Republic of Iraq

Ministry of Education
General Directorate of Curricula

Chemistry
Fifth Class of Science
Biological Branch

Authors
Prof.Dr.Ammar Hani Suhail Al-Dujaili Prof. Dr.Sarmed Bhagat Decran
Dr. Samir Hakim Kreem Dr. Saadi Mohammed Thaher
Khulood Mahdi Salim Dulaimi Akram Hanna Elia
Khalil Rahim Ali Majed Hussein Khalaf Aljasani
Kareem Abdl-Hussain Alkanani Basil Ibrahim Alshuk
Kadhim Rashid Musa

First Translated Edition 2020 / 1441

Scientific Supervisor : Dr. Huda Salah Kareem

Design Supervisor : Hiba Salah Mahdi

Translated from 8th Arabic Edution

‫استناداً الى القانون يوزع مجاناً ويمنع بيعه وتداوله في االسواق‬

 Introduction
Chemistry is a practical, theoretical science. These key pillars have grown
rapidly and the outcome has covered most aspects of our daily lives, mostly
meeting the needs of society Environment and living conditions. The con-
tinuous coordination with the rest of the other biological sciences, especially
physics, mathematics and life sciences.
The Authoring Committee of the book of chemistry for the fifth grade of
science; Interest in enriching the content of the vocabulary of the new cur-
riculum in chemistry to encourage the student and interested to follow up
on this vital science, taking into account the sensitive age for the student
and taught in previous years of this science starting from the first secondary
study. The committee emphasizes the need to support the theoretical mate-
rial in the book with two main pillars: conducting practical experiments
(including the practical needs of the teacher) and linking chemistry as much
as possible with daily life and its impact is growing in industry, agriculture,
nutrition and medicines cosmetics and others.
The chapters of the book included both horizontal and vertical expansion
that these chapters were modern and simple at the same time. Subject to the
use of the same symbols, formulas and references used in chemistry books
for previous years to facilitate the student and teacher and to be the chemis-
try curriculum in the secondary stage a coherent gradient (with the addition
of the update required by the various unit codes).
The committee assures of the chemistry teachers not to enter any symbols,
formulas or terminology, which is waged by confusing the minds of students
and adhering, as far as possible, by the quality of the questions and activi-
ties mentioned in end of each chapter - direct applied and deductive as well
as mentioned exercises In the margins and the accompanying educational
information illuminate the ideas of students and not be a stressful part In
the exam questions and in order to complete the benefit. The committee
considers that it organizes visits to qualitative and annual exhibitions and
the establishment of trips - scientific and entertainment at the same time - to
a factory near the school and informing students - on the ground - the suc-
cessive stages of manufacturing raw materials, through to the final product,
because it is a goal of great importance.
3
must be taken care of by teachers and School administrations.
Instruct students to prepare scientific reports for each visit or targeted
trip, be an extracurricular activity with an emphasis on investing local en-
vironmental resources some of us compensate for certain items that are not
available on time. Since this book a new composite needs to be reinforced
through feed back of the brothers teachers, educational specialists and par-
ents of students and all who relevant to the implementation of this curricu-
lum in the best way to provide the General Directorate of Curriculum / in
the Ministry of education and the books his proposal opinions produced by
the field of education to use them in the revision of the Subsequent editions.

Authers

4
 Contents

Chapter One
Development of the atomic concept 9

1.1 Introduction 10
1.2 Electron Discovery 10
1.3 Proton detection 11
1.4 Estimating the ratio of the charge of the electron to its mass 12
1.5 Determination of electron charge 12
1.6 Detection of the nucleus 13
1.7 Detection of atomic number 14
1.8 Detection of neutron 14
1.9 The configuration of electrons in an atom 14
1.10 Electromagnetic radiation 15
1.11 Quantum theory 15
1.12 Atomic Spectra 16
1.13 Linear emission spectrum for hydrogen 17
1.14 The wave nature of the electron 19
1.15 Wave mechanics 20
1.16 Quantum numbers 21
1.17 How to write an electronic configuration 25
Basic concepts 30
Chapter questions 31

5
 Content
Chapter Two
Bonding forces and the geometric shapes of molecules 35
2.1 Introduction 36
2.2 Types of chemical bonds 40
2.3 Resonance 44
2.4 Geometric shape of the molecules 45
2.5 Orbital hybridization 49
Basic concepts 59
Chapter questions 60

Chapter Three 61
Periodic Table and Chemistry of Transitional Elements
3.1 Periodic table 62
3.2 Transitional elements 77
3.3 Iron 90
Basic concepts 95
Chapter questions 96

Chapter Four
Solutions 97
4.1 Introduction 98
4.2 Solubility process 99
4.3 Types and properties of solutions 100
4.4 Raoult’s Law 110
4.5 Effect of non-volatile solute on some solvent properties 112
Basic concepts 119
Chapter questions 120
6
 Contents

Chapter Five
Chemical kinetics 121
5.1 Introduction 122
5.2 Chemical reaction speed 122
5.3 The reaction rate law 129
5.4 Theories of reaction rate 135
5.5 Heat of reaction 138
5.6 Activation energy 140
5.7 Factors affecting reaction rate 140
5.8 Reaction mechanism 142
Basic concepts 147
Chapter questions 148

Chapter Six
Acids, Bases and Salts 153
6.1 Introduction 154
6.2 Properties of aqueous solutions of acids and bases 154
6.3 Molecular concepts of acids and bases 156
6.4 Amphoteric substances 161
6.5 Acid and base reactions in aqueous solutions 162
6.6 Types of salts 163
6.7 Acids and bases indicators 166
6.8 Acids and bases solutions 166
6.9 Titration 168
Basic concepts 171
Chapter questions 172
7
14_00CO.JPG
 Contents
Chapter Seven
Polymer Chemistry 173

7.1 Polymers (plastics) 174

7.2 Polymers Types 174
7.3 Synthetic polymers 175
7.4 The world of plastics 187
Basic concepts 188
Chapter questions 189

Chapter Eight
191
Aromatic hydrocarbons

8.1 Introduction 192

8.2 Benzene composition 192
8.3 Mechanism of electrophilic substitution reactions 201
8.4 Phenols 202
8.5 Preparation of aspirin 205
8.6 Detection of phenols 205
8.7 Heterocyclic compounds 205
Basic concepts 208
Chapter questions 209

8
Chapter One
Development of the atomic concept 1
After completing this chapter, the student is expected to:
• Understand the importance of electrical discharge tubes experiments in electron
and proton discovery
• Characterize the properties of cathode ray and channel ray.
• Know the value of a single electron’s charge, its mass, and how it was found.
• Show the importance of nucleus discovery in the development of atomic structure.
• Know the phenomenon of electromagnetic radiation.
• Understand the meaning of quantum energy and the importance of quantum
theory.
• Distinguish the difference between linear and continuous spectrum.
• Show the dual nature of the electron.
• Differetiate between the perception of the principal energy level according to
Bohr’s theory and mechanics waveform.
• Recognize the four quantum numbers and their importance.
• Determine the four quantum numbers of any electron in an atom.
• Know the principle of exception to Pauli.
9
 Do you know 1-1 Introduction
Zinc sulfide is a fluorinated sub- It came in Dalton’s atomic theory in one of its clauses stat-
stance light flashes when light ed that “atoms are not able to divide or fragment” but the
falls on it. electrical discharge experiments through gases and radio
activity phenomenon particles, meaning that atom is di-
visble.
1-2 Electron Discovery
When supplied with an electric discharge tube, which con-
tains hydrogen gas under low pressure (Figure 1-1), with
a zinc sulfate-covered reagent barrier as well as a metal
plate near the cathode, with a narrow rectangular incision.
When the electrical current passes, we see a luminous line
Figure 1-1 across the barrier, and the explanation for that is beams or
Model for electric discharge electric rays emanate from the cathode, and are moving
tube. to anode, the metal plate stops most of them, but a rectan-
gular incision allows to this narrow beam runs through it,
and hits the barrier and produces a luminous line. Because
the beams appear to be coming from the cathode, they are
known as cathode – ray, when a cathode rounded a north
pole finder to a magnet from on the side of the tube, the
luminous line is curves down, either near south pole finder,
the luminous line curves upwards. The deviation trend in-
Figure 1-2 dicates that the cathode rays are negative electric charge
Cathode rays were affected by [Figure (1-2)]. The same conclusion can be made when
polarity magnetism. put charging plates electrically above and below the bar-
rier, the illuminated line curves towards the positive plate
because opposing charges attract each other [Figure (1-3)].
When studying different gases it was found that these
particles compose the luminous line. The luminance has a
negative charge of one value as well as its mass and these
particles are called electrons. The most important proper-
ties of cathode rays:
1 - Moving in straight lines emitted from the negative
pole towards the positive pole.
Figure 1-3 2 - It consists of particles with a very tiny mass and mov-
Cathode rays were affected by ing very fast.
electrical plates. 3 - Affected by the electric field and attracted towards the
positive electrode Which indicates it’s a negative charge
[Figure (1-3)].

10
 4 - Affected by the magnetic field [Figure (1-2)].
5 - Ionize the medium in which it passes.

1-3 Proton detection

The first experiments conducted by Goldstein in 1886, to
suggest that positive particles are also formed in the elec-
tric discharge tubes. In the Goldstein tube [Figure 1-4]
there is an positive pole (Anode) towards the right and
negative pole (cathode) on the left, which is formed from a
piece of metal engraved by a hole, a detector barrier can be
placed on the tube, to the left of the cathode. When pass-
Figure 1-4
ing an electric current is formed a luminous line, the for-
Goldstein tube
mation of this line can be explanied as follows:electrons
that are emitted from cathod are attracted towards anode
and because there is hydrogen gas in the tube, electrons
collide with Hydrogen neutral gas atoms, if the electrons
have enough energy, they can flush away other electrons
from the neutral atoms. positive particles remaining in its
place are capture mostly by electrons and become neutral,
a few numbers of these positive particles slip from the hole
to the area behind the cathode “so called a channel ray”.
It is formed a bundle of particles are affected by the elec-
tric field as it deflects toward the plate, which has negative
charges and is also affected by the magnetic field. Also,
they veer up toward the north finder pole looking up the Exercise 1-1
south pole and curve down the south finder pole. It also What are the properties of
found that these particles are usually heavier than electrons cathode rays?
where the mass depends on the type of gas in the tube.
These particles were later named proton.
From the study of the properties of these rays found the
following:
1- Attracted towards the negative pole, which confirms
that it is positively charged.
2- It has mass and speed which indicates that it is mat-
ter particles and are usually heavier than electrons and its
mass depends on the type of gas presents in the electric
discharge tube.
3- Affected by electric and magnetic fields.

11
 1-4 Estimating the ratio of the charge of the
electron to its mass
The first quantitative study on the deviation of electronic
beams by electric and magnetic fields was carried out by
the scientific Thomson in 1897 and Figure 1-5 illustrates a
model, where he used a cathode-ray tube and, by using the
properties of these rays. He found that these rays deviate
from their straight path under the influence of the mag-
netic field but returns to its original path when the electric
Figure 1-5
field is shed equal to the intensity of the magnetic field in
Model for Thomson tube.
a direction perpendicular to the direction of the magnetic
field. From know the intensity of the electric and magnetic
fields, it was possible to calculate the charge-to-mass ratio
of particles. Thomson shows that these constant percent-
ages do not change regardless of the metal which is used
to make the cathode or the nature of the gas used inside
the cathode ray tube. He has been found the ratio equal to
1.76 x 1011 C/kg (coulomb / kilogram).

1-5 Determination of electron charge

In 1909, Millikan assigned an electron charge by placing
small negative charge droplets between two electrodes as
in Figure 1-6, the positive electrode above tries to attract
oil droplets which is charged with negative charge and at
the same time droplets tend to descend down by the power
of gravity force. When the droplets stabilize in static state
between the poles and from the known values of radius,
density and the electric field intensity, Millikan enables to
calculate the carried charge on these droplets. The value is
1.6 × 10-19 C coulombs or one of their simple multiples be-
Figure 1-6 cause droplets can be charged with more than one negative
A model of the Melican experi- charge. The charge of one electron must be equal the value
ment. of the smallest charge carried by the droplet 1.6 × 10-19 C
and by using this value of the electron charge and the value
of the ratio of the charge of the electron to its mass which
is calculated by Thomson (paragraph 1 - 4 above) could
calculate the mass of an electron from the electron charge
found by Millikan:

12
Electron mass= (The electron charge found by Millican) / Do you know
(The ratio of the charge of the electron to its mass created Rutherford’s expression was
by Thomson) when he saw the results “It was
1.6×10-19 C something great, almost unbe-
Electron mass = lievable It was as if a shell was
1.76×1011 C/ kg
fired for range 15 inch on a thin
Electron mass = 9.1 × 10-31 kg translucent piece of paper then
she bounced again and hit you.
1-6 Detection of the nucleus
Thomson suggested that the atom could be considered a
positive spherical region, and negative electrons is embed-
ded in it ( such as a piece of cake filled with currant). The
bulk of the atoms mass must be bound by the positively
charged spherical region. It is a conclusion arise from a
note that positive parts of an atom are heavier than elec-
trons. In 1910 Rutherford conducted the traditional experi-
ment which tested Thompson model, he was examining
the dispersion of alpha particles by thin slice of metal. Ac-
cording to the Thomson model, the metal consists of atoms
as positively charged balls contains negative electrons and
because Alpha particles are too high energy, they will pen-
etrate through the metal strips as a straight line because the
positive charge and the mass is evenly distributed in the
metal (according to Thompson model) and there is little
chance of deviation from the original course. As expected,
99% of alpha particles penetrated the slice, some of them
Figure 1-7
skewed sharply and a few of them were reflected along her
A model for the Rutherford ex-
trajectory [Figure 1-7]. That was not true at all for Ruth-
permant
erford because Thomson model cannot explain these de-
viations and as a result Rutherford explained these devia-
Exercise 1-2
tions that the positive charge and mass in the metal slice
are centralized in a very small area. According to Ruther- Who attributes the discovery
ford opinion , the atom has a nucleus or center, where its of the nucleus of the atom?
positive charge and mass are centralized. The quantitative
results of dispersion experiments illustrate (such as that
Do you know
carried out by Rutherford) that the nucleus of the atom has
If the atom becomes bigger till
a diameter, which is equal to 10-13cm and atoms have di-
the nucleus size as a point (.)
ameters approximately 100,000 times the diameter nucleus
then the whole atom would be
(10-8cm).In other words, the nucleus occupies very little
bigger than a house.
space from the total size of the atom and most of its size is
a space occupied by electrons.
13
 1-7 Detection of atomic number
If the nucleus is containing protons, how many protons are
exist in a particular nucleus? This question remained un-
til the answer came from scientific observations that are
shown by Mossile in 1913 which led to the discovery of
the atomic number.

Do you know 1-8 Detection of a neutron

The atomic number (Z) is sig- At that time the question was how to interpret isotopes?
nificant as a special character- And how Different blocks of the same element can be ob-
istic In chemistry, since it gives tained, if they are all containing the same number of pro-
a number of protons the positive tons. The answer of this question is that there must be some-
charge in the nucleus which is thing else in the nucleus. In 1932 Chadwick threw a thin
equal to the number of electrons slice of beryllium by alpha particles, there were appeared
out the nucleus which make the rays resembling the rays of gamma which has very high
atom is neutral. And including energy from metal slice. Subsequent experiments showed
(Z) represents the number of that these rays represent the third model of the compo-
protons in a nucleus atom is a nents of the atom, called by Shadwick as Neutron because
nuclear property that provides its charge is neutral and its mass is roughly equal to that
us with important information of the proton. The nucleus is now composed of neutrons
about the rest of the atom. and protons. The neutron has a mass (atomic mass unit) of
1.00866 amu and charge value equal (0) and the proton
has a mass of 1.00727 amu and charge (+ 1). The symbol
(Z) of the atomic number, which represents the number
of protons in the nucleus of an atom of any element. The
symbol (A) represents the number of mass, which is equal
to the total number of protons and neutrons in the nucleus
of any element.

1-9 The configuration of electrons in an atom

The process of arranging the components of atom has
gone through several theories. These theories were fading
or evolving according to their ability to interpret physi-
cal and chemical phenomena. The model is considered to
be an atom is consisting of a positive nucleus which the
negative electrons are moving around it. This movement
of electrons counteract the gravitational force generated by
nucleus. Explanation of why these electrons do not radiate
energy as they are observed in all other cases.

14
Moving electric charges under the influence of gravita-
tional forces are losing energy. In the case of electrons,
a loss of energy will result in a slow motion of the elec-
tron, which will gradually attract it to the nucleus in a spi-
ral path approaching them to fall inside. In this spiral path
continuously gives energy as a continuous spectrum that
resembles a semi - ray solar spectrum but this does not
happen as the atoms are stable in their atomic construction.
This atomic stability led scientists to present other theories
about atomic construction.
1-10 Electromagnetic radiation
This expression includes different types of radiation, such
as warm that travels to us from a fireplace or light reflected
on bright surfaces, or radiology used in hospitals. They are
different but they are common in some intrinsic qualities,
all these types of radiation are transmitted in the space in
same speed equal to 3x108m/s, which is light speed. All of
these types are wave-nature and their waves are similar to
those which form above the surface of the water in which a
stone is thrown [Figure 1-8]. We notice repetition in these Figure 1-8
waves, that is, the wave is repeated at regular intervals and Model to form waves above the
the regular number of these waves is what is known as the surface water.
frequency of the wave that occurs per second ,and when ra-
diation passes at a specified point [Figure 1-9]. The waves
are small and their frequency is high and the opposite is
true and is related to length frequency wave in relation:
Where
C C : The speed of light in unit (m/s)
ʋ = υ : Frequency in unit (1/s)
λ Figure 1-9
λ : Wavelength in unit (m) A model of the wave nature of
1.11 Quantum theory
light.
Max Plank paved the way for the scientific revolution that
emerged at the beginning of the twentieth century, by his
hypothesis which is considered radiology electromag-
netism is like a bundle of small energy groups which he
called it Quantum. When he was studying the phenomenon
of light emission from hot objects, Planck suggested that
hot objects emit energy in small quantities is called (quan-
tum). It is the minimum amount of energy that can be lost
or gained by the body The process of losing or acquiring
energy is not a continuous process, but rather batches or
flows of energy so that the energy is lost or the acquired
multiplier is a true multiple of that quantum or quantity. 15
 Do you know The amount of quantum depends upon the frequency of
Impose a car moving at a certain radiation , and is associated with a constant (named Planck
speed, if we want to increase its constant) and its value 6.63 × 10-34 J.s . Planck’s energy is
speed to reach higher speed the known as: E=h υ Where E: Energy in Joule (J)
car must pass through all speeds h: Planck constant in unit (J.s)
that fall between lower speed and Einstein then came to expand Planck’s theory of interpreta-
higher speed. It cannot be jump tion of the photoelectric effect phenomenon, a phenomenon
from speed to higher speed, as of electrons emitting from a metal when the light is pointed
requires of the idea of quantum to, [Figure 1-10]. The emission of electrons depends on
leaps. the frequency of light, ie on its energy, if the frequency
of atomic radiation hits the metal surface in low value,
Exercise 1-3 we are not observe the emission of electrons. When using
What is a photon?
increased frequencies, the frequency suddenly reaches at
which it begins emitting electrons. Before that, however,
the radiation intensity of the low frequency was increased
the metal is not affected. If the radiation frequency exceeds
that the minimum value of emission does not increase the
number of electrons emitted but the energy carried by the
electron increases. Einstein has used Planck’s equation to
calculate the quantum energy of light, which he called (pho-
tons) and photon is a particle of electromagnetic radiation
has a mass of zero but holds a quantity of energy depends
on the frequency of the electromagnetic wave. The photon
energy has the lowest frequency needed to emit electrons
by overcoming its binding energy that electrons relate to
the atom. Einstein’s interpretation of this phenomenon was
the first application important for quantum theory. This in-
terpretation has also given light a trait particle along with
the wave characteristic

1-12 Atomic Spectra

The scientific Newton passed ordinary sunlight through
a glass prism , he found that the light decomposes into a
range of colors firstly from violet and finally to red (Fig-
ure 1-11). Because there are no separate areas between one
Figure 1-10 color to another, this spectrum is named with continuous
Photoelectric effect phenom- spectrum. But it was observed that if the atoms of a pure
enon. element exposure into the gaseous state to heat or in the
electric discharge tube in the case of low pressure emits ra-
diation (spectrum) of a glowing element atoms is not con-
nected or continuous but is consisted from a few lines of
16 light separated by relatively large dark distances.
 Figure 1-11
Continuous spectrum phenom-
enon.

call it (the line emission spectrum) because there are spac- Do you know
es between one color and another. It was later discovered Each element has a special emis-
that each element had a spectrum linearly distinguishes it sion spectrum. Emission spectra
from other elements [Figure 1-12]. linear emission spec- are used to designate Identities
trum shows that the emitted radiation from an atom emits of unknown samples and per-
a specific energy. In other words it is emitted in a quantum centage determination compo-
style rather than in a continuous manner. nents of stars.

Hydrogen lamp Figure 1-12

1.13 Linear emission spectrum for hydrogen Linear spectrum phenomenon.
Classical theory claims that a hydrogen atom can be ex-
cited by any amount of energy that receive. So scientists
predicted the emission of a series of continuous spectrum, Exercise 1-4
but the hydrogen atom emits only limited frequencies of That rainbow is a series of col-
light, why, when the current passes in the hydrogen gas ors?
under low pressure. The potential energy of some of its at- Discuss whether you consider
oms increases and the lowest atomic energy level is called this series of continuous or
(ground state). When the potential energy level of the atom linear emission spectrum.
is higher than the ground state led to atom to (excited state).

17
 When the atom returns from excited level to the stable
level, it lose the gained energy in form of electromagnetic
rays (photons) and the energy of this photon is equal to
the difference between first and second energy level [Fig-
ure 1-13]. Studies showed that hydrogen atoms release
only specific frequencies of light. The energy difference
between these energy levels of an atom is a specific differ-
ence and this means that the hydrogen electron in limited
Figure 1-13 energy level.
The transmission of an electron 1-13-1 Bohr’s theory of hydrogen atom
from a level of less energy to the After the discovery of the elements spectra, scientists tried
level of the highest energy and to explain them based on the movement of the electron
the reverse. in the atom, but the attempt failed. Suppose the electron
moves around the nucleus at a constant speed, in a curved
path and because of laws of physics that require any par-
ticle charged as an electron undergo a process acceleration
will lose energy on the image of electromagnetic radiation.
This slows down the speed of the electron and therefore
the electron will collide to nucleus and the atom collapses.
Because atoms do not collapse, scientists had to challenge
to explain how electrons rotate. The scientist Bohr used
ideas of Planck and Einstein mentioned earlier in his theo-
ry that he managed of calculating the energy of an electron
in hydrogen atom. Bohr was adopted in his theory on two
facts: first, the atoms do not collapse and the second is that
light emission from an atom is done at a certain frequency,
which means that changes in the atom energy are certain
and specific.
This shows that the electron exists in areas with specific
energy and cannot exist between them because it is forced
to specific energy levels in the atom.
In his theory, Bohr assumed that electrons rotate in spe-
cific orbits that has constant energy and volume [Figure
1-14]. Bohr theoretical hypotheses can be summarized as
follows:-
Figure 1-14 1. The electron rotate in specific orbit (with a specified
Major levels of energy with con- diameter) and a specific energy and do not radiate energy
stant volume and constant en- as a result of this rotation.
ergy. 2. Energy is emitted from the atom in one case, which is
the transmission of the electron from Specific orbit to an-
other orbit that has less energy than in first orbit.
18
Bohr’s theory has proved useful in the structure of a hy-
drogen atom, but when scientists tried to apply the ideas in
Bohr theory to atoms of the other elements was failed be-
cause the rest of the atoms contained more electrons so the
atomic spectra of these elements are more complex of the
atomic spectrum of hydrogen. That means Energy levels
is more complex and there are sub-levels of energy start-
ing from the second level so he start thinking about it again
and activate search for a more comprehensive theory.

1-14 The wave nature of the electron

We said that an electron is a particle with a specific mass
and negative charge and light has matter and wave length
nature. De Broley (1924) suggested the possibility of du- Exercise 1-5
ality nature (wavelength and particulate) in the case of
the particles also depends on what Einstein and Blank Why Bohr theory failed?
achieved.
E= mc2 ………………..(1) Einstein equation
E= hυ ………………..(2) Blank Equation
Since equation 1 and 2 are equal
hυ= mc2 ………………..(3)
Whereas υ=(c/λ), by substituting them into equation (3),
we obtain:
c
mc
2
=h λ ………………….(4)

By omitting the value of c from either side in equation (4)

it becomes:
mc = h …………………..(5)
λ
In order of equation (5) we get:
h
λ= …………………..(6)
mc

Momentum is known as the following relationship mc=p:

h‫ـ ـ ـ ـ‬
λ = ‫ـ ـ‬p …………………..(7)

Based on this derivation can a torrent of moving electrons

that leads to an interfering pattern, that is, a mobile elec-
tronic is similar to photons forming rays.
19
 1-15 Wave Mechanics
Despite the success of Bohr theory in the interpretation of
the atomic spectrum of hydrogen as well in the interpre-
tation of atomic spectra of systems in which one electron
such as( He+ and Li2+) but did not produce good results
when trying to explain the spectrum of atoms which has
more than one electron. In 1936 the scientific Schro-
Exercise 1-6
denger used the mathematics to study the hydrogen atom.
What does the excited state of A science called wave mechanics or quantum mechanics
the atom mean? and what hap- has begin. Schrodenger solved a mathematical equation
pens to the electron during it? named after by his name and also called the wave equa-
tion. Schrödinger’s idea is that instead of thinking about
an electron moving circularly in a constant orbit, we have
to assume a series of waves move within this stable orbit
and that the perimeter of the orbit must be equal a simple
multiple of the electron’s wavelength. According to the
assumption of Schrödinger electron wave behavior gives
an estimate of the probability of existence electron within
the wavelengths. This new concept is an alternative to the
system of the fixed orbits assumed by Bohr in his theory
rather than look of an electron as a particle orbiting in a
specified radius orbit wave mechanics describes the mo-
tion of an electron in terms of the wave function (Figure
1-15), which we call orbital and it depends on the total
energy and its potential energy and the coordinates of its
location (z, y, x). Nature of this electron wave behavior
makes our ability to determine the location of the electron
and its momentum (the amount of movement) at the same
time not possible. Heisenberg’s principle coined known by
his name (the principle of inaccuracy) which states it is not
possible to locate a particle and its momentum precisely at
the same time is able to measure one with greater uncer-
tainty in the accuracy of measuring the other.

Figure 1-15
The wave motion of an electron.

(a) (b)

20
Since it is not possible to draw a path for electrons, the
best thing to do is talking about the probability of finding
an electron at a specific location within an atom. The prob-
ability of finding an electron at multiple points within an
atom is mathmatical dilemma is very complicated. Quan-
tum mechanics solves this dilemma as an electron as if it
was a wave. In fact, the electrons wave properties appear
under some conditions, so that the electron beam moving
quickly, for example, appears diffraction, a characteristic
of motion waveform. To identify any electron at its atom,
solve the wave equation and know the values of the wave
function, it requires three quantum numbers. These are
real numbers relate to the potential energy and position
of electron and shape electronic cloud. Fourth electronic
quantum number must be set because the three theoreti-
cally quantum numbers were resulting from the solution
of the Schrödinger equation are not sufficient to explain
all observed properties of electrons in atoms. So the fourth
quantum number was entered to fill this deficiency and it is Exercise 1-7
called the electron spin quantum number because the elec-
How quantum mechanics was
tron can think of it as spinning around its axis as it moves
able to Explain the presence of
around the nucleus. The fourth quantum number shows the
an electron in a particular lo-
direction of the spinning.
cation in atom.
1-16 Quantum numbers
To know how electrons will be arranged we must examine
the energy levels in the atom by studying the following
quantum numbers:
1-16-1 Principal quantum number (n)
Energy levels in atom are determined by the principal
quantum number and the larger value of n the higher value
of energy level and farther the electron from the nucleus.
The value of n determines the size of the level and n take
integers 1, 2, 3, etc.

1.16.2 Momentum quantum number (ℓ)

Determines the shape of orbital in which the electron is
likely to be present and it is resulting from the movement of
the electron around the nucleus. That each principal level n
consists of one or more sub - (secondary) levels. The num-
ber of secondary levels in any orbital is equal to the num-
ber of principal quantum number, so n=1 consists of one
21
 Do you know secondary level of energy s. The second principal level n
The letters s, p, d, f were select- =2 contains two secondary levels is s and p, and the third
ed from the letters the first se- principal level, n=3, contains three secondary energy levels
quences The main Sharp, prin- are s, p and d as well as the fourth principal level of energy
cipal, diffuse, and fundamental n=4 contains four secondary levels of quantum numbers
these words represent the forms are s, p, d, and f. Each n value corresponds to a specific
of lines of linear spectra that are value for the quantum secondary number ℓ are integers
connected energy transitions. starting with zero and ending with ( n-1). If n=1, that ℓ will
start with zero and end with (1 - 1) so there is one value for
ℓ= zero but if n= 2 it is start with zero and end with (1 - 2)
by1 so there are two values for ℓ is zero and one. So when
n = 3 there are three values 0, 1, 2 and when n= 4 there are
four values: 0, 1, 2, 3 and so on (Table 1-1).
Table (1-1) The values of ℓ and the symbols of the secondary
levels indicated
ℓ value 0 1 2 3 4 5
The letter signifying them s p d f g h
In our study we will only consider secondary levels s, p, d,
f because they are only ones that are preoccupied with elec-
trons in atoms in ground state. To determine the secondary
level from any n principal level symbolically writes the
value of n for the n principal level then the character as-
signed to the secondary level, for example the secondary
level s from the second principal level has the symbol 2s
and has values(n=2, ℓ = 0). The secondary level d of the
third principal level is 3d and has quantum values (n = 3,
ℓ = 2) and so on.
Figure 1-16 1-16-3 Magnetic quantities number (mℓ)
Secondary level of S. Atomic orbitals can take the same shape around the nucle-
us but in different directions. Magnetic quantities number
indicates the direction of the orbital around the nucleus
where each secondary level consists of one or more orbit-
als. This number was used to explain the appearance of
extra lines in the spectrum when atom placed in a mag-
netic field. A spatial distribution of an electron, spherical
symmetric, ie, the probability of its existence is the same
in all trends from the nucleus [Figure 1-16]. On the other
Figure 1-17
hand the probability the presence of the electron p in some
Secondary levels of p
directions of the nucleus is more than in other. In fact, the
probability of distribution for the electron p of two diffuse
lobes to some extent, one on each side of nucleus [figure
22 1-17]. The p-level consists of three orbitals
The secondary level d consists of five orbitals and the
secondary level f consists of seven orbitals and their spa-
tial distribution is more complex to a large extent [Figure
1-18]. Each value of ℓ matched with the values of mag-
natic quantum numbers which are positive and negative
integers. When the value of ℓ=0 there is one value for mℓ
which is (0). If ℓ = 1, the values of mℓ are (+1, 0,-1) and
if ℓ= 2 the values of mℓ is (+2, +1, 0, -1, -2). When the
value of ℓ=3 then the value of mℓ are (+3, +2, +1, 0, -1, -2,
-3). Table (1-2) summarizes quantum numbers of primary, Figure 1-18
secondary and magnetic. Secondary levels of d

Table (1-2) Values of quantum numbers of primary, secondary and mag-

netic.
principal Secondary Symbol Magnetic quantum Number of
quantum quantum number (mℓ) orbitals at the
number number secondary level
(n) (ℓ)
1 0 1s 0 1
2 0 2s 0 1
1 2p +1, 0,-1 3
3 0 3s 0 1
1 3p +1, 0,-1 3
2 3d +2,+1, 0,-1,-2 5
4 0 4s 0 1
1 4p +1, 0,-1 3
2 4d +2,+1, 0,-1,-2 5
3 4f +3,+2,+1, 0,-1,-2,-3 7

1-16-4 Electron spin quantum numbers (ms)

We explained earlier that there is one orbital for the sec-
ondary level S, three orbitals for secondary level p, five for
secondary level d and seven for the secondary level f and
where these secondary levels can accommodates 2, 6, 10
and 14 electrons respectively. It follows that any orbital
can accommodate two electrons but electrons differ in the
same orbital by one important thing is that it has opposite
spindle rotation. The reason for talking about the spin of
the electron comes from sightings the magnetic behavior
of the material can be obtained from the behavior of mag-
netic for single atoms, by Ottostron experiment.

23
 In this experiment [Figure 1-19], a beam of silver atoms
Neutralization (resulting from the evaporation of silver)
passed between two magnetic electrodes. It was found that
the beam splits into two separate beams, that is, half the
atoms deflects in a certain direction and the rest deflects in
the opposite direction.
To explain this behavior that each electron behaves like a
beam of silver fine magnet. This magnetism is produced by the spin of the
atoms negative charge because it is known that the spindle rota-
hoied
silver tion of any charge is generated magnetic field. There are
two opposite directions of the spin; we expect each elec-
tron to attract each other but that disability of the attrac-
tion between the electron orbital repulsion in their charge.
two magnetic
electrodes Since the movement of the rotation of the two electrons is
board confined in only two directions so we have two values of
the spindle quantum ms are +1/2and -1/2.
Figure 1-19
It can summarize the characteristics of the electron in
Ottostron experiment.
the atom As follows:-
1. The principal quantum number n and this character-
istic indicates the order of the electrons configuration and
increase the distance from the nucleus.
2. The momentum quantum number mℓ describes the
type of orbital that the electron operates from, where its
spatial distribution is similar (for example, the electron s
has spherical distributions are symmetrical and the elec-
trons p have symmetrical distributions along separated di-
rections in the space).
3. The magnetic quantum number mℓ and this charac-
teristic defines any orbital from orbitals of the secondary
Do you know
level in which an electron is likely to be present.
To resemble the principal energy
4. The electron spin quantum number (ms) and this char-
levels around an atom by draw-
acteristic determines either of the two possible directions
er, that does not have a regular
of the spin of the electron.
dimensional drawer. The dis-
5. When the four characteristics of the electron are de-
tances are very high at first, then
termined in an certain atom, it will discover that it cannot
gradually diminishing whenever
exist in the same one atom another electron that has a set
you move away these levels from
similar to those of its the four characteristics. This fun-
nucleus.
damental definition is known as (the Puli exclusion prin-
ciple). This definition states that it cannot two electrons in
the same atom have the same values for all quantum four
number.
24
1-17 How to write an electronic configuration
For the purpose of writing the electronic configuration of
any atom, you must know the atomic number of that atom,
whereas the number of electrons of an atom is equal to
their atomic number (number of protons) if electrically
equivalent. Well it should we know that the basic process
in writing an electronic structure of an atom is that we start
by filling the orbitals with electrons of the less energy and
then the most energy, thus where they are arranged as fol-
lows: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p
7s 5f 6d , In the case of secondary levels containing more
than one orbital such as levels (p, d and f) electrons en-
ter individually in these orbitals to avoid electric repulsion
between charges according to the Hund’s rule. This rule
states that there is no duplication between two electrons in
the secondary energy level only after orbitals occupies an
individual firstly.
Exercise 1-8
Even if all orbitals of the secondary level now have an
one electron in each of them as at the secondary level p3 Write the electronic configu-
then the fourth electron enters to double ration of following elements:
with one electron that preceded it to make full orbitals 30
Zn , 20Ca
in the secondary level p4 and to overcome the
repulsion between the two electron charges within a single
orbital, the two electrons are represented by two opposite
arrows.
Example 1-1
Write the electronic configurations for the following ele-
ments: 23V ، 19K ، 18Ar
Solution
18
Ar 1s2 2s2 2p6 3s2 3p6
19
K 1s2 2s2 2p6 3s2 3p6 4s1
23
V 1s2 2s2 2p6 3s2 3p6 4s2 3d3

If we note in the previous example of the elements that 23V,

19
K That the electronic configuration of the first 18 elec-
trons of them is (1s2 2s2 2p6 3s2 3p6). This is the electronic
configuration of [Ar] so we can simplify this configuration.
So we will write the order of the first 18 electrons of po-
tassium 19K and vanadium 23V as well as for the elements
beyond which to [Kr]
As follows: 19K [Ar] 4s1 , 23
V [Ar] 4s2 3d3
25
 Example 1-2
Write the electronic configuration of the elements 29Cu , 24Cr
Solution
24
Cr [Ar] 4s2 3d4
Note that level d needs an electron to become half satu-
rated and atoms when their secondary levels are half full
or full be more stable so we write the configuration as
follows: 24Cr [Ar] 4s1 3d5
As for 29Cu
29
Cu [Ar] 4s2 3d9
So we write it as follows:
29
Cu [Ar] 4s1 3d10

Example 1-3
Write the electronic configuration for : 47Ag , 36Kr
Exercise 1-9
Solution
Set the four quantum values of Kr 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
36
the last electron of each of the or [Ar] 4s2 3d10 4p6
following atoms: Ag 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
47
I, Ba, 27Co
53 56 5s1 4d10
or
[Kr] 5s1 4d10

Example 1-4
Set the four quantum values of the last electron for each
of the following element atoms: 54Xe ، 23V ، 17Cl ، 3Li
Solution
3
Li 1s2 2s1
n=2 because the electronic configuration ended
in the second level 2
ℓ=0 because the electronic configuration ended
with the secondary level s
mℓ = 0 because the last electron is located in the
secondary level s
ms = +1/2 Because the last electron rotates clock
wise (the first electron of the Orbital).

