene-
solution)
chloroform
Negative
Acetone-
A-A, B-B < A-B
Negative
Increase
(exothermic)
A-A, B-B > A-B
Positive
water
Decrease
(endothermic)
Positive
Ethanolhexane
64
2.5.
CHEMICAL THERMODYNAMICS
2.5.1. Internal Energy
This is the total energy of the system-the total of all the energies that it possesses as a
consequence of the kinetic energy of its atoms, ions and molecules, plus all the potential
energy that arises from the binding forces between the particles making up the system.
A change in the internal energy is defined as,
∆E = Efinal - Einitial
When a system changes from one state to another, there is two ways for it to exchange
energy with its surroundings. One is for it to gain or lose heat energy. If the system
absorbs heat, its energy rises, and if it loses heat, its energy drops. The second way for
the system to exchange energy with its surroundings is to do work or have work done on
it. If the system does work, its energy drops. On the other hand if work is done on the
system, its energy rises.
The energy bookkeeping for both heat and work is taken care of by the equation
∆E = q – w
Where the symbol ∆ means “change in”, q is defined as the heat absorbed from the
surroundings by a system when it undergoes a change, and w is defined as the work done
by the system on its surroundings. This simply states that the change in the internal
energy is equal to the difference between the amount of energy gained by the system in
the form of heat and the amount of energy removed in the form of work that is performed
on the surroundings. Since the equation above deals with the transfer of amounts of
energy, it is necessary to establish sign conventions to avoid confusion in our
bookkeeping. Heat added to a system and work done by a system are considered positive
quantities.
Thus if a certain change is accompanied by the absorption of 50 J of heat and the
expenditure of 30 J of work, q = + 50 J and w = + 30 J. the change in internal energy of
the system is
65
∆Esystem = (+ 50 J) – (+ 30 J)
or
∆Esystem = + 20 J
Thus the system has undergone a net increase in energy amounting to + 20 J.
When the system gains 50 J, the surroundings lose 50 J; therefore q = -50 J for the
surroundings. When the system performs work, it does so on the surrounding. We say
that the surroundings have done negative work, and w = -30 J for the surroundings. The
change in the internal energy for the surroundings is thus
∆Esorroundings = (- 50 J) – (- 30 J)
∆Esorroundings = -20 J
The change in internal energy of the system is thus equal but opposite in sign, to ∆E for
the surroundings. This has to be so in order to satisfy the law of conservation of energy.
In summary,
q positive (q > 0); heat is added to the system.
q negative (q < 0); heat is evolved by (removed from) the system.
w positive (w > 0); the system performs work – energy is removed.
w negative (w < 0); work is done on the system – energy is added.
The internal energy is a state function. A state function is a property of a system that
depends only on its present state, which is determined by variables such as temperature
and pressure, and is independent of any previous history of the system. The magnitude of
∆E therefore depends only on the initial and final states of the system and not the path
taken between them. The sign and the magnitude of ∆E are controlled by the values of E
and the initial and final states. For any given change, ∆E, there are many different paths
that can be followed with their own characteristic values of q and w. however for the
same initial and final states, the difference between q and w is always the same.
Therefore, even though ∆E is a state function, q and w are not.
2.5.2. Enthalpy and Enthalpy Change
The heat absorbed or evolved by a reaction depends on the conditions under which the
reaction occurs. Usually, a reaction takes place in a vessel open to the atmosphere, where
66
it occurs at the constant pressure of the atmosphere. We will assume that this is the case
and write the heat of reaction as qp, the subscript p indicating that the process occurs at
constant pressure. There is a property of substances called enthalpy that is related to the
heat of reaction qp. Enthalpy (denoted H) is an extensive property of a substance that can
be used to obtain the heat absorbed or evolved in a chemical reaction. (An extensive
property is a property that depends on the amount of substance).
Consider a chemical reaction system. At first, the enthalpy of the system is that of the
reactants. But as the reaction proceeds, the enthalpy changes and finally becomes equal to
that of the products. The change in enthalpy for a reaction at a given temperature and
pressure (called the enthalpy of reaction) is obtained by subtracting the enthalpy of the
reactants from the enthalpy of the products. The change in enthalpy ∆H can be thus
represented as
∆H = Hfinal – Hinitial
Since you start from reactants and end with products, the enthalpy of reaction is
∆H = H(products) – H(reactants)
Because H is a state function, the value of ∆H is independent of the details of the
reaction. The key relation is that between enthalpy change and heat of reaction:
H = E + PV
For a change at constant pressure:
∆H = ∆E + P ∆V
If only PV work is involved in the change, we know that
∆E = q – P ∆V
Substituting this in the above equation, we have
∆H = (q – P ∆V) + P ∆V
∆H = qp
The enthalpy of reaction equals the heat of reaction at constant pressure.
To illustrate this concept, consider the reaction at 25°C of sodium metal and water,
carried out in a beaker open to the atmosphere at 1.00 atm pressure.
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
67
The metal and water react vigorously and heat evolves. Experiment shows that 2 moles of
sodium metal react with 2 moles of water to evolve 367.5 kJ of heat. Because heat
evolves, the reaction is exothermic, and we write qp = -367.5 kJ. Therefore, the enthalpy
of reaction, or change of enthalpy for the reaction, is ∆H = -367.5 kJ.
2.5.3. Entropy and Entropy Change
The thermodynamic quantity that is related to the randomness or statistical probability of
a system is called the entropy and is given the symbol S. The more random a system is,
the greater its entropy. Several things influence the amount of entropy that a substance
has in a particular state. For example, if we compare a liquid with a solid, we know that
the particles in a solid are highly ordered, while those in a liquid are disordered.
Therefore, for a given substance at a particular temperature, its liquid state has higher
entropy than the solid. Similarly, if we compare a liquid with a gas, we see that the
molecules in a liquid are confined to one region of the container (the bottom), but a gas
can spread its molecules randomly throughout the entire vessel. Thus the gas has higher
entropy than the liquid. So for a given substance at a given temperature
Ssolid < Sliquid < Sgas
Entropy is a state function just like E and H, which means that the magnitude of a change
in entropy, ∆S, depends only on the entropies of the system in its initial and final states.
∆S = Sfinal – Sinitial
One way that a change in entropy can be brought about is by the addition of heat to a
system. The more heat added to the system, the greater the extent of disorder afterwards.
It should not be surprising, therefore, to find that the entropy change, ∆S, is directly
proportional to the amount of heat added to the system. This heat is specified as qrev-the
amount of heat that would be added to the system if the change follows a reversible
path.
∆S  qrev
68
The magnitude of ∆S is also inversely proportional to the temperature at which the heat is
added. At low temperatures a given quantity of heat makes large change in the relative
degree of order. Near absolute zero, the addition of even a small amount of heat causes
the system to go from essentially perfect order to some degree of randomness – a very
substantial change, so ∆S is large. If the same amount of is added to the system at higher
temperature, however, the system goes from an already highly random state to one just
slightly more random. This constitutes only a very small change in the relative degree of
disorder and, hence, only small entropy change. Thus it can be shown that a change in
entropy is finally given as
S 
q rev
T
where T is the absolute temperature at which qrev is transferred to the system. Note that
entropy has units of energy/temperature, for example, calories per Kelvin (ca/K) or joules
per Kelvin (J/K).
2.5.4. The Second Law of Thermodynamics
The second law of thermodynamics provides us with a way of comparing the effect of
the two driving forces involved in spontaneous process-changes in energy and changes in
entropy. One statement of the second law is that in any spontaneous process there is
always an increase in the entropy of the universe (∆Stotal > 0). This increase takes into
account entropy changes in both the system and its surroundings,
S total  S system  S sorroundings
The entropy change that occurs in the surroundings is brought about by the heat added to
the surroundings divided by the temperature at which it is transferred. For a process at
constant P and T, the heat added to the surroundings is equal to the negative of the heat
added to the system, which is given by ∆Hsystem.
Thus
q surroundings   H system
The entropy change for the surrounding is therefore
69
S surroundings 
 H system
T
and the total entropy change for the universe is
S total  S system 
H system
T
or
S total 
TS system  H system
T
This can be rearranged to give
TS total  H system  TS system 
Since ∆Stotal must be a positive number for a spontaneous change, the product T∆Stotal
must also be positive. This means that the quantity in parentheses on the right, (∆Hsystem –
T∆Ssystem), must be negative so that –(∆Hsystem – T∆Ssystem) may be positive. Thus, in
order for a spontaneous change to take place, the expression (∆Hsystem - T∆S) must be
negative.
At this point it is convenient to introduce another thermodynamic state function, G, called
the Gibbs free energy. This is defined as
G = H – TS
For a change at constant T and P, we write
∆G = ∆H - T∆S
From earlier arguments we see that ∆G must be less than zero for a spontaneous process;
that is, ∆G must have a negative value at constant T and P. The Gibbs free energy
change, ∆G, represents a composite of the two factors contributing to spontaneity, ∆H
and ∆S. For systems in which ∆H is negative (exothermic) and ∆S is positive (increased
disorder accompanying the change), both factors favor spontaneity and the process will
occur spontaneously at all temperatures. Conversely, if ∆H is positive (endothermic) and
∆S is negative (increase n order), ∆G will always be positive and the change cannot occur
spontaneously at any temperature.
70
In situations where ∆H and ∆S are both positive, and both negative, temperature plays the
determining role in controlling whether or not a change will take place. In the first case
(∆H and ∆S both are positive), ∆G will be negative only at high temperatures, where T∆S
is greater in magnitude than ∆H. Therefore, the reaction will be spontaneous only at
elevated temperatures. An example is the melting of ice, which is non-spontaneous at low
temperatures (below 0°C) and spontaneous at high temperatures (above 0°C).
When ∆H and ∆S are both negative, ∆G will be negative only at low temperatures. An
example of this is the freezing of water. We know that heat must be removed from the
liquid to produce ice, so the process is exothermic with a negative ∆H. freezing is also
accompanied by an ordering of the water molecules as they leave the random liquid state
and become part of the crystal. As a result, ∆S is also negative. The sign of ∆G is
determined both by ∆H, which in this case is negative, and T∆S, which is also negative.
To compute ∆G we must subtract a negative T∆S from a negative ∆H. The result will be
negative only at low temperature. Therefore, at 1 atm we observe H2O to freeze
spontaneously only below 0°C. Above 0°C the magnitude of T∆S is greater than ∆H, and
∆G becomes positive. As a result, freezing is no longer spontaneous. Instead, the reverse
process (melting) occurs.
The effect of the signs of ∆H and ∆S and the effect of the temperature on spontaneity can
be summarized as follows.
∆H
∆S
Outcome
(-)
(+)
Spontaneous at all temperatures
(+)
(-)
Nonspontaneous regardless of temperature
(+)
(+)
Spontaneous only at high temperature
(-)
(-)
Spontaneous only at low temperature
71
2.6.
CHEMICAL KINETICS
Chemical kinetics, also referred to as chemical dynamics, is concerned with the speed, or
rates, of chemical reactions. The factors that control how rapidly chemical changes occur
include the following;
1. The nature of the reactants and products. All other factors being equal, some
reactions are just naturally fast and others are naturally slow, depending on the
chemical makeup of the molecules or ions involved.
2. The concentrations of the reacting species. For two molecules to react with each
other they must meet, and the probability that this will happen in a homogeneous
mixture increases as their concentrations increase. For heterogeneous reactionsthose in which the reactants are in separate phases-the rate also depends on the
area of contact between the phases. Since many small particles have a much large
area than one large particle of the same total mass, decreasing the particle size
increases the reaction rate.
3. The effect of temperature. Nearly all chemical reactions take place faster when
their temperatures are increased.
4. The influence of outside agents called catalysts. The rates of many reactions are
affected by substances called catalysts that undergo no net chemical change
during the course of the reaction. Because it is not consumed by the reaction, it
does not appear in the balanced chemical equation (although its presence may be
indicated by writing its formula over the arrow). A pure solution of hydrogen
peroxide, H2O2, is stable but when hydrobomic acid, HBr (aq), is added, H2O2
decomposes rapidly into H2O and O2.
2H 2 O 2 (aq) HBr(aq)
 2 H 2 O(l )  O2 ( g )
Here HBr acts as a catalyst to speed decomposition.
2.6.1. Definition of a Reaction Rate
The rate of a reaction is the amount of product formed or the amount of reactant used up
per unit of time. So that a rate calculation does not depend on the total quantity of
reaction mixture used, you express the rate for a unit volume of the mixture. Therefore,
the reaction rate is the increase in molar concentration of product of a reaction per unit
72
time or the decrease in the molar concentration of reactant per unit time. The usual unit
of a reaction rate is moles per liter per second, mol L-1 s-1.
Consider the gas-phase reaction discussed below
2N2O5 (g) → 4NO2 (g) + O2 (g)
The molar concentration of a substance by enclosing the formula of the substance in
square brackets. Thus, [O2] is the molar concentration of O2. in a given time interval ∆t,
the molar concentration of oxygen, [O2], in the reaction vessel increases by the amount
∆[O2]. The symbol ∆ means “change in”; you obtain the change by subtracting the initial
value from the final value. The rate of the reaction is given by
Rate of formation of oxygen 
O2 
t
This equation gives the average rate over the time interval ∆t. if the time interval is very
short, the equation gives the instantaneous rate-that is, the rate at a particular instant of
time. Because the amounts of products and reactants are related by stoichiometry, any
substance in the reaction can be used to express the rate of reaction. Therefore for the
above reaction, we can express the rate in terms of the rate of decomposition of N2O5.
Rate of decomposition of N2O5 = -
N 2 O5 
t
Note the negative sign. It always occurs in a rate expression for a reactant in order to
indicate a decrease in concentration and to give a positive value for the rate. Thus,
because [N2O5] decreases, ∆[N2O5] is negative and -∆[N2O5]/ ∆t is positive.
The rate of decomposition of N2O5 and the rate of formation of oxygen are easily related.