26
17
Cl 1s2 2s2 2p6 3s2 3p5
n=3 Because the configuration ended with the third prin
cipal level 3
ℓ=1 Because the configuration ended with the secondary
level p
mℓ = 0 Because the last electron is located in the orbital
which has a value of ℓ = 0
+1 0 -1

ms = -1/2 Because the last electron is the second electron

in the orbital.
V 1s 2s2 2p6 3s2 3p6 4s2 3d3
2
23 Exercise 1-10
n = 3 Because the highest principal level contains electrons
at Write the four quantum num-
the third bers of the penultimate elec-
ℓ=2 Because the electronic configuration ended with the tron for each of the following
secondary level d atoms:
mℓ = 0 Because the last electron is located in the orbital Cs, 36Kr, 26Fe
55
whose value is m ℓ = 0
+2 +1 0 -1 -2

ms = +1/2 Because the last electron is the first electron in

the orbital.
54
Xe 1s 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6
2

n = 5 Because the highest principal level contains elec

trons is the fifth Exercise 1-11
ℓ=1 Because the electronic configuration ended with the Write the electronic configu-
secondary level p ration of following ions
mℓ = -1 Because the last electron is located in the orbital O= , 3Li+1
8
whose value is m ℓ = -1
+1 0 -1
ms = -1/2 Because the last electron is the second electron in
the orbital.

Example 1-5

Write the four quantum numbers of the last electron for

each of the following ions: 20Ca2+ , 9F-
Solution
F- fluoride ion is a atom that has gained an electron and
its number of electrons becomes 10
9
F- 1s2 2s2 2p6
n=2 Because the highest principal level contains elec
trons is the second
ℓ=1 Because the electronic configuration ended
with the secondary level p

27
 mℓ = -1 Because the last electron is located in the orbital
whose value is m ℓ = -1
+1 0 -1

ms = -1/2 Because the last electron is the second electron

in the orbital.

The Ca2+ calcium ion is a calcium atom lost two electrons

Exercise 1-12
and the number of electrons become 18
Compare the four quantum 20
Ca2+ 1s2 2s2 2p6 3s2 3p6
numbers of an electron the last n=3 Because the highest principal level contains electrons
of both atomic elements. is the second
Si , 13Al ℓ=1 Because the configuration ended with the secondary
14
level p
mℓ = -1 Because the last electron is located in the orbital
which has a value of ℓ=-1
+1 0 -1

ms = -1/2 Because the last electron is the second electron

in the orbital.

Example 1-6

Compare the four quantum numbers of the last electron

to each of the two element atoms 3Li and 11Na
Solution
3
Li 1s2 2s1
n=2
ℓ=0
mℓ = 0
ms = +1/2
Exercise 1-13 11
Na 1s2 2s2 2p6 3s1
n=3
If the last electron of an ele- ℓ=0
ment atom has the following mℓ = 0
four quantum numbers: ms = +1/2
The difference is only in the principal quantum number.
n=4، ℓ= l، mℓ=0، = ms= -1/2
write the electronic configu-
ration of this atom and what Example 1-7
its atomic number.
If the last electron of an atom of an element has the fol-
lowing four quantum numbers
n = 3 ، ℓ = 2 ، m ℓ = +1 ، ms = -1/2
What is the atomic number of this element?
n=3 The principal level will be the third
ℓ =2 The secondary level is the level and contains five
orbitals

28
mℓ = +1
Located in the orbital indicator with the letter X
+2 +1 0 -1 -2

ms = -1/2 is the second electron

+2 +1 0 -1 -2

So from previous information the last secondary level in the Exercise 1-14
electronic configuration will end with a secondary 3d level Write the electronic configu-
which contains seven electrons then become an electronic ration of 5B atom
configuration of secondary levels in this atom as follows:
Then write the quantum num-
1s2 2s2 2p6 3s2 3p6 4s2 3d7
bers for all the electrons
adding the number of electrons above the secondary levels in in which .
the electronic configuration and the sum value represents the
atomic number of this element.
So Atomic number = 27

Example 1-8

Write the electronic configuration of the 25Mn atom and

write the quantum number of the electrons, for the last sec-
ondary level.
Solution
Mn
25
1s2 2s2 2p6 3s2 3p6 4s2 3d5
+2 +1 0 -1 -2

Electron Quantum number

n ℓ mℓ ms
First 3 2 +2 +1/2
Second 3 2 +1 +1/2
Three 3 2 0 +1/2
Four 3 2 -1 +1/2
Five 3 2 -2 +1/2

29
 Basic concepts
Cathode ray Bohr theory
A bundle of particles emerges from the cath- 1. Electron orbits in a specific orbit and a
ode to the anode direction that when passing specific energy.
a electric current in the electrode discharge 2. Energy is emitted from an atom when an
tube contains hydrogen gas under low pres- electron is transmitted from a higher energy
sure and affected electric and magnetic fields. orbit to a lower energy orbit.
It has a negative charge and mass. Uncertainty principle
Channel ray The location and momentum of an electron
A bundle of particles pops up when result- cannot be precisely located at the same time
ing particles slide that came from the expul- if measure one of them increased the lack
sion of electrons from gas neutralized atoms certainty in the measurement accuracy of the
through the hole in the metal into the area other.
behind the cathode. It affected by the elec- Principal quantum number
tric and magnetic fields and it has a positive The number determines the electron distance
charge. Its mass depends on the type of gas from the nucleus and the amount of energy of
used in the electric discharge tube. an electron at energy levels. The values of
Electromagnetic radiation these numbers will be positive integer values.
An expression used to identify different types Momentum quantum number
of radiation and share in similar properties The number determines the shape of the elec-
as transition in the space in same velocity. tronic cloud that probably having an electron
Each type of radiation has wave property and in them. This number comes from movement
special frequency of its own and also has a electron around the nucleus for each one
particle property. principal or more level of the secondary lev-
Quantum els are s, p, d, f etc.
The minimum amount of energy that can be Magnetic quantum number
lost or gained by the body and the process of The number determines the direction of the
loss of energy or gained in the form of fluxes orbital around the nucleus and shows the
or flows of energy. The amount of energy lost most probable for a position in any exists or-
or gained is a multiple integer to this quantum. bital and for each secondary level one orbital
Continue emission spectrum or more.
A group of decaying colors of sunlight which Spin quantum number
It starts from violet and ends in red and be The number determines the angular momen-
continuous and connected to each other. tum of the electron rotation around himself
Line emission spectrum and this movement is either towards a scor-
The decaying group of colors of teased pure pion clock or counterclockwise.
element atoms in gaseous state. Each color Pauli - excluded principle
is separated from other color by large dark Two electrons in the same atom cannot be
spaces and each element has a linear spec- they have one value for all four quantum
trum that distinguishes it from other ele- numbers.
ments.
30
 Chapter One Questions 1
1-1 Describe the model of electric dis- 1-12 What is the frequency wave?
charge tube with drawing, explain the What are the units of frequency, then
discovery of the electron? state the mathematical relationship be-
1-2 What are the characteristics of chan- tween frequency and wavelength.
nel rays. 1-13 Explain the experience of Auto-
1-3 What do the symbols (Z and A) mean? stron, and explain its importance.
1-4 Explain the effect photoelectric phe- 1-14 What quantum theory assumes.
nomenon. Inter-explanation Einstein to 1-15 Explain the following
this phenomenon. And what gave this ex- A- Non-repulsion of two electrons in
planation on the nature of light. the same orbital
1-5 What happens to an electron when it B- Millikan’s experience is comple-
gains energy. mentary to the Thomson experience.
1-6 What is the importance of exposing C- The secondary level s can accommo-
the electric discharge tube for magnetic date only two electrons either the sec-
and electric fields when studying proper- ondary level p accommodates just six
ties of electrons and protons. electrons.
1-7 Talk about the experience of Millikan D- The position and momentum of an
and what is the important of that have electron in an atom cannot be set in high
find out. resolution simultaneously.
1-8 How Rutherford discovered the nu- E- Secondary level 3p is filled with
cleus of an atom. What are the relation- electrons after secondary level 3s.
ship between the size of an atom and its F- The strongest attraction of the nucle-
nuclei. us on the electron closest to it.
1-9 What are the difference between 1-16 Define the four quantum numbers
A- Rutherford’s model and Thomson’s ( n, ℓ, mℓ, ms) What do they benefit
model of atomic structure. from each?
B- Rutherford’s model and Bohr model 1-17 Draw the shape of the orbital when
of atomic structure the value of ℓ = 0 and when they are
C- Forms of the principal levels of Bohr valuable ℓ = 1.
and quantum theory. 1-18 What we mean by all of what
D- Linear and continuous emission spec- comes.
trum. A- Photon.
1-10 Derive the mathematical relation- B- The wavelength.
ship λ=h/p based on Planck and Einstein C- The double character of the electron.
equations. 1-19 What benefits from Hund’s base in
1-11 What is the difference between the the electronic configuration.
secondary level and the orbital? What are
the numbers of orbitals in the first princi-
pal four levels.

31
1-20 show 1-28 Set the four quantum values for the
A- The Heisenberg principle. electrons located in the last major level of
B- Pauli’s exception principle. each The following atoms 4Be, 15P
C- How an atom can emit a photon. 1-29 Set the four quantum values for the
D- How Schrödinger was able to explain last electron only for each of the follow-
the spectrum of atoms which have a high- ing atoms 78Pt, 35 Br, 19K
er atomic number of hydrogen. 1-30 choose the correct answer for the
1-21 What are the values of the princi- following:
pal and secondary quantum numbers and 1- When the North Pole finder is rounded
magnetism in the following major levels to a magnet On cathode rays, the lumi-
(third and fourth). nous line is curved into:
1-22 If you know that the values of the A- Down
four quantum numbers of the last electron B- The highest
for the atoms (A, B, and C), respectively. C- Not affected
A 2- Mass of one electron equal:
n = 4 ℓ = 2 mℓ = +1 ms = -1/2 A- 1.76x1011 C/kg
B B- 1.6x10-19 e
n = 2 ℓ = 0 mℓ = 0 ms = + 1/2 C- 9.1x10-31 kg
C 3- The discovery of neutron is attributed
n=3 ℓ=1 mℓ = -1 ms = -1/2 to the scientist:
Write the electron configuration of atoms A- Thompson
(A, B, and C). What is the atomic number B- Henry Mosley
of each. C- James Chadwick
4- The light wavelength is proportional to
1-23 Mention at least two elements ends its frequency
up with distributed electron levels .s2 d6 A- directly
1-24 Write the electronic configuration of B- inversely
the following ions 29Cu + 2, 29Cu +1 C- equal
1-25 Write the electronic configuration 5- In the photoelectric effect phenomenon
of atoms and ions of the following 79Au, if the radiation exceeds minimum emis-
34
Se, 16S-2 , 37Rb, 50Sn sion value:
1-26 Compare the four quantum numbers A- The number of emitted electrons in-
of the last electron for each of the atoms creases
for following elements 38Sr, 21Sc, 20 Ca B- The number of electrons emitted de-
1-27 Write the electronic configuration creases
of the atom (9F) then write the quantum C- The energy carried by the electron in-
numbers of all the electrons. Show the creases
principle that corresponds through your
feedback to values quantum number for
electrons.

32
6- Quantum number that determines the C- n = 4 ℓ = 2 mℓ = -2 ms=+1/2
shape of the electronic cloud is : 12- If the values of the four quantum n
A- Principal. number for the last electron in the element
B- Secondary. atom is n = 3 ℓ = +1 mℓ = 0 ms=+1/2
C- Magnetic. The atomic number of the element is:
7- The diameter of the atom is greater A- 12
than the diameter of its nucleus B- 13
A-1000 times C- 14
B-10000 times 13- If the values of the four quantum
C-100000 times numbers of the penultimate electron in
8. If n = 2, there are: element atom is
A- one specific value for the secondary n = 2 ℓ = 1 mℓ = 0 ms=-1/2
quantum number = zero. The atomic number of the element is:-
B- two specific values for the number of A- 9
secondary quantum = 0 and 1 B- 10
C- three specific values for the number of C- 11
secondary quantum = zero, 1 and 2. 14- The electronic configuration of an
9. The principle which states that two atom of an element was:-
electrons cannot exist in the same atom 1s2 2s2 2p6 3s2 3p3
would have the same values, of the four The four quantum numbers of electrons
quantum is: are in secondary level 3p3 will only differ
A- The Heisenberg inaccuracy principle. in:
B- The Pauli exclusion principle. A- Secondary quantum number.
C- The principle of the rule of Hind. B- Magnetic quantum number.
10. If the electronic configuration ends at C- Electron spin quantum number.
the secondary level 2p6 there is a possibil- 15- The four quantum numbers of elec-
ity: trons are in secondary level 4p2 will only
A- The atom should be only one element differ in:
which is an element.10Ne A- the secondary quantum number.
B- That there are two element atoms, B- the magnetic quantum number.
namely 9F and 10Ne C- the electron spin quantum number.
C- That there is more than one atom el- 16- The four quantum numbers of elec-
ement in addition to 10Ne that these ele- trons are in secondary level 5p6 will only
ments are lost or acquired electrons to be differ in:
rearranged like an atom element 10Ne A- the secondary and electron spin quan-
11- If the electronic configuration of an tum number.
element’s atom ends 4s2 3d5, so The four B- the secondary and magnetic quantum
quantum numbers of the last electron be: number.
A- n = 3 ℓ = 2 mℓ = -2 ms=+1/2 C- the magnetic and electron spin quan-
B- n = 4 ℓ = 2 mℓ = +2 ms=+1/2 tum number.

33
17. In his theory, Bohr assumed that elec- 22. Quantum number which indicates the
trons orbit at : direction of the orbital around the nucleus
A- specific size and energy orbitals. is:
B- specific size and variable energy orbit- A- Secondary quantum number.
als. B- Magnetic quantum number.
C- variable size and specific energy orbit- C- Electron spin quantum number.
als. 23. Principle quantum number is always
18. In an electrostatic discharge tube pos- equal for number:
tive particles that slides through the hole A-Secondary levels.
into an area behind the cathode are called B- Orbitals.
A- cathode rays. C- Electrons.
B- channel ray. 24. The electron orbits in a specific orbit
C- x - ray. with specific diameter and energy (ac-
19. If we assume that the atom includes cording Bhor assumption) and as a result
only protons in its nucleus, it means: of its rotation:
A- There is no difference in the atomic A- Emit energy.
number of all elements atoms. B- Absorb energy.
B- There is no difference in the number of C- Does not emit energy.
mass of all elements atoms.
C- There is no difference in the mass
number of all atoms for one element.
20. Plank assumed when he was studying
a phenomenon of light emission of hot
objects:
A- Hot objects release electromagnetic
energy in the form of waves.
B- Hot objects release electromagnetic
energy as specific small amounts.
C- Hot objects release electromagnetic
energy as continuous small amounts.
21. The atom is excited:
A- when the energy of the atom is at a
stable level.
B- when the potential energy level of atom
becomes higher than the stable level.
C- when energy is lost in the form of elec-
tromagnetic radiation (Photon).

34
 More polarity Less polartiy

Chapter Two
Bonding forces and geometric shapes
of molecules
2
After completing this chapter, the student is expected to:
• Understand why atoms combine together and the meaning of chemical bonds and
their types.
• Understand the factors that determine the type of bonding and know how the
atoms combine among them.
• Determine the properties of ionic compounds and distinguishes between their
properties and those of covalent compounds
• Realize the absence of molecules in ionic compounds.
• Understand the concept of hybrid orbitals and their types.
• Recognize the principle of resonance and its effect on the effectiveness of some
covalent compounds.
• Draw the geometric patterns of some molecules.
• Differentiate the bonds of sigma and the bonds of π.
35
 2-1 Introduction
Before we get into the details of how chemical reactions
Do you know
occur among the atoms of the elements in nature around
us, we must review chemical properties of noble or inert
The number of elements already
elements, this group that we have already studied in pre-
in nature is 94 elements, As for
vious levels, which include elements of helium He, Neon
the elements prepared industri-
Ne, Argon Ar, Krypton Kr and xenon Xe has a special and
ally are 26 elements due to chem-
important characteristic that distinguishes it from other el-
ists’ efforts in world, the number
ements in nature.
of elements are 120 elements.
But the resulting compounds
All elements of this group are chemically inert and on most
within the various branches of
are ineffective in normal circumstances. because the outer
chemistry are in millions. These
energy level has saturated with electrons, unlike the rest of
compounds are increasing with
the existing elements in nature. So the formation of elec-
the days and the main and basic
trons saturated orbitals would be the goal of any atom to
reason of that is nature of the at-
achieve chemical stability.
oms that are bonded together to
form independent molecules or
It is worth to mention that there are a number of elements
crystalline structures. Chemical
found in nature as a single unit (molecule) like oxygen O2
bonds are the ladder, which if
and nitrogen N2 and sulfur S8, where the atoms are bound-
the chemist guided in his labora-
ed together by bonds to arrived their external orbitals to
tory to create new materials as
saturation (which resembles noble gases).
we see it in our daily lives.

All atoms of the elements except those of noble gases pos-

sess uneven chemical activity is under normal conditions
and can enter into chemical reactions to saturate the last
energy level through loss or acquire or share electrons to
access the electronic configaration to the nearest noble (in-
ert) gas.

The tearm (chemical bonding) are mainly used to know

how atoms or ions are bonded together to form covalent
or ionic molecules or others, which is mainly used to un-
derstand and follow chemical reaction process that lead to
creation of new chemical compounds [Figure 2-1].

Here we have to go through our imagination on some logi-

cal scientific questions we need to answer, in order to com-
plete the general idea of how these bounds occur:

36
* How and why? Atoms are bound together to form mul-
tiple molecules atoms were simple or complex?
* Why different chemicals materials possess different
physical properties from each other such as color, melt-
ing point, boiling point, acidity and thermal conduction,
electrical conductivity and solubility in different liquids
.....etc?
* Why chemicals materials have different chemical quali-
ties in terms of resistance to the effect of acids, bases or
various chemical reagents and affected by temperature?

Hydrogen Carbon Oxygen

atom (H) atom (C) atom (O)

Water molecule Carbon dioxide Ethyl alcohol

(H2O) molecule (CO2) molecule (C2H6O)
Figure 2-1
Many other questions need appropriate scientific answers Chemical bonding
to them to explain causes, and thus extrapolate other cas-
es where extrapolation is an investigation to the desire of
man and his curiosity to uncover the secrets of scientific
knowledge. For the purpose of remind students by some of
the concepts discussed in the previous year we will review
some of them in the following items:

2-1-1 Chemical reaction

The chemical reaction between two elements is an in-
teraction between atoms of two elements to form a new
compound molecule has chemical and physical properties
completely different from the characteristics of the ele-
ments involved in the reaction. Its purpose is to form orbit-
als saturated with electrons to achieve chemical stability.
There are many scientific facts and notes must be in the
thinking of students, when they are studying and learning
of chemical reaction project as follows:

37
 1. The nuclei of elements atoms does not enter the chemical
reaction, the same is true for internal electronic orbitals which
are saturated and stable.
2.The chemical reaction is limited to the sharing of valence
electrons located in the outer shell of unsaturated orbitals.
3. The atoms of the elements involved in the compound com-
position are strongly associated with abinding on the elements
involved. Stability or weakness of bonds in the resulting com-
pound depends on the nature of the bonding force between the
atoms; the most important is the electronegativity for involved
atoms.
4. When partition the compound into its initial elements, ie sep-
arating the two elements from each other with, various chemi-
cal or physical methods dismantling of the bonding force (ie,
bond breaking) are require. Bond breaking which needs energy
equal to that emitted energy to bond formation.
5. When elements are combined, the electronic configuration of
external unsaturated orbitals are changed to reach stable state
as the atoms which has saturated molecular orbitals.

2.1.2 Lewis symbol and the Octet Rule

Exercise 1-2
You knew from your previous study that electrons in the outer
Write Lewis symbol for shell of atom element’s is configured in Lewis order, so that the
atoms: 16S, 11Na, 10Ne symbol of the chemical element is written surrounded by dots
each point representing one electron and every two adjacent
points are an electronic pair. These points are distributed to the
four sides surrounding the symbol so that do not exceed two
points on each side.

Example 2-1
Write the Lewis symbol for magnesium (12Mg) and chlorine (17Cl)
Solution
Electronic configuration of magnesium atom:
12
Mg : 1s2 2s2 2p6 3s2

So Louis’ symbol for the magnesium atom is:

Electronic configuration of chlorine atom:

17
Cl : 1s2 2s2 2p6 3s2 3p5

So Louis’ symbol for the chlorine atom is:

38
These electrons determine the nature of bonds between at-
oms and also determine chemical formulas for compounds
resulting from the atoms union when covalent bonds forms
between atoms, they share a number of electrons to be ar-
ranged similar to the noble gas closest to them which has
eight electrons in its outer shell (except helium who has
only two electrons). This can be illustrated by writing
Lewis to a number of molecules as follows:
Water molecule (H2O) the electronic configuration of an
atom (H) is 1H: 1s1
Lewis symbol of atom (H) is H.

And atom(O) is 1s2 2s2 2p4

Lewis symbol of atom (O) is

Water molecule is

Ammonia molecule (NH3) The electronic configuration of

an atom (N) is N: 1s2 2s2 2p3
and the hydrogen atom as mentioned above.

Lewis symbol for the ammonia molecule is

You may notice that each of the atoms O and N in these

compounds, is surrounded by eight electrons and this rule
applies to atoms in many compounds. This rule is called
the (octet rule), where the central atom in the molecule
is surrounded by eight electrons. This rule is not applica-
ble for all the atoms in the molecules as in the phosphorus
pentachloride PCl5 and Boron trifluoride molecule BF3

Exercise 2-2
Which one of these molecules
apply octet rule to Its central
atom the CH4 or BeF2. Note
It notes that the central phosphorus atom was surrounded
that the atomic number of
by ten electrons, as for the boron atom is surrounded by
F = 9, Be = 4, H = 1, and C = 6
six electrons, so it does not agree with rule of eight. Au-
thorized rule of eight is not general in all cases as there are
many atoms that do not agree with this rule.
39
 2-2 Types of chemical bonds
Chemical bonds are the phenomenon of the presence of atoms
coherent together in molecule or crystal. The material atoms
are bonded together by chemical bonds, the type and strength
of the chemical bond depends on the electronic configuration
of the constituent atoms that form bond. There are several types
of chemical bonds can form between the atoms of different ele-
ments, which are:
2-2-1 Ionic bond
Ionic bonds emerge between a metallic element and a non-me-
tallic element through loss and gain electrons. It results from
the interaction of two atoms, one of which have High electro-
negativity (such as atoms of halogen group elements) and oth-
ers have low electronegativity (e.g., alkaline earth metals group
and alkaline metals group elements). In this case, the valence
electron will move completely from the atom of the low elec-
tronegative element into the atom of the high electronegative
Figure 2-2 element, and then we have two ions, the first is positive charge
Ionic bond of NaCl. ions as a result of the loss of an valence electron , the second is
negative ion as a result of the reception of this electron[Figure
2-2]. These different charged ions are then bound as a result
of the electrostatic attraction to form the complex and form
neutral crystal lattice [Figure 2-3] like; sodium chloride NaCl,
potassium chloride, KCl, magnesium chloride MgCl2 and po-
tassium fluoride KF and calcium chloride CaCl2 , as well as all
hydrides of alkaline metals group elements and alkaline earth
metals group elements, such as sodium hydride NaH.
The most important properties of ionic compounds are:
1. It exists in form of a crystalline structures, a regular geomet-
ric arrangement of negative and positive ions.
Sodium ion 2. It has very high melting and boiling point, to overcome the
(Na )
+
attraction forces between the negative and positive ions and to
Chloride ion break the crystalline structures.
(Cl-) 3. Inability of conductive electrical in solid state due to ions
connection and their inability to move within the crystalline
Figure 3-2 structures while becomes conductive electrical, when melted
Crystalline structure of sodium or dissolved in water (then the ions will be free to move in the
chloride. melted and aqueous solution).
4. Soluble in polar solvents such as water and do not dissolve in
non-polar organic solvents such as gasoline or ether.

40
 2-2-2 Covalent bond
Covalent bond is formed when the valence electron is dif-
ficult to transmit a complete transition from one atom to
another, in this case the pair is formed from electronic con-
tribution or participation of both atoms. Charges are not
shown on atoms, covalent bonds often occur between non-
metals. The difference in the electronegativity value (see
electronegativity values for some elements in Table 2.1)
plays an important role in the formation of covalent bonds.
It can lead to two types of covalent bonds:
A- Pure covalent bond:
This bond emerge between two atom of non-mitallic ele-
ments, where they are similar in electronegativity, or be-
tween two elements atoms that are similar in electronega-
tivity, where the difference in electronegativity is zero for
the two case. The pair of electrons will spend equal time in
the acquisition of both atoms [Figure 2-4].
An example of this type of covalent bond is the bond in the Figure 2-4
nitrogen molecule N2, the chlorine molecule Cl2 and the Pure covalent bond.
oxygen molecule O2 and in the fluorine molecule F2.
B- Polar covalent bond:
This bond emerge between two elements atom where they
are similer electronegativity but in this case the difference
must be greater then zero and less than 1.7 to participation
with one or more electrons pair [Figure 2-5].
Examples of this type of covalent bonds found in mol-
ecules H2O, and ammonia NH3 of water H2O, ammonia
NH3, Hydrogen fluoride HF, aluminum chloride AlCl3, and
brominde HBr, In this case one of the atoms carry partial
negative change (negative delta δ-) and the second atom Figure 2-5
carry partial positive change (positive delta δ+). Polar covalent bond.
Table (2-1) Electronegative values of some elements of the pe-
riodic table.
Element Symbol Electro- Element Symbol Electro-
negative negative
Fluorine F 4.0 Silver Ag 1.9
Oxygen O 3.5 Iron Fe 1.8
Nitrogen N 3.0 zinc Zn 1.6
Chlorine Cl 3.0 Manganese Mn 1.5
sulfur S 2.5 Aluminum Al 1.5
Hydrogen H 2.1 Magnesium Mg 1.2
Nickel Ni 1.9 Lithium Li 1.0
Copper Cu 1.9 Sodium Na 0.9
Carbon C 2.5 Potassium K 0.8 41
 There are many types of covalent bonds that differ in number
of electronic couplings bonding between atoms. most com-
mon covalent bonds is a single bond, which share only one
electronic pair, such as molecule F2. When participating in two
electronic pairs, they are called double covalent bonds, In the
case of participation in three electronic pairs makes it a triple
covalent bond. An example of double bond is what we find in
the oxygen molecule O2. An example of triple bond is what we
Single covalent bond. find in the nitrogen molecule N2. Covalent bonds compounds
are characterized by the following:
1. Low melting and boiling point, so do not need high heat en-
ergy, because the forces of attraction between their molecules
are weak.
2. Do not conducted electric current because they dont form
negative or positive ions in their melts or aqueous solutions.
Double covalent bond. 3. Mostly don’t dissolve in polar solvents as water while dis-
solve in organic solvents such as ether and benzene.
2-2-3 Coordinate bond
Coordinate bond is formed when one of the atoms provides
a pair of electrons to another atom have the ability to receive
this electronic pair to form bond. Then this pair will be shared
between two atoms. The electron donor atom is the Lewis base
and contains a pair of free electron such as oxygen atom in a
Triple covalent bond. water molecule or a nitrogen atom in the ammonia molecule.
The receiving atom is often a transitional metals (Lewis acid)
because they have empty orbitals of the type d such as nickel or
hydrogen atom ion.
We can say that coordinate bond Is a special type of covalent
bond, except that the source of the electron pair is only from
one atom, and the coordinate bond is longer and weaker than
Exercise 2-3
covalent bond.
Use the Lewis symbol for at-
oms F, O and N to illustrate Example 2-2
the formation of covalent Graphically illustrate the emergence of the coordinate bond
bonds single, double and tri- in the ammonium ion NH4+
ple. Solution
This ion consists of the binding of ammonia NH3 to the hy-
drogen ion H+ in aqueous solution:
H H +
H +
H N +H
H
+
[H N H
H
] [ H N→H ]
H
42
Note from the previous example that the nitrogen atom has a Exercise 2-4
pair of electrons not involved in the NH3 molecule which can Represent the two ions H3O+ ,
be involved, The hydrogen ion has an empty orbital that can re- BF-4 using Lewis symbol and
ceive this pair. When the ammonia molecule is close enough to interpreted the formation
hydrogen,an attraction occurs and the pair of electrons shared of coordinate bond between
between, as in covalent bond exactly and form NH4+ ion. This
them.
type of bond is called coordinate bond and is referred to in the
order of the Lewis symbol with an small arrow rather than a
line that represents a covalent bond.
2-2-4 Metallic bond
A chemical bond that happens between the atoms of an ele-
ment of metals, this bond is formed due to the metal atoms
possessing electrons in their outer shells contribute to the for-
mation of a crystal and these atoms has free electrons move-
ment in this crystal. When metal atoms are binding together,
they do not reach the electronic configuration of noble gases.
Atoms of metals such as sodium and potassium, they are easy
to lose their equivalent electrons and become positive ions be-
cause their electronegativity is low. The strength of the metal
bonds is affected by several factors, the most important is the
charge density which is equal to the ion charge / ion size (pro-
portional to the number of orbits), where the ion charge is the
charge that the metal gains after losing electrons in the last orbit
(+1,+2,+3).
Therefore, the force of the metallic bond depends on the num-
ber of the electrons valence beam in the metal atoms, the more
electrons of the valence beam, the greater the coherence of
the metal to be more solid and higher in boiling point. Higher
charge density on the ion increased bond strength and as a re-
sult higher melting point obtained. Many of the properties of
natural metals depends on the nature of this bond, electrical
conductivity and thermal conductivity of metals caused by the
movement of free electrons between atoms.
2-2-5 Hydrogen bonding
Hydrogen bonds arise due to the attraction that occurs between
the positive terminal (hydrogen atom) and the negative termi-
nal atom possess electronic pair or more. These three properties Hydrogen bond in water
are limited to three elements only are ; oxygen, fluorine and molecule.
nitrogen atoms. Therefore we find water, ammonia, hydrogen
fluoride molecules and others are agglomerated by effect of hy-
drogen bonds.
43
 The hydrogen bond is the cause of the high boiling point
of water, the melting point of ice and the expansion of the
volume of frozen water that led to ice floats on the water.
Hydrogen bond is a weak physical bonding force between
molecules and not a precise chemical bond: therefore its
strength is much lower than other.

2-3 Resonance
Some types of bonds can have more than a dot shape such
as ozone (composition I, II) in which the central atom has a
single bond with one atom and a double bond with another.
The dots shape cannot tell us which of the atoms has a
double bond. Both atoms have the same chance of a double
bond. These two structures are likely to be called resonance
Hydrogen bond.
structure or resonance. In fact, the structure of ozone is a
hybrid resonance structure between its resonance structure
(composition III). Rather than having a double bond and
a single bond, in fact, the two bonds are equal, which is
a middle state between the single bond and double bond,
Composition I where three electrons are at each all time. Carbonate ion
(CO3-2) can also be drawn by any of the following formu-
las:

Composition II

It is clear from the structures above that the central atom

(C atom) is connected with the other atoms (O atoms) with
two mono bonds and one double bond alternating their
Composition III position on the three atoms and these potential structures
Hybrid resonance structure are called resonance. In fact, the ion structure (CO3-2) is a
between the two structures hybrid resonance structure between the above three struc-
I,II tures as following
Resonance of ozone
Exercise 2-5
Draw the resonant shapes
of the sulfate ion(SO4-2) and
phosphate ion(PO4+3).

44
2-4 Geometric shape of the molecules
The molecules of chemical compounds take certain geo-
metric forms that control in its composition a number of
factors, the same that control the type of chemical bond
(whether ionic, covalent or metallic), which are:
1. Number and type of atoms associated in the molecule.
2. Electronic configuration of element atoms involved in
the formation of the molecule.
3. The ability of atoms to acquire, lose or share valence
electrons.
4. The presence or absence of an empty secondary outer
shell in the atom.
The atom reaches a more stable state and a minimum en-
ergy state so that the repulsion between its electrons is
minimums in atoms and attracting as much as possible.
Theories explain the formation of the bonds and the forms
of molecules and recall the most important now, these the-
ories:
2-4-1 Valence Shell Electron-pair Repulsion Theory (VSEPR)
This theory explains the configuration of atoms around a
central atom based on the repulsion between pairs of elec-
trons involved or not involved in the valence shell of the
central atom. The repulsion between these pairs in mini-
mum, when they are as far as possible from each other so
as to achieve this greater gravitational attraction of the at-
oms makes them more stable and less energy. To illustrate,
the geometry of some molecules can be studied:
BeF2 molecule:
When Lewis symbol is written for both of 4Be beryllium
atom and the 9F fluorine atom as follows:
The Lewis symbol shows that the beryllium atom shares
the two fluorine atom by pair of electrons atoms to have
a two pairs of electrons shared around beryllium atom. In
order for the electrons to be in the lowest state of repulsion,
they are distributed on both sides of the beryllium atom
and the fluorine atoms are distributed according to the fol-
Figure 2-6
lowing:
Linear form of beryllium fluo-
So you notice that the three atoms are arranged on a straight
ride molecule.
line so that beryllium falls in the middle and the spatial
shape of the molecule is linear and the angle between the
two bonds are 180ْ [Figure 2-6].
45
 BCl3 molecule:
The Lewis symbol for the boron chloride molecule BCl3 is
written as follows:

It is clear from this symbol that there are three pairs of

electrons surrounded borons atom. In order to be less re-
pulsive between pairs of electrons as possible, the chlorine
Figure 2-7 atoms are distributed around the boron atom in the form
Triangle planar. of a triangle planar. The angles between its bonds are 120°
(Figure 2-7).
CCl4 molecule:
The Lewis symbol for CCl4 is as follows:

It notes that there are four sets of electron pairs surround-

ing carbon atom, the repulsion between them at its lowest
then chlorine atoms are distributed towards the heads of a
regular tetrahedral so that the angle between each pair is
Figure 2-8 109.5 ° (Figure 2-8).
Regular tetrahedral.
H2O molecule

It is clear from Lewis symbol of the water molecule that

there are four pairs of electrons surround the oxygen atom,
further note the composition of the oxygen atom in water
molecule is similar to the composition of a carbon atom in
CCl4 molecule. The pairs of electrons are distributed to-
ward a angular shape. The presence of non-participating
electronic pairs pushes two hydrogen atoms because they
Figure 2-9
are spread over two vertical sides in a curved shape at an
Geometric shape of water (angular).
angle of magnitude (Approx.104.5°) [Figure 2-9].

NH3 molecule

It is clear from the Lewis symbol that there are four sets of
electron pairs surrounded nitrogen atom and it is expected
to be distributed as in the case of the CCl4 molecule toward
the form a trigonal pyramidal, but the presence of a non-
participating electron pair push the three hydrogen atoms
Figure 2-10 to be distributed on the heads of the three-base pyramid.
Geometric shape of ammonia The angle between the three bonds is about (107.3 °) [Fig-
(trigonal pyramidal). ure 10-2].
46
 2-4-2 Valence bond theory
The theory of repulsion of electrons pairs of valence shell
was able to explain different shapes of molecules, but they
could not explain how the electrons were distributed be-
tween the valence shells of the two constituent atoms of
bond. One of the theories that contributed to this clarifica-
tion theory of the valence bond, which was adopted in its
interpretation of the formation of the covalent bond on the
interference of atomic orbitals of the valence of the atoms
where the electrons move around the nuclei and increase
the electronic density in the area of interference between
the two atoms. This leads to the two nuclei approaching
each other, reducing their energy and increasing their at-
traction towards the overlap area of the orbitals, thus form-
ing a covalent bond between them. This can be illustrated
in the formation of covalent bonds of some simple mol-
ecules such as H2, HF,Cl2, O2, N2. Atomic orbitals overlap
as follows:
A- Orbital overlap S with two hydrogen atoms in the hy-
drogen molecule
When writing the electronic configuration of the hydrogen
atom, we notice that it has one electron at the secondary
level (1s): 1H : 1s1
Bond forms in the hydrogen molecule consists of the over-
lap of the orbital 1s in the first atom with the orbital 1s of
the second atom as shown in Figure 2-11:

Axis of the bond

Figure 2-11
It is noted from the figure that the electrons of the bond are Vertical overlap of Orbitals S.
concentrated between the nuclei of the two atoms, where
the electronic density increases in the area of interference
around the axis connecting the two nuclei. This bond is
called sigma (σ)
B- Orbital overlap s with orbital p from other atom as in
HF molecule
When writing the electronic configuration of hydrogen
atom, we note that it has one electron at the secondary lev-
el 1s 1H : 1s1
When1 writing the electronic arrangement of the fluores-
cence atom is as follows: 9F : 1s2 2s2 2p5
47
 It notice that there is a single electron in orbital in the fluo-
rine atom, so a sigma bond is created, when the orbital p
from the hydrogen atom overlaps with the orbital p (lo-
cated on the vertical axis) where the electronic density is
distributed around the axis connecting the two nuclei. This
bond is also named sigma bond [Figure 2-12].

orbital type P orbital type S Hybrid orbital

Figure 2-12 C- Orbital overlap p of the first atom with orbital p of the
Two orbitals p,s vertical second atom
overlap. Orbitals p overlap in two ways as follows:
1- Orbital p overlaps with orbital p on the same axis
When orbital type p of atom is vertical overlap with orbit-
al type p of another atom (at the same axis), then electronic
density is distributed between two nuclei of the two atoms
symmetrically along the axis connecting them. This bond
is called sigma bond as in Cl2 molecule.

Atomic orbital 3p approaches, which contains one electron

in both of the two atoms and it overlap between them and
arises cloud of electronic density. It represents the sigma
bond between the chlorine atoms [Figure 2-13].

Vertical overlap

Figure 2-13 Bond axis

Two orbitals p vertical over-
lap. 2. Orbital p overlap with orbital p on the side axis
When an orbital p of an atom overlaps with an orbital p of
another atom, the lateral interference occurs vertically, the
electronic cloud will be distributed over the bond axis that
bonds of the two atoms. These overlap will form a cova-
lent bond type pi (π), as oxygen molecule.

48
When lateral interference occurs horizontally, the electron-
ic cloud distributes on the right and left axis of the bond
axis that connected the two atoms, that gives bond type pi.
Three cases can be found from overlap (vertical, lateral
vertical, lateral horizontal) as shown between the two at-
oms of nitrogen molecule in [Figure 2-14].
bonds

N atom N atom bonds molecule

2-5 Orbital hybridization Figure 2-14

The process of overlap can occur between orbitals of one Formation of πbond and
atom which are close in energy to produce a number of sigma bond in nitrogen
atomic hybrid orbitals which are equal to the number of or- molecule.
bitals involved in the interference. This condition is called
orbital hybridization, called orbitals that are result from
hybrid orbitals. Thus, the hybridization process represents
interference atomic orbitals have to produce hybrid orbit-
als similar in shape and size and equal in energy and more
stable and lead to get stronger covalent bonds between at-
oms involved in molecule formation. Regarding following
are some observations regarding the hybridization process
and its products:
1. The hybridization process occurs in the same atom or-
bital after excitation the electrons that fall within the main
outer plane and their task reducing the repulsion between
electrons of resulted molecules.
2. Hybridization is done between atomic orbitals converg-
ing in energy together like (2s,2p), (3s,3p), (4s,4p).
3. The holographic geometry of hybrid atomic orbital dif-
fers from Stereoscopic geometry of atomic orbital before
hybridization. The hybrid atomic orbital form consists of
two lobes, one of which is large; where the cloud is rela-
tively concentrated and the other is small and often ne-
glects while drawing [Figure 2-15].
4. The name of the hybrid orbital is derived from the
names and number of pure orbitals, as example the hybrid
orbital (sp3) which is involved in the hybridization process. Figure 2-15
It means the participation of three atomic orbitals of type Hybrid orbital sp.
(p) with one atomic orbital type within the same principal
electronic shell.
49
 5. The number of hybrid atomic orbitals formed is equal to
the number atomic orbitals involved in the hybridization
process. Based on this is when the hybridization of three
atomic orbitals of the type (p) with atomic orbital one of
type (s) to form four orbitals of type (sp3).
6. The energy of atomic hybrid orbitals formed is equiva-
lent or equal.
7. The ability of hybrid orbitals to interfere with atomic
orbitals of another atom is greater than the susceptibility
of not hybridized atomic orbitals that participate in the
hybridization process, because atomic orbitals hybrids or-
bitals are more extensible in space than the participated
atomic orbitals not hybridized.

2-5-1 Types of hybrid orbitals

It can get types of hybrid atomic orbitals according to num-
ber and type of atomic orbitals involved. The most impor-
tant will be addressed at the current school level.
A-Hybrid orbitals type (sp): According to the above obser-
vations, it know that this type of hybrid atomic orbital may
be as a result of the participation of two atomic orbitals
one of type (s) and the other of type (p).
The overlap of these two atomic orbitals will lead to for-
mation two hybrid orbitals type sp located on a straight
line and at an angle of ˚180 in order to have the least re-
pulsion between the two orbitals like the beryllium hydride
molecule BeH2. For the purpose of clarifying how that is
happen, the following steps are followed:
1. The electronic configuration of beryllium atom:
4
Be 1s2 2s2
The orbitals of the casings in the stable state are expressed
as follows:

Energy level

50
2. One of the electrons will rise from the orbital 2s to the
orbital (2px)

Energy level

3. Then the interference occurs to generate two hybrid or-

bitals type sp in equivalent energy less than the energy
atomic orbitals 2px, 2py, 2pz and higher than the energy of
atomic orbital 2s2 and on the following figure.

Energy level

4. These two orbitals have the ability to form molecular or-

bitals with two hydrogen atoms orbitals (1s1) to form two
covalent bonds between Be and H and form the beryllium
hydride molecule. The four steps above can be illustrated
by the representation of stereotypes geometric shapes of
atomic orbitals before and after hybridization. In this case Figure 2-16
we have two hydrogen atoms each one has an orbital (1s1), Composition of hybrid orbit-
They interfere with the two hybrid orbitals sp of beryllium als type sp from atomic or-
atom. This overlap is done on one of the coordinates to bital interference type s with
give the linear shape of the resulting molecule as shown in two orbitals type px, py of the
Figure (2-16). s-type .

51
 B- Hybrid orbitals of type (sp2): This type of atomic orbit-
als hybridized by the participation of one atomic orbital
type (s) with two atomic orbitals type ( p) within the same
main orbital. The overlap of these three atomic orbitals will
lead to formation three hybrid atomic orbitals type sp2 fall
at the same level between them is an angle centered at the
center of the central atom of 120°. This will be led to less
repulsion between the orbitals as it occurs in a trifluoride
boron molecule BF3. Below is an explanation of how this
hybridization was achieved the following steps:
1. The electronic configuration of boron is:
5
B 1s2 2s2 2p1
Orbital (1s2) is considered internally and will not partici-
pate in hybridization, so the representation of the elec-
tronic configuration of the outer shell of this atom is repre-
sented as follows:

Energy level

2. One of the two electrons in the orbital (2s2) goes up to

the orbital py with the approximate energy:

Energy level

3. Then the interference occurs to generate three hybrid

atomic orbitals of type sp2 has less equivalent energy than
non-hybrid atomic orbitals
52
2px, 2py, 2pz and higher than non-hybrid atomic orbital 2s2.
These orbitals are distributed around the nucleus in level
that shares two coordinates, and they have equal angles.

Energy level

4. These three orbitals have the ability to form three mo-

lecular orbitals with three orbitals for three fluoride atoms
to form a triple boron fluoride molecule.
The four steps above can be illustrated by the represen-
tation of stereotypes geometric shapes of atomic orbitals
before and after hybridization. In this case we have three
fluoride atoms interfering with sp2 hybrid orbitals of boron
atom. This overlap is done at one level as shown in Figure
(2-17).

three hybrid orbitals type sp2 three hybrid orbitals type sp2
around atom B

C- Hybrid orbitals type (sp3): This type of atomic orbitals Figure 2-17
hybridized by one atomic orbital interference of type (s) Composition of hybrid orbit-
with three atomic orbitals of type (p). als type sp2 of atomic orbital
The overlap of these the four atomic orbitals will produce overlap of type 2s with two
four hybrid atomic orbitals type (sp3), these orbitals form a orbitals px and py.
regular tetrahedral shape around the nucleus of the central
atom with vertical angles of 109.5o,
In order to have less repulsion between the orbitals as in
the methane molecule CH4. Below is an explanation of
how the hybridization takes place according to the follow-
ing steps:

53
 1. The electronic configuration of carbon atom is:
6
C 1s2 2s2 2p2
Orbital (1s2) is considered internally and will not partici-
pate in hybridization. So the representation of the electron-
ic configuration of this atom is as follows:

Energy level

2. One electron from orbital (2s2) rises to orbital (2pz)

Energy level

3. The interference is happened to generate four hybrid

atomic orbitals from sp3 with equivalent energy less than
energy of non-hybrid atomic orbitals 2px, 2py, 2pz and
higher atomic orbital energy of 2s. These hybrid orbitals
are distributed in a regular tetrahedral shape around the
central carbon atom with equal angles.

Energy level

54
4. These four orbitals have the ability to form four molecule
orbitals with four atomic orbitals for four hydrogen atoms Exercise 2-6
(1s1) to form a methane molecule. The four steps above can Explain how hybrids get in
be illustrated by the representation of stereotypes geomet- molecule for SiCl4 according
ric shapes of atomic orbitals before and after hybridization to the four steps, that we fol-
[ Figure 2-18]. In this case we have four hydrogen atoms lowed in the hybridization
each one have orbital (1s1) has an overlap with the hybrid process (atomic numbers Cl =
orbital sp3 of carbon atom as shown in Figure (2-18). 17 and Si = 14).

four hybrid orbitals sp3

around carbon atom
four hybrid orbitals sp3
Figure 2-18
Composition of hybrid orbit-
als sp3 of orbital overlap from
2-5-2 Examples of hybridization and shape geometry 2s with three orbitals px, py,
1. Ammonia molecule: pz.
The regularity of hybrid orbitals has a significant impact
on the geometry shape of molecule, in other words on the
configuration of the nuclei of united atoms. An important
example, So what we see happens in the orbitals of some
atoms and their formation For molecules with a known
geometric shape as we find it in a molecular state of am-
monia NH3. The nitrogen atom has the following electron-
ic configuration:
7
N 1s2 2s2 2p3
↼

↼
level

level

2px 2py 2pz 2px 2py 2pz

‫م�ستوى الطاقة‬

‫م�ستوى الطاقة‬

hybridization
↼

‫تهجين‬
↼↼
Energy

Energy

sp3 sp3 sp3 sp3

↼
↼

2s 2s

55
 When a nitrogen atom approaches three atoms of hydrogen
Exercise 2-7 each has an atomic orbital of type 1s1, then it can the ex-
ternal atomic orbitals of the nitrogen atom hybridize with
Write the steps of hybridiza-
each other. This hybridization occurs after the orbital (2s)
tion of atom orbitals of oxygen
interferes with the orbital 2p, to give four hybrid orbitals
in water molecule. What is
(sp3),They are equivalent in form, energy and possess ir-
geometry shape and why?
regular four-sided geometric shape. The three atomic or-
bitals ( 1s1) of the three hydrogen atoms overlap with hy-
brid atomic orbitals (sp3) which is containing one electron
in each orbital to form three σ covalent bonds The hybrid
orbital does not participate, that contains an electronic pair
in this interference. The geometric shape that this molecule
will take it will be an irregular tetrahedral pyramid, formed
of a central nitrogen atom surrounded by three hydrogen at-
oms and an unrelated free electronic duplex [Figure 2-19].

Figure 2-19
Geometry shape of ammonia
molecule

2. Ethylene molecule C2H4:

The carbon atom has an electronic configuration:
6
C 1s2 2s2 2p2
When one of the 2s2 electrons is upgraded to the orbit pz,
the interference occurs between atomic orbitals we get
three hybrid orbitals sp2 leaving the fourth orbital pz non-
hybridized.
Energy level

Energy level

hybridization

Now, if two hybrid carbon atoms approach the hybrid or-

bitals sp2 each other, and at the same time, four hydrogen
atoms each one have one electron on s [Figure 2-20A] are
overlapped and electrons are pairs then five sigma cova-
lent bonds one (C-C) and four (C-H). This is followed by
the overlap of non-hybrid orbitals pz in two C atoms are

56
 lateral interference and the electrons are duplexed to form
bond between the two carbon atoms that form with
Signal bond
the sigma bond double bond between carbon atoms [Fig-
ure 2-20B]. This means that the bond of the sigma type is
resulted from the interfering of hybrid orbitals sp2 of two
contiguous carbon atoms. Pure pz (non-hybridized) orbit-
als of adjacent carbon atoms, thus, the ethylene molecule
Stereosco image of ethylene
is raised. The pi bond was produced by interference pure
pz (non-hybridized) orbitals of contiguous carbon atoms,
where ethylene molecule is created.

Figure 2-20
A. Overlap of orbitals to form
non-hybrid orbitals A non-hybrid orbitals sigma bonds in the ethylene
molecule.
B. Orbitals overlap to form
bonds in the ethylene
molecule.

non-hybrid orbitals B Ethylene molecule

bonds
3.Actylene molecule:
In the case of the acetylene molecule, that 2s and 2px suffer
hybridization, while pz and py orbitals remain non-hybrid-
ized, meaning that hybridization will be of a type sp.
Energy level

Energy level

hybridization

When two carbon atoms approach this kind of hybridiza-

tion and two atoms approach at the same time, hydrogen
interferes with atomic orbitals, it has three σ covalent bonds
of one type C-C and two type C-H, and be connected to a
single straightness.

57
 As a side overlap occurs for a non-hybrid orbitals pz, py of
each C atom and creates two ( ) bonds to form with the
(σ) bond, triple bond and thus creating a straight molecule
of acetylene with triple bonds as shown in Figure 2-21.

B First bond second bond

Figure 2-17
A. Overlap of orbitals to form
(σ) bonds in acetylene mol-
ecule.
B. Orbitals overlap to form
bonds (σ) bonds in acetylene
molecule

58
 Basic concepts
Chemical reaction Ionic Bond
Is the interaction that occurs between at- Is the power of electrical attraction between
oms of two elements to form the molecule two ions different in charges, one of which is
of a new compound possesses chemical and a metal with a positive charge for the loss of
physical properties completely different from one or more electrons and the other is non-
the characteristics of the elements involved metal that carries a negative charge to ac-
in the reaction. The purpose of the chemi- quire one or more electrons.
cal reaction is to form saturated orbitals with Covalent Bond
electrons through loss or gain or participate It is the bonding force between two equal
in electrons to achieve chemical stability. atoms in electronegativity or a slight differ-
Chemical bonding ence between them, so that each atom shares
It is the phenomenon of the presence of atoms an electron from its outer shell the two atoms
coherent together in a molecule or crystal share by this electronic pair by force of at-
within an element or compound, the strength tachment is called covalent bonding.
that binds them is called the chemical bond. Polar Covalent Bond
Valence Shell Electron-pair Repulsion They are covalent bonds of two atoms the
Theory (VSEPR) difference between electronegativity is rela-
This theory explains the configuration of tively significant , the double electronic link
atoms around central atom depending on is attracted between them towards the high-
repulsion between participating and non- est electronegative atom gain, a second nega-
participating electrons pairs existing in the tive charge have partial positive charge.
cavity casing of the central atom, so that it is Hydrogen Bond
the repulsion between these pairs is minimal It is the power of electrical attraction between
when be as far as possible from each other. the ends of different charge of molecules
Valance bond theory with polarized covalent bonds, provided that
Adopted the theory of parity insistence in its one part hydrogen.
interpretation on the theory of hybridization Coordinate Bond
of orbital where it follows hybrid orbitals of It is a special covalent bond between two at-
the central atom so that The disharmony be- oms, one of them is granted the electronic
tween these orbitals is minimal when be as pair of its outer shell to the second atom con-
far as possible from each other. taining empty orbital in its outer shell.
Orbital hybridization Metallic Bond
It is the process of overlapping atomic orbit- It is the force that binds the metal atoms to-
als to produce hybrid orbitals are equal to the gether, it results from the sharing of each
number of orbitals the atomic involved in the metal atom with equivalents electrons, and
hybridization process are similar form, size, returns them to the metallic properties.
equal in energy and more stability than non-
hybrid atomic orbitals and lead to stronger
covalent bonds between atoms involved in
the formation of the molecule.
59
 Chapter Two Questions 2
2-8 A- What is the ionic bond? What are
2-1 Reasons for each of the following: the conditions of its composition?
1. The boiling point of water (100° C) is B- Why molecules are not formed in ionic
higher than the boiling point of hydrogen compounds?
sulfide gas (-60 ° C). C- What are the most important character-
2. The NH4Cl ammonium chloride molecule istics of ionic compounds?
contains three polar covalent bond, coordi- 2-9 What do we mean by hydrogen bond-
nate bond and ionic bonds. ing? Explain this in example, draw hydro-
3. The angle between the two hybrid orbitals gen bonds between Methanol (CH3OH)
sp2 in C2H4 Molecule is equal to 120° C. molecules.
4. Ionic compounds do not conduct electrical 2-10 Covalent bonds may be polarized,
in solid state but for melting or solutions of when is that?
ionic compound in water has a good ability 2-11 What are the factors that determine
of electrical conductivity. the bonding between two atoms covalent
5. When you put a piece of ice in the water, or covalent polarized? or Ionic?
float however when you put a frozen piece of 2-12 What are the metal bonds? what ef-
gasoline in gasoline immerse. fect it has on the properties of pure metals?
2-2 Compare between each of the following: 2-13 Rewrite the following statements as
1. Covalent bond and coordinate bond. corrected (may) you find it from scientific
2. The sigma (σ) bond and pi(π) bond. mistakes:
3. Hybrid orbital and non-hybrid orbital. A- All compounds with covalent bonds
2-3 Draw a figure showing a excited carbon didn’t dissolve in water.
atom and an hybrid Carbon atom sp3 B- Protons and neutrons are involved in the
2-4 What is the hybridization situation in composition ionic bond.
each of the following. Mentioning forms of C- (π) bond is less energy than (σ) bond of
the resulting compounds: the same molecule.
1. Oxygen in water. D- The triple bond in (C2H2) contains three
2. Nitrogen in HN = NH types of (π) bond.
3. Nitrogen in Ammonia E- The type of hybiridization in C atom
2-5 Draw resonance formulas for: orbitals in Methan is similar to N atom in
CH3COO-, O3, CO3-2 ammonia which is (sp3).
2-6 Why are the atoms of elements combined 2-14 Draw the geometry of the following
together? are the combinations of atoms of particles using hybridization once and us-
elements always produce compounds? Dis- ing the (VESPR) theory again:-
cuss your answer scientifically with at least BF3 , CH4 , BeCl2
two examples. 2-15 Draw and compare the shapes of the
2-7 A- What is chemical bond? Identify it ethylene and acetylene molecules in terms
accurately. of the hybridized orbitals which of it the
B- Numbering only types of bonds you bond between the two carbon atoms is
know. stronger.

60
Chapter Three
Periodic Table and Chemistry of
Transition Elements
3
After completing this chapter, the student is expected to:
• Recognize the historical sequence of the emergence of the periodic table and
attempts made to divide the elements.
• Arrange elements in a modern periodic table based on atomic numbers instead of
their atomic masses.
• Able to determine the number of periods included in the periodic table and on the
group numbers.
• Recognize the periodic table and can know its parts.
• Understand the periodic properties of elements in the periodic table.
• Understand when an element has an atomic spectrum and that the atomic spectrum
is a characteristic for the element.
• Distinguish between a transitional element and a non-transitional element as
recognize the internal transition elements.
• Distinguishe between the elements that are attracted towards the magnetic field
and those that are not attracted.
61
 3-1 Periodic Table
3-1-1 Introduction
The emergence and maturation of the periodic table has
Do you know undergone a series of developments synchronized with the
The word periodic means that development of scientific concepts in each period, where
The properties of the elements it came from simple ideas to become a scientific model
change periodically due to an in- proud of all Who contributed to the arrangement of these
crease in their atomic numbers. ideas. Those attempts had begun a simple idea for a chem-
ist to arrange the elements in a given table This arrange-
ment then evolved into what it is now.
It seems there is no need in the old to classify items in a
special table because the number of elements at that time
did not exceed the fingers. After chemistry development
and the discovery of more elements, They are beginning to
need classification
the elements and scheduling to facilitate their study and
facilitate dealing with them were the first Attempts to sort
and arrange items in a table are split Items into two groups:
A- Metals group
B- Nonmetals group
Because there is no boundary dividing the elements into
metals and nonmetals, where many share properties be-
tween them. There have been other attempts to classify
the elements on other bases and most important , those
are built on the relationship between the properties of ele-
ments and their atomic masses. Following is a summary of
these attempts:
1. Dobriner’s trilogy: Doeberiner observed in 1817 The
differences between the atomic masses of elements such as
calcium(Ca), Strontium (Sr) and barium (Ba) are fixed dif-
ferences as the difference in the atomic mass in these trip-
lets of any element is almost constant for the element that
precedes and is attached to it. That the physical and chemi-
cal properties of the element in these triplets is the average
properties of the two previous and next elements. In other
cases such as iron, cobalt and nickel, the elements have
approximately equal atomic masses as shown in Table 3-1.

62
Table (3-1) Atomic mass of some elements and the difference between them
Element Atomic mass Difference
1 Calcium (Ca( 40.07
47.56
2 Strontium (Sr( 87.63
49.74
3 Barium (Ba( 137.37
4 Iron (Fe( 55.84
3.10
5 Cobalt (Co( 58.94
0.25
6 Nickel (Ni( 58.69
The previous groups, each of which is composed of three, are
called elements are similar in many of their chemical and physi-
cal properties are called Dobriner›s trilogy.
2. Second attempt by the English scientist Newlands
In 1864, he arranged the elements that were known to him and
which number 63 elements by increasing their atomic masses in Table 3-2
H F
Eighties of scientific Newlands
Cl Co/Ni Br Pd I Pt/Ir

the form of groups, each group has eight elements. It is found Li Na K Cu Rb Ag Cs Ti

Ga Mg Ca Zn Sr Cd Ba/V Pb
if it starts with an element, the eighth element is similar in its Bo Al Cr Y Ce/La U Ta Th
chemical properties to the element that began in it. The similar- C Si Ti In Zn Sn W Hg

ity between this observation and the musical scale it is called of N

O
P
S
Mn
Fe
As
Se
Di/Mo Sb
Ro/Ru Te
Nb
Au
Bi
Os
these remarks;(the Law of Eighties of scientific Newlands) the
law, is true in the case of the first seventeen elements either the
following elements there have been some contradictions due to
the lack the accuracy of the atomic masses on the one hand and
on the other hand it left no place empty in his schedule for many
elements that were not known in it’s time as shown in Table 3-2.
3. The attempts of Doeberiner and Newlands ended when the
scientific Russian Mandeleeff (1869) was developed his known
the periodic table it is worth mentioning that Mandeleeff and
Meyer worked independently of each other and came to the
same conclusion. They arranged the elements ascending ac-
cording to their atomic masses, They were summarized each
the two results of his research in this subject in the following
phrase (elements properties are periodically dependent on their
atomic masses). This phrase is defined as (Cyclical law). It was
taken by Mandeleeff mainly to classify the elements. Man-
deleeff spent several years collecting relevant scientific find-
ings elements before he publishes his periodic law, which relies

63
 mainly on practical facts and not on theoretical grounds. It
is worth mentioning number the elements discovered at the
time were limited and therefore left empty places for those
elements for possible future discovery. The possibility of
Mendeleev in high guess places those elements that have
not been discovered. The name of the periodic table have
been associated with scientific Mandeleeff because he was
able to determine the properties of some undiscovered ele-
ments of his time, including the Germanium and named it
Similar to silicon.
Mandeleeff also relocated some elements to its new loca-
In 1869 the Russian Dmitry Iva-
tion better aligned with the new neighboring elements. It
novich Mandeleeff developed
has been corrected some errors in the development of cer-
the first periodic table by ar-
tain elements depending on the values of atomic masses.
rangement the elements discov-
As for the weaknesses in a table Mandeleeff are:
ered in his time in view tabular
A-Hydrogen Position: The position of hydrogen in the ta-
by increasing atomic masses
ble remains controversial as it is placed sometimes in the
relative (atomic weight).
first group above the alkali metals and sometimes above
the group seventh (halogens) and that there are some simi-
larities between him and these elements in the two groups
mentioned.
B-Position of internal transition elements (lanthanides):
These fourteen elements are similar in many properties,
especially in equivalence (often 3) and their place in the
Mandeleeff table in the third group, place element is usu-
ally placed at the bottom of the table as well as the case in
Actinides.
C- Disparate arrangement of some elements:
It is known that the elements are arranged in the Mandeleeff
table in ascending order atomic masses, even if this rule in
any case strictly followed some elements have occurred in
places where their properties are inconsistent with those
of the elements in the other group had to be overcome this
difficulty of change put these elements regardless of the
value of their atomic masses. For example, atomic mass
of cobalt Co is 58.94 and the nickel 58.69 is nevertheless
placed cobalt Co before nickel as well as in the following
pairs of elements:

64
 Elements Atomic mass
Arcon 39.94
Potassim 39.09
Tellurium 127.61
Iodine 126.93

Mandeleeff arranged the elements in ascending order ac-

cording to their atomic mass so that similar elements in the
properties are located below each other. Mandeleeff table
included horizontal rows called periods and vertical rows
called groups.

Table 3-3 Mandeleeff table

I II III IV V VI VII VIII

1 H=1

2 Li=7 Be=9.4 B=11 C=12 N=14 O=16 F=19

3 Na=23 Mg=24 Al=27.3 Si=28 P=31 S=32 Cl=35.5

Fe=56,Co=59
4 K=39 Ca=40 ?=44 Ti=48 V=51 Cr=52 Mn=55 Ni=59

5 Cu=63 Zn=65 ?=68 ?=72 As=75 Se=78 Br=80

Ru=104,Rh=104
6 Rb=85 Sr=87 ?Yt=88 Zr=90 Nb=94 Mo=96 ?=100 Pd=106

7 Ag=108 Cd=112 In=133 Sn=118 Sb=122 Te=127.6 J=127

8 Cs=133 Ba=137 ?Di=138 ?Ce=140

9
Os=195,Ir=197
10 ?Er=178 ?La=180 Ta=182 W=184 Pt=198

11 Au=199 Hg=200 Tl=204 Pb=207 Bi=208

12 Th=231 U=240

65
 3-1-2 Modern Periodic Table of Elements
After the discovery of the electron and the appearance of
the concept of atomic number by Moseley in 1914, the
elements were arranged in ascending order by increasing
their atomic numbers rather than by increasing the atomic
mass as arranged by Mandeleeff. That is, each element in
the modern periodic table exceeds the element preceded
by a single electron known as a distinct electron. This ar-
rangement according to the increase in atomic numbers
corresponds with the order of elements by increase energy
levels from least to most energy. Thus, the image of the
modern atomic table, [Figure 3-1] in the form of 7 horizon-
tal periods with groups arranged by columns, their number
(18) groups which are:
1. The first short period includes hydrogen and helium.

2. The second and third short period, each consisting of 8
elements. They are arranged in the periodic table as fol-
lows:

3. The fourth long period consists of 18 elements and they
are arranged in the periodic table as follows:

Do you know
Transitional elements
Due to the difficulty of putting
lanthanides with lanthanum in
the third group, is placed, in a
4. The fifth long period consists of 18 elements and they
separate row below the periodic
are arranged in the periodic table as follows:
table. The same is the case with
actinides which is relative to the
actinium element, they are also
Transitional elements
placed in another row below pe-
riodic table .

66
5. The six long period consists of 32 elements and they are
arranged in the periodic table as follows:

Lanthanides

6. The seventh long period consists of 24 elements and

they are arranged in the periodic table as follows:

Actinides

Main group Main group

elements elements
Atomic number metals (main group)
lA 1 metals (transitional elements) VlllA
)1( H Symbol metals (Internal transitional elements) )18(
1.008 metalloids
1 Atomic mass nonmetals 2
1 H llA lllA lVA VA VlA VllA He
1.008 )2( )13( )14( )15( )16( )17( 4.003
3 4 5 6 7 8 9 10
2 Li Be transition B C N O F Ne
6.941 9.012 elements 10.81 12.01 14.01 16.00 19.00 20.18
11 12 13 14 15 16 17 18
3 Na Mg lllB lVB VB VlB VllB VlllB lB llB Al Si P S Cl Ar
22.99 24.31 )3( )4( )5( )6( )7( )8( )9( )10( )11( )12( 26.98 28.09 30.97 32.07 35.45 39.95
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.39 69.72 72.61 74.92 78.96 79.90 83.80
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
85.47 87.62 88.91 91.22 92.91 95.94 )98( 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn
132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 )209( )210( )222(
87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds
)223( )226( )227( )261( )262( )266( )262( )265( )266( )269( )272( )277( )285( )289(

Internal transitional elements

58 59 60 61 62 63 64 65 66 67 68 69 70 71
6 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
140.1 140.9 144.2 )145( 150.4 152.0 157.3 157.3 162.5 164.9 167.3 168.9 173.0 175.0
90 91 92 93 94 95 96 97 98 99 100 101 102 103
7 Actinides Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
232.0 )231( 238.0 )237( )242( )243( )247( )247( )251( )252( )257( )258( )259( )260(

Figure 3-1
Modern periodic table.

67
 The groups in the periodic table and its number (18 groups)
divided into Group A (8 groups) and Group B (10 groups)
as shown in Figure (3-1). These groups are arranged in
vertical columns based on equal number of electrons in the
outer shell (external valence electrons). Thus, the elements
of a single group are similar in their chemical properties,
despite the great difference in their atomic masses so as
they equal by number of electrons, these atom can lose or
gain, or contribute to during the entry into the chemical
reaction. While in a single period the elements where ar-
ranged by increase of the atomic number, therefore these
elements has convergent in their atomic masses, yet they
differ in their physical properties. For example, Nitrogen
is near carbon and oxygen in the second period, despite
their approximate in the atomic mass, they have different
in physical properties, each of the energy shell in elements
atoms contain secondary energy shell (sub), its number
equal to the number of principal shell. These principal
shells are filled with electrons as it increases in the energy
where fills the lowest energy level(s) and then the higher
energy level. This configuration in filling shell is similar to
the configuration of elements in periodic table by increas-
ing atomic number.

3-1-3 Periodic Table Parts

The periodic table includes four parts arranged as follows
[Figure (3-2)]:
First part: includes the first group lA and the second group
llA and both are ends with (ns) shell known as alkali met-
als and alkaline earth metals.
Second part: include the elements in the groups:
lllA, lVA, VA, VIA, VIIA, VIIIA (the last group is called
0). The elements of this area (Part I and Part II) are full the
electrons in the sub-energy levels (s and p). The first and
the second part are called the represented elements.
Third part: includes all the elements in subgroups B, name-
ly:
lB, llB, IIlB, IVB, VB, VIB, VIIB, VIllB (Group 0) which
include (8B, 9B, 10B ) and the outer shell of these elements
is type s, d and d is not full of electrons is called tran-
sitional elements because they are transite by properties
68
between elements with the outer shell s (group, lA, IlA)
and elements with outer shell p group (lllA, lVA, VA, VIA,
VIIA, VIIIA). They mediate the periodic table.
Fourth part: comprises two series of internal transitional
elements:which are: Lanthanides and Actinides
First part Second part

Third part

Fourth part

Lanthanides

Actinides

Figure 3-2
Parts of periodic table.
3-1-4 Periodicity of properties in the periodic table
Many of the physical properties of the elements change pe-
riodically accordingly to position of these elements in the
periodic table in terms of group and period, some of these
features are discussed below.
A-Atomic size Atomic size is a difficult characteristic for
several reasons one of the most important reasons is that
the probability of electronic distribution is influenced by
atoms adjacent in the chemical compound and thus the
size of the atom changes somewhat when moving from
one case to another. Therefore, when checking any table of
atomic radiuses, it should be remembered that tabulated
values may be a meaning only when considering a relative
comparison of volumes, Figure (3-3) shows

69
 Method of measurement of atomic radii for elements de-
rived from distances measured from the centers of adja-
cent atoms in pure elements and on this basis atomic size is
defined as half distance between the centers of two identi-
cal atoms in a crystal and it is measured using X-rays.
In general, atomic radii are reduced per one period when
moving from left to right (ie, atomic number increases)
in the periodic table. This behavior can be explained by a
table showing the change of atomic radii along the second
period.
In the case of the group, the radius increases in one group
Figure 3-3 from top to bottom as the atomic number increases.
Method of measurement of The reason for this is the addition of electronic shells with
atomic radii of elements. increasing quantum numbers farther from the nucleus as
shown in the table of atomic radii[Figure(3-4)].
In the transitional elements
1. The radius decreases gradually in each transition chain
until the fifth element, ie to half of the period then gradu-
ally increases until the end of the series, where the size de-
creases due to the increase of attraction force, as the elec-
tron, which is added by increasing the number of atomic
number form one element to another enters the secondary
shell orbitals (d).
2. Observations have shown that adding half of this num-
ber, ie 5 electrons, is accompanied by a state of stability
and this electronic system block the effect of the nucleus
so the attracted force will be less to the electrons that are
added , which it explains the increase in size slightly after
the fifth element.

Figure 3-4 In the internal transitional elements, the size also decreases
Changing atomic radii (ie, the gradually by increasing the atomic number up to the seventh
sizes of atoms) of some elements element (ie to half of the chain) and then gradually increas-
in the period and group. ing again and on the same basis of interpretation in point 2
70
 For the radius ions of atoms it is known that atoms can lose or
gain one or more electrons to form ions. Electrons have nega-
tive charge so atoms gain an extra charge when gain or lose
electrons. So the ion is an atom or an atomic group has positive
or negative charge. a
When an atom loses electrons and is a positive ion is formed,
the atom size becomes smaller. This is due to two factors:
firstly, the electron that the atom loses often it’s an equivalent
electron, and the loss of electron led to form an empty external
orbit, causing radius decreases. Secondly, electrostatic repul- sodium atom sodium ion
sion decreases between the remaining electrons and increases
the attraction between them. The nucleus has a positive charge,
allowing the electrons to get closer of the nucleus and figure
(3.5a) shows the decrease in the ion radius of sodium atom b
when it formed a positive ion.
Conversely, when an atom acquires electrons and became neg-
ative ions,their size increases because the addition of an elec-
tron to the atom generates electrostatic repulsion bigger with
the outer-level electrons, and pushes them strongly outward. Chlorine atom Chlorine ion
An increase in the distance between external electrons results or
in an increase in the amount of radius. Figure 3.5b shows how Figure 3-5
the radius of the chlorine atom increases, when it forms nega- A- positive ions smaller than
tive ion. Figure (3.6) shows the ion radius of some elements. their neutral atoms.
B- negative ions are larger than
their neutral atoms.
Period

B- Ionization energy Figure 3-6

It is the energy needed to extract an electron from a neutral Ion radius of ideal elements
atom in its gaseous state and to form positive ion as in the fol- measured in pico meters.
lowing equation: M + Energy M+ + e-
(1pm=10-12m)
This energy is consumed in extracting an electron from external
valance electrons. The ionization energy is measured in units of
electron volts (ev) and electron volts is a small energy equal 1.6
x10-19 J. Energy is needed to remove the first electron is called
first ionization energy. The result of that ion with one positive
charge. 71
 The energy needed to extract a second electron is also called a
second energy ionization. Thus, the second ionization energy
is always greater than the first ionization energy because the
positive nucleus charge attracts the second electrons strongly
larger.
If the relationship between ionization energy and the atomic
number of elements is plotted, it is noted that change occurs
periodically Figure( 3-7).
The noble elements are located at the great ends of it in order
to stabilize their system. The alkaline elements also located on
the lower ends due to the large size of the atomic volumes and
that the quantum layer penultimate contains (8) electrons are
characterized by a high degree of stability and work as a barrier
obscures the effect of the charge of the nucleus on the valence
electron is easy to extract.