Two moles of N2O5 decompose for each mole of oxygen formed, so the rate of
decomposition of N2O5 is twice the rate of formation of oxygen. To equate the rates, you
must divide the rate of decomposition of N2O5 by 2 (its coefficient in the balanced
chemical equation). Rate of formation of O2 = ½(Rate of decomposition of N2O5)
O2 
N 2 O5 
= -½
t
t
73
2.6.2. Dependence of rate on concentration
Experimentally, it has been found that a reaction rate depends on the concentration of
certain reactants as well as the concentration of catalyst, if there is one. A rate law is an
equation that relates the rate of a reaction to the concentrations of reactants (and
catalyst) raised to various powers. Consider the reaction
2NO2 (g) + F2 (g) → 2NO2F (g)
The rate law for the reaction is:
Rate = k[NO2][F2]
Note that in this rate law both reactant concentrations have an exponent of 1. Here k,
called the rate constant, is a proportionality constant in the relationship between rate and
concentrations. It has a fixed value at any given temperature, but it varies with
temperature. Whereas the units of rate are usually given as mol L-1 s-1, the units of k
depend on the form of the rate law. From the rate law above we have
k
rate
NO2 F2 
from which we get the following unit for k:
molL1 s 1
 lmol 1 s 1
2 2
mol L
As amore general example, consider the reaction of substances A and B to give D and E,
according to the balanced equation
C
aA + bB 

dD + eE
C = catalyst
You could write the rate law in the form
Rate = k[A]m[B]n[C]p
The exponents m, n, and p are frequently, but not always integers. They must be
determined experimentally and cannot be obtained simply by looking at he balanced
equation. For example, note that the exponents in the equation Rate = k[NO2][F2] have no
relationship to the coefficients in the balanced equation 2NO2 + F2 → 2NO2F.
Once you know the rate law for a reaction and have found the value of the rate constant,
you can calculate the rate of a reaction for any values of reactant concentrations.
74
2.6.3. Reaction Order
The reaction order with respect to a given reactant species equals the exponent of the
concentration of that species in the rate law, as determined experimentally. For the
reaction of NO2 with F2 to give NO2F, the reaction is first order with respect to NO2
because the exponent of NO2 in the rate law is 1. Similarly, the reaction is first order with
respect to F2. The overall order of a reaction equals the sum of the orders of the species in
the rate law. In this example, the overall order is 2; that is, the reaction is second order
overall.
2.6.4. Determining the rate law
The experimental determination of the rate law for a reaction requires that you find the
order of the reaction with respect to each reactant and any catalyst. The initial-rate
method is a simple way to obtain reaction orders. It consists of doing a series of
experiments in which the initial or starting concentrations of reactants are varied. Then
the initial rates are compared, from which the reaction orders can be deduced.
2.6.5. Change of concentration with time
A rate law tells you how the rate of a reaction depends on reactant concentrations at a
particular moment
Concentration-time equations
2.6.6. First Order Rate Law
Let us look at first order rate laws. Consider the decomposition of nitrogen pentoxide.
2N2O5 (g) → 4NO2 (g) + O (g)
Has the following rate law;
Rate = -
N 2 O5 
= k[N2O5]
t
Using calculus, one can show that such a first-order rate law leads to the following
relationship between N2O5 concentration and time;
75
ln
N 2 O5 t
N 2 O5 0
  kt
or
log
N 2 O5 t
N 2 O5 0

 kt
2.303
Here [N2O5]t is the concentration at time t, and [N2O5]0 is the initial concentration of
N2O5 (that is, the concentration at t=0). The symbol “ln” denotes the natural logarithm
(base e = 2.718. . .), and “log” denotes the logarithm to base 10. This equation enables
you to calculate the concentration of N2O5 at any time, once you are given the initial
concentration and the rate constant. Also you can find the time it takes for the N2O5
concentration to decrease to a particular value.
More generally, let A be a substance that reacts to give products according to the
equation
aA → products
where a is the stoichiometric coefficient of reactant A. suppose that this reaction has a
first-order rate law
Rate = 
 A
 k  A
t
Using calculus, you can get the following equation:
ln
At
A0
log
or
At =  kt
A0 2.303
Here [A]t is the concentration of reactant A at time t, and [A]0 is the initial concentration.
The ratio [A]t/[A]0 is the fraction of reactant remaining at time t.
2.6.7. Second Order Rate Law
Consider the reaction
aA → products
and suppose it has the second-order rate law
Rate = 
 A
 k  A2
t
An example is the decomposition of nitrogen dioxide at moderately high temperatures
(300°C to 400°C).
2NO2 (g) → 2NO2(g) + O2(g)
76
Using calculus, you can obtain the following relationship between the concentration of A
and the time
1
1
 kt 
At
A0
Using the equation, you can calculate the concentration of NO2 at any time during its
decomposition if you know the rate constant and the initial concentration. At 330°C, the
rate constant for the decomposition of NO2 is 0.775 Lmol-1s-1. Suppose the initial
concentration is 0.0030 molL-1. What is the concentration of NO2 after 645 s? By
substituting into the previous, you get
1
1
= 0.775 Lmol-1s-1 × 645 s +
= 8.3 × 102 Lmol-1
1
NO2 t
0.0030molL
If you take the inverse of both sides of this equation, you find that
[NO2]t = 0.0012 molL-1. Thus, after 645s, the concentration of NO2 decreased from
0.0030 molL-1 to 0.0012 mol L-1.
2.6.8. Half-lives
An important quantity, particularly for first-order reactions, is the half-life, t1/2- The
length of time required for the concentration of the reactant to be decreased to half of its
initial value. At this point t=t1/2 and
At

1
A0
2
Therefore
In
A0
1
A0
 kt 1
2
ln2 = kt½
0.693 = kt½
Solving for t½ gives
t½ = 0.693
k
77
2
The half-life of a second order reaction differs from that of first-order reaction by being
concentration dependent. Following the same procedure as above, we find that a secondorder reaction whose rate law is
Rate = k B 
2
has a half-life given by
t½ =
1
k[ B ] 0
This means that if we cut the concentration of B in half, the value of t ½ will double.
Therefore, during the course of the reaction, each successive half-life is twice as large as
the preceding one. If a second order reaction such as this starts with a half-life of 20
minutes, the concentration at the beginning of the second half-life is half of the initial
value, so the second half life will be twice as long. Similarly, the third half-life will be
twice as long as the second, and so forth.
2.7.
HESS’S LAW OF HEAT SUMMATION
Since enthalpy is a state function, the magnitude of ∆H for a chemical reaction does not
depend on the path taken by the reactants as they proceed to form the products. Consider,
for example, the conversion of 1 mol of liquid water at 100°C and 1 atm to 1 mol of
vapor at 100°C and 1 atm. This process absorbs 41 kJ of heat for each mole of H2O
vaporized and, hence, ∆H = +41 kJ. We can represent this reaction as
H2O(l) → H2O(g). An equation written in this manner, in which the energy change is also
shown, is called a thermochemical equation and is nearly interpreted on a mole basis.
Here, for instance, we see that 1 mol of H2O(l) is converted to 1 mol of H2O(g) by the
absorption of 41 kJ.
The value of ∆H for this change will always be +41 kJ, provided that we refer to the same
pair of initial and final stages. We could even go so far as to first decompose the 1 mol of
liquid in to gaseous hydrogen and oxygen and then recombine the elements to produce
H2O(g) a 100°C and 1 atm. The net change in enthalpy would still be the same, +41 kJ.