Figure 3-7
It is evident that the ionization energy increases in one period
The relationship of ionization
with increasing atomic number because of their small atomic
energy with the atomic number.
sizes (radii) except atoms whose last shell is saturated or half-
saturated, where ionization energy is greater than ionization en-
ergy of the next atom. For example, that The ionization energy
of 7N is greater than the ionization energy of 8O and the reason
for this is due to that the last nitrogen shell is half-saturated
in which three electrons and to be more stabile than oxygen
despite being the largest atomic number. As well in the case
of manganese 25Mn and iron 26Fe the Mn ionization energy is
greater than Fe ionization energy for the same previous reason.
In one group, ionization energy decreases as atomic number
increases due to large atomic size which facilitates extraction
the outer electrons of the atom, for example, the ionization en-
ergy reduce from lithium to Cesium and from beryllium to ra-
72 dium as shown in Figure (3-8).
Ionization energy (KJ/mol)

rgy
ne
ne
io
z at
Increas oni
ed ioni
zation e di
energy eas
cr
Figure 3-8In
C-Electron affinity Ionization energy changes in
An electronic affinity is known as the amount of energy one period and group.
released from an neutral atom in the gaseous state when
it acquires an electron forming negative ions, according to
the following equation:
Fg + e- F-g + 328kJ/mol

Electronic affinity increases in periods by increasing the Do you know

atomic number, the small atomic sizes make it easier for The electronic affinity measures
the nucleus to attract the electron. Electronic affinity of the severity of additional elec-
elements in a single group decreases as atomic number tron binding to the atom.
increases because of increasing atomic size. Just as there
is first and second ionization energy, so is there first and
second electronic attraction whilst some energy is released
upon acquisition the first electron often absorbs some en-
ergy when second electron is acquired because there is a
repulsive force between the negative ion and the acquired
electron.
The emission of energy when an electron is added to an
atom leads to the transition to lower energy case, ie to a
more stable state and this explains the tendency of some
atoms to acquire electrons as elements in the VII group
in chemical reactions to reach a more stable and minimal
state of energy.
The low values in the electronic affinity of noble gases and
the elements to the far left of the periodic table are attrib-
utable to their inability on the formation of negative ions
compared to fluorine (F), chlorine (Cl) and bromine which
form easily negative ions as shown in Figure (3-9).

73
Figure 3-9
The relationship of electronic af-
finity with the atomic number. D-Electronegativity
Is the ability of atoms in molecules to attract electrons to-
ward them of other atoms associated with them by chemi-
cal bond. There are values numerical elements as shown in
Figure (3-10) these numbers describe the relative ability of
an atom to form a bond , ie turns negative state and shared
electron can then attract to it. Fluorine element has high
Do you know electronegativity relative to any element in the periodic
And now we ask the benefit of table because it located at the end of its period and on the
the value of electronegativity top of its group. The noble gases elements are not formed
the answer to this question that chemical bonds and did not agreed on their values yet. In
one of these benefits is to predict general electronegativity increases in a single period from
which bonds are ionic and which left to right as the atomic number increases. Existing ele-
covalent . It can also benefit of ments at the far left of the periodic table (group IA, IIA)
electronegativity in the polar have low electronegativity.
prediction where two elements The elements on the far right except the group (0) has high
apart for electronegativity values electronegativity and it is attributed for the elements group
whenever necessary, the bonds (VII) the following electronegativity to:
are more polar and the bond F=4.0, Cl=3.0, Br=2.8, I=2.5
between H and Cl is more polar The order of decrement in the electronegativity shall be
than those between (Cl and Br). uniform, other than the order of affinity in one group, the
electronegativity decreases with increasing atomic number
as we move from the top to down of the group.

74
 Figure 3-10
E- Metallic and non-metallic properties: Electronegative values for some
Metals have a number of properties including metallic lus- elements periodic table.
ter and electrical and thermal conductivity and high melt-
ing and boiling points such as iron, copper and zinc, while
nonmetals, are characterized by no luster and gloss. They
are often fragile, with low melting and boiling points such
as sulfur, carbon, phosphorus and gases. The metalloid are
elements their properties between metals and nonmetals
such as boron and silicon. Their properties are graded in
the periodic table are as follows:
In one period the metallic properties decrease and the non-
metal properties increase as they increase atomic number.
We find that the elements of the beginning of the period
are all metals and then less this property and a non-me-
tallic property starts to appear as we go to the right of the
period, that is, an increase atomic number for example in
the second period lithium (3Li) and beryllium (4Be) me-
tallic properties, while boron (5B) shows the properties of
semi-metals. The rest comes second period elements such
as carbon, nitrogen, oxygen and fluorine to appear proper-
ties of nonmetals where the atomic number increases. In
one group metallic properties increase and Non - metal-
lic properties decrease with increasing atomic number. All
elements of groups (IA and IIA) are metals while elements
of groups (VII and 0) are nonmetals. The rest of the groups
all elements are not in one class, for example, in the fifth
group shows (N) metallic properties, while (S) shows the
behavior of semimetals and Bismuth comes the last ele-
ment in the fifth group with metallic properties.

75
 In periods, the elements of the first period (H and He) are non-
metallic for metals in the next four periods there is a gradual
transition from metallic properties to nonmetallic properties of
properties. In the sixth period all its elements are metals except
the last two elements are non-metallic elements of the seventh
period are all metals and show transitional elements and ele-
ments Lanthanides and actinides have metallic properties, as
shown in Figure (3-11).
Reduction of metal properties
lA metals (main group) VlllA
Increase of metallic transition elements properties

metals (transitional elements)

1 metals (Internal transitional elements) 2
1 llA metalloids lllA lVA VA VlA VllA
H nonmetals He
3 4 5 6 7 8 9 10
2 Li Be transitional B C N O F Ne
elements
11 12 13 14 15 16 17 18
3 lllB lVB VB VlB VllB VllB lB llB
Na Mg )8( )9( )10( Al Si P S Cl Ar
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
4 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54
5 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
55 56 57 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
6 Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn
87 88 89 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118
7 Fr Ra Ac Rf Db Sg Bh Hs Mt Ds

Internal transitional elements

58 59 60 61 62 63 64 65 66 67 68 69 70 71
6 Lanthanides Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
90 91 92 93 94 95 96 97 98 99 100 101 102 103
7 Actinides
Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

F-Flame Spectrum:
Figure 3-11
Change of metallic and nonme- The spectrum of an element can be obtained by placing the
tallic properties in one period element or its vapor in an electric discharge tube under low
and group.
pressure and high voltage that irritates its atoms. It emits linear
spectra and each element has a distinct linear spectrum, there
is a relationship between the linear spectrum and the atomic
structure of the element, that’s called the atomic spectrum. For
example, the hydrogen atom is in a stable state if the electron
is found at the first energy level and when increased energy
of an electron moves to a higher energy level and is said to be
Sodium flame Potassium flame an atom is excited or irritated when an electron falls from an
higher energy level to a lower energy level it loses energy equal
to the difference between energy levels.
This energy appears as an electromagnetic radiation (Spec-
trum) accompanied by a color and has a specific wavelength
and frequency for each element, linear spectrum is character-
istic of it. when heating a calcium metal on a flame, it flames
Copper flame Lithium flame brick red and strontium is crimson, barium is yellowish green,
sodium yellow and potassium violet, cesium blue and rubidium
76 dark red.
 3-2 Transitional elements
3-2-1 Introduction
Transitional elements appear in the fourth, fifth, sixth and
seventh periods of the periodic table and has an electronic
configuration, orbitals electrons d and f play an important
role. These elements can be traditionally divided into two
groups:
The set of elements d and the set of elements f
The set of d elements consists of three complete series and
a fourth series incomplete. Each series of the third series Ions of salts of transition elements
have ten elements, these are: and their solutions from left to
1.The first transition series: starting from the element scan- right.
dium (21Sc) to zinc element (30Zn). Mn2+,Fe2+,Co2+,Ni2+,Cu2+,Zn2+
2. The second transition series: starting from the element
yttrium (39Y) to cadmium element (48Cd).
3.The third transition series: starting from the element lan-
thanum (57La) to gold element (30Au).
The fourth transition series starts with actinium element
89
Ac and ends with element daramstadtium 110Ds, which is
composed of (8) elements can be observed these serieses
of looking at the periodic table Figure (3-1). The f group
consists of two series and in each series 14 elements are
called internal transition elements, the first series is called
lanthanides and the second series is the actinides, and these
two series are:
1. The lanthanide series starts from the cerium Ce atomic
number 58 to the lutetium Lu atomic number 71.
2. The actinides series begins with the element thorium Th
atomic number 90 to the element lorticium Lr atomic num-
ber 103.
In each series, the electronic configuration of the full en-
ergy levels remains the same with higher main quantum
number is constant while internal energy levels are gradu-
ally filled d(n-1) and f(n-2) with increasing atomic number
Table (3-4). In the series of transition elements, orbitals 3d,
4d, and 5d are gradually filled, while the internal transi-
tions are gradually filled with orbitals 4f and 5f. The tran-
sition element can thus be defined as the element, which
has an electronic distribution in which orbitals d or f is
partially full in the case of neutral or chemically unified
atoms in their compounds.
77
 Table (3-4) Electron configuration of transitional elements
First transitional series
Atomic number Symbol Name Electron configuration
21 Sc Scandium [18 Ar] 3d1 4s2
22 Ti Titanium [18 Ar] 3d2 4s2
23 V Vanadium [18 Ar] 3d3 4s2
24 Cr Chromium [18 Ar] 3d5 4s1
25 Mn Manganese [18 Ar] 3d5 4s2
26 Fe Iron [18 Ar] 3d6 4s2
27 Co Cobalt [18 Ar] 3d7 4s2
28 Ni Nickel [18 Ar] 3d8 4s2
29 Cu Copper [18 Ar] 3d10 4s1
30 Zn Zinc [18 Ar] 3d10 4s2
Second transitional series
39 Y Yttrium [36 Kr] 4d1 5s2
40 Zr Zirconium [36 Kr] 4d2 5s2
41 Nb Niobium [36 Kr] 4d3 5s2
42 Mo Molybdenum [36 Kr] 4d5 5s1
43 Tc Technetium [36 Kr] 4d5 5s2
44 Ru Ruthenium [36 Kr] 4d7 5s1
45 Rh Rhodium [36 Kr] 4d8 5s1
46 Pd Palladium [36 Kr] 4d10
47 Ag Sllvar [36 Kr] 4d10 5s1
48 Cd Cedmium [36 Kr] 4d10 5s2
Third transitional series
57 La Lanthanum [54 Xe] 5d1 6s2
72 Hf Hafnium [54 Xe] 4f14 5d2 6s2
73 Ta Tantaium [54 Xe] 4f14 5d3 6s2
74 W Tungsten [54 Xe] 4f14 5d4 6s2
75 Re Rhenium [54 Xe] 4f14 5d5 6s2
76 Os Osmium [54 Xe] 4f14 5d6 6s2
77 Ir Iridium [54 Xe] 4f14 5d7 6s2
78 Pt Platinum [54 Xe] 4f14 5d9 6s1
79 Au Gold [54 Xe] 4f14 5d10 6s1
80 Hg Mercury [54 Xe] 4f14 5d10 6s2

78
 First internal transitional series
Atomic number Symbol Name Electron configuration
58 Ce Cerium [54 Xe] 4f 1 5d1 6s2
59 Pr Praseodymium [54 Xe] 4f 3 6s2
60 Nd Neodymium [54 Xe] 4f 4 6s2
61 Pm Promethium [54 Xe] 4f 5 6s2
62 Sm Samarium [54 Xe] 4f 6 6s2
63 Eu Europium [54 Xe] 4f 7 6s2
64 Gd Gadolinium [54 Xe] 4f 7 5d1 6s2
65 Tb Terbium [54 Xe] 4f 9 6s2
66 Dy Dysprosium [54 Xe] 4f 10 6s2
67 Ho Holmium [54 Xe] 4f 11 6s2
68 Er Erblum [54 Xe] 4f 12 6s2
69 Tm Thulium [54 Xe] 4f 13 6s2
70 Yb Ytterbium [54 Xe] 4f 14 6s2
71 Lu Lutatium [54 Xe] 4f 14 5d1 6s2
Second internal transitional series
90 Th Thorium [86 Rn] 6d2 7s2
91 Pa Protactinium [86 Rn] 5f 2 6d1 7s2
92 U Uranium [86 Rn] 5f 3 6d1 7s2
93 Np Neptunium [86 Rn] 5f 4 6d1 7s2
94 Pu Plutonium [86 Rn] 5f 6 7s2
95 Am Americium [86 Rn] 5f 7 7s2
96 Cm Curium [86 Rn] 5f 7 6d1 7s2
97 Bk Berkelium [86 Rn] 5f 9 7s2
98 Cf Californium [86 Rn] 5f 10 7s2
99 Es Elnsteinium [86 Rn] 5f 11 7s2
100 Fm Fermium [86 Rn] 5f 12 7s2
101 Md Mendelevium [86 Rn] 5f 13 7s2
102 No Nobelium [86 Rn] 5f 14 7s2
103 Lr Lawrencium [86 Rn] 5f 14 6d1 7s2

79
 The definition of the transitional element by the electronic
configuration of neutral atom excludes copper, silver and
gold whose electronic configurations are in stabilization
ns1 (n-1)d10 state as well as zinc, cadmium and mercury,
ns2 (n-1)d10 on the other hand if the presence of electrons
in chemically united atoms is the only property excluding
the elements Y, La and Ac.
3-2-2 General characteristics
1 - Physical properties
All transition d elements are generally highly dense met-
als and low atomic volumes and high melting and boiling
points. It is fused and boils the last elements of the group
distinctly at lower temperatures as compared to other el-
ements of the group. Mercury is the last element of the
transition series represents the observed exception of met-
als for being fluid under normal condition. Elements of
the lanthanide series possesses metallic properties which
show a metallic luster and good conductivity of electrical
and heat, as well as that the degrees of melting and boiling
points as expected.
Table (3-5) melting and boiling point and atomic radii for transition d elements
First transitional series
Sc Ti V Cr Mn Fe Co Ni Cu Zn
melting point /K 1539 1998 1988 2103 1520 1801 1763 1725 1356 693
boiling point /K 2727 3533 3773 2573 2363 3008 3313 3113 2853 1180
atomic radii / pm 162 132 122 117 117 116 116 115 117 125
Second transitional series
Y Zr Nb Mo Tc Ru Rh Pd Ag Cd
melting point /K 1509 2373 2223 2870 2140 2670 2240 1828 1233 594
boiling point /K 2927 3870 5370 5070 4470 4170 3443 2453 1040
atomic radii / pm 182 145 134 129 135 124 125 128 134 141
Third transitional series
La Hf Ta W Re Os Ir Pt Au Hg
melting point /K 920 2570 3289 3670 3420 2970 2727 2047 1336 234
boiling point /K 3469 5470 6270 6070 4870 4770 4070 2970 630
atomic radii / pm 187.7 159 147 141 137 135 136 139 144 155

80
2.Chemical properties
A - Elements of a group d:
In general, group d elements are relatively ineffective with
oxygen, halogens, sulfur, nitrogen, hydrogen and water
vapor at normal conditions but at high temperatures are
reacted with these reagents more easily. The group of tran-
sition elements reacts with hydrogen under certain con-
ditions to form substances with structures called internal
hydrides. This name originally implied that the arrange-
ment of the metal atoms is roughly the same as that of the
There are no zero oxidants in the
metal crystallization while the hydrogen atoms enter the
first series of transition elements as
distances. Although this is not the reality, the term remains
shown In the larger circles in the
traded. These hydrides occupy a volume larger than the
graph.
metal volume, where it was formed from it and has a me-
tallic appearance. It is prepared by direct union of metals
and hydrogen at high temperatures. For example, plati-
num, palladium and iron are permeable metals to hydrogen
at high temperatures. Tantalum absorbs hydrogen, creating
an easily broken product. The elements of group d react
directly when the carbon-element mixture is heated at tem-
peratures above about 2200ºC produced carbides. The car-
bides have high melting degrees and are very hardened, which
is on two groups with a general formula MC and M2C such as
titanium carbides, zirconium, hafnium, vanadium and others.
These carbides are characterized by chemical inertness, such
as titanium carbide TiC, it is not affected by water or aqueous
solutions of hydrochloric acid even when 600ºC.
B. Elements of group f lanthanides:
Lanthanide metals are soft and more effective with known re-
agents of group d elements, so lanthanides react slowly with
halogens forming MX3 compounds and with oxygen forming-
M2O3 at room temperature but easily ignite with these reagents
at temperatures above 200ºC and react with sulfur at boiling
point to form M2S3 and nitrogen at a temperature higher than
1000ºC to form MN. At temperatures above 300ºC, lanthanides
react rapidly with Hydrogen to form a kind of hydride and at
high temperatures this reaction with boron and carbon gives
borides and carbides respectively.

81
 3-2-3 The first transition series
The elements of this series (from scandium to zinc) (Zinc)
in the fourth period of the periodic table between calcium
Ca in group (IIA), and Galuim Ga in group (IIIA). Table
(3-4) shows the electronic configuration of these elements.
These elements are called transition elements or group of
d elements because they possess internal energy level 3d.
Transitional elements

It is proposed that the transitional elements of the fourth

period or the first transition series should end with nickel
element because the 3d energy level of the next two ele-
ments (copper and zinc) is full, and fact that copper shows
multiple characteristics of the transition elements, while
zinc shows intermediate characteristics between the tran-
sition elements and the main group elements. For this, it
is appropriate to include both copper and zinc in the first
series of transition elements. That the electronic configu-
ration differences that distinguish transitional elements of
other elements, lead to the emergence of special physical
and chemical properties of the transition elements. These
properties are not necessary unique behavior to the tran-
sitional elements. but combined they give transitional el-
series of the first transitional ements have a behavior that is distinct from that of any
elements. other type of element. These characteristics can be limited
to the elements of the first transition series:
A- Metallic Characteristics:
One of the most prominent properties of the elements of
the first transition series is that they are all of metals with
high melting and boiling point and good conductors for
heat and electrical. They are generally hard materials and
are strong alloys with them and the possession of these
qualities gives them a unique technological importance of
its kind. Although the transition elements are denser and
stiffer and have higher boiling points than the main groups,
there is no systematic increase in the degree of these prop-
erties as the atomic numbers increase. That metal elements
of this series are divided into two groups; first groups from
Sc to Mn and the second from Mn to Zn with peaks at Ti
and V and at Co, and Ni. The division of the series into
two groups has to do with the fullness of orbitals d. The
energy level 3d of manganese is half full and then it be-
comes occupied by individual electrons fills by electrons
82 pairs to become complete fullness at copper and zinc.
 3500

3000
K / ‫درجة الغليان‬

2500

2000

1500

Figure 3-12
1100
Sc Ti V Cr Mn Fe Co Ni Cu Zn The relationship between
‫االعداد الذرية‬

That electronic configuration 3d5 in chrome and 3d10 in boiling point and atomic
copper gets by removal of an electron from the 4s level number for elements of the
for the purpose of obtaining the most stable ranking. These first transition series.
elements are intended to release the largest number of ex-
ternal valence electrons for metal bonding, that it gives the
highest binding energies and is reached in the middle of
the series.
B - Oxidation cases:
There are more than one oxidation state in all transition
elements, for example, iron has two oxidation states, +2
and +3, and an element atom Cobalt has two oxidation
states of +2 and +3 and Cr also has numerous oxidation
states are +2, +3, +4, +5, and +6 in their ionic and covalent
compounds. The oxidation state of the transition elements
reaches +7, in the case of manganese as in the compound
potassium permanganate KMnO4 Table(3-6). The cause of
the oxidation state is multiple for atoms of the transition
elements are due to the number of electrons in the outer
shell ns and (n-1)d to the atom of that element, where the
loss of electrons from ns begins first and then in terms of
the loss of these electrons one after the other, that the num-
ber of electrons in d does not exceed five electrons. Each
electron loses an oxidation state appear, so that manganese
has an oxidation state of +1, +2, + 3, +4, +5,+ 6 and +7 due
to the difficulty of losing all electrons in (n-1)d because
of its need for high ionization energy so prefer to compose
coordinate .The resulting ion is a strong oxidizing agent
that draws electrons from neighboring atoms. The highest
oxidation state of the atoms of the transition elements in
the first series depends on:
1. Power of oxidizing agent.
2. Nature of the resulting compound 83
Table 3-6 Oxidation states reported by the first transition series
Multiple oxidative states Highest oxidation The most stable
Element Electronic configuration of the
case
symbol two orbitals (ns), (n-1) 3d oxidation state

Sc 4s2 3d1 +3 +3 +3
Ti 4s2 3d2 +3 , +4 +4 +4
V 4s2 3d3 +2 , +3 , +4 , +5 +5 +4
Cr 4s1 3d5 +2 , +3 , +6 +6 +3
Mn 4s2 3d5 +2 , +3 , +4 , +5 , +6 , +7 +7 +2
Fe 4s2 3d6 +2 , +3 +6 +3
Co 4s2 3d7 +2 , +3 +4 +2
Ni 4s2 3d8 +2 , +3 +4 +2
Cu 4s1 3d10 +1 , +2 +3 +2
Zn 4s2 3d10 +2 +2 +2
From the Table (3-6) you can see the following:
1-The presence of familiar oxidative state (+2) when elec-
tronic loss 4s2
2- The increase in the number of oxidation state of scan-
dium (Sc) to (Mn) and in the latter element the oxidation
state is consistent with the loss of electrons 4s2 , 3d5
3- The sharp decrease in the number of oxidants after man-
ganese due to the difficulty of removing electrons after
doubling.
C-Acid and base properties
The acidic and basic properties of the transition elements
are based on Lewis’ concept of oxidation state, so the high-
er the oxidation number of element, the lower the basic
and the more acidity as shown in Table 3-7 for manganese
in its oxides.
Table 3-7 Acid and base properties for magnesium transition element in its oxides
Oxidation number property formula Oxide name
+2 basal MnO Manganese oxide

+3 Weak basic Mn2O3 Trioxide dimagnesium

+4 Amphoteric MnO2 Dioxide magnesium

+6 Acidity MnO3 Trioxide magnesium

+7 Rich acidity Mn2O7 Heptoxide dimagnesium

84
D-Formation of coordinate Complexes
The positive ions of the transition elements shed strong electro-
static attraction on molecules or ions that contain one or more
pairs of non-bonding electrons are known as such molecules or
ions by (ligands). The result of this attraction is the so-called
complex coordinate. Thus, we can say that the coordinate com-
plex is an centralized atom is often a transitional element sur-
rounded by a group of atoms, molecules, or ions are called li-
gands. The central atom often, they are transition metals, but
ligands may be a negative ion of signal atom, such as a halide
ion, or may be a polyatomic molecule, or an ion containing
a donated atom belonging to oxygen or nitrogen group such
as H2O or NH3 or CN or NO2 and others. The term coordinate
number uses to refer to the number of atoms that donated elec-
trons related to the central atom. The coordinate number of the
iron atom in the coordinate ion [Fe(CN)6]-4 equals 6, and the
oxidation state of iron equals +2. The number -4 represents the
complex ion charge, which is equal to (algebraic sum of the A) octahedral complex.
coordinate charge number and the central ion charge), where
the sum of coordinate charge number equal (-6) and central ion
charge equal (+2) so, the difference between them is (-4) repre-
senting the complex ion charge. The coordinate number varies
from one element to another but in the complex compounds of
the transition elements in oxidation states +2 and +3 is usually
4 or 6. Examples of such complexities are [Co(NH3)6] +3 and
[Ni (CO)4]+2.Common forms of such complexes are octahedral
of the coordinate number (6), and tetrahedral of the coordinate B) tetrahedral complex.
numbers 4. There are other forms such as linear, binary, tri-
angular, and others. These complex shapes will the student is
informed about it in his university studies.
H- Color
All complexes of the transition elements are characterized by
their distinctive bright colors, for example CuSO4.5H2O cop-
per sulphate has a light blue color, while Ni (OH)2 light green
color and copper fourth ammonia (II) [Cu (NH3)4]+2 dark blue
and others.
I- Magnetic properties
Electrons have a charge and their motion creates two magnetic
effects one is associated with the quantum number (l) and the
other is accompanied by the quantum number (s) and these two
effects give each electron the characteristics of a small mag-
netic pole. 85
 Types of magnetic A magnetic moment in electron-filled shell is equivalent
behavior to the magnetic moment of each other and the atom as a
whole will possess the resultant magnetic moment only if
there are individual electrons in the valence orbit. When
this happens it will make the outer magnetic field projected
to direct the magnetic moment of the atoms toward that
this field and this such behavior is called (paramagnetism).
When all electrons are double in external orbitals, in this
case a magnetic fields induced that reflect the magnetic
field that generated it. This makes material repel with the
external magnetic field, and this behavior is known as (dia-
magnetism). For example, we find iron as an element is
Diamagnetism attracted towards the magnet field because of its existence
individual electrons in its outer shell where the outer shell
is for the iron atom (3d).
3d
↼
↼
↼
↼
↼
↼

If the iron is within the coordinate ion as in [Fe(CN)6]-4

(Hexaciano ferrate(II)) it is not attracted to the magnetic
field because the orbitals the outer shell of the iron ion (II)
has been saturated with electrons granted by ligands. For
the component [Fe(CN)6]-3 (Hexaciano ferrate(III)), it is
where the triple iron ion gravitates toward the magnetic
field (paramagnetism) due to the presence of a single elec-
tron in the outer orbital shell 3d of triple iron ion as shown
in the following figure:
Paramagnetism 3d 4s0 4p
↼
↼
↼
↼
↼

Information
The green color represents the At the approach of the ligand(coordinate group), such as
electron pairs And the yellow (CN-1) has a high rang of orbits 3d of triple iron ion works
color represents single elec- on duplex electrons in 3d orbitals for iron ion as shown in
trons. following figure:
3d 4s0 4p
x x x x x x
↼
↼
↼
↼
↼

↑→

↑→

↑→

↑→

↑→

↑→

- - - - - -
CN CN CN CN CN CN

The electrons given by the coordinate groups are placed in

the form of (X) differentiated from electrons not granted
by coordinate groups ( )

86
A complex ions that have a paramagnetic property is
[NiCl4]-2 ,where the outer shell (3d) of the Ni+2 ion con-
tains two single electrons, so it shows a single paramag-
netic property

28
Ni2+ 1s2 2s2 2p6 3s2 3p6 4s0 3d8

in orbitals 3d 4s0 4p
↼
↼
↼
↼
↼
↼
↼
↼

If we take the zinc component (30Zn) we observe from its

electronic configuration that all the electrons in its outer ferromagnetism
shell are in pairs and so it is not attracted to the magnetic
field.
30
Zn 1s2 2s2 2p6 3s2 3p6 4s2 3d10
3d 4s2
↼
↼
↼
↼
↼
↼
↼
↼
↼
↼
↼
↼

A third type of magnetic behavior is ferromagnetism,

which is actually very rare but it is extremely importance.
a special case of paramagnetism occurs in compounds that
contain a large part of atoms or ions containing single elec-
trons.
For this type of components and in appropriate conditions,
the single electrons of each atom interact and regulate
themselves with electrons. This process is repeated dur-
ing all atoms in the component. Therefore, the effect leads
to the construction of permanent magnets. Ferromagnetic
behavior can be observed mainly among metals, alloys and
transition elements oxides and less impact in internal tran-
sition elements oxides. Permanent magnets made of
materials ferromagnetism
Gouy balans
The material with dia or paramagnetic properties can be
identified using a sensor called (Gouy balance). This de-
vice has a very sensitive balance attached in one hand the
material whose magnetic properties are to be defined and
the equivalent weights are placed the material in the sec-
ond cuff as shown in Figure (3-13).

87
 Material
There is no magnetic
Diamagnetism field Paramagnetism
material
where the material is placed in a strong magnetic field, if
Figure 3-13
it has paramegnetism properties, it will gravitate towards
Gouy balans.
the magnetic field and increase the scale reading, but if
the material diamegnetism it will repet with the feild and
decrease the scale reading
I-Effectiveness as catalysts:
The catalysts derived from the transition elements has great
importance in many biological processes and systems and
are indispensable in chemical industries too. Almost all of
the transition elements ability to behave as catalysts either
in their free states or in component forms. This ability is
Lu
71

likely to come from the use of d orbital or from the forma-

Yb
70

tion of compounds that can absorb and activate reactants.

Tm

The ability of these compounds to behave as catalysts by

69

ability to create low-energy pathways for reactions by ei-

Er
68

ther oxidation-induced changes or by the formation of suit-

Ho
67

able intermediate compounds.

Dy
66

3-2-4 Lanthanides and Actinides

Tb
65

1.Lanthanides
Gd

Lanthanum (La) (atomic number 57) shows the first ele-

64

ment in the third series of transition elements which has

Eu
63

electronic configuration [36Kr] 4d10 5s2 5p6 5d1 6s2 .

Pm Sm
62

The next element is cerium and its atomic number is 58

61

and electronic configuration [36Kr] 4d10 4f1 5s2 5p6 5d1 6s2.
Nd
60

Where electrons begin to full the energy level (peak level)

4F until we get to the element lutetium (Lu) has an atomic
Pr
59

number (71) and then the 4d level returns to fullness. Ele-

Ce
58

Lanthanides series ments from lanthanum to lutetium are called lanthanides

the first internal transition series consists of 14 elements.

88
These elements are called a Rear earth because they ex-
ist in uncommon mixtures as they were thought to be ele-
ments of soil or oxides.
The elements of lanthanides are very similar to each other,
and their separation is a major problem because all their
compounds are very similar, where the oxidation state (+3)
appears and this state shows the predominant ionic proper-
ties are similar to the alkaline earth metal ions except that
they are positive trilogy and not positive dualism. There is
no aerobium (Eu) (one of the elements in this series) free
in nature and can only be discovered when obtained within
an nuclear fission output.
2. Actinides:
Similarly to the lanthanide series, the actinides begin
with an element actinium (Ac) has an atomic number of 89
and has an electronic configuration: [54Xe] 4f14 5d10 6s2 6p6
6d1 7s2 Before we know about the possibility of the pres-
ence of elements after uranium has developed heavier nat-
ural elements such as thorium, erotectinium and uranium
in the sixth period of periodical classification in tandem

Lu
71
elements of hafnium, tatalium and tungsten. It was inferred

Yb
that these elements were the beginning of a new series of 70
four ten elements in which the sixth quantum level is filled Tm
69

in exactly the same way that the fifth quantum level is full
Er
68

by hafnium, tantalum and tungsten elements. The discov-

Ho
67

ery of several post-uranium elements and the study of their

Dy
66

qualities in fact demonstrate the emergence of a new inter-

nal transitional series that begins after actinium. So post-
Tb
65

actinium elements are now called actinides. Whatever the

Gd
64

extent of oxidants possessed by oxidants, they always have

Eu
63

an oxidative state (+3) within oxidants.

Sm

In the actinides, the element curium is one of the ele-

62

ments, its inner subshell casing is probably half-filled, and

Pm
61

the vast majority, for its compounds curium (Cm) is a posi-

Nd
60

tive trilogy, while the elements are above americium have

Pr
59

many oxidative states behave like +2, +3, +4, +5, +6 and
Ce

an Berkelium element (Bk) after curium exhibits oxida-

58

tive states +3, +4. Polymorphism in oxidative states of ac- actinides series.
tinides elements to extent amersium elements makes the
chemistry of the elements of this series very complex.

89
 In addition, in cadolinium (Gd) the subshell level is
filled in half with electrons. This is known particularly
represents electronic configuration and ionization. Cado-
linium forms Gd+3 ions only, (with the loss of three exter-
nal electrons) and does not show inclination to add or lose
electrons in a half-filled inner level. This behavior can be
compared with the element that comes before cadolinium
is an aerobium (Eu), and this element exhibits an oxida-
tion state of +2 and also +3,the next element, terbium (Tb),
exhibits oxidative states +3 and +4. We will take iron as an
example of transition elements.
magnetite 3-3- Iron
3-3-1 Introduction
Iron is the second metal after aluminum and the fourth
element after oxygen, silicon and aluminum in proportion
to the crust of the earth, believed to be the center of the
earth, consists mainly of iron and nickel. Iron is spread in
the earth’s crust, united with other elements forming mul-
tiple ores; hematite which contains Fe2O3 and magnetite
contains Fe3O4 and limonite contains FeOOH and siderite
contains FeCO3.
Iron is also the most important transitional element, it is
one of the elements involved in the synthesis of blood he-
moglobin and is also found with molybdenum in the syn-
limonite thesis of the nitrogen fixing enzyme. It is one of the neces-
sary elements that are Chlorophyll for the food industry in
the plant.
3.3.2 The location of iron in the periodic table
Iron falls in the periodic table and in the fourth period
the eighth group B, and has the following electronic con-
figuration: 26Fe : 1s2 2s2 2p6 3s2 3p6 4s2 3d6
The fact that the iron element of the transitional valence
fall within the two outer shell [ns, (n-1) d] but it does not
show an oxidation state equal to this number (8).
The most common oxidation states in iron are +2 and +3,
the process of losing two electrons from the iron atom to
hematite form an iron ion (II) or the loss of three electrons to form
iron ion (III) depends on the reactants nature . The relation-
ship between the oxidation states can be represented the
following equation below: Fe2+ Fe3+ + e-

90
The equation above shows that the iron ion(II) can oxidize
(lose an electron) to become an iron ion(III), or the iron
ion(III) reduces (acquires an electron) to the iron ion(II).
3-3-3 Properties of iron
Iron is a shiny white metal that is pure and is character-
ized by all the propertied of metals such as hardness, good
thermal and electrical conductivity, and ability of hammer-
ing and , extrusion and other properties. It is a magnetized
metals with melting point 1528°C, boiling point 2861°C
and density 7.86 g/cm3 at 25°C
3-3-4 Iron reactions
1. Iron does not react in normal temperatures with dry air
and does not react with water free from dissolved air, but
reacts with oxygen in moist air to form a brownish reddish
layer called rust.
This layer of iron is separated in the form of crusts, which
is aqueous iron oxide and does not stop rust at the outer
surface of the iron piece (due to permeability and incoher-
ence of the oxygen layer formed)
This layer of iron is separated in the form of crusts, which
is aqueous iron oxide (III) and does not stop rust at the
outer surface of the iron piece (due to permeability and
incoherence of the oxygen layer formed)
4Fe + 3O2 + nH2O 2Fe2O4.nH2O
2. Iron is oxidized when heated to high temperatures (de-
gree of redness) and in the presence of air forming iron
oxide magnetic Fe3O4 as in the following equation:
3Fe + 2O2 ∆ Fe3O4
3.Iron reacts with water vapor, forming magnetic iron ox-
ide and realizing hydrogen gas as in the following equa- Do you know

tion: 3Fe + 4H2O ∆ Fe3O4 +4H2 that the most famous iron sul-
4. Iron reacts easily with dilute acids such as hydrochloric phate is iron pyrite FeS2, which
acid and sulfuric acid forming iron salt (II) and realizing is known as gold false because
hydrogen gas as in the following equations: of the color similar to the color
Fe +2HCl FeCl2 +H2 of gold, which is used in cheat-
ing gold without easily detected.
Fe + H2SO4 FeSO4 + H2
It reacts with concentrated acids such as hot concentrated
sulfuric acid and nitric acid, forming iron salts and water,
and rising SO2 gas as in the following equation.
3Fe + 8H2SO4 FeSO4 +Fe2(SO4)3 + 4SO2 + 8H2O
Iron sulfate (II) Iron sulfate (III) 91
 5. When heating a mixture of iron filings and sulfur pow-
der form Iron sulfide (II) as in the following equation:
Do you know Fe + S ∆ FeS
That iron enters into a lot of 6.Iron reacts with halogens (fluorine, chlorine and bro-
the most famous cyanide com- mine) to produce iron halides (III) according to the equa-
pounds Fe4 [Fe(CN)6] is called tion:
a blue tincture Prussia which is
2Fe + 3X2 2FeX3 ; X = F,Cl, Br
used to remove yellowing water
For example, iron chloride (III) is produced from the pas-
from sediment salts iron, called
sage of chlorine gas on the heated iron filings to the red-
locally ( Jweet ).
ness temperature as in the following equation:
2Fe + 3Cl2 ∆ 2FeCl3
3-3-5 Iron extraction
Iron is produced industrially in the bloating oven by re-
ducing iron ores, especially hematite and magnetite with
carbon at 2000°C.
The method involves mixing iron ore with coke and
limestone(mostly calcium carbonate) and add the mixture
from the top hole of the bloating oven [Figure (3-14)].
Then the hot air from the nozzles is blown. In a result of
the high temperature of the mixture, many reactions hap-
pen which can be summarized. Firstly coal combustion to
heat the oven and form a carbon monoxide:
2C + O2 2CO
Carbon monoxide reduces iron ore to produce fusible iron
and carbon dioxide:
Fe2O3 + 3CO ∆ 2Fe + 3CO2
The oven heat also caused the decomposition of calcium
carbonate into calcium oxide and carbon dioxide accord-
ing to the equation:
CaCO3 ∆ CaO + CO2
Then calcium oxide combines with sand to be molten cal-
cium silicate according to the equation:
CaO + SiO2 CaSiO3
The calcium silicate molten floats above the iron molten,
because its density less than the density of iron, forming a
layer called slag that prevents mixing of iron fusion with
the material above it. The molten iron is drawn from time
to time from the bottom of the oven, where it is poured into
special molds and then called cast iron, which is used as
iron ore in the following stages of iron manufacturing.
92
The slag formed is also pulled from special holes below
oven, which is used in cement, concrete or concrete roads
and other construction. Due to environmental concerns
about the use of coke, alternative methods of iron process-
ing have emerged, one of which is the reduction of iron
using natural gas. Iron is produced in the form of a powder
called spongy iron, which used in steel industry. The pro-
cess consists of two main interactions namely: oxidation of
natural gas with the help of auxiliary agent and heat.
2CH4 + O2 2CO + 4H2
the resulting H2 and CO gases then at 800-900°C reduction
of iron ore to produce iron spongy.
Fe2O3 + CO + 2H2 ∆ 2Fe + CO2 + 2H2O
The sand is then removed by adding calcium carbonate in
the next step to form slag as in the previous method.
This method of iron production is characterized by:
1. Does not require expensive coke and is not available.
2. The cost of its creation is much lower than the cost of
setting up the blower oven.
3. Technology is simple and easy to use.
4.The iron produced is carbon-free while the cast iron is
produced from the lower oven contains approximate
ly 4% carbon.
Ore, coke and limestone

Hot gases existing

out of the oven

Fire brick

Compressed
air Figure 3-14
Bloating oven.

slag Fused iron

93
 3-3-6 Iron types
We learned that the iron produced from the bloating oven
is called cast iron contains impurities ranging from (6% to
8%) mostly carbon and silicon. Percentage change these
impurities can get different types of iron ones:
A- Casting Iron
When re-melting cast iron and pouring into molds to make
the required tools then it is called cast iron and it has two
types depending on casting and cooling method. If you use
metal molds for casting the casting process is quick and
the output is called white cast iron. If sand molds are used
then the solidification process is slow and the product is
called gray cast iron. In general, cast iron, known locally
as ‘Aheen’, is characterized very rigid but brittle and does
not withstand strong shocks. It is often used in making
some parts for fireplaces, pipes and sewage covers.
B) casting (steel):
Iron produces steel from cast iron after oxidation of exist-
ing impurities in a special oven, add the required amount
of carbon (0.2% to 1.5%) and other elements according to
the desired qualities of the desired steel quality. The prop-
erties of iron and steel depend on the ratio of carbon and
other additives.

94
 Basic concepts
Periodic Table Parts Electronegativity
The modern periodic table consists of four It is the ability of atoms in molecules to
parts, the first part includes the elements attract electrons towards it from other
that ends Its outer shell with s type orbit- atoms connected with it chemical bond.
als. The second part includes elements Electronegativity is increased by increas-
whose outer shell ends with orbitals of ing the atomic number in one period and
type s and p while elements are whose decreased by increasing atomic number.
outer shell ends with a type of orbital s Transition element
or d the third part of the periodic table. An element that has an electronic config-
Internal transitional elements series (lan- uration that is orbitals d or f are partially
thanides and actinides) fourth part of the full in the neutral or chemically bonding
periodic table, whose outer shell ends in their compounds.
with s and f orbitals Coordinated Complexes
Atomic Volume It is a central atom and is often for a tran-
It is half the distance between the centers sitional element surrounded by a group
of two atoms two consecutive crystals are of atoms or molecules or ions are called
measured using radiology X- ray. Atomic actinides. Actinides may be a single atom
radii are lower in the one period when you negative ion like a halide ion or it could
move from left to right but they are in- be polyatomic molecule or an ion that
crease when moving from top to bottom contains donation atom and belong to the
in one group. atom of the Oxygen or Nitrogen group
Ionization Energy such as H2O, NH3, CN and NO2 and oth-
Is the energy needed to extract an elec- ers.
tron from an neutral atom in its gaseous Coordinated number
state and positive ion formation. The en- It is the number of atoms, molecules or
ergy needed to extract the first electron is ions that donor of the electrons related to
called the first ionization energy is called the central atom. The number of symmet-
the energy needed to extract the second ric elements changes to another but in the
electron is the second ionization energy transition elements is equal to commons
and so on. 4 or 6.
Electron Affinity Magnetic Properties
It is the amount of energy released from The electron has properties as the result
the neutral atom in gaseous state when of the movement of qualities small mag-
acquired electron creating negative ions. netic rod with magnetic momentum. In
This trait increases in the one period by the case of materials contain full orbitals
increases the atomic number while de- with electrons, the magnetic moment of
creasing in the group by increasing atom- one electrons are equivalent to the mag-
ic number. netic moment of the other electron. Thus
is not attracted to the outer magnetic field,
and it is said to be a diamagnetic material.
95
 Chapter Three Questions 3
3-1 After trying to divide the elements A- Electronic affinity.
into metals and nonmetal, attempts were B- Increasing electronegativity
made to classify them. Mention these at- C- Atomic radius increasing
tempts and on what basis those attempts D- Increasing ionization energy
were built. 3-10 Explain
3-2 Indicate the periodic law reached by A- There is no need in the old to clas-
each From Lands and Mandeleeff. sify the elements in periodic table.
3-3 What are the weaknesses of the scien- B- The division of elements into metals
tific Mandeleeff table and nonmetals did not last long.
3-4 What is the difference between the C- High ionization energy 17Cl com-
Mendeleev table and the new periodic pared to ionization energy 12Mg
table. D- Energy is released when the first
3-5 How the atomic size of the elements electron is acquired, but many often ab-
of the single period and single group sorb some energy when gaining the sec-
change by increases the atomic number ond electron.
and why. E- Fluorine ionization energy is greater
3-6 Which atoms in each of the follow- than oxygen ionization energy.
ing pairs are larger atomic size (3Li, 4Be) 3-11 How to obtain a spectrum for the
(16S, 8O), (17Cl,35Br) . element. Does the spectrum of the ele-
3-7 Arrange the following according to ment appear when it absorbs energy or
the ionaization energy increasing, men- when it emits energy.
tioning reason. Al, Al+, Al+2, Al+3. 3-12 What is the coordinate number
3-8 If you have the following elements of the central atom and complex Ion
(17Cl, 15P, 11Na) answer the following charge of each of the following:
questions: A- Hexaciano ferrate [Fe(CN)6](III)
A- Any of the elements have a larger B- Quaternary ammonia copper
atomic size and which the smallest atom- [Cu(NH3)4](II)
ic size. C- Quaternary chloro Nickel [NiCl4](II)
B- Arrange these elements as increasing 3-13 Mention methods of industrial
electronegativity, mentioning the reason extraction of iron? Which is better and
for this arrangement. why?
C- Arrange them according to the in- 3-14 Define each of the following
creasing attraction electronic affiliation, 1- Transitional element.
mentioning the reason 2- d elements group.
D- Any of these elements in which the 3- Coordinated number.
metallic properties are expected. 4- Ligand.
3-9 You have atoms 34Se, 16S, 8O ar-
ranged them according to the following
with the reason.

96
 4
Chapter Four
Solutions

After completing this chapter, the student is expected to:

• Identify solutions and understand the meaning of solubility and the factors
affecting it.
• Identify the types of solutions and some methods of classification and distinguish
between saturated solutions, unsaturated and supersaturated solutions.
• Recognize some methods of expressing solution concentrations.
• Distinguish between real solutions, plankton and colloidal systems.
• Distinguish between ideal and non-ideal (real) solutions and defines Raoult’s
law.
• Recognize the reason of solute some effection on some of solvent properties.
• Realize the concept of collective properties of solution and the quantitative
relation ships of solute concentration.

97
 4-1 Introduction
The solution is generally defined as a homogeneous mix-
ture of substances consisting Solvent and solute of one or
more, mixing proportions are different from solution to
solution. The largest material quantity In the mixture is the
solvent and the least amount is the solute.
The term solutions include different types of mixtures in
which Solid, liquid or gaseous materials exhibit either
solvent or solute behavior. But the solvent is usually liq-
uid such as water, alcohol, gasoline and ether chloroform,
carbon tetrachloride, acetone, etc. For example, sea water
contains a number of dissolved substances in addition to
other substances suspended in it. Water saturated with car-
bon dioxide is an example of a gas solution dissolved in
a liquid. The blood also contains a plasma solution with
suspended substances are blood cells.
On the other hand there are solutions that liquids doesn’t
formed solvent. The dental fillings that the doctor attends
is an example of a solid amalgam, in other words a solution
in which liquid mercury is formed liquid substance in the
metallic solid phase as zinc, whereas metallic alloys are an
example of solid solutions. Atmospheric air is one of the
solutions for gases in nature, oxygen as well as other gases
dissolved in formed solute while the solvent is nitrogen.
Table (4-1) shows different types of solutions and some
examples for each type
Table (4-1) different types of solutions and some examples for each type
Solution status Solute status Solvent status Examples
Gas Liquid Oxygen in water
Liquid Liquid Liquid Ethanol in water
Solid Liquid Sugar in water

Gas Gas Oxygen in air

Gas Liquid Gas water droplet in atmosphere
Solid Gas Dust particles in air

Gas Solid Hydrogen in platimim

Solid Liquid Solid Mercury in zinc
Solid Solid alloys

98
 4-2 Solubility process
The speed and ease of solubility process for any solute in a particu-
lar solvent depends on two important factors: Firstly, the change in
the temperature of the product mixture (solution) and secondly the
extent of dissolution of the solute (in the form of ions or molecules)
between the solvent molecules, associated with the solubility pro-
cess. It can be said that the solubility process takes place more
easily:
1. If the solubility process is accompanied by an increase in the
temperature of the resulting solution( hottest the solution).
2. If the diffusion of dissolved components in the solvent widely
and completely.
The change in energy associated with the solubility process is
called heat of solution, The increase (or decrease) in the solution
temperature in the melting process depends mainly on the reac-
tion forces (attraction or repulsion) between the solute and solvent
molecules. When mixing a solvent with a solute to prepare a solu-
tion, each of the solvent molecules will be affected with other sol-
vent molecules surrounding them, in addition to being affected by
solute molecules. The same can be said for solute molecules. It is
the amount of force of these effects that determines the amount of
solubility and ease with which the solubility process takes place. It
can be said that the ease of solubility is determined by the follow-
ing conditions:
1. weak affect between the solute molecules among them.
2. Weak effect between solvent molecules among themselves.
3. Large effect between solute molecules with solvent molecules.
Figure 4.1 shows the energy change associated with each step of
the solubility process. For the solubility process to take place, forc-
es must be overcome the attraction force between solute molecules
(step A) is accompanied by a heat absorption. (step B) represents
the separation and faraway of solvents particles of each other to
be ready to receive solute particles. This step needs to be absorbed
by the heat, ,but thermal energy is emitted as a result of the effect
between the solute and solvent molecules (step C) and thus the
solubility process in this case produces heat and the solution after
preparation is hot.
On the other hand, the solubility of a number of solids in liquid
solvents is accompanied by heat absorption (cold output solution).
this can be explained by widespread and complete diffusion of
the solute components between the solvent molecules during the
99
 solubility process and become free to move and randomly
Figure 4-1
in the solution after they were restricted movement in a
Change in system energy
solid state. Solvent molecules are also more free to move.
content accompanying the
They become in an environment that contains solute com-
steps involved in the melting
ponents that surround each molecule, including a mixture
process.
of solute and solvent molecules [Figure (4-2)].

4-3 Types and properties of solutions

Solutions in which the solvent is liquid can be clas-
sified in several ways Depending on the concentration or
type of solvent or mixture type that forms the solution as
follows:

4-3-1 Diluted and concentrated solutions

This classification only depends on the difference in the
amount of solute in the solution, ie on the relative masses
of solute and solvent, Therefore, it is a descriptive expres-
sion of the quantities of solute and solvent. A solution that
contains a larger amount of solute is called or described
as a concentrated solution, while a solution with a small
amount of solute is called with diluted solution.

4-3-2 Expression of concentration

Chemists use the term concentration to refer to the
Figure 4-2 amount of solute in an amount of solvent. The concentra-
Sodium chloride solution tion of the solution is expressed as is known amount of
(dissolved) in water (sol- solute in a given weight or volume of solvent. There are
vent). many ways to express the concentrations of solutions and
these are in two ways:
100
Firstly. Expression of concentration in terms of mass of
solute in mass of solvent (mass in mass) or amount of sol-
ute in amount of solvent (amount in amount). most impor-
tant are:
1. Mass percentage of solute
See Figure 4-3, What does the percentage written on the
bottle mean?
Percentage (20%) refers to the concentration of sodium hy-
droxide solution. Each 100g solution contains 20 g sodium
hydroxide and the remaining 80 g water. The solution can
be prepared with the same concentration if we dissolve 2g Sodium hydroxide
of sodium hydroxide in 18g of water. The mass percentage NaOH
of solute can be defined as the number of grams of solute 20%

in 100g solution and the mathematical relationship to cal-

culate the mass percentage of solute in a solution :
Figure 4-3
Percentage of mass to express
of the solution concentration

Exercise 4-1
What is the mass of potassium
chloride needed to prepare
Where m1represent solvent mass and m2 represent solute
250 g of a 5% concentration
mass
Example 4-1 solution by mass?

Calculate the mass percentage of sodium hydroxide in

a solution prepared by dissolving 8 g NaOH of 50g of Exercise 4-2
water.
How many grams of sulfuric

Solution acid H2SO4 are contained in

one litter of aqueous solution
m
% NaOH = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬NaOH
‫ × ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬100% of this acid if the percentage
mNaOH+mH O
2
of acid in it is 34% and the
density of the aqueous solu-
8 g
% NaOH = ‫ × ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬100% tion of this acid is equal 1.24
8 g + 50 g
g/ml (Note: the density of the
solution is defined as the ratio
% NaOH = 13.8 %. between mass to volume)

101
 2. Molality (m*)
Molality is the number of solute moles that is contained in
one kilogram of solvent. If one mole of sugar is dissolved
in one kg of distilled water you get a concentration of solu-
tion (1mol / kg) or one molali 1m Molality is known as the
following mathematical relationship:

Molality(m) ( ‫ ـ ـ ـ‬mole
‫ = )ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬solute moles number (mole)
kg solvent mass (kg)
n (mol)
m (mol/kg) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m (kg)
This method of expressing concentration is useful as this
expression refers to the ratio between the number of solute
particles (i.e., number of moles) to the solvent particles(ie
its mass) in solution.
The concentration of the solution expressed in molarity is
not affected by temperature change. As previously know,
Moles are expressed as follows:

m (g)
n (mol) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
M (g/mol)
Where m is the mass of solute in grams and M is the sol-
vent atomic molar
Example 4-2
Exercise 4-3 Calculate the molarity of the prepared solution by dis-
Dissolve 6.2 x 102 g ethylene solving 36 g of glucose sugar.
Molar mass = 180 g / mol in 360 g distilled water.
glycol(C2H6O2) used as an
Solution
antifreeze in the auto radia-
Calculate molality of solution through the relation:
tor in 4kg of distilled water.
m (g)
Calculate the concentration n (mol) = ‫ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬36
‫(ـ ـ ـ ـ ـ ـ ـ‬g)
‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.2 mol
M (g/mol) 180 (g/mol)
of the ethylene glycol solution
Calculate the number of moles of solute (glucose sugar)
expressed as the molar con-
m = m (g) × ‫ـ ـ‬1‫(ـ ـ ـ ـ‬kg) ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬360 g × ‫ـ ـ ـ‬1‫(ـ ـ ـ ـ‬kg) ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.36 kg
centration. 1000 (g) 1000 (g)
Change solvent mass to unit (kg)
n (mol) 0.2 mol
m (mol/kg) = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.556 mol/kg = 0.556 m
m (kg) 0.36 kg
* The student should distinguish between the mass sym-
bol (m) and the symbol used to express the molality
concentration(m)
102
3. Molar fraction x
Some properties of the solutions depend on the relative
quantities of all the components of the solution expressed
in terms of the number of moles (molar fractions). A meth-
od to express the components of the solution. For two
component solution solute (A) and solvent (B), the molar
fraction of the component (A) known, as the ratio of the
number of moles of this component nA to the total number
of moles of all the components of the solution (nA + nB), it Exercise 4-4
can be expressed as follows: n
xA = ‫ ـ ـ ـ ـ ـ ـ ـ ـ‬A‫ـ ـ ـ ـ ـ ـ ـ‬ Prepare a solution of sodium
nA + nB
Thus, the molar fraction of the solvent (xB) is the ratio be- hydroxide by dissolving 4 g of
tween the number of solvent (nB) to the total number of solid base at 16.2g of distilled
solutions (nA + nB) n water. Calculate the molar
xB = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ‬B‫ـ ـ ـ ـ ـ ـ‬
nA + nB fraction of the components of
The molar fractions of the components of the solution are
the solution.
always equal to one integer as described below:
n n
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ‬A‫ ـ ـ ـ ـ ـ ـ ـ ـ‬+ ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ‬B‫ = ـ ـ ـ ـ ـ ـ ـ‬1 ⇒ ⇒⇐⇐) xA + xB = 1 (
nA + nB nA + nB
Example 4-3
Prepare sucrose solution C12H22O11 (molar mass=342 g/
mol) by dissolving 34.2 g of it in 180g of distilled water
(molar mass=18 g/mol). Express the concentration Exercise 4-5
of sugar and water in the solution in terms of molar Calculate the molar fraction of
fraction for each one. water in a mixture consists of
solution 9.0 g of water and 120g of
Calculate the number of molars of sugar and water acetic acid (CH3COOH)
from the relationship:

m (g)
n (mol) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
M (g/mol)
nC H O = ‫ـ ـ ـ ـ ـ‬34.2 ‫( ـ ـ ـ ـ ـ ـ ـ ـ ـ‬g)
‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.1 mol
12 22 11
342 (g/mol)
180 (g)
nH O = ‫= ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬10 mol
2 18 (g/mol)
n H O
xC H O = ‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬C‫ ـ ـ‬12 ‫ـ ـ ـ ـ ـ ـ‬22‫ـ ـ ـ ـ ـ‬11‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.1
‫ ـ ـ ـ ـ ـ ـ ـ‬mol
‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.01
12 22 11 nC H O + nH O 0.1 mol + 10 mol
12 22 11 2
10 mol
xH O = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.99 In the same way
2 0.1 mol + 10 mol
103
 Second. Expression of concentration in terms of solute
Exercise 4-6 mass in solvent volume(mass in volume) or amount of sol-
ute in solvent volume (amount in volume), the most impor-
What the molarity concentra-
tant of these expressions are:
tion of chloride sodium NaCl
Molarity M*
solution prepared by dissolv- Mole denotes the amount of matter, and often there is a
ing 4.39 g of salt in distilled need to deal with the substance when is dissolved in a solu-
water to get a solution of 250 tion, so it is appropriate use of the Molarity concentration
mL? to determine the number of moles of material is dissolved
in a certain volume of solution. The Molarity are number
of the solute that is contained in one liter of solution, ie
moles number of solute (mole)
Molarity (M) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
solotion volume (liter)
n (mol)
M = ‫ـــــــــــــــــ‬
V (L)
Therefore, the unit of molarity is (mol/L). For example, if
one mole (62g) of ethylene glycol (C2H6O2) in sufficient
quantity from distilled water to obtain one liter of solution,
the concentration of the resulting solution is equal to 1mol
/ L or written in the simplest form 1M.
Example 4-4
Calculate the molarity concentration of a solution
prepared by dissolving 2.3g of ethanol (C2H6O) (molar
mass = 46 g/mol) in 3.5L of distilled water.
Solution
Calculate the number of ethanol moles
m (g) 2.3 (g)
n = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.05 mol
C2H6O
m (g/mol) 46 (g/mol)
Accordingly, his molarity can be calculated as follows:

M=n ‫( ـ ـ ـ ـ‬mol)
‫ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.05
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬mol
‫= ـ ـ ـ ـ ـ ـ‬0.015 mole/L = 0.015 M
* The student should V (L) 3.5 L
distinguish between the In general, a solution can be prepared for any substance
molar mass symbol M in two ways, firstly as it has already been explained in the
and the symbol used previous examples by dissolving a known weight of soluble
to express the molarity solid or liquid in a known weight of solvent or secondly
concentration M. by dissolving the known weight of the solute in a given
volume of solution.

104
 4-3-4 Dilution law
We often need to prepare a dilute solution (low concen-
Exercise 4-7
tration) from concentrated solution (high concentrated),
when diluting the solution (by adding an amount of sol- A solution of 2L and a concen-
vent to it) its volume increases and the concentration of the tration of 1.5M of Na2CO3,
dissolved material decreases (reduced concentration), but how many grams of carbon-
the amount of dissolved material remains constant. That is, ate sodium you need to pre-
the number of solute moles does not change [Figure (4-4)].
pare this solution?
And this is mean:

(A) (B)

Figure 4-2
Dilute solution:
(A) Volumetric flask fills
(C) with 100 mL solution of 0.10
Number of solute moles before dilution = Number of sol- M potassium chromate.
ute moles after dilution (B) The solution was com-
pletely converted to 1L vol-
umetric flask. (C) Distilled
Return to the general relationship to express molarity con-
centration can say that: water was added until the
beaker reaches the final vol-
Number of solute moles (n) = solution volume (V) (L) × ume of (1L). The new solu-
concentration (M) (mol /l) tion is 0.01 M and its volume
is 1L and contains the same
If we symbolize the volume and concentration of the solu- amount of K2CrO4 as if con-
tion before dilution (V1) and (M1) respectively and its size tained in the original solu-
and concentration after dilution (V2) and (M2) so: tion before dilution.

V1× M1 = V2× M2

105
 Example 4-5
Calculate the molarity concentration of sodium
hydroxide in a solution prepare it by adding 150 mL
of distilled water to 100 mL of NaOH solution with a
concentration of 0.2 M.
Exercise 4-8
solution
How much distilled water
V1 = 100 mL ‫ و‬M1 = 0.2 M
should be added to 250 mL of
V2 = 100 mL + 150 mL = 250 mL
0.5 M barium chloride solu-
M2 = ?
tion to get this salt solution at
V1× M1 = V2× M2
a concentration of 0.25 M
V×M
∴M2= ‫ـ ـ ـ ـ‬1‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1‫ = ـ‬100
‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬mL
‫×ــــــــ‬
‫ـ ـ ـ ـ ـ‬0.2
‫ـ ـ ـ ـ ـ ـ ـ‬mol/L
‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.08 mol/L = 0.08 M
V2 250 mL
4-3-4 Aqueous and non-aqueous solutions
The solution in which water is a solvent is called aqueous
solution such as water-sugar solution and sodium chloride
in water solution, while a solution in which a liquid other
than water is soluble is called a non-aqueous solution, such
as sulfur solution in carbon disulfide and naphthalene solu-
Exercise 4-9 tion in gasoline.
Calculate the volume of the
concentrated sulfuric acid so- 4-3-5 Saturated and unsaturated solutions
The solution that cannot be dissolved an additional amount
lution 18M required to pre-
of solute at a certain temperature, is called a saturated solu-
pare a dilute acid solution its
tion, [Figure (4-5a)], while the solution in which a greater
volume of 250 mL and 1.8M amount of solute can be dissolved, at a given temperature,
is called an unsaturated solution [Figure 4.5b]. It should be
noted here that the saturated solution at a certain tempera-
ture can become an unsaturated solution if the temperature
increases (considering that solubility of most substances
increases with increasing temperature) and vice versa.
Figure 4-5
a : saturated solution.
b : unsaturated solution.
c : supersaturated solution.

a b c

106
A condition that the solution can reach is called an over-
saturation state, the solution is therefore called the super-
saturated solution [Figure (4-5c)]. It can be defined as a
solution that contains an additional amount of solute from
that calculated at equilibrium (saturation state) at a certain
temperature. The condition above the saturation that the
solution may reach, its unstable state in which the solution
attempts to reach a stable state (saturation). This can be as-
sisted by adding a crystal of the solute or any other solid
or by moving the solution.
4-3-6 Solubility
Solubility of a substance is defined as the largest quantity
of the matter (grams or moles ... etc.) that can dissolve in
a certain amount (volume or mass) of solvent or solution at
specific temperature and pressure. Therefore solubility is
often defined as a substance dissolved in a given solvent at
any temperature as the number of grams solute which can
dissolve to reach the saturation state at 100 g of solvent.
4-3-7 Factors affecting solubility
There are a number of factors affecting solubility:
1. Temperature
Increased temperature often leads to increased solubility
and vice versa true, but this rule does not necessarily apply
in general. For this reason you should follow the follow-
ing:
If the process of solute dissolved in specific solvent to low
temperature solution (cooler solution) The rate of solu-
bility increases with temperature, an example is that the
solubility of potassium nitrate salt increases with tempera-
ture increasing. If soluble solubility in a particular solvent
leads to overheating of the resulting solution (the resulting
solution is hotter), the solubility will decreases by increas-
ing the temperature such as the process of soluble calcium
oxide in water. Gas solubility is generally greater in cold
solvents than in hot solvents.
2. The nature of the solute and the nature of the solvent
Solubility depends mainly on the nature of both solute and
solvent. A general rule of solubility is like dissolves like.
The polarized solute dissolves in a polarized solvent (such
as solubility of potassium chloride in water), while the
non-polarized solute dissolves in a non-polarized solvent
(such as iodine solubility in carbon tetrachloride). 107
 3. Pressure:
Pressure change has no noticeable effect on solubility solid
or liquid materials however, increased pressure leads to in-
creased solubility of gases.
4-3-8 Factors that accelerate the solubility process
There are a number of factors that affect the speed of solu-
bility of the material (but it is does not affect solubility):
1. The volume of solute particles
When the solute is dissolved, the solubility process takes
Do you know place on the outer surface of each particle of the solute,
Increasing water temperature which is in contact with the solution. Therefore, increasing
will reduce solubility gases such the total surface area of the precipitate (contact area with
as CO2 and O2. Decreases O2 the solution) will increase the solubility. Solute particles
solubility in water and then grinding and making them smaller sizes increases the total
drain it from water led to death surface area for these particles which increases the rate of
of living things, this is called solubility.
thermal pollution. 2. Stirring and shaking
Stirring or shaking exposes the surface of the solute par-
ticles to new batches of solvent continuously and this leads
to increased solubility rate.
3. Temperature
Temperature increases led to solubility increasing of the
solids and liquids, which have been described previous-
ly, and to increase solubility rate because it increases the
movement of the components of the solution by increas-
ing its kinetic energy and creating convection currents in
it and thus have the same effect as the mechanical stirring
process. On the other hand, the effect of increasing tem-
perature is adversely affecting the solubility rate of gases.
Increasing the temperature increases the kinetic energy of
the gas molecules leading to overcoming the interconnect-
ed forces that bind gas molecules to components, so it can
escape the solution.
4-3-9 Real solutions, plankton and colloidal systems
1.Real Solution:
Is a homogeneous solution in which the solute particles
have volumes ranging from (0.1-1.0nm) (1.0nm=10-9m)
that is within molecular volumes, so their solubility in the
solution form a homogeneous system. They can be seen
even with the strongest types of microscopes or separated
from the solution by filtration such as sodium chloride so-
108 lution in water or sugar water solution.
2. Colloidal systems
Also called colloidal solutions, colloidal state or colloidal
diffusion, its a case between real solutions and plankton.
Solute particles size are ranged between (1.0-100nm). So
they form a heterogeneous two-phase system (solid and
liquid) in which the solute particles aren’t separated by
sedimentation. It seems to the naked eye or the use of
a normal microscope as the real solution and solute can
not be separated by filtration and examples of this type are
milk and blood. However, if a high beam of light is pro-
jected onto the solution and then viewed from a vertical Figure 4-6
angle relative to the direction of incident light , it is able It enables us to see the beam
see the light dispersed by the solution. This phenimena is of ablution falling on the col-
called the “Tendle effect” and is characteristic of colloidal loidal solution from a vertical
solutions [Figure 4-6]. angle relative to the direction
3. Plankton of the rays. Because of their
They are solid particles stuck in a liquid medium and reflection from the surface of
therefore they are systems consisting of two distinct phas- the suspended particles in the
es (solid and liquid) with a solute particles size more than solution.
(100 nm). The solute particles in these systems may be
invisible, but they can be seen with a microscope. It form a
heterogeneous and unstable system (i.e., a temporal state)
because of the ability of the solute particles to sediment is
stagnated by gravity and can be easily separated by filtra-
tion such as sodium chloride is stuck in gasoline and sand
is stuck in water. Figure (4-7) and Table (4-2) show the
relationship between the solutions properties and solute
particles diameter dissolved in them.
Table (4-2) Relationship between the solutions and diameter of solute particles
Mixture Particle diameter Properties
The solute particles pass through the precise filter, and
Real solution Less than 1.0nm cannot distinguish them with a microscope and have a
high diffusion capacity.

The solute particles pass through the usual filter papers

Colloids Between 1.0-100nm but it does not pass through precise filters and can identify
by a microscope and has a weak ability to spread.

Plankton granules do not pass through the pores of

normal filter papers and has no ability to spread The
Plankion More than 100nm usual filtration has no propagation potential. It can
see with a normal microscope and sometimes with the
naked eye.

109
 Particles size particles size between particles size
less than 1.0nm 1.0- 100nm more than 100nm

Real solution Colloids solution Plankton

Figure 4-7
Comparing volumes of sol- 4-3 Raoult’s law
ute particles in various solu- For a solution containing two components similar in prop-
tions. erties such as toluene and benzene or heptane and octane
blends, each component behaves if it were pure. This type
of solution is approaching behavior of the ideal solution
and it is called the ideal solution. If the ideal solution con-
tains volatile components A and B, Raoult low provides
as follows:
“The vapor pressure of any component in an ideal solu-
tion is directly proportional with the molar fraction of that
component in the solution ” this proportionality can be ex-
pressed as the following:
PA∝ xA
or for a relationship of equality
PA = xA P°A
where PA represents the vapor pressure of component A
in the solution and fraction xA represent the molar of this
component. P oA represents a constant proportional ( vapor
pressure of pure component).The same applies to the other
component B in the solution.
PB∝ xB

PB = xB P°B
The total vapor pressure of the solution (PT) is equal to the
total sum of the two components partial vapor pressure ac-
cording to Raoult Law:
PT = PA + PB
This relationship can be represented by the graph shown in
Figure 4-8. All solutions that obey Raoul‘s law are called
ideal solutions, however in fact, most solutions do not ex-
hibit ideal behavior. While called
110
solutions not obey Raoul’s Law are not ideal, which are
called real solutions.
P =P +P
T A B
Vapor pressure

Figure 4-8
Raoul’s law of an ideal solu-
tion consisting of two volatile
components.
mole fraction
Exercise 4-10
Example 4-6
Steam pressures for pure ben-
At 40 °C, the vapor pressure of the pure heptane and oc- zene (C6H6) and pure toluene
tane are 92 Torr and 31 Torr respectively. Calculate the (C7H8) is 44.5 Torr and 88.7
pressure the vapor for each component in the solution Torr at a certain tempera-
and the total vapor pressure of the component solution ture. An ideal solution was
for solution of 1 mole of heptane and 4mole of octane? prepared at the same tempera-
solution ture by mixing 60 g of benzene
Raoult’s law with 40 g of toluene. Calcu-
PA= xA P°
A late the partial vapor pressure
we symbolize heptane with the letter H and octane with of benzene and toluene in the
the letter O. First, the mole fraction heptane (xH) and solution and total vapor pres-
octane (xO) in solution sure (atomic masses H = 1 and
C = 12)
n 1 mol
xH = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ‬H‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ‬0.2
nH + nO 1 mol + 4 mol
Since the sum of the molar fractions of the components
of the solution is equal to one, xO is equal to:
XO = 1 - XH = 1 - 0.2 = 0.8
By applying the Raoult’s law for volatile components,
we obtain
PH = xHP°H = (0.2)(92 Torr) = 18.4 Torr
PO = xOP°O = (0.8)(31 Torr) = 24.8 Torr
PT = PH + PO = 18.4 Torr + 24.8 Torr = 43.2 Torr

111
 4-5 Effect of non-volatile solute on
some solvent properties
The physical properties of the solution, which depend only
on the number of particles of the solute (molecules or ions),
and not on the type of those particles, in a certain amount
of solvent are called the colligative properties. There are
four important proporties of this type of solution that are
directly proportional to their value with the number of sol-
ute particles present, namely:
1.Lowering vapor pressure of solution.
2. Increasing boiling point of solvent.
3. Decreasing freezing point of solvent.
4. Osmosis pressure.
All of these properties depend on the total concentration
of the solute particles regardless of the ionic or molecu-
lar or charge or volume nature. The effect of solubility of
non-electrolyte and non-volatility materials is discussed
(which, when it dissolves, that it will produce one mole of
soluble species per mole of solute) as follows:
1.Lowering vapor pressure of solution
Many experiments have shown that a solution containing a
nonvolatile liquid or a solid as a solute always has a lower
vapor pressure than the pure solvent (Figure 4-9). The va-
por pressure of a liquid depends on the ease with which the
Figure 4-9
molecules are able to escape from the surface of the liquid.
Lowering vapor pressure. If
When a solute is dissolved in a liquid, some of the total
no air is present in the ap-
volume of the solution is occupied by solute molecules,
paratus, the pressure above
and so there are fewer solvent molecules per unit area at
each liquid is due to water
the surface. The number of particles escaping from the sur-
vapor. This pressure is less
face of the liquid decreases and the vapor pressure of the
over the solution of sugar
solution decreases.
and water.

water sugar
solution

Mato meter
112
2. Increasing boiling point of solvent
The definition of boiling point of a liquid is the tempera-
ture at which its vapor pressure equals the applied pressure
at its surface. For liquids in open containers, this is atmo-
spheric pressure. We have seen that the vapor pressure of
a solvent at a given temperature is lowered by the presence
of a nonvolatile solute. Such a solution must be heated to
a higher temperature than the pure solvent to cause the va-
por pressure of the solvent to equal atmospheric pressure
(Figure 4-10). In accord with Raoult’s Law, the elevation
of the boiling point of a solvent caused by the presence
of a nonvolatile, nonionized solute is proportional to the
number of moles of solute dissolved in a given mass of
solvent. Mathematically, this is expressed as
∆Tb∝ m
∆Tb = Kbm
The term ∆Tb represents the elevation of the boiling point
of the solvent, that is, the boiling point of the solution mi-
nus the boiling point of the pure solvent. The m is the mo-
lality of the solute, and Kb is a proportionality constant
called the molal boiling point elevation constant. This con-
stant is different for different solvents and does not depend
on the solute. Kb corresponds to the change in boiling point
produced by a one-molal ideal solution of a nonvolatile
nonelectrolyte. The units of Kb are C°/m.
Table (4-3) shows Some properties of common solvents