78
Consequently, it is possible to look at some overall change as the net result of a sequence
of chemical reactions. The net value of ∆H for the overall process is merely the sum of all
the enthalpy changes that take place along the way. These last statements constitute
2.7.1. Thermochemical Equations
Thermochemical equations serve as a useful tool for applying Hess’s law. For example,
the thermochemical equations that correspond to the indirect path just described for the
vaporization of water are
H2O(l) → H2(g) + ½O2(g)
∆H = +283 kJ
H2(g) + ½O2(g)→ H2O(g)
∆H = -242 kJ
Notice that fractional coefficients are allowed in thermochemical equations. This is
because a coefficient of 1/2 mol. (In ordinary chemical equations, however, fractional
coefficients are avoided because they are meaningless on a molecular level; one cannot
have half an atom or molecule and still retain the chemical identity of the species.). The
two equations above tell us that 283 kJ are required to decompose 1 mol of H2O(l) in to
its elements, and that 242 kJ are evolved when they recombine to produce 1 mol of
H2O(g). The net change (the vaporization of 1 mol of water) is obtained by adding the
two chemical equations together and then canceling any quantities that appear on both
sides of the arrow.
H2O(l) + H2(g) + ½O2(g) → H2O(g) + H2(g) + ½O2(g)
or
H2O(l) → H2O(g)
We also find that the heat of the overall reaction is equal to the algebraic sum of the heats
of reaction for the two steps.
∆H = +283 kJ + (-242 kJ)
∆H = + 41 kJ
Thus, when we add thermochemical equations to obtain some net change, we also add
their corresponding heats of reaction. To describe the nature of these thermochemical
changes, we can also illustrate them graphically.
79
H2(g) + 1 O2(g) (0.0 kJ)
2
0
 H = + 283
 H = - 242
Enthalpy
(Energy absorbed)
(Energy released)
H2O (g) (- 242 kJ)
 H = + 41 kJ
H2O (l) (-283 kJ)
This type of figure is frequently called an enthalpy diagram. Notice that we have chosen
the enthalpy of the free elements at the zero point on the energy scale. This choice is
entirely arbitrary because we are interested only in determining differences in H. in fact,
we have no way at all of knowing absolute enthalpies, just as we have no way of knowing
absolute internal energies. We can only measure ∆H.
2.7.2. Heat of Formation
A particularly useful type of thermochemical equation corresponds to the formation of
one mole of a substance from its elements. The enthalpy changes associated with these
reactions are called heats of formation or enthalpies of formation and are denoted as ∆Hf .
For example, thermochemical equations for the formation of liquid and gaseous water at
100°C and 1 atm are, respectively,
H2(g) + ½O2(g) → H2O(l) ∆Hf = -283 kJ/mol
H2(g) + ½O2(g) → H2O(g) ∆Hf = -242 kJ/mol
We can use these equations to obtain the heat of vaporization of water by reversing the
first equation and then add it to the second. When we reverse this equation, we must also
remember to change the sign of ∆H. if the formation of the H2O(l) is exothermic, as
indicated by a negative ∆Hf, the reverse process must be exothermic
(Exothermic)
H2(g) + ½O2(g) → H2O(l)
∆H = ∆Hf = -283 kJ
(Endothermic)
H2O(l) → H2(g) + ½O2(g)
∆H = ∆Hf = +283 kJ
80
When this last equation is added to that for the formation of H2O(g), we obtain
H2O(l) → H2O(g)
And the heat of reaction is
H  H fH 2O ( g )  H fH 2O ( l )
∆H = -242kJ – (-283kJ) = +41 kJ
Notice that the heat of reaction for the overall change is equal to the heat of formation of
the product minus the heat of the formation of the reactant. In general, we can write that
for any overall reaction
∆Hreaction = (sum of ∆Hf of products) – (sum of ∆Hf of reactants)
81
CHAPTER THREE
ORGANIC CHEMISTRY
Organic chemistry is the chemistry of compounds containing carbon. Some organic
compounds are ethanol (grain alcohol), ethylene glycol (automobile antifreeze), and
acetone (nail polish remover). A unique feature of carbon is its ability to bond to other
carbon atoms to give chains and rings of various lengths. Petroleum, for example,
consists of molecules that have up to 30 or more carbon atoms bonded together, and
chains of thousands of carbon atoms exist in molecules of polyethylene plastic. The
tetravalence of carbon-that is, its covalence of four-also makes possible the branching of
chains and the fusion of several rings. Moreover, other kinds of atoms, such as oxygen,
nitrogen, and sulphur, may be attached to the carbon atoms by single or multiple bonds.
Thus, great variety can be found, even among the smaller organic molecules.
3.1.
HYDROCARBONS
The simplest organic compounds are hydrocarbons, compounds containing only carbon
and hydrogen. All other organic compounds are considered for classification purposes to
be derived from hydrocarbons.
3.1.1. Alkanes
Hydrocarbons are classified into two main types: aliphatic and aromatic. Aromatic
hydrocarbons are hydrocarbons that contain benzene rings or similar structural
features. (Benzene consists of a ring of six carbon atoms with alternating single and
double carbon-carbon bonds.) Aliphatic hydrocarbons are all hydrocarbons that do not
contain benzene rings. The simplest hydrocarbon is methane.
3.1.2. Methane, the Simplest Hydrocarbon
A carbon atom has four valence electrons and forms four covalent bonds. Therefore the
simplest hydrocarbon consists of one carbon atom to which four hydrogen atoms are
bonded. The C—H bonds in this simplest hydrocarbon, called methane, are formed from
82
tetrahedrally directed sp3 hybrid orbital on the carbon atom. You can represent the
structure of methane by its molecular formula, CH4, which gives the number of different
atoms in the molecule, or by its structural formula, which shows how these atoms are
bonded to one another:
H
CH4
C
H
H
H
molecular formula
of methane
structural formula
of methane
3.1.3. The alkane series
Methane is one of the saturated hydrocarbons—that is, a hydrocarbon in which all
carbon atoms are bonded to the maximum number of hydrogen atoms. (There are no
carbon-carbon double or triple bonds.) One series of saturated hydrocarbons is the alkane
series. The alkanes, also called paraffins, are saturated hydrocarbons with the general
formula C n H 2n  2 . For n = 1, you get the formula CH4 ; for n = 2, C2H6; for n = 3, C3H8;
and so on. The structural formulas of the first four alkanes are
H
H
H
H
C H
C C H
H
H
H
H
H
propane
methane
H
C C C
H
H
H
H
H H H
H
H C C C
H
H H H
ethane
H
C H
H
butane
Structures of organic compounds are often given by condensed structural formulas,
where
H
CH3 means H
C
H
and
CH2 means
C
H
H
Condensed formulas of the first four alkanes are
CH4
methane
CH3CH3
ethane
CH3CH2CH3
propane
83
CH3CH2CH2CH3
butane
Note that the formula of one alkane differs from that of the preceding alkane by a —
CH2— group. A homologous series is a series of compounds in which one compound
differs from the preceding one by a —CH2— group. The alkanes thus constitute a
homologous series. Members of a homologous series have similar chemical properties,
and their physical properties change throughout the series in a regular way. The table
below lists the melting points and boiling points of the first ten straight-chain alkanes.
(These alkanes have all carbon atoms bonded to one another to give a straight chain;
hydrogen atoms fill out the four valences of each carbon atom.) They are also called
normal alkanes. Note that the melting points and boiling points generally increase in the
series from methane to decane. This is a result of increasing intermolecular forces, which
tend to increase with molecular weight.