Table (4-3) shows Some properties of common solvents

Boiling point (C°) Increasing molal boiling Freezing point Decreasing molal
solvent
(pure solvent) constant Kb C°/m (C°)(pure solvent) freezing point Kf C°/m

water 100 0.512 0.00 1.86

benzene 80 2.530 5.48 5.12

acetic acid 118 3.070 16.60 3.90

nitrobenzene 210 5.240 5.70 7.00

phenol 182 3.560 43.00 7.40

camphor 207 5.601 178.40 40.00

113
 We have already identified molality and knowledge of the
following relationship with the symbol for solvent 1 and
solute 2
n (mol)
m (mol/kg) = ‫ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m1 (kg)
As is known, the number of moles of solute is equal to

m (g)
n2 = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
M2 (g/mol)
Where m2 is the mass of the solute in unit (g) and M2 is the
Exercise 4-11 molar mass of the solute (g/mol). The mass of the solvent
When 0.15 g of a substance is (m1) is converted from the unit (g) to (kg)
dissolved in 15 g of a solvent,
the boiling point of the sol- m1 (kg) = m1 (g) × ‫ ـ ـ ـ ـ‬1(kg)
‫ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬m
‫ ـ ـ ـ ـ‬1‫( ـ ـ ـ‬g)
‫ـــــــــــــ‬
vent is increased by 0.216 oC
1000 (g) 1000 (g/kg)
at boiling point of pure sol- Substitute in molality formula, we get
vent. What is the molar mass
m (g)
of the solute if you know that ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
the constant of the increasing M (g/mol) 1000 (g/kg) × m (g)
m (mol/kg) = ‫ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m (g)
boiling point of the solvent is ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬m1 (g) × M2 (g/mol)
equal to 2.16 oC/m ? 1000 (g/kg)
Therefore, increasing boiling point equal to

1000 (g/kg) × m (g)

∆Tb (°C) = Kb (°C/ m) × ‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m1 (g) × M2 (g/mol)
Example 4-7
Predict the boiling point of the 1.25 m sucrose solution?
Note that The boiling point of pure water is 100 °C and
increasing molal boiling point constant for water is
0.512 °C/m.
solution
Use the relationship ∆Tb = Kbm to find the amount of
increase first and then add this increase to the boiling
point of pure water:
∆Tb = (0.512 °C/m) ( 1.25 m) = 0.640 °C
This means that the solution will boil at a temperature
higher than the boiling point pure water by a value
0.640 °C, so the boiling point of the solution will be
100 °C + 0.640 °C = 100.640 °C

114
 Figure 4-10
Increasing boiling point (by a
value ∆Tb) Decreasing freez-
ing point (by a value ∆Tf ).
Because a nonvolatile solute
lowers the vapor pressure of
3. Decreasing freezing point of solvent
a solvent.
Molecules of liquids move more slowly and approach one
another more closely as the temperature is lowered. The
freezing point of a liquid is the temperature at which the
forces of attraction among molecules are just great enough
to overcome their kinetic energies and thus cause a phase
change from the liquid to the solid state. Strictly speaking,
the freezing (melting) point of a substance is the tempera-
ture at which the liquid and solid phases are in equilibrium.
When a dilute solution freezes, it is the solvent that begins
to solidify first, leaving the solute in a more concentrated
solution. Solvent molecules in a solution are somewhat
more separated from one another (because of solute par-
ticles) than they are in the pure solvent. Consequently, the
temperature of a solution must be lowered below the freez-
ing point of the pure solvent to freeze it.
The freezing point depressions (∆Tf ) of solutions of non-
electrolytes have been found to be equal to the molality
(m) of the solute multiplying constant called the molal
freezing point depression constant, Kf . The values of Kf
for a few solvents are given in Table 4-3. Kf knows as the
freezing point depression of a one-molal ideal solution of
a nonelectrolyte in that solvent.
The relationship is as follow and similarly to the relation-
ship of the increasing in boiling point:
1000 (g/kg) × m (g)
∆Tf (°C) = Kf (°C/ m) × ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m1 (g) × M2 (g/mol)
where ∆Tf represent freezing point depression (pure sol-
vent freezing point- solution freezing point)
115
 Example 4-8
When 15.0 grams of ethyl alcohol(atomic mass=46g/
gmol), is dissolved in 750 grams of formic acid, the
freezing point of the solution is 7.21 °C. The freezing
point of pure formic acid is 8.40 °C. Evaluate Kf for
formic acid.
Solution
We find the value of ∆Tf from difference between the
freezing point of pure solvent and pure freezing point of
solution:
∆Tf = 8.4 °C - 7.21 °C = 1.19 °C
Apply the relationship
1000 (g/kg) × m (g)
∆Tf (°C) = Kf (°C/ m) × ‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m1 (g) × M2 (g/mol)
Exercise 7-3
So Kf value equal
What is the freezing point of
∆T (°C) × m (g) × M (g/mol)
aqueous solution of 0.05m for Kf = ‫ـ ـ ـ ـ ـ ـ ـ ـ‬f‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
non-electrolyte solute? Note
1000 (g/kg) × m2 (g)
that the constant of molal Kf = ‫ـ‬1.19 ‫ـ ـ ـ ـ×ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬750
‫(ـ ـ ـ ـ ـ ـ ـ ـ ـ‬°C) ‫ـ ـ ـ ـ×ـ ـ ـ ـ ـ ـ ـ‬46
‫(ـ ـ ـ ـ ـ ـ ـ ـ ـ‬g) ‫( ـ ـ ـ ـ ـ ـ‬g/mol)
‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2.74 °C/ m
freezing point depression of 1000 (g/kg) × 15 (g)
water is 1.86 °C/ m.
Example 4-9
Soluble 1.2 g of an unknown covalent compound in 50
g of benzene, and the freezing point of the solution was
found to be 4.92 °C. Calculate the molar mass of the
unknown compound, if you know that the freezing point
of pure benzene is 5.48 °C and the value of Kf is 5.12
°C/m.
Solution
Firstly, we find the value of ∆Tf which is equal to the
freezing point of the pure solvent - the freezing point of
the solution:
∆Tf = 5.48 °C - 4.92 °C = 0.56 °C
Use the relationship
1000 (g/kg) × m (g)
∆Tf (°C) = Kf ( °C/ m) × ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
m1 (g) × M2 (g/mol)
M2 value will be
5.12 (°C/m) × 1000 (g/kg) × 1.2 (g)
M2 (g/mol) = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬219.4 g/mol
50 (g) × 0.56 °C

116
4. Osmosis pressure
Osmosis is the spontaneous process by which the solvent
molecules pass through a semi – osmotic (semipermeable
membrane from a solution of lower concentration of sol-
ute into a solution of higher concentration of solute). A
semipermeable membrane separates two solutions. Sol-
vent molecules may pass through the membrane in either
direction, but the rate at which they pass into the more con-
centrated solution is found to be greater than the rate in the Figure 4-11
opposite direction. The initial difference between the two Laboratory instruments to
rates is directly proportional to the difference in concentra- display the osmotic proper-
tion between the two solutions. ty, the solute molecules can
After a period of time
not permeate through the
start semipermeable membrane.
solution level after
period of time of osmotic
pressure

solution level at
start sugar molecules do not
penetrate the membrane
sugar solution

Rubber some water molecules penetrate

out of the solution

semipermeable water water molecules cross

membrane (solvent) over the membrane

Solvent particles continue to pass through the membrane

(Figure (4-11)).The column of liquid continues to rise until
the hydrostatic pressure due to the weight of the solution
in the column is sufficient to force solvent molecules back
through the membrane at the same rate at which they enter
from the dilute side. The pressure exerted under this condi- Figure 4-12
tion is called the osmotic pressure of the solution. Osmotic The pressure that is just suf-
pressure depends on the number, and not the kind, of solute ficient to prevent solvent
particles in solution; it is therefore a colligative property. flow from the pure solvent
The osmotic pressure of a given aqueous solution can be side through the semiperme-
measured by shedding an equal external pressure to it able membrane to the solu-
(sufficient pressure to prevent osmosis) as shown in Fig- tion side is a measure of the
ure (4-12). The external pressure to be shed is equal to the osmotic pressure of the solu-
solution osmotic pressure. tion.
117
 The osmotic pressure can be calculated using a law-like
mathematical relationship
general gases

P = ‫ ـ ـ‬nRT
‫ـــــــــــ‬
V

Replace the pressure symbol P with symbol π which indi-

cates the osmotic pressure,
as follows:

π = ‫ ـ‬nRT
‫ـــــــــــ‬
V

In this equation n is the number of moles of solute in vol-

ume, V, (in liters) of the solution. R is the gas constant
(0.0821 L. atm/mol. K) and T temperature in degeree Kel-
vin. The term n/V is a concentration term. In terms of mo-
larity, M,
π = MRT

Osmotic pressure increases with increasing temperature
because T affects the number of solvent–membrane colli-
sions per unit time. It also increases with increasing molar-
ity because M affects the difference in the numbers of sol-
vent molecules hitting the membrane from the two sides,
and because a higher M leads to a stronger drive to equal-
ize the concentration difference by dilution and to increase
disorder in the solution.

118
 Basic concepts
Solution of solvent or solution at specific temper-
In general the solution is defined as a ature and pressure.
homogeneous mixture of substances
consisting of solvent and one solute or Raoult’s Law
more. Mixing ratios among them are dif- The vapor pressure of a solvent in an
ferent from solution to solution. The larg- ideal solution is directly proportional to
est amount in the mixture is the solvent the mole fraction of the solvent in the so-
and the least amount is solute. lution.
PA∝ xA⇒PA =xA P°A
Molality m
Molality is the number of solute moles, PB∝ xB ⇒ PB =xB P°B
which it contains in one kilogram of sol-
vent Osmotic Pressure
m (mol/kg) = ‫ـ ـ‬n‫(ـ ـ ـ ـ‬mol)
‫ــــــــــــ‬ Osmosis is spontaneity process by which
m (kg) solvent molecules pass through a semi-
permeable membrane from a dilute so-
Mole Fraction x lution into a more concentrated solution
Defined as the ratio between the num- in solute. Also solvent molecules pass
ber of components of a component (nA) through the membrane in both directions
to the total number of components of all and The shed pressure at this condition
components (nA + nB). They can be ex- are called the osmotic pressure.
pressed as follows:

n nB
xA = ‫ـ ـ ـ ـ ـ ـ ـ ـ‬A‫ـ ـ ـ ـ ـ ـ ـ‬ x
or B n ‫ ـ ـ‬+‫ـ ـ ـ‬n‫ـ ـ ـ ـ ـ‬
= ‫ـ‬ ‫ـ‬ ‫ـ‬ ‫ـ‬ ‫ـ‬
nA + nB A B

Molarity M
Molarity is the number of solute moles,
in one liter of solution

M = ‫ـ ـ‬n‫(ـ ـ ـ ـ‬mol)
‫ـــــــــــــ‬
V (L)

Solubility
Solubility of a substance is defined as
the largest amount of matter (number of
grams, number of moles,…. etc.) that can
dissolve in in certain (volume or weight)

119
 Chapter Four Questions 4
4-1 Give an example of a solution that um hydroxide, by dissolving 1g of KOH
contains each of the following: in 100 mL of ethyl alcohol (C2H5OH) (al-
(a) a solid dissolved in a liquid. cohol density = 0.789 g / mL), expressing
(b) a gas dissolved in a gas. the concentration of the resulting solution
(c) a gas dissolved in a liquid. by:
(d) a liquid dissolved in a liquid. 1.Molarity concentration.
(e) a solid dissolved in a solid. 2.Molal concentration.
4-2 Discuss the following statement: that 3.Mole fraction of sodium hydroxide.
the heat produced or consumed during the 4.Mass percentage of sodium hydroxide.
dissolution process are important factor 4-9 Calculate the boiling point of the
in determining the solubility or dissolu- aqueous solution of ethylene glycol
tion of solute. What is the other important compound (non-volatile substance) of
factor? How does it affect the process? concentration 2.5 m. (The boiling point
4-3 Define all of the following: the mo- of pure water is 100 °C). Note that the
lar fraction of a component in a solution, molal constant of water increasing boil-
Raoult’s law, the osmotic pressure, Satu- ing point is (Kb = 0.512 °C/ m)
rated solution. 4-10 Dissolve 3.75 g of non-volatile sub-
4-4 How many grams of table salt are stance in 95 g of acetone, the boiling point
needed to dissolve in 40 mL of water to increased to (56.50 °C), compared to the
obtain a solution salt where the mass per- boiling point of pure acetone (55.95 °C).
centage of salt is 15%? (Water density = Note that the molal constant of acetone
1 kg / L). increasing boiling point is (Kb = 1.71 °C/
4-5 How to prepare 250 mL of diluted m). what is the molar mass of a solute?
hydrochloric acid solution a concentra- 4-11 When grinding 0.154 g of sulfur
tion of 0.5 M from concentrated acid so- smoothly and dissolved in 4.38 g of cam-
lution of 12M? Explain this with the re- phor, freesing point of camphor is de-
quired calculations. creased by 5.47 °C. What is molar mass
4-6 A 60 mL sample of diethyl ether, of sulfur and its molecular formula? If
(C2H5)2O, is dissolved in enough metha- you know that molar constant of camphor
nol, CH3OH, to make 300 mL of solution. depression freezing (Kf = 40 °C / m)
The density of the ether is 0.714 g/mL. 4-12 Using the Table (4.3), what is the
material that caused the greatest depres-
What is the molarity M of this solution?
sion in the degree of freezing?
4-7 Explain, using the necessary calcula-
A. Benzene B. Camphor
tions, how you can prepare one liter of
C. Acetic acid D. Phenol
sodium chloride solution (NaCl) has a
4-13 Using Table 4.3, find the freezing
concentration of 0.215 m, if you know the
point of a solution prepared by dissolving
resulting solution density is 1.01 g/mL.
1.5 g of matter has a molar mass of 125 g
4-8 A solution was prepared for potassi-
/ mol in 30 g of nitrobenzene.

120
 5
Chapter Five
Chemical Kinetics

After completing this chapter, the student is expected to:

• Calculates the reaction rate in terms of the change in the concentration of material
over time.
• Determine the rate of reaction and deduces the law of reaction speed.
• Explaine chemical reactions by collision theory and transition state theory.
• Determine the factors influencing reaction rate and how to control it to increase
reaction speed.
• Distinguish between exothermic and endothermic reactions.
• Propose an acceptable reaction mechanism that conforms to the specified
conditions.

121
 5-1 Introduction
We have learned from our previous study of chemical
changes and simplified that they occur between the re-
actants to form the products, and expressed by a bal-
anced chemical equation, which is useful in calculating
the number of moles and masses of materials and the
volumes of reactive and resulting gases. But the chemi-
cal equation did not tell us other details regarding the
speed at which the reactants are converted into products
as well as the number of steps that the reaction pass to
create the products. All these changes are studied by
chemical kinetics and concern by the following points:
1.The rate of the chemical reaction and how to measure
it and the factors affecting it.
2. The reaction mechanism, ie the initial reactions that
the reactants pass through to reach the final product and
how they are expressed in balanced chemical equations.
Chemical kinetics is an important science because knowl-
edge speed of the reaction and the factors that affect it,
such as concentration, pressure, temperature, and the na-
ture of the reactants, allow students to predict how quickly
the reaction will reach equilibrium. Also and that study
of the reaction mechanism (the initial steps that the reac-
tion passes through to the reaction of the reactants into the
products), enables them, when they know, to control the
flow of the reaction to obtain the desired material in the
required quantities and in appropriate economic ways.
5-2 Chemical reaction speed
The definition of speed is familiar in our daily lives. If a
car drives 70 kilometers in one hour, we say it has a speed
of 70 Kilometer / hour, or student reads twenty pages of
a book in 30 minute, or the people of Baghdad consume
one million cubic meters of potable water per day. These
expressions of speed share a common explanation that a
change occurs over a certain period of time. The speed of
the car expresses the change in place measured in kilome-
ters per hour. For a student reading a book, the number
of pages read increases twenty pages every 30 minutes.
As for the people of Baghdad who consume one million
cubic meters of potable water per day, water tanks will de-
crease by one million cubic meters during this time period.
122
Similar to chemical reactions, they occur at varying rate
according to the change in the quantities of substances en-
tering or leaving the reaction per unit of time. Chemical
reactions vary in rates. Some of them are very fast when
mixing the reactants in a fraction of a second, such as neu-
tralization reactions between acids and bases and combus-
tion reactions. Others have moderate rates that take several
minutes to several months, such as iron rust and reactions
that lead to the ripening of fruits and vegetables. There are
very slow reactions that take several years or millions of
years to occur, such as reactions that lead to human growth
and age and turn dead plants into coal. Note the Figure (5-
1).
5-2-1 Measurement of chemical reaction rate
The rate of a chemical reaction is measured by knowing
the concentration change in reactive or product material
in unit of time (unit of time may be second (s), minute
(min), hour (hr), day (day) or any other units of time). In
this chapter we will use the molar concentration exclusive-
ly, which is defined as the number of moles of solute in
one liter of solution (mole /L) expressed by a square arc [
],where any symbol inside the arc means molar concentra-
tion. For example [X] means the molar concentration of
material X. So we will express the rate in the mathematical
relationship as the following:

rate of reaction=( Change in concentration of one of the

reactants/change in time)
Where ∆means change, and [ ] means molar concentration
in unit (mol/ L). Therefore have the chemical reaction rate
Figure 5-1
unit: Differential reaction rate, from
(mole / L.t )(or).( mole L-1 t-1) top to bottom: combustion is
For the following general reaction: very fast reaction, ripening fruits
and vegetables Moderate reac-
tion rate, repel a iron reaction
Where R represents any (reactant) and P represents any moderate rate and lower than
product substance. The rate of this general reaction is cal- rate of fruit ripening, human
culated by growth reaction is very slow.

123
 The relationship above and in terms of reactants R and as
follows:
Rate of reaction in term R = (-)concentration of R/change
in time
molar concentration (R)

If the reaction starts with a certain concentration of the re-

actant, let [R]1 be in the time of (t1) and as the reaction
progresses, the concentration of the reactant material de-
creases, while concentration of product material will be in-
creased. In time (t2) concentration of the reactant material
Time (t)
[R2], Note Figure (5-2), rate of reaction in term of R as
The concentration of reac- follows:
tants ICl and H2 decreases,
while the concentration of
the resulting substances
HCl and I2 increases over
time.

Figure 5-2
The concentration of reactant R
decreases over time.

Time (t)
The negative signal is added in the law when expressing
the reaction rate in terms of the change in the concentration
of a reactant, because the reactant is consumed during the
reaction and its concentration is lowered, so that the sec-
ond concentration is less than the first concentration and
the change in concentration is negative. so the negative
signal was added to make the reaction rate positive signal
note Figure (5-2).
When the reaction rate is measured in terms of the change
in the concentration of the substance produced by the reac-
tion P, the relationship is as follows:
Rate of reaction in term P = change in P concentration/
change in time

124
The change in the concentration of the resulting substance
Δ [P] is positive because its concentration increases over
time and the second concentration is [P]2 greater than the
first concentration [P]1, so no negative sign is added in the
law. As shown in Figure (5-3).

Figure 5-3
Time (t) The concentration of the result-
ing substance P increases over
time.
Example 5-1

The concentration of material R changed from 1.2 mole

/ L to 0.75 mole / L over 125s depending on the follow-
ing reaction:
R P
Calculate the reaction rate S as a function R.
Solution time(s(

∆ [R] [R] - [R] Exercise 5-1

RateR = (-) ‫) = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬-( ‫ـ ـ ـ ـ ـ ـ‬2‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1
∆t t2 - t1 Assume the following reac-
(-) ( 0.75 - 1.20 ) mol/L tion:
RateR = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.0036 mol/L.s
( 125 - 0 ) s 3H2 + CO CH4 + H2O
The concentration of H2 var-
ies according to the above
diagram. Find the rate of this
reaction and the time needed
to decrease the concentration
of H2 to 0.300 mol / L.

125
 5-2-2 The relationship of reaction rate with the
number of moles
The reaction rate varies in terms of the change in the con-
centration of the substance depending on the number of
moles of the substance in the balanced reaction equation.
For example, the reaction between hydrogen and iodine
to form hydrogen iodide as in the following equation:

When one molecule of hydrogen is consumed (one mole),

one molecule of iodine is also consumed (one mole), while
two molecules (two moles) of hydrogen iodide is formed.
ie rate of consumed of hydrogen or I2 equal to half the rate
Exercise 5-2
of HI formation.
The rate of formation of NH3
in the following reaction is
0.15mol / L.min. or the formation rate of HI is twice rate of consumption
N2 + 3H2 2NH3 H2 or I2
So the rate in terms of N2
consumption is equal n unit
mol/L. min the equation can be write as follows
A- 0.150
B- 0.075
C- -0.175 Rate(HI) means the rate in term product HI concentration,
D- 0.200 and so on for the rate in terms of the reactants H2 and I2, we
E- 0.300 note that we are not put signals when rate is used because
rate is always positive, while change in concentration is
negative for reactants and positive for products substances.
For general reaction

Where (a, b, g and h) are the number of moles of substanc-

es in the balanced reaction equation, and the rate of this
reaction is expressed in terms of the reactants produced as
follows:

1 ∆ [A] 1 ∆ [B] 1 ∆ [G] 1 ∆ [H]

Rate = - ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ × ـ ـ ـ ـ ـ‬- ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ × ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ × ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ × ـ ـ ـ ـ ـ‬
a ∆t b ∆t g ∆t h ∆t

or
1 1 1 1
‫ ـ ـ ـ ـ ـ‬Rate(A) = ‫ ـ ـ ـ ـ ـ‬Rate(B) = ‫ ـ ـ ـ ـ ـ‬Rate(G) = ‫ ـ ـ ـ ـ ـ‬Rate(H)
a b g h

126
 When the rate is expressed in terms of one mole of any
substance involved in the reaction overall rate of reaction
is used.It is the overall rate of reaction in terms of one mole
of reactants or products, and as follow:

Where nJ represents the number of moles of J with its nega-

tive signal of the reactants and the positive of the products,
it was –a or –b or g or h in the general reaction above. Exercise 5-3
Which one of the expressions
below do not represent the
Example 5-2
suitable expression for the
For the following reaction general rate of reaction
2H2(g)+ O2(g) 2H2O(g) 2A + 3B F + 2G

1.Express the rate in terms of the change in concentra- A- -∆ ]A[

tion of H2, O2 and H2O. ∆t

2. Calculate the H2O formation rate if the O2 consump- B- -∆ [B]

tion rate is equal 0.023 mol / L.s 3∆t
3. Calculate the amount of H2O formed after 18s C- -∆ ]F[
Solution ∆t

(1)Rate(H ) = (-) ‫∆ ـ‬ [H ]
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ‬ D- -∆ [G]
2 ∆t 2∆t

Rate(O ) = (-) ‫∆ ـ‬
‫[ـ ـ ـ ـ ـ‬O
‫ـ ـ ـ ـ ـ‬2‫ـ ـ]ـ ـ‬ H- -∆ [A]
2 ∆t 2∆t

Rate(H O) = ∆ ‫ـ ـ ـ ـ ـ ـ‬2‫ـ‬O]
‫[ ـ ـ ـ ـ‬H ‫ـــــــ‬
2
∆t
(2) ‫ ـ ـ‬1‫∆ ـ ـ ـ‬ ‫ ـ ـ ـ ـ ـ‬2‫ـ ـ‬O]
‫[ ـ ـ ـ ـ ـ‬H ‫) = ـ ـ ـ ـ‬-( ‫[ـ ـ ـ ـ∆ـ ـ‬O
‫ ـ ـ ـ ـ ـ‬2‫]ـ ـ‬
2 ∆t ∆t
Rate(H O) = 2 Rate(O )
2 2

Rate(H O) = 2 × 0.023 mol/L.s = 0.046 mol/L.s

2

(3)Rate(H O) = ‫[ـ ـ ـ ـ∆ـ ـ‬H O] [H O] - [H O] ([H O] - 0) mol/L

‫ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ‬1‫ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
2
∆t t2 - t1 (18 - 0 )s
[H2O]2 = Rate(H O) mol/L.s × 18 s = 0.046 mol/L.s × 18 s
2

[H2O]2 = 0.828 mol/L

127
 Example 5-3
The reaction between ethylene gas and ozone gas is ex-
pressed by:
C2H4(g) + O3(g) C2H4O(g) + O2(g)

It was found that the concentration of O3 changed as the

reaction progressed in the table below

Exercise 5-4
For the following reaction: A- Calculate the reaction rate during the first 10s of the
reaction.
4NO(g) +3 O2(g) 2 N2O5(g)
B- Calculate the reaction rate during the last 10s of the
reaction.
A- Express the reaction rate in
C- What is the reason for the difference between the rate
terms of change in the concen-
of reaction in both cases?
tration of each substance over
time.
Solution
B- Calculate the rate of O2
A- Reaction rate during the first 10s of the reaction.
consumption if consumption
∆ [O ] [O ] - [O ]
rate of NO was equal Rate(O )= - ‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3‫ = ـ ـ ـ ـ‬- ‫ـ ـ ـ ـ ـ ـ‬3‫ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3‫ـ ـ ـ‬1‫ـ‬
3
∆t t2 - t1
1.60 × 10-4mol / L. s
(2.42 × 10-5 - 3.20 × 10-5) mol/L
= - ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬7.8 × 10-7 mol/L.s
(10 - 0) s

B-The reaction rate during the last 10s of the reaction.

∆ [O ] [O ] - [O ]
Rate(O )= - ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3‫ = ـ ـ ـ ـ‬- ‫ـ ـ ـ ـ ـ ـ‬3‫ ـ ـ‬2‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3‫ـ ـ ـ‬1‫ـ‬
3
∆t t2 - t1

‫ـ ـ ـ ـ×ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬10
‫ـ ـ ـ ـ ـ ـ ـ ـ‬-‫ ـ ـ ـ‬1.23
‫ـ ـ ـ ـ ـ× ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬10
‫ـ ـ)ـ ـ ـ ـ ـ ـ ـ‬mol/L
-5 -5
=-(1.10 ‫= ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1.3 × 10-7 mol/L.s
(60 - 50) s
C- The reaction rate during the first 10s is equal to six
times as reaction rate during the last 10 s, indicating that
the reaction rate is not constant. It decreases over time
in terms of the concentration of the reactant as a result
of its consumption.

128
 5-3 The reaction rate law
In general, the reaction rate is directly proportional to the
product the concentrations of the reactants and each con-
centration are raised to a given exponent.
The following general reaction:

The reaction rate is proportional to the concentrations of

the reactants as follows:

Or in the form of equality

Exercise 5-5
The student should note that the number of moles of the re-
For the following gaseous re-
actants a, g and h is not related to the ranks of the reactants
action:
α, β, and γ, but are derived from practical experiments
only. The general rate of reaction is represented by Rate A+B C
for any of the reactants on the prodacts as the following: Experimentally found that the
law of reaction rate
Rate∝ = K [A]2 [B]
So the order of this reaction
A- First order
B- Second order
C- Third order
D- zero order
E- half order

Where [A], [G] and [H] are the molar concentrations of A,

G and H respectively.
α is called the reactant order A and β is the order of the
reactant G and γ is the order of the reactant H. The general
order of the reaction n is defined as being equal to the sum
of the orders of the reactants as follows:

n values are zero, 1, 2, or 3 and may be fractional. K is

called the reaction rate constant and the relationship
above framed by the law of reaction rate.

129
 If the reaction is zero, then the values of α, β and γ are
equal to zero, meaning that the reactant concentration do
not affect the reaction rate and therefore the law of reaction
rate for zero-order reactions is as follows:
Rate= K
*The student notes that the If the reaction is a first order, the value α = 1
rank of the reactant in the law the value of β and γ is zero, the law of velocity is written
of reaction rate may not cor- as a
respond to number of moles Rate= K[A]
of that reactant in the balanced For example, the following reaction is of the first order *
equation, so it is necessary to
2N2O5 2N2O4 + O2
create the term molecular reac-
tion and compare it with the or-
The law of reaction rate is:
der of reaction.
Rat= K[N2O5]
The reaction molecule is defined
The general order of the reaction is first (n = 1)
as the number of molecules or
If the second order reaction, the law of rate can be written
ions of reactants in the balanced
as following forms:
equation that can theoretically
be found from the number of
Rate= K[A] [G]
moles, Either reaction order is
or
a value that can only be known
Rate=K[A]2
by experience, and there is an-
other difference that the order
So for other orders. The following examples illustrate the
of reaction may be zero while
law of rate and orders of some reaction.
the reaction molecule cannot be
The law of reaction rate:
like this.
2NO(g) + 2H2(g) N2(g) + 2H2O(g)
Is
Rate=K [NO]2 [H2]
* The reaction is of the first order for the H2 reactant and
the second for the NO reactant. So the general reaction rate
order is n=3
For the following reaction:
(CH3)3CBr + H2O (CH3)3COH + HBr

The law of reaction rate is:

Rate=K [(CH3)3CBr]

Reaction rate is first order for the reactant (CH3)3CBr and

it is zero order for the H2O reactor and the general order of
the reaction is firstly, ie n = 1
130
The reaction between acetone and iodine has the presence
of MnO2 as a catalyst .
MnO2
CH3COCH3 + I2 CH3COCH2I + HI
It was observed that its rate does not depend on the con-
centrations of reactants and the reaction rate law for this
reaction is
Rate=K Exercise 5-6
Thus, this reaction is zero order From the above example,
Determine the orders of the
it is clear that the law of reaction rate and its order cannot
reactants and the general Re-
to be concluded by just considering the chemical equa-
actions orders for each of the
tion of the reaction, indeed this should be determined by
following reactions:
experiment only.
A-
The values of α, β and γ are not determined by the number
of moles of the reactants a, g and h as in the general reac- CH3CHO(g) CH4(g) + CO(g)
tion: aA + g G + h H pP Rate = K[CH3CHO]
3/2

B-
5-3-1 Determination of reaction orders
- +
To illustrate how to set the values of α, β, γ experi- H2O2(aq) + 3I (aq) + 2H (aq)
mentally. It is used several experimental methods, we will -
I3 (aq) + 2H2O(l)
look at one here. This method involves making a number -
of experiments in which the concentration of a reactant is Rate = K[H2O2] [I ]
altered and maintained the concentrations of the other re- C-
actants are constant, so we find the order of the variable
2NO(g) + Cl2(g) 2NOCl
reactant, and so on for the other reactants. This method can
2
be explained by the following examples: Rate = K[NO] [Cl2]

Example 5-4
The first reaction at a given temperature:
2N2O5 2 N2O4 + O2
The reaction rate was measured at different concentra-
tions of the reactant N2O5. Data were obtained in the
table below.

Exp.No. [N2O5] / mole/ L Rate / mole / L.s

1 0.0113 6.7 × 10-6
2 0.0084 5.0 × 10-6
3 0.0042 2.5 × 10-6

Determine the order of the reactant and deduce the law

of reaction rate and its general reaction rate order.
131
 Solution
Exercise 5-7
1 - We write the reaction rate law
For the following gaseous
Rate = K [N2O5]α
reaction:
2- To determine the value (α) we choose any two
A+B C
experiments such as 1 and 2 and compensate the results
Experimentally found that the
law of his rate of each experiment in the law of reaction rate and divide
Rate= K [A]2 [B] one equation by the other and simplify of fraction we
If the concentration of A is get (α) value
three times greater than it’s α
concentration. B is twice it’s Rate∝ K [N O ]
‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ‬2‫ـ ـ ـ ـ ـ ـ‬5‫ ـ ـ‬α1‫ـ‬
concentration. The reaction Rate∝2 K [N2O5]2
rate is increased by:
A- 6 B- 9 C- 12
6.7×10-6 (0.0113)
D- 18 E- 36 ‫ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬K
‫ــــــــــــــــــــــــــ‬
5.0×10-6 K (0.0084)

(1.34)1 = (1.34)α ⇒⇒ α =1
Therefore the value of α is equal to the integer one, so
Exercise 5-8
the reaction is of the first order and its rate law is written
as follows: Rate = K [N2O5]
For the following gaseous
reaction: 5-3-2 Calculating the value of the reaction rate constant
2NO + 2H2 N2 + 2H2O The reaction rate constant is a constant amount that
It has the following rate law: does not change unless the temperature changes. Its value
Rate = K [NO]2 [H2] is calculated from the compensation of the results of an
If the NO concentration experiment in the law of reaction rate, as in the following
reduced it’s concentration in example:
half and the H2 concentration Example 5-5
is tripled added, the change
Calculate the value of the rate constant of the reaction in
in reaction rate relative to the
Example 5.4.
first rate will be:
Solution
A- Decreases by 3/4
To find the value of rate constant we compensate the re-
B- Increases by 3/4
sults of one experiment in the law of reaction rate, that
C- increases by 2/3
we got in the previous example as follows:
D- decreases by 2/3
Rate = K [N2O5]
E- Remains the same
Suppose we have chosen experiment number 1 and off-
set its values in the relationship above
6.7×10-6 mol/L.s∝ = K (0.0113) mol/L
K = ‫ـ‬6.7×10
‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬mol/L.s
-6
‫ = ∝ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬5.95 × 10-4 s-1
(0.0113) mol/L
132
 5-3-3 Rate constant units
The units of rate constant vary depending on the general
order of the reaction and the unit of time used to express
the reaction rate. It can be derived mathematically from
compensation in the law of reaction rate, or from the ap- Exercise 5-9
plication of the following law: 1-n 1-n
‫ ـ‬mol [ ]
Use the constant rate value
= ‫ـ‬ ‫ـ‬ ‫ـ‬ ‫ـ‬ ‫ـ‬
‫ـــــــــ‬
‫ـ‬ ‫ـ‬ ‫ـ‬
[ ] M
L‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ‬ calculated in example 5-5 to
Unit of rate constant t t calculate the consumption
Where n represents the general reaction order and M
rate of N2O5. When its
molarity. Table (5-1) shows units of rate constant for dif-
concentration is equal to
ferent orders if time is expressed in (s)
0.1 mol/L at the same set
Table 5-1 Units of rate constant corresponding to the general reaction order
temperature.
Order (n) Units of rate constant corresponding to the general reaction rate

0 M. s-1 or mol L‫او‬-1 s-1 or mol/L. s

1 s-1 or 1/s
2 M-1. s-1 or L.mol-1s-1 or L/mol. s
3 M-2. s-1 or L.2mol-2s-1 or L2/mol2.s
Example 5-6
Exercise 5-10
For the following reaction
For the following reaction
O2(g) + 2NO(g) 2NO2(g)
X + 2Y P
Determine the orders of the reactants and deduce the law
It is found first order for X
of reaction rate, general reaction rate, and rate constant
and second order to Y.
of experiments results in the table below:
The units of rate constant of
Exp.No. [O2] / mole/ L [NO] / mole/ L Rate / mole / L.s
this reaction if time in seconds
1 1.10×10-2 1.30×10-2 3.20×10-3
is:
2 2.20×10-2 1.30×10-2 6.40×10-3 A- M. s-1
3 1.10×10-2 2.60×10-2 12.8×10-3 B- M-2. s-1
4 3.30×10-2 1.30×10-2 9.60×10-3 C- M-3. s
5 1.10×10-2 3.90×10-2 28.8×10-3 D- M2. s-1
Solution E- M-1. s-1
Write reaction rate law Rate= K [O2]α [NO]β
To determine the value of (α), select two experiments
such as 1 and 2, where [NO] is constant in the two ex-
periments while [O2] variable and compensate for the
results of each experiment in the law of rate. Divided
one equation to the other and simplify the fraction we
get α value
133
 α β
Rate∝2 K [O2]2 [NO]2
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬α‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
β
Rate∝1 K [O2]1 [NO]1
α β
6.40 × 10 -3 K (2.20 × 10 -2
) (1.30 × 10 -2
)
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬-3‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ‬α‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬β
3.20 × 10 K (1.10 × 10-2) (1.30 × 10-2)
Exercise 5-11 (2)1 = (2)α ⇒⇒ α =1
For the following reaction:
CH3COOCH3 + OH- So the reaction is first order for O2
CH3COO- + CH3OH To determine the value (β), we select experiments 1 and 3
From the results of the three where [O2] is constant in both experiments, while [NO] is
experiments in the table below:
a variable and we substitute the results of each experiment
in the rate law and divided one equation over the other
]C3H6O2[ ]OH[ Rate/ mol/L. s
and simplify the fraction, we get a (β) value.
0.040 0.040 0.000225
0.040 0.080 0.00045 α β
Rate [O ] [NO]
0.080 0.080 0.00090 ‫∝ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3 = K ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬2‫ـ ـ ـ‬3‫ ـ‬α‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬3‫ـ ـ ـ‬
β
A- Determine the ranks of the Rate1∝ K [O2]1 [NO]1
α β
-3 K (1.10 × 10-2) (2.60 × 10-2)
reactants. 12.80 × 10
‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬α‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬β
B- Deduce the law of reaction 3.20 × 10-3 K (1.10 × 10-2) (1.30 × 10-2)
rate. β β

C- Find the value of the

(4) = (2) ⇒⇒ (2)2 = (2) ⇒ β = 2
constant rate.
So the reaction is second order for NO.
The general order of the reaction is equal
n=α+β=1+2=3
so the reaction rate law is Rate= K[O2] [NO]2
We use the information in one of the experiments in the
table above and replace it In the law of reaction rate to get
the value of the rate constant. We used information from
experiment 2
2
Rate∝ = K [O2] [NO]
6.4 × 10-3 mol/L.s = K(2.20 × 10-2 ) mol /L (1.30 × 10-2 )2 mol2 /L2

6.4 × 10-3 mol/L.s

K = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬1721.4 L2 /mol2.s
(2.20 × 10-2 ) mol /L (1.30 × 10-2 )2 mol2 /L2

Note that the unit of rate constant for third order reactions
is L2 / mol2.s

134
5-4 Theories of reaction rate
A number of theories have been developed to explain how
the reaction occurs. The most important of these theories
is the theory of collision and the theory of the transition Exercise 5-12
state or called the theory of activated complex. The first What the texts below are
explained how the reaction occurs on the basis of the vis- erroneous with regard to
ible, while to the second explained the reaction in a way hypotheses collision theory:
that happens inside molecules subject to interaction and A- Occurrence of collision
we will briefly explain both theories. between the reactant molecules
5-4-1 Collision Theory is a basic condition of the
The rate of chemical reactions varies depending on the reaction.
properties of the reactants particles (atoms, molecules, or B- All intermolecular collisions
ions) with varying reaction conditions. lead to the formation of
To illustrate how chemical reactions occurs and why they products.
differ their rates. Collision theory had been submitted, the C- In order for the products
hypotheses of which are: to be formed, the collision
1. The chemical reaction occurs as a result of the colli- molecules must be in a suitable
sion of the reactants’ particles, this theory assumed that the appropriate space.
shape of the colliding particle was spherical.
2. The reaction rate is directly proportional to the num-
ber of collisions between the reactants particles per unit of
time. Where you can say it whenever the number of col-
lisions between the reactants particles increased the prob-
ability of the reaction to occur.
3. Not all collisions are effective and lead to the formation
of products, in that the number of collisions between par-
ticles are enormous and about 1×1027 collisions per second
between gas particles of 1 liter volume under normal con-
ditions. Therefore, if such a large number of collisions lead
to the formation of products to complete all the reactions at
the moment of mixing of reactants. So most collisions are
ineffective and does not lead to products.
To be effective, the collision must meet the following two
conditions:
1. Collision molecules have a minimum of potential energy
to overcome the power of repulsion among themselves
when colliding and breaks the bonds of materials and turn
them into products. This minimum energy Necessary to
reaction is called activation energy.
2. The direction of the colliding particles is appropriate,
that the particles are in an appropriate geometric space and
in the right direction when colliding. This it creates the 135
desired products. To illustrate this, consider Figure (5-4).
 products

products

products
From the above we find that in order for reaction to occur
Figure 5-4
according to this theory, the molecules must possess the
Some possible collisions be-
minimum activation energy and that their collision is spa-
tween N2O and NO molecules in
tially oriented in the right direction. That’s known that the
the gas phase.
internal (total) energy of any moving molecules are equal
(a) collision that could be effec-
to the sum of the kinetic energy and the potential energy
tive in producing the products.
it possesses, and the molecules moving very quickly have
(b, c) Collisions that would be
high kinetic energy and low potential energy. When two
ineffective orientation.
moving molecules approach too fast, part

136
Kinetic energy of each one will convert to potential energy.
Because of The repulsion between them is caused by the
presence of electrons in their outer shells. At the moment
of the collision, the two molecules stop moving and the
kinetic energy of each is converted to the potential energy,
if that energy is less than energy of the activation energy of
the reaction will bounce off each other without any reac-
tion (Non effective Collision) [Figure( 5-4b and c)].
If they have enough kinetic energy, they will rush with
great power and high speed, they can overcome the power
of repulsion and entry the reaction(Effective Collision)
[Figure (5-4a)].
In fact, the collision theory failed to give the true rate of
reactions when tested, the reason for the failure of this the-
ory is to assume that the molecules are all spherical. This,
in fact, applies only to a small number of molecules that
are monatomic. Therefore, other scientists have proposed
a new theory to explain how the reaction occurs, namely
the theory of the transition state or called the theory of ac-
tivated complex.
5-4-2 Transition State Theory
This theory showed that in all chemical reactions cannot
resulting material consists directly, but must pass through
the so-called state active transition, which is where the so-
Activated complex
called activated complex. It is a non-static active compound
suggested that its synthesis is a compromise between the Transition state
Activation
energy

reactants and the resulting substances, so this theory called energy

reactant
activated complex theory. it is also in equilibrium with the
reactants [Figure( 5-5)]. The energy of this compound is
product
always greater than that of the reactants and the resulting
reaction progress
substances. It should be noted that the activated complex
is often inseparable but can disintegrate either to produce Figure 5-5
products or to give reactants according to reaction condi- Reaction progress accord-
tions. ing to the theory of transition
An important example to illustrate the idea of collision state.
theory and transition state theory is the reaction between
the I- ion and methyl chloride CH3Cl.
I- + CH3Cl CH3I + Cl-
This reaction occurs according to the method shown in
[Figure (5-6)]. The reaction begins with the collision of
the iodide I- ion with the methyl chloride molecule CH3Cl.
137
 From behind the C-Cl bond through the center of the three
hydrogen atoms, where the bond begins between carbon
and iodine while elongating and weakening between car-
bon and chlorine and the transition state of the reaction
(the activated complex) is formed. Activated complex has
three standard C-H bonds and two bonds weak (partial
bonds) between carbon and iodine, C-I, carbon and chlo-
rine C-Cl. As the bond strength increases between carbon
and iodine, the bond is established between them, while
the bond breaks down between carbon and chlorine and
reaction is done [Figure (5-6)a].
If the collision between the iodide ion and the methyl
chloride molecule is in the wrong direction, the reactants
remain unchanged. The reaction doesn’t occurs, note the
[Figure (5-6)b].

Figure 5-6
Some types of molecular colli-
sions in the gaseous state.
A- Effective collision leading
to the formation of products. 5-5 Heat of reaction
B- Collisions are ineffective The rate of the chemical reaction is related to the activation
because the direction of the energy, wherever the activation energy value is large, the
colliding molecules is incor- reaction is slow, and the reaction is quicker as the activa-
rect in addition to lack of en- tion energy is smaller.
ergy Sufficient for an effective The chemical reaction is accompanied by a change in ener-
collision. gy due to the absorption or emission of a quantity of heat.
This quantity represents the difference between the energy
of the products and the energy of the reactants. They are
called reaction temperatures and are calculated as follows:

Heat of reaction = Energy of products - Energy of reactants

138
The reaction is endothermic when the reaction temperature
is a positive value, ie the energy of the resulting material is
greater than that of the reactant. As shown in Figure (5-7).

Activation Figure 5-7

energy The following reaction scheme
(+) Absorption
heat A + B2 AB + B
When the energy of the
products is greater than the
energy of the reactants have
the reaction endothermic so
signal heat value (+)

If the amount of reaction temperature is negative value,

that is the, energy products > energy of the reactants, so the
reaction is Exothermic. As shown in Figure (5-8).

Figure 5-8
The following reaction scheme
Ea
Activation A + B2 AB + B
energy
When the energy of the re-
sulting material is smaller of
(-) Emitting the energy of the reactants
heat have the reaction is exother-
mic, so heat signal heat val-
ue (-)

139
 5-6 Activation Energy
Activation energy is defined as the minimum energy re-
quired that the reactants must possess them in order for the
collision to be effective.
5-7 Factors affecting reaction rate
Reactions vary in rate, some slow and others fast and under
the same conditions, the reaction rate can be changed by
controlling the factors affecting the reaction rate, namely:

5-7-1 Concentration
We observed from our study of the reaction rate law that
Figure 5-9 increasing the concentration of a reactant often increases
Increase in the speed of igni- the reaction speed. Figure (5-9) shows effect of increasing
tion of the splinter by increas- oxygen concentration on reaction rate where the burning
ing the concentration of oxy- fragment lights in the air, containing oxygen gas 20%, but
gen. flame glow when inserted into a filled bottle with Oxygen
(bottom) ignition of the gas it caused by an increase in the concentration of oxy-
splinter in the air (top) burn- gen, which leads increased ignition rate.
ing flame glow inside a bottle It also concentration has the effect of increasing the rate
filled with oxygen gas. of reactions involving on reactants in the liquid state, the
pressure has a similar effect on reactions involving reac-
tants in the gaseous state. That increased pressure reduc-
es the volume of gas and thus increases its concentration
which increases of the chemical reaction rate.

5-7-2 Physical nature and reactant material nature

The rate of chemical reactions depends on the physical
state of the reactants. If, for example, we had three piec-
es of metal, sodium, zinc and tin have the same size and
we put each piece in a hydrochloric acid solution has the
Figure 5-10 same concentration. We find that the sodium piece reacts
Phosphorus has two forms. strongly, the zinc piece will react at a lower rate for the tin,
White Phosphorus (above) ig- it will react very slowly. So it’s a physical state of the re-
nites and burns quickly when actants are very important to determine their effectiveness
exposed to oxygen air there- and their rate of reaction, as well as the effectiveness of
fore, it should be stored under white and red phosphorus [Figure (5-10)]
water.
Red phosphorus (under) re-
acts very slowly severe with air
so it can be stored with open
bottles.

140
The surface area of a given mass of material increases as
it is small the size of the constituent particles, and as the
surface area increases,. As this area increases, the number
of collisions then, reaction rate increases, [Figure (5-11)].
The surface area of the solid can be increased, either by
soluble it in a solvent, the solvent separates the particles
from each other or grinds them and turn it into a fine pow-
der.
5-7-3 Temperature
The reaction rate increases as the temperature rises and Figure 5-11
it decreases with temperature decreasing. This is because The chalk powder (pure car-
high temperature increases the speed of the reacted mole- bonate calcium CaCO3) re-
cules, then collision of molecules increases. So the number acts quickly with diluted hy-
of molecules has kinetic energy equal to or greater than the drochloric acid because it has
activation energy of the reaction, which can react to form a large surface area, while the
the product, as shown in Figure 5-12. chalk finger with a small sur-
face area reacts much slower.

Figure 5-12
Temperature increasing leads
to a marked increase in the
number of molecules with
high energies equal to or
greater than activation energy
which will increase the num-
Most reactions rates are doubled if temperture increases by ber of collisions and the rate of
10 degrees (10°C) because of the increasing proportion reaction.
of molecules involved in Reaction as shown in Table 5-2.
Do you know
The mixture of coal dust and air
Table 5-2 The reaction rate doubles as the reaction temperature is a mixture capable of explo-
increases (10 oC) sion at high temperature and
Number of molecules involved in the reaction Temperature may cause the explosion of coal
1.70 ×10-9 25 oC (298 K) mines, as well as the mixture of
flour dust and air may cause an
3.29 ×10-9 35 oC (308 K)
explosion for the same reason.
6.12 ×10-9 45 oC (318 K)

141
 5-7-4 Catalyst
Catalyst is a substance that increases the rate of reaction
without to consume through it. The effect of adding cata-
lyst is stronger than the effect of high temperature on reac-
Do you know tion rate. Since the catalyst reduces the activation energy
Enzymes are vital catalysts that level of the reaction, the number of colliding particles with
increase the rate of reactions in kinetic energy is equal to or greater the energy of activation
the human body. When eating energy, so that it can react to form the products, as shown
a meal containing protein, the in Figure (5-13). Note from the figure that the activation
digestive enzymes act on break- energy is reduced for reaction also leads to the opening of
ing down their molecules within a new line that differs from the first line before adding the
a few hours, but without en- catalytic by creating a new activated complex with low
zymes, the process takes several energy. The effect of the catalyst on increasing the reaction
years. rate of hydrogen with oxygen can be illustrated:
2H2 + O2 2H2O
The reaction is hardly happening at the normal tempera-
ture, but is taking place quickly by add a little platinum
powder as an catalyst.

Activation energy
without catalyst

tivation energy
with catalyst
Energy

Reactants

Reaction without catalyst

Figure 5-13 Reaction with catalyst
The catalyst reduces the acti- products
vation energy and increases
reaction progress
the reaction rate.
5-8 Reaction mechanism
Most chemical reactions occur through a number of se-
quential steps represented by so-called reaction mechan-
ics. Reactions occur in more than one step called complex
reactions, while one-step reactions are called primary reac-
tions.
A complex reaction involves more than one initial reac-
tion. For example, the interaction between NO and O2 oc-
curs as follows: 2NO + O2 2NO2
142
In practice it’s found that the reaction takeplace in two
steps:-
1) 2NO N2 O2 Do you know

2) N2O2 + O2 2NO2 In 1998 a new science was intro-

In practice, this reaction was found to occur in two step duced is called the femtosecond
Each of these steps is an primary reaction. If we combine science, by a group of scientists,
equations (1) and (2), we would get the above general re- including the Arab scientific,
action. Dr. Ahmed Zewail. This science
N2O2 is the intermediate compound, which is defined as a enters in many areas including
compound produced in one step the initial reaction and it medicine, electronics and sci-
is consumed in another step of these reactions. So the in- ence space, chemistry, physics
termediate compound does not appear in the general equa- etc.
tion of the reaction but it appears in the mechanism steps As a first time and by using this
of the reaction or the so-called primary reaction steps. This science scientists were able to
is because it is an unstable compound, it is more stable observe what happens during
than the activated complex because its atoms are bound chemical reaction and slow im-
together with normal bonds and in some cases can be sepa- aging and see the movement of
rated from the reaction. Mechanism steps include the re- atoms using super speed camera
action on the specified step for speed (Rate-determining which helped to suggest a reac-
step), defined as the slowest step of reaction mechanism tion mechanism.
steps. When an primary reaction occurs in the mechanism
in a lower rate than any other primary reactions, so this
reaction is the rate determined step of reaction rate. The
rate of this reaction, it determines the general reaction rate
whether this step is at the beginning, end, or intermediate.
For example, the reaction of nitrogen dioxide NO2 and car-
bon monoxide CO was experimentally found that it is an
reaction of second order for NO2 compound and zero order
for CO compound.
NO2(g) + CO(g) NO(g) + CO2(g)

This is contrary to what can be deduced at first glance from

the above chemical equation as a first order reaction for
NO2 and CO on this it was suggested that this reaction fol-
lows the following mechanics:

NO2 + NO2 NO3 + NO slow step

NO3 + CO NO2 + CO2 fast step

NO2 + CO NO + CO2

143
 Since the reaction rate is the speed of the slow step, the law
of the reaction rate law is:
Rate∝ = K [NO2] [NO2]
Exercise 5-13 or
Rate∝∝ = K [NO2]2
A reaction mechanism has
been proposed for a given reac- This is consistent with the law of reaction rate of empiri-
tion; constitute of five steps are cally concluded. Observations of the reaction mechanism
as follows: is forming intermediate compound NO3. Deduce from pre-
viously that any proposed mechanism for any chemical re-
2NO(g) N2O2(g) action must be agreed with the general chemical equation
2H2(g) 2H(g)) of reaction. To propose a reaction mechanism must apply
the following conditions:
N2O2(g) +H(g) N2O(g)+HO(g)
1. When combining the equations of the primary reactions
2(HO(g) + H(g) H2O(g))
in the proposed reaction mechanism, we obtain the general
H(g) +N2O(g) HO(g) + N2(g) equation for the reaction.
2.The order of reaction for any primary reaction must be
1.Write the balanced reaction a first or second order and primary reactions are excluded
equation. from the third order due to the difficulty of effective colli-
2. Write the law of reaction sion occurring between three molecules at once.
rate for each step. 3. The law of reaction rate for rate-determining step the
3. Set the reaction order for specified step of velocity must be identical to the law of
each step. general reaction rate because the reaction rate will depend
4. determine the intermediate on the “rate-determining step” rate.
materials. The followning examples illustrate a mechanical steps for
reactions according to above point:

Example 5-8
The law of reaction rate of an reaction: Rate=K [NO2]2
Suggested reaction mechanism:

1) NO2(g) + NO2(g) NO3(g) + NO(g)

2) NO3(g) + CO (g)
NO2(g) + CO2(g)

A. Write the general balanced equation for the reaction

and what is the intermediate compond.
B. Write the law of rate for each step.
C. Set the rate-determining step.

144
Solution:
A - Combine steps 1 and 2 and cut down similar items to
get reaction equation
NO2(g) + NO2(g) NO3(g) + NO(g) Exercise 5-14
NO3(g) + CO (g) NO2(g) + CO2(g)
What is the proposed reaction
mechanism
NO2 + CO NO + CO2 NO2Cl(g) NO2(g) + Cl(g)
The intermediate compound is NO3 because it did not ap-
NO2Cl(g)+ Cl(g) NO2(g)+Cl2(g)
pear in the general equation of the reaction.
B-The law of reaction rate for the first step Rate= k[NO2]2
Reaction rate law is:
The law of reaction rate for the second step Rate= k[NO3]
Rate= K[NO2Cl]
[CO]
1. Write the balanced general
The first and second steps are second order.
reaction equation.
C - The first step is the rate-determining step for rate, be-
2. Write the law of reaction
cause the law of rate of it is matched with the law of reac-
rate for each step.
tion rate.
3. What a general reaction rate
4. which one of the two steps
Example 5-9
rate-determining step and
For the following reaction:
why.
2NO2 + F2 2NO2F
Reaction rate law is : Rate= K [NO2] [F2]
The reaction mechanism has been proposed in four dif-
ferent ways, explain with the reason being given the right
reaction mechanism, and false one.
NO2+ NO2 N2O4 slow step (1)
N2O4 + F2 2NO2F quick step

NO2+ F2 NO2F2 slow step (2)

NO2F2 + F2 NO2F4 quick step

2NO2 + F2 2NO2F (3)

NO2+ F2 NO2F + F slow step (4)

NO2 + F NO2F quick step
Solution
(1) False , because reaction rate law of rate-determining
step Rate=K [NO2]2 does not matched to the law of reac-
tion rate
145
 (2)False, because the sum of the two steps does not pro-
duce the reaction equation:
NO2+2 F2 NO2F4
(3) False, because the reaction mechanism is from one
step and third order
Exercise 5-15
(4) True, because it meets the following three conditions:
Suppose the following reaction A- When we combine the two steps we get the equation of
3H2 + CO CH4 + H2O reaction.
Follows the following mechanism B- The reaction order for each step is of the second gen-
eral order and it is acceptable.
H2+CO H2CO slow step
C- The law of the rate-determining step is matched to the
H2+H2CO CH4+O fast step
law of reaction rate: Rate= [NO2][F2]
H2 + O H2O fast step

1. Rate∝ = K [H2]2 [CO] 2 Example 5-10

2. Rate∝ = K [H2]2 [CO] Proposed the following reaction:
3. Rate∝ = K [H2] [CO] 2 A + 2B AB2
4. Rate∝ = K [H2] [CO] And suppose the following mechanism:
5. Rate∝ = K [H2]2 [CO] 3 B+B B2 slow step
B2 + A AB + B fast step
B + AB AB2 fast step

A - What is the law of the rate of reaction of this reaction.

B - Does this mechanism satisfy this reaction.
C- What is the general order of reaction.
D- What are the intermediate compounds in this reaction.

solution

A - Yes, this mechanism achieves this reaction because the

combination of the initial reactions leads to the general
equation of the reaction.
B-Reaction rate law is Rate=K[B][B] or Rate= K[B]2
C-Reaction is of second order
D- Intermediate compound is B2 and AB

146
 Basic concepts
The Rate of Chemical Reaction Rate theories
The amount of decrease in the molar Developed theories are to explain how
concentration of reactant materials Reac- that reactions occur. There are two theo-
tion or increase in molar concentration of ries, collision theory and theory of acti-
product materials with the corresponding vated complex or transition state.
change in unit time. Activation Energy
General reaction rate Is the minimum energy needed for reac-
The Overall Rate of Chemical Reaction tant molecules to form products
Rate in terms of one mole of any reactant Effective Collision
material product material or resulting. Those collisions in which the two condi-
The Rate Low tions are met: The two basic molecules
Relationship between rate of reaction possess the minimum of activation ener-
with the concentrations of reactants A, G gy and be geometric vacuum situation for
and H are powered to their orders re- colliding molecules in the right direction.
spectively and with the constant reaction Heat of reaction
rate K. For the general reaction: Heat absorbed (positive value) or emit-
aA + g G + h H pP ted (negative value) during the chemical
Rate law is written as follows: reaction Their values depend on the dif-
α β
Rate∝ = K [A] [G] [H]
γ ference between the energy of the reac-
tant material energy and product material
energy.
The Rate constant
Reaction mechanism
The general rank of the reaction is equal
Number primary steps (reactions) that re-
to the sum of the reactants in the law of
action passes to convert reactant material
velocity:
to product.
n=α+β+γ Rate-Determining step
The slowest step of the reaction mecha-
Rate Constant Units nism (primary reactions) and the rate of
The unit of reaction rate depends on the this step limits the law of reaction rate ie
order of reaction and can be derived by the number of reactants moles in rate-de-
the following relationship: termining step are orders of the reactants
1−n themselves.
⎡ mol ⎤
1−n
⎢ L ⎥⎦
Unit of rate constant = ⎣ =
[Mee ]
t t

147
 Chapter Five Questions 5
5-4 for the following reaction:
5-1 for the following reaction: Pt(NH3)2Cl2+H2O pt(NH3)2(H2O)
N2O5 2NO2 + ‫ـ ـ ـ‬1‫ ـ ـ ـ ـ‬O2 Cl++Cl-
2
Reaction rate law
The results in the following table: Rate = K [Pt(NH3)2Cl2]
t / hr [N2O5] / mol/L Its value K= 0.090 hr-1
0.00 0.849 A-Calculate reaction rate when concen-
0.50 0.733 tration is
1.00 0.633 pt(NH3)2Cl2 = (0.040M,0.020M,0.010M)
B-How the consumption rate changes
2.00 0.472
of Pt(NH3)2Cl2 with its concentration
Calculate the rate of reaction in hour unit changes.
of the following . C-What is the effect of concentration
A- From 0.00 to 0.50 change of Pt(NH3)2Cl2 on formation rate
B-From 0.50 to 1.00 of Cl-.
C-From 1.00 to 2.00

5-2 Explain how the following factors af- 5-5 What effect does the catalyst on the
fect chemical reaction rate following:
A-Temperature. A- Heat of reaction.
B- Surface area of reactants. B- Activation energy.
C- The nature of the reactants. C- Reactant energy and product energy.
D- Concentration of reactants.
5-6 The reaction below is found that
5-3 The following reaction is a first order the formation rate at a given time is
for Br2 reactant and it is second order for 0.036M.s-1 for C.
NO. Rate in terms of change in A,B and D and
2NO(g) + Br2(g) 2NOBr(g) general reaction rate (all in unit M.s-1) is
A- Write the reaction rate law 2A+3B 4C+2D
B- How the reaction rate changes when
the concentration of Br2 is to be three A: 0.018,0.027, 0.018, 0.009
times the initial concentration. B: -0.018, -0.027, 0.018, 0.009
C- How the reaction rate changes when C: -0.072, -0.048, 0.072, 0.144
the NO concentration is doubled only. D: -0.036, -0.036, 0.036, 0.009
D- How the reaction rate changes when E: -0.018, -0.012, -0.018, -0.018
the concentration of both Br2 and NO is
doubled.

148
5-7 Methyl acetate decomposes in a basic
solution to ion acetate and methyl alco- A- Conclude the law of reaction rate.
hol, according to the following equation: B- Calculate the value of the rate con-
stant.
CH3COOCH3(aq)+OH−(aq) C- Calculate the reaction rate when the
CH3COO−(aq)+CH3OH(aq) concentration NO, H2 equal to 8.0x10-3M

Reaction rate law Rate = K B - Calculate the reaction rate when the
[CH3COOCH3] [OH-] , K=0.14L/mol.s concentration of each O3 and C2H4 are
equal to 2.0x10-7M
A-What is the dissociation rate of meth-
yl acetate when concentration of both
CH3COOH3 and OH- are 0.025M. 5-9 For the following interaction and the
results of experiments in the table below:
B-What rate of appearance CH3OH in so-
lution. C2H4(g)+O3(g) 2CH2O(g)+ ‫ـ ـ ـ‬1‫ ـ ـ ـ ـ‬O2(g)
2
Reaction rate law
Rate = K [CH3COOCH3] [OH-] Exp. Rate
No. ]C2H4[ ]O3[ mol/L.s
K=0.14L/mol.s
1 1.0×10-8 0.5×10-7 1.0×10-12

2 1.0×10-8 1.5×10-7 3.0×10-12

3 2.0×10-8 1.0×10-7 4.0×10-12

5-8 The following reaction
2NO(g)+2H2(g) N2(g)+2H2O(g) A- Deduce the law of reaction rate, then
calculate K value.
Exp. Rate
No. ]H2[ ]NO[ mol/L.s B- Calculate the reaction rate when the
1 2.5×10-3 5.0×10-3 3.0×10-3 concentration of each O3 and C2H4 are
equal to 2.0x10-7M
2 2.5×10-3 15.0×10-3 9.0×10-3

3 10.0×10-3 15.0×10-3 36.0×10-3

From the results of the experiments in the

following table:

149
5-10 Write the law of reaction rate for 5-13 Methyl format is decomposed in
the following primary reactions: an acid solution Made up of methyl al-
A- O(g)+ O3(g) 2O2(g) cohol and formic acid according to the
B- Cl(g)+ H2(g) HCl(g) + H(g) following equation:
C- 2NO2(g) N2O4(g)
HCOOCH3 + H2O + H3O+
HCOOH(aq)+CH3OH(aq)
5-11 The following reaction
2NO2(g)+ F2(g) 2NO2F(g) Rate = K [HCOOCH3]
Why ion H3O+ does not appear in the
From the results of the experiments in law of reaction rate though it is found
the following table: in the reaction equation ?

Exp. Rate 5-14 The following reaction was done

No. ]NO2[ ]F2[ mol/L.s to a 600K
2NO2(g)+ O2(g) 2NO2(g)
1 0.001 0.005 2.0×10-4

2 0.002 0.005 4.0×10-4 The results of the experiments in the

following table:
3 0.002 0.002 1.6×10-4
Exp. Rate
No. ]O2[ ]NO[ mol/L.s
A- Conclude the law of reaction rate.
1 0.01 0.01 1.2×10-5
B- What is the order of the reactants for
each substance in the law of rate? 2 0.01 0.02 4.8×10-5
C- Calculate the reaction rate constant.
3 0.02 0.01 2.4×10-5

5-12 The following reaction:

NO2(g)+CO(g) NO(g) + CO2(g) A- Conclude the law of reaction rate.
From the results of the experiments in
the table below, specify the reactants or- B- Calculate the reaction rate when the
ders, deduced the law of reaction rate concentration of each of NO and O2 are
and general reaction rate. equal to 0.025M and 0.05M respective-
ly
Exp. Rate
No. ]NO2[ ]CO[ mol/L.s C-Calculate the rate of consumption
1 0.10 0.10 0.005 NO, and the composition rate of NO2
if O2 consumption rate was 1.0x10-
2 0.40 0.10 0.080 4mol/L
3 0.10 0.20 0.005

150
5-15 Select the right statement and the false show what are the letters a, b and c mean
statement, then correct the error in all of the and whether the reaction is endothermic re-
following: action or exothermic reaction and why?
A-The reaction rate constant varies with the
reaction temperature and over time.

B - The reaction rate constant varies with the

concentration of the reactants. 5-17 The reaction comes in second order
for the NO2 reactant and zero order for CO
C-The reaction rate remains constant over reactant:
time for a first order reaction when the tem-
perature is constant. CO(g) + NO2(g) CO2(g) + NO2(g)

D- The reaction rate is doubled for the reac- 1- Write down the rate law of reaction.
tion of zero order when the concentration of
the reactant is doubled. 2- How does the rate of the reaction chang-
es when the NO2 concentration is halved.

5-16 In the scheme below is plotted between 3-How does the rate of the reaction chang-
energy es when the CO concentration is doubled.
(Y axis) and reaction progress (x axis) for the
following reaction:
Energy

reaction progress

151
5-18 The following reaction and results of the
experiments are show in the table below:

CH3COCH3(aq) +Br2(aq)+ H+(aq)

CH3COCH2Br(aq)+HBr(aq)

Exp. Rate
No. ]CH3COCH3[ ]Br2[ [H+] mol/L.s

1 0.30 0.05 0.05 5.7×10-5

2 0.30 0.10 0.05 5.7×10-5

3 0.30 0.05 0.10 12.0×10-5

4 0.40 0.05 0.05 7.6×10-5

1-Write down the rate law of reaction.

2- Calculate the rate order constant value

of reaction.

3- How does the rate of the reaction change

when:
[H+] = 0.050M [CH3COCH3]=[Br2]= 0.10M

152
 6
Chapter Six
14_00CO.JPG
Acids, Bases and Salts

After completing this chapter, the student is expected to:

• Recognize the properties of aqueous solutions of acids and bases.
• Identify the different molecular concepts of acids and bases according to Arrenius
and Bronshted-Lauri and Lewis theories.
• Characterize the auto-ionization of water and the amphoteric properties of some
substances.
• Recognize the acids and bases reactions in aqueous solutions.
• Distinguish between the types of salts, the method of forming each and the
properties of thier aqueous solutions.
• Recognize the indicators that used in neutralization reaction.
• Recognize the titration process and understand its importance.

153
 6-1 Introduction
In nature, there are many acids, bases and salts that are
used in different fields, for example, digestive juices in the
human body contain a solution of hydrochloric acid at a
concentration of about 0.1 molar and in the blood of human
and also water components in body cells have a moderate
acid action. The solution is in the car battery is composed of
40% by mass of sulfuric acid solution. On the other hand,
sodium hydroxide is used in the manufacture of soap, pa-
per industry and a number of other chemical industries.
The salt also has extensive uses, baking soda is one of the
salts of carbonic acid. Table salt (sodium chloride) uses
to give a salty taste to food on the one hand and on the
other hand it is used to preserve many types of food. An-
other example of the use of these substances is the appli-
cation of calcium chloride salt to dissolve snow accumu-
lated on public roads. This salt is also used in the treatment
of people who has heart attacks. Ammonium salts are
also used as nitrogen fertilizers to increase soil fertility.
Most organic acids and their derivatives are found natu-
rally, the vinegar we use have about 4% of acetic acid in
its contents. The pain that you have feel when ants bite
you is caused by the formic acid that these insects secrete
into the body. Amino acids form protein modules in liv-
ing organisms. Many other examples illustrate the im-
portance and uses of these substances in everyday life.
6-2 Properties of aqueous solutions
of acids and bases
The aqueous solutions of most acids have distinctive char-
acteristics attributed to the presence of positive hydrogen
ion (H+) (proton) or crosses about hydronium ion (H3O+) in
its aqueous solution. The properties are:
1. Has an acidic taste.
2. Change the color of a number of indictors (pigments),
for example change the pigment of a sunflower tint from
blue color to red color
3. Acids react with most metals and release hydrogen gas
(H2).

154
4. Reaction with metal oxides and metal hydroxides to
form salts and water.
5. Its aqueous solutions have the ability to conduct electri-
cal current because of their full or partial ionization. The
Table (6-1) shows some common acids.

The aqueous solutions of most bases also have distinc-

tive characteristics, attributed to the presence of hydroxide The acid solution changes the
ion ( OH-) in its aqueous solution and qualities. The prop- color of the pigment sunflow-
erties are: er to red color.
1.Has a pungent taste.
2. It has a viscous soapy texture as in the case of aqueous
solution of sodium hydroxide.
3. Change the color of a number of indictors, for example
change the pigment of the sun flower tint from
red color to blue color.
4. Reacts with acids (neutralize) to form salts and water.
5. Its aqueous solutions have the ability to conduct electri- The basic solution changes
cal current to their ability to be ions. the color of the pigment sun-
flower to blue color.

Table (6-1) Some of the common acids and their uses.

Acid name Acid name Uses

Purification of minerals and purification of mineral ores

hydrochloric acid HCl and preparation of some food. It is the main component of
stomach acids.
Manufacture of fertilizers, explosives, glues and liquid used in
Sulfuric acid H2SO4
the cars batteries.

Nitric acid HNO3 Manufacture of fertilizers, explosives and adhesives.

Plastic and rubber industry and in food preservation is the

Acetic acid CH3COOH
main component for vinegar.

It is present in all carbonated drinks due to the reaction of

Carbonic acid H2CO3
carbon dioxide with water.

Hydrofluoric acid HF Clean the metal and polish the glass and engraving it.

155
 6-3 Molecular concepts of acids and bases
In this section we will look at the molecular concepts of
acids and bases according to the main theories are:
6-3-1 The Arrhenius Theory
In 1884 Arrhenius presented his theory of electrolysis
dissociation, which later resulted in a theory for naming
acids and bases which states on the following:
Acid is the substance that contains hydrogen which is ion-
ized to given hydrogen ions (H+) in the aqueous solution.
By the definition of Arrhenius, HCl is considered to be Ar-
rhenius acid because it produces H+ ions in aqueous solu-
tion:
HCl(aq) H+(aq) + Cl-(aq)
The base is the substance that contains the hydroxide group
(OH-) which ionizes the data of the negative hydroxide
Ionization of hydrochloric ions in the aqueous solution.
acid is in water. According to Arrenius’ definition of the base, NaOH is the
Arenius base because it produces hydroxide ions OH- in
the aqueous solution:
NaOH(aq) Na+(aq) + OH-(aq)
The process of neutralization according to this theory is
defined as a union H+ ions and OH- ions to form water
molecules:
H+(aq) + OH-(aq) H2O (l)

Arrhenius theory successfully explained the acids reac-

tions that have protons with metal hydroxides (hydroxide
bases), although the application of this theory is limited, it
has led to the development of more comprehensive theo-
ries to describe the properties of acids and bases.

Ionization of sodium hy- Hydronium ion (hydrated hydrogen ion)

droxide in water. Arrhenius described hydrogen ions in water as H+ (H2O)n,
(where n is an integer) this is because the attraction between
H+ ions and oxygen atom carrying a negative partial charge,
delta δ- in the polarized water molecule. Although we do not
know exactly how hydrogen ions hydrolyze in most aqueous
solutions, we usually represent hydrogen ion as hydronium
H3O+. Thus, it is argued that the hydrated ion of hydrogen

156
is to give aqueous acid solution its specific acidity prop-
erties. The hydrogen ion can be expressed as H+(aq) or
hydrolyzed H3O+ as follows:

Hydrogen ion Water ion hydronium

6-3-2 The Bronsted – Lowry Theory

In 1923, scientists Bronsted and Lowry introduced the
individually developed the theory of Arrhenius and thus
emerged new theory known as Bronsted-Lowry theory.
According to this theory, acid is defined as the donor of the
proton, and the base is the proton receptor.
These definitions of acid and base are highly general as
any molecule or ion contains hydrogen and has the capa-
bility to release a proton is acid, while any molecule or ion
can receive a proton is a base. So an acid reaction with a
base can be defined as that the reaction involves the trans-
fer of a proton from the acid to the base. For this reason,
the ionization process of hydrochloric acid (strong acid) in
water, It is an acid-base reaction in which the water mol-
ecule behaves as a base because it acquires a proton. This
can be expressed as follows:
First step (Arrhenius description)
HCl(aq) H+(aq) + Cl-(aq)
Second step
H2O(l) + H+(aq) H3O+
____________________________________________
Overall reaction (Bronsted-Lowry description)
HCl (aq) + H2O (l) H3O+ + Cl-(aq)

The reactions of acids and bases can be described by the

Bronsted-Lowry concept in terms of pairs (conjugate acid- Ionization of hydrochloric
base) and each pair is two different classes in a proton. acid is in water.
In the previous equation, HCl (acid) with Cl- (conjugate
base) is in the sense of a pair (conjugate acid-base),while
H2O (base) with H3O+ (conjugate acid) ie is another pair of
(conjugate base-acid ) is as in the following figure:

157
 On the other hand, the ionization of hydrogen fluoride
(weak acid). It occurs similarly but to a lesser extent and
therefore represents the following:
-

-
In this equation HF is acid with F conjugate base, while
H2O represents a base with H3O+ conjugate acid. In the
↼
forward reaction (↼) HF and H2O exhibit acid and base
↼
behavior respectively, while in the reverse reaction ( ↼ )
H3O+ exhibits acid behavior (ie, a proton donor) and fluo-
-
ride ion F exhibits base behavior (ie, a proton receptor).
We note from the examples above that when a weak acid
(HF) dissolves in water, HF molecules will produce a
small amount of H+ ions. It can be acquired by one of the
two basic classes in the solution F- or H2O. In fact, HF dis-
sociates slightly, indicating that F- is a stronger base than
H2O. When a strong acid (HCl) is dissolved in water, the
HCl molecules will produce H+ ions that can be gained
-
by one of the two basic classes in the solution Cl or H2O.
Since HCl dissociates completely in diluted aqueous solu-
-
tion, this means that Cl is lower base than H2O. That all
means that the weak acid has strong conjugate base and
the strong acid has weak conjugate base. It can be general-
ized by saying: the greater the strength of acid, the weaker
the conjugated base strength of his context and vice ver-
sa. These concepts must be used correctly; a strong acid
or a weak acid (as well as a base) are qualities that you
will use to describe a relative situation. When we say (in
-
the previous example) that F is a strong base, it does not
-
mean that it is so by comparing it with OH for example,
158
But we mean it is a stronger base relative to the conjugate
base of the strong acid in the example (H2O). In the aque-
ous solution of ammonia, the ammonia molecules behave
as a weak Bronsted base behavior while the water mol-
ecules behave as a acidic behavior and can be expressed
as follows
+ -

conjugated conjugated
acid1 base2

As is evident in the reverse reaction, the ion NH4+ be-

-
haves as conjugate acid of ammonia while the OH
exhibits the behavior of a conjugate base of water.
It is noted from the above that water exhibits acid-
ic (proton-donor) behavior in its reaction with NH3
while its behavior is basic (proton-receptor) at its re-
action with HCl and HF, therefore, the behavior of wa-
ter as acid or as a base depends on the other class con-
tained in the solution, and this can be described that water
has an amphoteric behavior and will explain this later.