Number of
Melting
point Boiling
Name
Carbons
Formula
(°C)
(°C)
Methane
1
CH4
-183
-162
Ethane
2
CH3CH3
-172
-89
Propane
3
CH3CH2CH3
-187
-42
Butane
4
CH3(CH2)2CH3
-138
0
Pentane
5
CH3(CH2)3CH3
-130
36
Hexane
6
CH3(CH2)4CH3
-95
69
Heptane
7
CH3(CH2)5CH3
-91
98
Octane
8
CH3(CH2)6CH3
-57
126
Nonane
9
CH3(CH2)7CH3
-54
151
Decane
10
CH3(CH2)8CH3
-30
174
point
In addition to the straight-chain alkanes, branched-chain alkanes are possible. For
example, isobutene (or 2-methylpropane) has the structure
84
H
H
H
H
C
C
C
H
H
H
H
C
or
CH3CHCH3
CH3
H
H
isobutane
(2-methylpropane)
Note that the molecular formula for isobutene is C4H10, the same as for butane, the
straight-chain hydrocarbon. Butane and isobutene are structural isomers of one another
(compounds with the same molecular formula but different structural formulas). Because
they have different structures, they have different properties. The number of structural
isomers corresponding to the molecular formula C n H 2n  2 increases rapidly with n.
3.1.4. Nomenclature of Alkanes
The nomenclature is formulated in rules agreed upon by the International Union of Pure
and Applied Chemistry (IUPAC). The first four straight-chain alkanes (methane, ethane,
propane and butane) have long-established names. Higher members of the series are
named from the Greek words indicating the number of carbon atoms in the molecule,
with the suffix –ane added.
The following four IUPAC rules are applied in naming the branched-chain alkanes:
1. Determine the longest continuous (not necessarily straight) chain of carbon atoms in
the molecule. The base name of the branched-chain alkane is that of the straight-chain
alkane corresponding to the number of carbon atoms in this longest chain. For
example, in
85
H
CH3CH2CH2CH2
C
CH3
CH2
CH3
The longest continuous carbon chain, shown in color, has seven carbon atoms, giving
the base name heptane. The full name for the alkane includes the names of any
branched chains. These names are placed in front of the base name, as described in
the remaining rules.
2. Any chain branching off the longest chain is named as an alkyl group. An alkyl
group is an alkane less one hydrogen atom. When a hydrogen atom is removed from
an end carbon atom of a straight-chain alkane, the resulting alkyl group is named by
changing the suffix –ane of the alkane to –yl. Thus, removing a hydrogen atom from
methane gives the methyl group, —CH3. The structure shown in Rule 1 has a methyl
group as a branch on the heptane chain.
Some alkyl groups
Name of Alkyl
Original Alkane
Structure of Alkyl Group
Group
Methane, CH4
CH3—
Methyl
Ethane, CH3CH3
CH3CH2—
Ethyl
Propane, CH3CH2CH3
CH3CH2CH2—
Propyl
Propane, CH3CH2CH3
CH3CHCH3
Isopropyl
Butane, CH3CH2CH2CH3
CH3CH2CH2CH2—
Butyl
Isobutane, CH3CHCH3
CH3
Tertiary-butyl
CH3CCH3
(t-butyl)
CH3
3. The complete name of a branch requires a number that locates that branch on the
longest chain. For this purpose, you number each carbon atom on the longest chain in
86
whichever direction gives the smaller numbers for the locations of all branches. The
structural formula in Rule 1 is numbered as shown in the following structure;
H
H
7
6
5
4
1
3
CH3 CH2CH2CH2
C
CH3
not
2
3
4
5
CH3 CH2CH2CH2
C
6
2CH2
CH3
CH2
7
1
CH3
CH3
Thus, the methyl branch is located at carbon 3 of the heptane chain (not carbon 5).
The complete name of the branch is 3-methyl, and the compound is named 3methylheptane. Note that the branch name and the base name are written as a single
word, with a hyphen following the number.
4. When there are more than one alkyl branch of the same kind (say two methyl groups),
this number is indicated by a Greek prefix, such as di-, tri-, or tetra- used with the
name of the alkyl group. The position of each group on the longest chain is given by
numbers. For example,
H
7
6
5
CH3CH2CH2
4
C
H
3
C
CH3
7
CH3
6
5
4
CH3CH2CH2CH2
3
C
2
2
1
1
CH3 CH2
CH3
CH2
CH3
CH3
3,4-dimethylheptane
3,3-dimethylheptane
Note that the position numbers are separated by a comma and are followed by a
hyphen.
When there are two or more different alkyl branches, the name of each branch, with its
position number, precedes the base name. The branch names are placed in alphabetical
order. For example,
87
CH3
CH3CHCHCH2CH3
CH2CH3
3-ethyl-2-methylpentane
Note the use of hyphens.
3.1.5. Alkenes
Unsaturated hydrocarbons are hydrocarbons that do not contain the maximum number
of hydrogen atoms for a given carbon-atom framework. Such compounds have carboncarbon multiple bonds and, under the proper conditions, add molecular hydrogen to give
a saturated compound. For example, ethylene adds hydrogen to give ethane.
H
H
C
C
H
+
H2
Ni catalyst
H
H
H
H
C
C
H
H
H
ethane
ethylene
Alkenes are hydrocarbons that have the general formula CnH2n and contain a carboncarbon double bond. (These compounds are also called olefins). The simplest alkene,
ethylene, has the condensed formula CH2=CH2. Ethylene is a gas with a sweetish odor. It
is obtained from the refining of petroleum and is an important raw material in the
chemical industry. Plants also produce it, and exposure of fruit to ethylene speeds
ripening. In ethylene and other alkenes, all atoms connected to the two carbon atoms of a
double bond lie in a single plane. This is due to the need for the maximum overlap of 2p
orbitals on carbon atoms to form a pi bond.
You obtain the IUPAC name for an alkene by finding the longest chain containing the
double bond. As with the alkanes, this longest chain gives you the stem name, but the
suffix is –ene rather than –ane. The carbon atoms of the longest chain are numbered from
the end nearest the carbon –carbon double bond and the position of the double bond is
88
given the number of the first carbon atom of that bond. This number is written in front of
the stem name of the alkene. Branched chains are named as in the alkanes. The simplest
alkene, CH2=CH2, is called ethene, although the common name is ethylene.
Rotation about a carbon-carbon double bond cannot occur without breaking the pi bond.
This requires energies comparable to those in chemical reactions, so rotation does not
occur. This lack of rotation about the double bond gives rise to isomers in certain alkenes.
For example, there are two isomers of 2-butene.
CH3
CH3
C
C
C
H
H
CH3
C
H
H
CH3
trans-2-butene
b.p. 0.9oC
cis-2-butene
b.p. 3.7oC
The different boiling points indicate that these are indeed different compounds.
Cis-2-butene and trans-2-butene are geometric isomers. Geometric isomers are isomers
in which the atoms are joined to one another in the same way but that differ because
some atoms occupy different relative positions in space. The cis isomer has identical
groups (in the case of the butenes, alkyl groups) attached to the same side of the double
bond, whereas the trans isomer has them on opposite sides. An alkene with the general
formula
A
D
C
C
B
E
has a geometric isomer only if groups A and B are different and groups D and E are
different. Thus, there is no isomer of propene, CH2=CHCH3.
89
3.1.6. Alkynes
Alkynes are hydrocarbons that have at least one triple covalent bond between two
adjacent carbons. The general formula for the alkyne is: CnH2n-2.
Alkynes are similar to alkenes in that the triple bond is between two carbon atoms that
have undergone sp hybridization using one 2s and one 2p atomic orbital to produce two
sp hybrid orbitals. The other 2p orbitals that did not enter the hybridization process are
double lobed pure "2p" orbitals that are 90 degrees apart. These pure p orbitals can
overlap with p orbitals on another sp carbon to form a Pi bond. Since there are two such p
orbitals on each SP carbon, then there can be two Pi bonds. The Chemistry is very similar
to alkenes in that both are formed by elimination reactions, and the major chemical
reactions that alkynes undergo are addition type reactions. There are some differences but
not many.
Nomenclature of Alkynes
1. Determine the longest continuous chain of carbons that have the triple bond
between two of its carbons. By "longest continuous chain" is meant to be able to
trace through the carbons without raising the tracer (or finger) off the surface. The
chain does not necessarily have to be straight.
2. Number the carbons in the chain so that the triple bond would be between the
carbons with the lowest designated number. This means that you have to decide
whether to number beginning on the right end or left end of the chain. If it makes
no difference to the triple bond then shift attention to the branched groups.
3. Identify the various branching groups attached to this continuous chain of carbons
by name
4. Name the branched groups in alphabetical order attaching (hyphenating) the
carbon number it is attached to along the continuous chain of carbons to the front
of the branch name. If more than one of the same kind of branched group is
attached to the chain, identify the number carbon each group is attached to as a
series of numbers separated by commas between each number then a hyphen and
finally use a greek prefix attached to the branch name.
5. Attach a numerical prefix indicating the lowest carbon number the triple bond is
between onto the normal alkane name
90
6. Drop the "ane" ending and add the "yne" ending associated with the Alkene
family
Let's try an example. Determine the IUPAC name of the following structure:
H
CH3 C
C
H2C CH3
H
C
C
C
CH3
H
Cl
CH3
1. Determine the longest continuous chain of carbons that have the triple bond
between two of its carbons. By "longest continuous chain" is meant to be able to
trace through the carbons without raising the tracer (or finger) off the surface. The
chain does not necessarily have to be straight.
That would be seven carbons long. Actually there are two alternate continuous
chains of carbon that have seven carbons, but the rules say that we choose the one
that results in the greatest number of substituents.
2. Number the carbons in the chain so that the triple bond would be between the
carbons with the lowest designated number. This means that you have to decide
whether to number beginning on the right end or left end of the chain. If it makes
no difference to the triple bond then shift attention to the branched groups.
We would number from the left end and go toward the right end of the continuous
chain.
3. Identify the various branching groups attached to this continuous chain of carbons
by name
There is a methyl group on carbon #4, an ethyl group on carbon #5, and a Chloro
group on carbon #6
4. Name the branched groups in alphabetical order attaching (hyphenating) the
carbon number it is attached to along the continuous chain of carbons to the front
of the branch name. If more than one of the same kind of branched group is
attached to the chain, identify the number carbon each group is attached to as a
series of numbers separated by commas between each number then a hyphen and
finally use a greek prefix attached to the branch name.
6-Chloro-5-ethyl-4-methyl
91
5. Attach a numerical prefix indicating the lowest carbon number the triple bond is
between onto the normal alkane name
6-Chloro-5-ethyl-4-methyl-2-heptane
6. Drop the "ane" ending and add the "yne" ending associated with the Alkene
family
6-Chloro-5-ethyl-4-methyl-2-heptyne
Here are some examples for you to try:
Identify the IUPAC name for the following structures:
CH3
H3C
C
H
Cl H
C
CH2
C
C
C
H
H
CH3
H3 C
C
C
C
Br
H
C
CH3
CH3
CH3
(b)
(a)
Answers (a) 6-Chloro-3,3-dimethyl-4-octyne (b) 4-Bromo-5-methyl-2-hexyne
Dehydrohalogenation of Alkyl Dihalides
This reaction is particularly useful since the dihalides are readily obtained from the
corresponding alkenes by addition of halogen. This amounts to conversion by several
steps of a double bond into a triple bond.
Dehydrohalogenation can be carried out in two stages. The halides thus obtained, with
halogen attached directly to double bonded carbon, are called vinyl halides, and are very
unreactive. Under mild conditions, therefore, dehydrohalogenation stops at the vinyl
halide stage; more vigorous conditions, use of stronger base is required for alkyne
formation. If only the first step of this reaction is carried out, it is a valuable method for
preparing unsaturated halides.
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3.2.
REACTIONS OF HYDROCARBONS
3.2.1. Oxidation
All hydrocarbons burn in an excess of O2 to give carbon dioxide and water.
7
C 2 H 6 ( g )  O2 ( g )  2CO2 ( g )  3H 2 O(l ); H   1560kJmol 1
2
C6 H 6 ( g ) 
15
O2 ( g )  6CO 2 ( g )  3H 2 O(l ); H   3267 kJmol 1
2
The large negative ∆H’s explain why the hydrocarbons are useful as fuels.
Unsaturated hydrocarbons are oxidized under milder conditions than are saturated
hydrocarbons. For example, when aqueous potassium permanganate, KMnO4(aq), is
added to an alkene or an alkyne, the purple color of KMnO4 fades and a precipitate of
brown manganese dioxide forms.
3C4H9CH=CH2 + 2MnO4-(aq) + 4H2O
1-hexene
H
H
3C4H9C
C
O
O
H
H
H + 2MnO2(s) + 2OH-(aq)
Saturated hydrocarbons are unreactive with KMnO4(aq), so this reagent can be used to
distinguish them from unsaturated hydrocarbons (this is called the Bayer test of
unsaturation).
3.2.2. Substitution reactions of alkanes
Alkanes react with the halogens F2, Cl2, and Br2. reactions with Cl2 or Br2 requires
sunlight (indicated hv) or heat.
H
H
H
C
H +
Cl-Cl
hv
H
C
H
H
93
Cl +
H-Cl
This is an example of a substitution reaction. A substitution reaction is a reaction in
which a part of the reagent molecule is substituted for an H atom on a hydrocarbon
group. All of the H atoms of an alkane may undergo substitution, leading to a mixture of
products.
+ Cl2
hv
CH2Cl2 + HCl
CH2Cl2 + Cl2
hv
CHCl3 + HCl
+ Cl2
hv
CCl4 + HCl
CH3Cl
CHCl3
Fluorine is very reactive with saturated hydrocarbons and usually gives complete
substitution. Bromine is less reactive than Cl2 and often requires elevated temperatures
for substitution.
3.2.3. Addition Reactions of Alkenes
Alkenes are more reactive than alkanes because of the presence of the double bond.
Many reagents add to the double bond. A simple example is the addition of a halogen,
such as Br2, to propene.
CH3CH
CH3CH2
CH2 + Br2
Br
CH2
Br
An addition reaction is a reaction in which parts of a reagent are added to each carbon
atom of a carbon-carbon multiple bond, which then becomes a C-C single bond. The
addition of Br2 to an alkene is fast. In fact, it occurs so readily that bromine dissolved in
carbon tetrachloride, CCl4, is a useful reagent to test for unsaturation. When a few drops
of the solution are added to an alkene, the red-brown color of the Br2 is immediately lost.
There is no reaction with alkanes, and the solution retains the red-brown color of the Br2.
Unsymmetrical reagents, such as HCl and HBr, add to unsymmetrical alkenes to give two
products that are isomers of one another. For example,
CH3CH
CH2 + HBr
3
2
CH3CH
1
CH2
CH3CH
Br
H
2-bromopropane
CH2 + HBr
3
2
CH3CH
1
CH2
H
Br
1-bromopropane
94
In one case, the hydrogen atom of HBr adds to carbon atom 1, giving 2-bromopropane; in
the other case, the hydrogen atom of HBr adds to carbon atom 2, giving 1-bromopropane.
(The name 1-bromopropane means that a bromine atom is substituted for a hydrogen
atom at carbon atom 1.)
The two products are not formed in equal amounts; one is more likely to form.
Markownikoff’s rule is a generalization stating that the major product formed by the
addition of an unsymmetrical reagent such as H-Cl, H-Br, or H-OH is the one obtained
when the H atom of the reagent adds to the carbon atom of the multiple bond that already
has the more hydrogen atoms attached to it. In the preceding example, the H atom of HBr
should add preferentially to carbon atom 1, which has two hydrogen atoms attached to it.
The major product then is 2-bromopropane.
3.2.4. Addition Reactions of Alkynes
Because of the unsaturated nature of ethyne addition reactions can occur across the triple
bond.
Addition of Hydrogen
When acetylene and hydrogen are passed over a nickel catalyst at 150°C, (or over
platinum black catalyst at room temperature) ethene is first formed and then this is further
reduced to ethane.
HC
CH + H2
Ni, 150°C
Ethyne
H2 C
CH2
Ni, 150°C
Ethene
C2 H 6
Ethane
Addition of Halogens
Ethyne reacts explosively with chlorine at room temperature, forming hydrogen chloride
and carbon. To control the reaction, acetylene and chlorine (also bromine) are added in
retorts filled with kieselguhr (hydrated silica) and iron filings.
HCCH + Cl2
ClCH
CHCl + Cl2
95
Cl2HCCHCl2
Addition of Hydrogen Halides.
Ethyne reacts with the halogen acids. Hydrogen iodide adding on the most readily, at
room temperature. A similar reaction occur with hydrogen bromide at 100°C. Reaction
with hydrogen chloride occurs very slowly.
HCCH + HCl
CH3CHCl2
CHCl + HCl
H2 C
Addition of Water (Hydration)
Hydration of ethyne occurs when the gas is passed into dilute sulphuric acid at 60°C.
Mercuric sulphate is used as a catalyst for the reaction, and the product formed is ethanal
(i.e. acetaldehyde).
HCCH + H2O
HgSO4, 60°C
CH3CHO
3.2.5. Oxidation of Alkynes
Ethyne is oxidised by a dilute aqueous solution of potassium permanganate to form
oxalic acid. Thus, if ethyne is bubbled through a solution of potassium permanganate the
solution is decolourised. This is Baeyer's test for unsaturated organic compounds.
CH3 C
CH
KMnO4/OHH+
CH3 C
OH
3.3.
O + H
C
O
OH
FUNCTIONAL GROUPS
The following is a brief introduction to some of the common functional groups in
Organic Chemistry. No attempt has been made to delve into their respective reactions for
that will require an entire course altogether. The aim here is to make the student aware of
the general features of these important groups in the field of organic chemistry body of
knowledge.
96
Alkyl group
Alkyl, and occasionally aryl (aromatic) functions are represented by the RExamples:
Methyl: CH3–
Ethyl: CH3CH2–
Propyl: CH3CH2CH2–
Isopropyl: (CH3)2CH–
Phenyl: C6H5–
Alkyl halide group
R
X
X = F, Cl, Br, I
Alkyl halides (haloalkanes) consist of an alkyl group attached to a halogen e.g. F, Cl, Br,
I.
Chloro, bromo and iodo alkyl halides are often susceptible to elimination and/or
nucleophilic substitution reactions.
Alcohol group
Primary alcohols
H
R
C
OH
H
Primary alcohols have an -OH function attached to an R-CH2- group.
Primary alcohols can be oxidised to aldehydes and on to carboxylic acids. (It can be
difficult to stop the oxidation at the aldehyde stage.)
Primary alcohols can be shown in text as: RCH2OH
Secondary alcohol
R
R
C
OH
H
Secondary alcohols have an -OH function attached to a R2CH- group.
Secondary alcohols can be oxidized to ketones.
Secondary alcohols can be shown in text as: R2CHOH
Tertiary alcohol
Tertiary alcohols have an -OH function attached to a R3C- group.
Tertiary alcohols are resistant to oxidation with acidified potassium dichromate (VI).
97
Tertiary alcohols can be shown in text as: R3COH
Carbonyl group
O
C
The carbonyl group is a super function because many common functional groups are
based on a carbonyl, including: aldehydes, ketones, carboxylic acids, esters, amides, acyl
(acid) chlorides, acid anhydrides
Aldehyde group
O
C
R
H
Aldehydes have a hydrogen and an alkyl (or aromatic) group attached to a carbonyl
function.
Aldehydes can be shown in text as: RCHO
Aldehydes are easily oxidised to carboxylic acids, and they can be reduced to primary
alcohols.
Aldehydes can be distinguished from ketones by giving positive test results with
Fehling’s solution (brick red precipitate) or Tollens reagent (silver mirror).
Aldehydes give red-orange precipitates with 2,4-dinitrophenyl hydrazine.
Ketone group
O
C
R
R
Ketones have a pair of alkyl or aromatic groups attached to a carbonyl function.
Ketones can be shown in text as: RCOR
Ketones can be distinguished from aldehydes by giving negative test results with
Fehling’s solution (brick red precipitate) or Tollens reagent (silver mirror).
Ketones give red-orange precipitates with 2,4-dinitrophenyl hydrazine.
98
Carboxylic Acid group
O
C
R
OH
Carboxylic acids have an alkyl or aromatic groups attached to a hydroxy-carbonyl
function.
Carboxylic acids can be shown in text as: RCOOH
Carboxylic acids are weak Bronsted acids and they liberate CO2 from carbonates and
hydrogen carbonates.
Ester group
O
C
R
O R
Esters have a pair of alkyl or aromatic groups attached to a carbonyl and linking oxygen
function.
Esters can be shown in text as: RCOOR or (occasionally) ROCOR.
ester + water
carboxylic acid + alcohol
This is an acid catalyzed equilibrium.
Amide group
O
C
R
NH2
Primary amides (shown) have an alkyl or aromatic group attached to an amino-carbonyl
function.
Primary amides can be shown in text as: RCONH2
Secondary amides have an alkyl or aryl group attached to the nitrogen: RCONHR
Tertiary amides have two alkyl or aryl group attached to the nitrogen: RCONR2
Amines group
Primary amine
R
NH2
Primary amines have an alkyl or aromatic group and two hydrogens attached to a
nitrogen atom.
Primary amines can be shown in text as: RNH2
Primary amines are basic functions that can be protonated to the corresponding
ammonium ion.
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Primary amines are also nucleophilic.
Secondary amine
R
NH
R
Secondary amines have a pair of alkyl or aromatic groups, and a hydrogen, attached to a
nitrogen atom.
Secondary amines can be shown in text as: R2NH
Secondary amines are basic functions that can be protonated to the corresponding
ammonium ion.
Secondary amines are also nucleophilic.
Tertiary amine
R
R
N
R
Tertiary amines have three alkyl or aromatic groups attached to a nitrogen atom.
Tertiary amines can be shown in text as: R3N
Tertiary amines are basic functions that can be protonated to the corresponding
ammonium ion.
Tertiary amines are also nucleophilic.
Acid chloride group
O
C
R
Cl
Acid chlorides, or acyl chlorides, have an alkyl (or aromatic) group attached to a carbonyl
function plus a labile (easily displaced) chlorine.
Acid chlorides highly reactive entities are highly susceptible to attack by nucleophiles.
Acid chlorides can be shown in text as: ROCl
Acid anhydride group
R
O
O
C
C
O
R
Acid anhydrides are formed when water is removed from a carboxylic acid, hence the
name.
Acid anhydrides can be shown in text as: (RO)2O
100
Nitrile
R
C
N
Nitriles (or organo cyanides) have an alkyl (or aromatic) group attached to a carbontriple-bond-nitrogen function.
Nitriles can be shown in text as: RCN
Note that there is a nomenclature issue with nitriles/cyanides. If a compound is named as
the nitrile then the nitrile carbon is counted and included, but when the compound is
named as the cyanide it is not.
For example: CH3CH2CN is called propane nitrile or ethyl cyanide (cyanoethane).
Carboxylate ion or salt
O
C
O-
R
Carboxylate ions are the conjugate bases of carboxylic acids, i.e. the deprotonated
carboxylic acid.
Carboxylate ions can be shown in text as: RCOO–
When the counter ion is included, the salt is being shown.
Salts can be shown in text as: RCOONa
Ammonium ion
+
R
N
R
R
R
Ammonium ions have a total of four alkyl and/or hydrogen functions attached to a
nitrogen atom.
[NH4]+
[RNH3]+
[R2NH2]+
[R3NH]+
[R4N]+
Quaternary ammonium ions are not proton donors, but the others are weak Bronsted acids
(pKa about 10).
Amino acid
COOH
R
C
H
NH2
101
Amino acids, strictly alpha-amino acids, have carboxylic acid, amino function and a
hydrogen attached to the same carbon atom.
There are 20 naturally occurring amino acids. All except glycine (R = H) are chiral and
only the L enantiomer is found in nature.
Amino acids can be shown in text as: R-CH(NH2)COOH
Ether
O
R
R
Ethers have a pair of alkyl or aromatic groups attached to a linking oxygen atom.
Ethers can be shown in text as: ROR
Ethers are surprisingly unreactive and are very useful as solvents for many (but not all)
classes of reaction
Alkoxide ion
O-
R
Alkoxide ions an alkyl group attached to an oxyanion.
Alkoxide ions can be shown in text as: RO–
Sodium alkoxides, RONa, are slightly stronger bases than water and so cannot be
prepared in water. Instead they are prepared by adding sodium to the dry alcohol.
Hydroxynitrile
CN
R
C
OH
R
Hydroxynitriles (also called cyanohydrins) are formed when hydrogen cyanide, H+ CN–,
adds across the carbonyl function of an aldehyde or ketone.
Carbocations
Primary carbocation
H
C+
R
H
Primary carbocations have a single alkyl function attached to a carbon centre with a
formal positive charge.
Carbocations - also and more correctly called carbenium ions - are important reactive
intermediates implicated in electrophilic addition reactions and electrophilic aromatic
substitution reactions.
Stability: primary << secondary << tertiary
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Secondary carbocation
R
C+
R
H
Secondary carbocations have a pair of alkyl functions attached to a carbon centre with a
formal positive charge.
Tertiary carbocation
R
C+
R
R
Tertiary carbocations have three alkyl functions attached to a carbon centre with a formal
positive charge.
Acyl cation
O
C+
R
Acyl cations have an alkyl (or aromatic) group attached to a carbonyl function with a
formal positive charge.
Acyl cations are important reactive intermediates and are implicated in electrophilic
addition reactions and electrophilic aromatic substitution reactions.
Acyl cations are commonly formed from the corresponding acyl/acid chloride plus
aluminium chloride.
Polymer
X
C
CH2
H
n
Polymers consist of small monomer molecules that have reacted together so as to form a
large covalently bonded structure.
There are two general types of polymerization: addition and condensation.
Linear chain polymers are generally thermoplastic, while three dimensional network
polymers are not.
103
Diol or polyol
OH OH
R
C
C
H
H
R
Diols and polyols are alcohols with two or more -OH functions.
Diols and polyols are very soluble in water. They are used as high temperature polar
solvents.
3.4.
ELIMINATION REACTIONS
3.4.1. E1 and E2 Reactions
An elimination reaction is a type of organic reaction in which two substituents are
removed from a molecule in either a one or two-step mechanism. Either the unsaturation
of the molecule increases (as in most organic elimination reactions) or the valence of an
atom in the molecule decreases by two, a process known as reductive elimination.
Important classes of elimination reactions are those involving alkyl halides or alkanes in
general, with good leaving groups, reacting with a Lewis base to form an alkene in the
reverse of an addition reaction. The one and two-step mechanisms are named and known
as E2 reaction and E1 reaction, respectively.
Elimination reactions are important as a method for the preparation of alkenes.
The two most important methods are:

Dehydration (-H2O) of alcohols, and

Dehydrohalogenation (-HX) of alkyl halides.
There are three fundamental events in these elimination reactions:
1. removal of a proton
2. formation of the C=C pi bond
3. breaking of the bond to the leaving group
Depending on the relative timing of these events, different mechanisms are possible:

Loss of the leaving group to form a carbocation, removal of H+ and formation of
C=C bond: E1 reaction
104

Simultaneous H+ removal, C=C bond formation and loss of the leaving group: E2
reaction

Removal of H+ to form a carbanion, loss of the leaving group and formation of
C=C bond (E1cb reaction)
In many cases the elimination reaction may proceed to alkenes that are constitutional
isomers with one formed in excess of the other. This is described as regioselectivity.
Zaitsev's rule, based on the dehydration of alcohols, describes the preference for
eliminations to give the highly substituted (more stable) alkene, which may also be
described as the Zaitsev product. The rule is not always obeyed; some reactions give the
anti-Zaitsev product. Similarly, eliminations often favor the more stable trans-product
over the cis-product (stereoselectivity)
3.4.2. Carbocations.
These are important species in elimination reactions. Let’s have a look at some of their
most important characteristics.
Stability:
The general stability order of simple alkyl carbocations is: (most stable) 3 o > 2o > 1o >
methyl (least stable)
This is because alkyl groups are weakly electron donating due to hyperconjugation and
inductive effects. Resonance effects can further stabilize carbocations when present.
Reactivity:
As they have an incomplete octet, carbocations are excellent electrophiles and react
readily with nucleophiles (substitution).
Alternatively, loss of H+ can generate a pi bond (elimination).
105
Rearrangements:
Carbocations are prone to rearrangement via 1,2-hydride or 1,2-alkyl shifts provided it
generates a more stable carbocation. For example:
Notice that the "predicted" product is only formed in 3% yield, and that products with a
different skeleton dominate. The reaction proceeds via protonation to give the better
leaving group which departs to give the 2o carbocation shown. A methyl group rapidly
migrates taking its bonding electrons along, giving a new skeleton and a more stable 3o
carbocation which can then lose H+ to give the more stable alkene as the major product.
2o carbocation to 3o carbocation
This is an example of a 1,2-alkyl shift. The numbers indicate that the alkyl group moves
to an adjacent position.
Similar migrations of H atoms, 1,2-hydride shifts are also known.
106