6-3-3 Lewis Theory

In 1939 Lewis proposed the most comprehensive theory
among other theories for the definition of acid and base.
The base, according to Lewis’ concept is any substance
that can donate a non-bonded pair of electrons in thier
chemical reactions, while the acid is any substance has
a blank orbital and can accept the electronic pair of an-
other class. This theory does not stipulate that the elec-
tronic pair must move completely from one atom to an-
other but instead assume that the electronic pair which
one atom has, becomes shared between the two atoms.
So the neutralization process (acid-base reaction) iden-
The dissolution of ammo-
tify as it is to form coordinate bond, and the reaction
nia in its aqueous solution.
between trichloride boron with ammonia is an typi-
cal example of the reaction between Lewis acid-base.
159
 (g) (g)

Self-ionization of water
Careful measurements showed that pure water ionizes ever
so slightly to produce equal numbers of hydronium ions
and hydroxide ions. Where one water molecule donates a
proton to another water molecule, and can to express this
according to Bronsted concept:

conjugated conjugated
acid1 base2

The transition of the proton to the base involves the forma-

tion of a coordinate bond. It is clear that in the process of
self-ionization, one of the molecules of water behaves acid
and other molecule behaves as base behavior so it is said
that water behavior is amphoteric behavior.

160
 6-4 Amphoteric substances
As we already know, a certain substance can behave as
acid or base behavior depending on the medium it is in.
Amphoteric is general term describing the reactivity of a
substance either as an acid or as a base. The amphoteric
behavior describes the state in which the substance has an
amphoteric property by gaining or losing a proton (H+).
For a number of the metallic hydroxides that low soluble in
water are an amphoteric property as they react with acids
to form salts dissolved in water but at the same time they
can dissolve by reacting with an excess of strong base. For
example, aluminum hydroxide is a typical example of me-
tallic hydroxide amphoteric as it exhibits a basic behavior
by reacting with nitric acid to form salt as in the following
equation:
Al(OH)3 (s) + 3HNO3 (aq) Al(NO3)3 (aq) + 3H2O
When a solution of any strong base (such as NaOH) is
added to the solid aluminum hydroxide powder, Al (OH)3
exhibits acidic behavior and begins to dissolve, forming
dissolved sodium aluminate as in the following equation:
Al(OH)3 (s) + NaOH(aq) NaAl(OH)4 (aq)
A number of other metal hydroxides have the same behav-
ior, as shown in Table (6-2).
Table (6-2) Some amphoteric hydroxides
metallic ion metalloids ion Insoluble amphoteric The complex ion is formed
hydroxide with an excess of a strong base
2-
Beryllium ion Be2+ Be(OH)2 [Be(OH)4]
-
Aluminum ion Al3+ Al(OH)3 [Al(OH)4]
Chrome ion Cr 3+
Cr(OH)3 [Cr(OH)4]-
Zinc ion Zn 2+
Zn(OH)2 [Zn(OH)4]-2
Tin (II) ion Sn 2+
Sn(OH)2 [Sn(OH)3]-
Tin (IV) ion Sn4+ Sn(OH)4 [Sn(OH)6]2-
Lead ion Pb2+ Pb(OH)2 [Pb(OH)4]2-
Arsenic(III) ion As3+ As(OH)3 [As(OH)4]-
Antimony(III) ion Sb3+ Sb(OH)3 [Sb(OH)4]-
Silicon ion Si4+ Si(OH)4 SiO44-‫و‬SiO32-
Cobalt ion Co 2+
Co(OH)2 [Co(OH)4]2-
Copper ion Cu 2+
Cu(OH)2 [Cu(OH)4]2-
161
 6-5 Acid and base reactions in aqueous solutions
The reaction between acid and base which results salt and
water is called neutralization reaction, most salts are ionic
compounds. Table (6-3) shows common strong acids and
bases
Table (6-3) Common strong acids and bases
Base Acid

Lithium hydroxide LiOH Hydrochloric acid HCl

Sodium hydroxide NaOH Hydrobromic acid HBr

Potassium hydroxide KOH Hydroiodic acid HI

Rubidium hydroxide RbOH Perchloric acid HClO4

Cesium hydroxide CsOH Chloric acid HClO3

Calcium hydroxide Ca(OH)2 Nitric acid HNO3

Strontium hydroxide Sr(OH)2 Sulfuric acid H2SO4

Barium hydroxide Ba(OH)2 Chromic acid H2Cr2O7

When any acid reacts with a base in stoichiometric will

result a normal salt (equivalent). It does not contain hydro-
gen atoms or ionized hydroxide groups.
For example, the complete neutralization of phosphoric
acid (H3PO4) with sodium hydroxide (NaOH) is produced
a normal salt sodium phosphate (Na3PO4) according to the
following equation:
H3PO4 (aq) + 3NaOH(aq) Na3PO4 (aq) + 3H2O(l)

If base is added less than the equivalent amount required to

completely neutralize the acid, it will result acidic salts due
to the ability of these salts to react with the base, as shown
in the following equations:

H3PO4 (aq) + NaOH(aq) NaH2PO4 (aq) + H2O(l)

H3PO4 (aq) + 2NaOH(aq) Na2HPO4 (aq) + 2H2O(l)

162
There are many examples of this type of salts, including
hydrogenated sodium carbonate (acidic), also called so-
dium bicarbonate (NaHCO3).
When a polyhydroxide base (a base containing more than
one hydroxide group in its chemical formula) is reacted
with an equivalent amount of acid, a normal salt is formed
as described in the following equation:

Al(OH)3 (s) + 3HCl(aq) AlCl3 (aq) + 3H2O(l)

When this type of base reacts with less amount of acid

than is necessary to equivalent it, then strong base salts
are formed (i.e. salts contains non-reactive OH groups (as
shown in the following equations:

Al(OH)3 (s) + HCl(aq) Al(OH)2Cl (s) + H2O(l)

Al(OH)3 (s) + 2HCl(aq) Al(OH)Cl2 (s) + 2H2O(l)

It should be noted that the basic salts is not necessary to

have a basic character, but they can react and neutralize
acids as described in the following equation:

Al(OH)2Cl + 2HCl AlCl3 + 2H2O

6-6 Types of salts

As we learned earlier, when a solution of acid is neutral-
ize with the base completely and accurately, the resulting
solution is a salt derived from a pair acid-base. That such a
situation often occurs in chemical reactions used in chemi-
cal analysis processes which will be necessary to calculate
the acidic function (pH) or in other words the concentra-
tion of hydrogen ion in the solution. Salts are is a strong
electrolyte, therefore salt will completely dissociated in the
solution. If the acid and base are strong, the acidic value
of aqueous salt solution remains constant does not change.
If one or both pairs (the acid and the base that formed the
salt) is weak, then the calculation of the acidic function of
solution becomes more complecated, so it is appropriate
to address this issue by dividing the salt into four main
sections:
163
 6-6-1 Salts derived from strong acids and strong base
reactions
Example of reaction of hydrochloric acid HCl with sodium hy-
droxide NaOH
HCl + NaOH NaCl + H2O
The solubility of this group of salts does not affect the equilib-
rium process between hydrogen and hydroxide ions in water:
↼
H3O+ + OH- ↼ 2H2O
Therefore, the solution remains neutral.
6-6-2 Salts derived from weak acids (HA) and strong
bases (MOH)
The weak acid HA reacts with the strong base of MOH as fol-
lows:
HA + MOH MA + H2O
The salt (MA) of this type completely dissociated in its aque-
ous solution:
MA M+ + A-
As is known, the aqueous solution also contains small amounts
of hydrogen and hydroxide ions resulting from dissociation of
water molecules. Because acid HA is weak acid, it is partially
ionized, so the amount of A- ions that can be present in the solu-
tion with H+ ions is also small, and to maintain the equilibrium
state in the solution, A- ions combine with H+ ions to form the
weak acid HA
↼
H++A- ↼ HA
The only source of hydrogen ions in the solution is the disso-
ciation of more water molecules. That the continued dissocia-
tion of water molecules will produce quantities of hydroxide
ion and hydrogen ion consumed from the solution to form the
weak acid HA. All that leads finally to increase in the concen-
tration of hydroxide ions in the solution at the expense of the
concentration of hydrogen ions, and the solution becomes ba-
sic. This condition in which an ion (or ions) of salt reacts with
water ions is called hydrolysis process. The hydrolysis process
of a salt derived from a weak acid and a strong base can be ex-
pressed by the following equation:

MA + H2O ↼
↼ HA + MOH

164
The property of the solution depends on the relative
strength of the acid and the base resulting from the hydro-
lysis process, an example of this type of salts is sodium
acetate (CH3COONa) and format potassium (HCOOK).
6-6-3 Salts derived from strong acids (HA) and weak
bases (B)
This type of salts is formed according to the following
equation:
HA + B (BH)A
This salt is fully ionized in the aqueous solution:
(BH)A BH+ + A-
In the aqueous solution of salt (BH) A, the concentration
of BH+ ion (formed by the complete dissolution of the salt)
decreases, due to its association with the hydroxide ion (re-
sulting from the dissolution of water molecules) to form
the weak base B and reach to the equilibrium state, as in
the following equation:
BH+ + OH-
↼
↼ B + H2O
↼
BH+ + H O ↼ B + H O+
2 3
Therefore the concentration of hydrogen ion in the solu-
tion will increase at the expense of the hydroxide ion (con-
sumed to form the weak base), so the solution becomes
acidic.
An example of this type of salts is ammonium chloride salt
(NH4)Cl and ammonium sulfate salt (NH4)SO4.

6-6-4 Salts derived from weak acids (HA) and weak

Information
bases(B)
Dissociation constant of weak
In the aqueous solution of this type of salt, the following
acid and base means (ability of
two instantaneous reactions occur:
dissociation of weak acid and
A-+ H2O
↼ HA + OH-
↼ weak base to produce their ions
↼
BH+ + H2O ↼ B + H3O+ in the solution).
The property of the acidic or basic solution depends main-
ly on this case on the relative strength of acid and base
(whichever is stronger than the second). If they are the
same strength (acid dissociation constant = base dissocia-
tion constant) the aqueous solution of salt is neutral. If the
acid dissociation constant is larger than the base dissocia-
tion constant. The aqueous solution of the salt is acidic and
vice versa.
165
 6-7 Acid and bases indicators
Acid and base indicators are organic dyes (organic compounds),
the color of which depends on the concentration of H3O+ ions
in the solution. The color of the indicators shows the amount
of acidity or basic of the solution. The first indicators that used
for this purpose was plant dyes such as sunflower (Litmus), but
most of the indicators used now is laboratory-made materials.
Most of the indicators for acid-base reactions is weak organic
acids (symbolized as HIn), which have a color for its non-dis-
sociated formula differ than the color of dissociated formula
In-. For example, the non-dissociated blue bromophenol dye in-
dicator ( is a very weak organic acid) have yellow color, while
the dissociated dye have a blue color as shown in the following
equation for the indicator HIn:
HIn + H2O H3O+ + In-
Color 1 yellow Color 2 blue
The color of the solution is determined by the ratio of the quan-
tities of both dissociated In- and non- dissociated HIn available
in the solution. A type of indicator used in neutralization reac-
tions is called general or comprehensive indicators (consisting
of a combination of a number of acid-base indicators). These
indicators show a gradient in color for a wide range of acidic
function.
6-8 Acid and base solutions
Aqueous solutions of acids and bases are considered to be the
most important solutions. These solutions are vitally important,
as the biological reactions that occur within our bodies take
Figure 6-1 place in aqueous acidic or basic media, and any difference,
The colors of the three solu- even if a slight concentration of these solutions may alter these
tions are common indicators reactions, it may result in impaired physiology. Acids and bas-
at values for the acidic func- es prepare in the laboratory in different concentrations accord-
tion of the solution (pH) rang- ing to what we have already learned about the preparation of
ing from 3-11 (A) red methyl, solutions previously. For example, a solution of hydrochloric
(b) blue bromothymol, (c) acid (HCl) can be prepared by dissolving hydrogen chloride
phenolphthalein. gas HCl in water, and solutions of sodium hydroxide can also
be prepared by dissolving solid NaOH in water. The prepara-
tion process each time is done by dissolving a known quantity
(mass or number of moles) of the solute, and precisely, in a
specific amount (mass or volume) of the solvent according to
calculations made for this purpose.

166
 Example 6-1
Calculate the molarity of a solution containing 3.65 g HCl
in 2 L of solution (molar mass 36.5g/mol).
Solution
According to the law of the molar

M=n
‫( ـ ـ ـ ـ‬mol)
‫ـــــــــــ‬
V (L)
Calculate the number of moles of HCl from the following
relationship:
m (g) 3.65 (g)
nHCl = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.1 mol
M (g/mol) 36.5 (g/mol)
So molarity is equal to:
n (mol) 0.1 (mol)
M = ‫ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ = ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬0.05 mole/L = 0.05 M
V (L) 2.0 (L)

Example 6-2

Calculate the mass of Ba(OH)2 (molar mass = 171 g Exercise 6-1

/ mol) needed to prepare 2.5 L of barium hydroxide Calculate the molarity of
solution at a concentration of 0.06 M. concentrated sulfuric acid
Solution solution in one liter (molar
Based on the law of molarity: mass = 98 g/mol) if you know
the mass percentage of the
n (mol) acid in the solution is 96.4%
M (mol/L) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
V (L) and the mass of the solution is
From the definition of moles numbers equal 1.96 kg / L .
m (g)
nHCl = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
M (g/mol)
Combining the two relationships above we get:
m (g)
‫ـــــــــــــــــــــــــــــــــــــــــ‬

M (g/mol)
M (mol/L) = ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬
V (L)
and then we get the mass of Ba(OH)2
m(g) = M(mol/L) × V(L) × M (g/mol)
m(g) = 0.06 (mol/L) × 2.5(L) × 171.3(g/mol)
m(g) = 25.695 g

167
 6-9 Titration
Laboratory (acid-base) reactions are usually used to mea-
sure the volume of a solution (known concentration) to
react with a certain volume of another solution with un-
known concentration and then the measured volume is
used to calculate the concentration of the first solution, this
process is called titration. Titration is the process in which
the solution of one of the two reactants is added gradu-
ally from a cylindrical glass instrument inserted in such a
way that it can accurately measure the volume of the solu-
tion and it is called the burette, to a solution of the other
reactant typically found in a conical flask called a (Erlen-
meyer flask) until the reaction between the two substances
is complete. Then measured the volume of the solution that
should added to complete the reaction.
To determine the point at which the titration process is
stopped (the point at which the reaction between the acid
and the base ends) is called the (end point) or the actual
equivalence point where a substance color-changing is
added that its color-changing at this point called the indi-
cator.
For example, an acid solution with unknown concentration
in the conical flask can be titrate by gradual addition of a
standard sodium hydroxide solution (with a known con-
centration) from the burette [Figure (6-2)] and using the
phenolphthalein indicator.

Figure 6-2
Titration of a solution of un-
known concentrated acid
against a standard solution of
the base where the reaction
endpoint is determined when
the color of the added indi-
cator to the titrated solution
changes.

168
 Example 6-3
What is the molar concentration of an acid solution for
hydrochloric acid if know that 36.5 mL of it was necessary
to react with 43.2 mL of 0.236M of sodium hydroxide
solution?
Solution
HCl + NaOH NaCl + H2O

The reaction equation shows that one mole of acid is

equivalent to one mole of the base, or one mlimole of acid
is equivalent to one mlimole of the base, so it calculates
the number of mlimoles of sodium hydroxide involved in
the reaction firstly. Then it can be calculated the molar of
hydrochloric acid concentration because the volume of
the acid solution is known.
Information

VHCl = 36.5 mL
MHCl = ?
VNaOH = 43.2 mL
MNaOH = 0.236 M

Using the dilution law of the solution

MHCl × VHCl = MNaOH × VNaOH

MHCl × 36.5 mL = 0.236 M × 43.2 mL

MHCl = ‫ـ ـ‬0.236 ‫ـ ـ ـ ـ×ـ ـ ـ ـ ـ ـ‬43.2

‫ــــــــــــ‬M ‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬mL
‫ = ـ ـ ـ ـ ـ ـ ـ‬0.278 M
36.5 mL

169
 Example 6-4
In the process of titration of sulfuric acid solution with
standard sodium hydroxide solution. It was found that
43.2 mL of a solution of 0.236 M sodium hydroxide
Exercise 6-2
consumes 36.5 mL of acid solution to reach the end
(Adjusted oncentration) Potas- reaction point the reaction (changes the color of the blue
sium hydroxide solution was bromophenol indicator from blue color to yellow color.
titrated by titrating 25 mL of it What is solution molarity of H2SO4?
with the standard hydrochloric Solution
acid solution (0.08 M).
H2SO4 + 2NaOH Na2SO4 + 2H2O
If you know that 42 mL of acid
has been consumed in the cor-
The chemical equation of the reaction shows that the
rection process until the color
reaction ratio is 1 mol of H2SO4 per 2 mol of NaOH, so
of the red methyl indicator used
the number millimoles of sulfuric acid are equal to half
changes from yellow to red.
of the mlimoles of sodium hydroxide, so we write the
what is the molar concentration
dilution law as follows:
of potassium hydroxide solu-
tion? 1
MH SO × VH SO = ‫( × ـ ـ ـ ـ ـ ـ ـ‬MNaOH × VNaOH)
2 4 2 4 2

Information
VH SO = 36.5 mL
2 4
MH SO = ?
2 4
VNaOH = 43.2 mL
MNaOH = 0.236 M

Substitute in the above relationship

1
MH SO × 36.5 mL = ‫( × ـ ـ ـ ـ ـ ـ ـ‬0.236 M × 43.2 mL)
2 4 2

MH SO = ‫ـ ـ‬0.236 ‫×ـــــ‬
‫ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬M ‫ـ ـ ـ ـ ـ‬43.2
‫ ـ ـ ـ ـ ـ ـ ـ ـ ـ‬mL
‫ = ـ ـ ـ ـ ـ ـ ـ ـ‬0.139 M
2 4
2 × 36.5 mL

170
 Basic concepts
Hydronium ion(H3O+) Arrhenius base
Is a common expression of full hydrogen Substance produce ions OH- in aqueous
ion. solutions. Strong bases are dissolved in
Self – Ionization water and completely dissociated, while
Ionization reaction occurs between simi- weak bases are partially deionized.
lar molecules (For the same substance). Base Bronshted Laurie
Ionization in aqueous solution Proton - acquired class.
A process that involves the reaction of a Lewis Base
compound molecule with water to form Any item can contribute by providing
ions. (honor) a pair of electrons to form a co-
Titration ordinate bond.
A process by which the standard solution Amphoteric substance
volume (with a known concentration) is The substance that shows the amphoteric
required to react with specific amount of property and that of during their ability
material to be estimated. to acquire or lose protons, i e, any sub-
Neutralization stance that has the potential to behave
Acid reaction with base to form salt and either acid or base behavior.
water Hydrogen ions usually interact Equivalence
with ions hydroxide to form water mol- The point at which equivalent amounts
ecules. chemical reactants have reacted.
Arrhenius acid Standard solution
A substance that produces H +
ions in
(aq)
A solution whose concentration is pre-
aqueous solutions. Strong acids ionize cisely known.
completely or almost completely in di- Calibration
lute aqueous solutions, while weak acids A process by which the concentration of
are partially deionized. the solution is determined accurately, it
Acid Bronshted Laurie is usually corrected against known con-
Proton donates class trol quantity of standard material.
Lewis acid Hydrolysis
Any item can contribute to receiving a Ionization or dissocaition of the sub-
pair of electrons to form a symmetric stance in water.
nucleus.
Indicator
An organic substance that can show dif-
ferent colors when present in different
acid solutions, so it is used to determine
the point at which the reaction between
two substances is finished( one of them
act as acid and the other as a base).

171
 Chapter Six Questions 6
6-8 Calculate the mass of HNO3 con-
6-1 How did Arrenius defind the acid, tained in 5 mL of concentrated acid so-
base and neutralization process? Explain lution( Its mass percentage = 69.8% and
this by example. its density = 1.42 g / mL).
6-2 Using the Bronshted and Laurie the- 6-9 How much ammonium sulfate salt
ory, explain what is meant by each of the (NH4)2SO4 (molar mass = 132.8 g/mol)
following with example: required to prepare 400 mL of solution
A- acid B- conjugate base in a concentration of 0.25 M?
C- base D- conjugate acid 6-10 How much the volume of concen-
E- pairs of conjugate acid-base trated sulfuric acid( Its mass percentage
6-3 Distinguish between acids and bases = 98% and its density =1.84 g / mL) is
in each of the following using the theory required to prepare 100 mL of diluted
of Lewis, indicating the donor and ac- acid solution( Its mass percentage 20%
quired of electron pair: and its density = 1.14 g / mL).
H2O + H2O H3O+ + OH- 6-11 Put (✓) in front of the correct
HCl + H2O H3O+ + Cl- statement and a sign ( × ) in front of in-
correct statement as follows:
NH3 + HCl NH4+ + Cl-
1. All Bronshted-Laurie acids and bases
6-4 What is acid salt? Write balanced
are considered Arrhenius acids and bas-
chemical equations to show how to get
es.
the following acidic salts from appropri-
2. Ammonia reacts with water as Bron-
ate acids and bases:
shted base.
A- NaHSO3 B- KHCO3
3. The aluminum chloride (AlCl3) is an
C- NaH2PO4 D- NaHS
acid Louis.
6-5 Explain why
4. The conjugate base of strong acid be-
A- Salts solutions derived from strong ac-
have as strong behavior base.
ids and strong bases will be neutral.
5. Boron trifluoride is considered Lewis
B- Salts solutions derived from strong ac-
acid.
ids and weak bases are acidic.
6-12 Which of the following statements
C- Salts solutions derived from weak ac-
relates to Arrhenius theory are correct?
ids and strong bases are basic.
1. Acid-base reaction according to
6-6 Calculate the molarity of the nitric
Bronshted-Lauri theory includes pro-
acid solution if you know that 35.7 mL
tons transmission.
from it neutralized, in titration process,
2. One of the essential requirements for
0.302g from Na2CO3.
the Bronshted-Lauri statement is the
6-7 How much the volume of concentrat-
presence of hydroxide ion in its chemi-
ed HCl acid(Its mass percentage = 36%
cal formula.
and its density = 1.18 g/mL) is required
to prepare 500 mL of dilute solution of
the same acid at concentration of 2 M?

172
 7
Chapter Seven
Polymer Chemistry

After completing this chapter, the student is expected to:

• Recognize molecules (polymers) as large molecules built from small units called
monomers.
• Understand polymerization of monomers process for the formation of polymers.
• Recognize some types of natural and synthetic polymers.
• Explain the effect of different general structural composition of polymers on their
properties.
• Specify the types of polymers (polymers of addition and condensation polymers)
and how they are prepared in laboratory.
• Describe the formation of polymers by addition polymerization as well as by
condensation polymerization .
• Recognize rubber (as a type of polymer) and distinguish between natural rubber
and synthetic polymer.
• Know some of common uses of synthetic polymers like plastics synthetic and fibers.

173
 7-1 Polymers (plastics)
Plastics, synthetic fibers, rubber, cellulose and proteins,
the clothes we wear all these materials are polymers, starch
and fatty are other examples of polymers. The word poly-
mer consists of two segments, (poly) the Greek word means
multi and the word (mer) is a molecule, and so on, polymer
means polymorphic(multi-molecules). Polymers are huge
molecules composed of a large number of small molecules
called (monomers). These monomers were connected by
mostly covalent bonds in the form of long chains accord-
ing to organic chemical reactions. One of the polymers
contains approximately 1000 to 200,000 atoms, mostly
carbon and hydrogen. The monomers are small molecules
that may be organic or inorganic compounds that combine
with each other by chemical reaction to form huge chains,
which is polymer. A polymer consisting of units of these
duplicate molecules is called duplicated unit, and the reac-
tion process of these monomers of the formation of poly-
mers is called polymerization process. So polymerization
is a bonding process of small molecules called monomers
come together to form huge molecules called polymers. If
the monomer is denoted by the letter M and the number of
monomers is n (n values are between 5 and 50,000), the
polymerization process can be represented to form poly-
mer (M)n as in the following equation:
Polymerization
nM process
[ M ]n
n monomers polymer with n duplicate units

7-2 Polymers types

Polymers were initially classified into two types:
7-2-1 Natural Polymers
They are found in nature such as starch, cellulose, proteins
and natural rubber.
7-2-2 Synthetic Polymers
Which was prepared laboratorial or synthetically by hu-
man such as polyethylene, polyamide (nylon), polyvinyl
chloride and other thousands of polymers are known now-
adays.Polymers may be classified according to the effect
of temperature on:
174
A-Thermoplastic polymers
This type of polymer is softened when it is heated
where it can be reformed several times and then it hardens
at a low temperature. These polymers are named plastics.
The most famous examples are: polyethylene, polypropyl-
ene, polyvinyl chloride and polystyrene.
B-Thermosetting polymers
Do not soften when heated, but maintain its original
shape. One of the famous example melamine, synthetic
rubber. Polymers may also be classified according to their
structural composition and effect of heating on, it is either
linear or branched or crosses - linked as shown in Figure
7-1. When linear polymer is heating the molecules are free
to move and slide easily forward and backward on top of
each other as they are thermally unstable. As for heating the
branching polymer and its molecules contain side chains
that prevent the particles slide off each other easily but are
likely still not thermally stable. The cross-link polymer has
molecules adjacent are interrelated so it cannot slide over
each other by heating. It remains in this state to maintain
its shape thermally.

Figure 7-1
7-3 Synthetic polymers
Types of polymer according to
synthetic polymers (which are made by Human) can be
their structural composition.
made by two types of polymerization, namely:
7-3-1 Addition Polymerization
A chemical reaction in which too many small unsatu-
rated molecules (monomer containing double bond) are
added to form one huge molecule, which is a polymer
without any by-product (those polymers may be made of
one type monomer). Examples of addition polymers are
polyethylene (PE), polyvinyl chloride (PVC), polystyrene
(PS) and polypropylene (PP), and we will look at the meth-
ods of preparation and compositions in a brief summary.
175
 1. polyethylene (PE)
Polyethylene is prepared from the addition of ethylene
molecules(CH2=CH2) for some of them to be a polyethyl-
ene molecule. As shown in the following equation:

Catalysts
Heat

Ethylene polyethylene
The letter (n) represents the number of polymerized eth-
ylene molecules which may be a number repeated units
of approximately 50,000 or more. The addition reaction is
repeated many times to form a polymer with length n of
monomers (very long molecular chain).This reaction can
Ethylene Molecule be repeated hundreds or even thousands of times, the eth-
ylene unit in the polymer chain is called the repeated unit,
where one of the two bonds breaks between the two car-
bon atoms in the ethylene molecule ( π bond). Each atom
is bound to a carbon atom of another molecule, and this
process is repeated until thousands of ethylene molecules
are bonded to form polyethylene as follows:

This equation can be written briefly as in the previous

equation and figure 7-2 shows how the number of units of
a molecule is known polymer (polyethylene), consisting of
three repeated units.

Figure 7-2
Number of units of polymer
molecule (polyethylene). Monomer (3) Monomer (2) Monomer (3)
Forms of ethylene
Polyethylene is usually prepared in various forms ac-
cording to the conditions of the polymerization reaction.
There are three forms of polyethylene according to tem-
perature, pressure and catalysts used in the preparation:
176
A-High-Density polyethylene (HDPE)
It is formed when ethylene is heated to 100 ° C in a hy-
drocarbon solvent at normal atmospheric pressure a high
density linear polymer is formed. Because linear molecules
are very closely aligned, their size is small this makes the
density is high, so the polymer remains strong and solid,
making it plastic containers such as milk containers and
juices.
B-Low-Density polyethylene (LDPE)
It is formed when ethylene is heated to 200 ° C at high
pressures (approximately 1000 atm) with little oxygen as
the initiator of the reaction. Branched polymer is formed Do you know
when removal of hydrogen atoms from the molecule and The polymer chain is in adhesive
the addition of ethylene molecules instead in these sites, tapes, low density polyethylene
unlike the linear molecules because their molecules cannot with a value of n about 10,000
convergence. So the density of branching chain is less than units each series has an average
the density of the linear chain. So the polymer is less rigid molar mass of about 300,000.
than high density polyethylene and therefore used in mak-
ing ordinary plastic bags.
C- Cross-linked polyethylene (CPE)
This type of polyethylene can be obtained when re-
moved hydrogen atoms of polyethylene molecules, where
two adjacent molecules in the chain are bonded and form
an cross-link bonding between two molecules. The cross-
type is known to be stiffer and stronger than high-density
polyethylene, making things that require extreme stiffness.
Figure (7-3) shows the different forms of polyethylene
with their different uses.

Figure 7-3
Properties of various forms of
polyethylene and some special
uses to them.

177
 Figure 7.4 shows some of the different uses of polyethylene.
Beauty Tools

Housewares Garden benches

Plastic bags Waste containers

polyethylene

Figure 7-4
Different uses of polyethylene.
2. polypropylene (pp)
It forms of the union of a large number of propylene
molecules(CH2=CH-CH3) with catalysts in the following
equation written briefly as follows:

Catalysts
Heat

Propylene polypropylene
This process can be repeated until thousands of propyl-
ene molecules bind to the polypropylene as follows:

Catalysts
Heat
Propylene Propylene Propylene polypropylene

Polypropylene has the following characteristics:

1. Easily formed, poured and resistant to heat and chemi-
cals.
2. Unbreakable
3. Transparent and odorless
4. Used in the manufacture of medical instruments, toys,
tubes and Figure (7-5)
Shows some different uses of polypropylene.
178
 Paper clip plastic cups Plastic chair

Plastic bags Plastic ropes

polypropylene

CD Covers Meat cutting boards

Figure 7-5
3. Polyvinyl chloride (PVC) Some different uses of
It consists of the bonding of a large number of vinyl polypropylene.
chloride molecules (CH2 = CH-Cl) with catalysts, as the
following simplified reaction:

Catalysts
Heat
Vinyl chloride polyvinyl chloride

PVC has the following qualities:

1. More durable, heat resistant and chemicals than poly-
ethylene and propylene.
2. Cheaper
3. It is used instead of metal pipes in the manufacture of
water pipes.
4. Water resistant and insulator so it enters the car brush-
Molecular structure of
less and raincoats industry.
vinyl chloride
Figure (7-6) show some uses of polyvinyl chloride.

179
 In sports airframes

In the artificial skin In the stickers

In plastic pipes In the water pipes

polyvinyl
cholride

In window frames

Figure 7-6 4. Polystyrene (PS)

Various uses of polyvinyl Polystyrene consists of the polymerization of a polymer
chloride. molecule called styrene with the presence of suitable cata-
lysts and the structural formula of styrene is as follows:

or

Polymer styrene to form polystyrene according to the fol-

lowing formula:

Catalysts

180
Polystyrene is a white solid that is easy to form and resist
for acids and alkalis, used in the manufacture of synthetic
sponge (cork), insulators, pipes, some household utensils
and vegetable containers. Figure 7-7 illustrates some uses
of polystyrene.
In cork and plastic cups

In the packaging In cork dishes

materials polystyrene

Figure 7-7
Different uses of polystyrene.

In thermal insulation of buildings

Cork is made of polystyrene, which is small balls as shown
in Figure (7-8). It size varies depending on the type of cork
and there is a monolith together. When it exposed to flame
in an atmosphere saturated with oxygen (at least 30%), it
ignites and because of this characteristic cork is used in the
manufacture of high-temperature resistant materials, note
Figure (7-9), which made polystyrene to compete with
many other polymers prepared.

Figure 7-8
Cork is a small balls.

Figure 7-9
Cork resistance to heat in the
normal atmosphere.

181
 5. Poly tetrafloroethylene (Teflon)
This polymer is prepared from the tetrafluoroethylene
monomer (CF2 = CF2), note the structural formula adjacent
to the polymer.
Where molecules bonded to this monomer in the form
of a long chain to form a substance known as Teflon. This
substance contains a non-reactive and stable carbon fluo-
Structural formula of rine bond at temperature is 325 °C. It has a very low fric-
tetrafluoroethylene. tion coefficient and this it means that the material slides
easily off the surface, making it an important material in
the manufacture of parts of heat-resistant machines in ad-
dition to use in the manufacture of kitchen utensils that do
not stick food (Tefal) Figure (7-10).
Teflon has the following characteristics:
1. Its extreme resistance to heat and chemicals
2.More fixed than any polymer material, whether natural
or manufactured.
3. It does not burn and is not corroded by weathering.
The most important uses are:
1. In the manufacture of tools exposed to heat.
2. In coating cooking utensils to prevent food sticking to
Figure 7-10 them.
Frying pan coated with 3. In the manufacture of heat resistant clothing.
Teflon. Note from the examples above that added polymers are
all from hydrocarbons, which they are very similar in form
and chain the method of composition, differ in terms of
groups that are related to the carbon atom; for example,
the hydrogen atom in ethylene is replaced by the methyl
group -CH3 as in propylene, chlorine atom -Cl as in vinyl
chloride, and the phenyl ring -C6H5 as in styrene.The four
atom of hydrogen may be replaced by fluorine atoms -F
as in tetrafluoroethylene. This causes a difference in the
qualities of plastics produced from them.
Table(7-1) illustrete the monomer structures and refind
units of addition polymers and some ot their uses.

Teflon uses

182
 Table 7-1 Structures of monomers and refined units of addition polymers and their names
The repeated unit
Monomer name Polymer name Monomer form uses
form in the polymer

plastic bags
Polyethylene
Ethylene ,slices, strips
.etc.

Glasses, Plastic
Propylene Polypropylene
bottles ,etc

Poly chloride Waterproof

Chloro ethylene
vinyl (PVC) materials
(vinyl chloride)
isolators, cylinders

Canning, roofing
Styrene Polystyrene
of buildings

Tetrafluoroethylene Teflon Non - stickutensils

7-3-2 Condensation Polymerization

Monomers are bonded together with the removal of a simple mol-
ecule (by-product) such as water. Each polymer may contain two types
of monomer molecules. It contains two functional groups which make
each monomer associate with another and withdrawn a molecule as a
by-product. One of the most important examples of this polymers type
are polyamides (nylon) and polyester polymers (Terylene) and will
show the preparation and properties of these polymers in brief.
1.Polyamide
Nylone (nylone) is the first common manufactured condensation
polymer derived from New York and London. It was made of two
monomers, one of which has the amine group(- NH2) at both ends, The
other monomer has a carboxyl group(-COOH) at both ends. These two
monomers are easily linked together with removal molecule of water
and they are formed polyamide and the bond roup between the two
O H
monomers for the nylon chain -C-N- is called amide group. The follow-
--
-

ing equation shows the nylon industry:

183
 Hexagonal methylene Diamine Adipic acid

6,6 nylon water

Figure 7-11 shows nylon 6,6 wrapping around the glass
leg. We conclude from the equation that the reaction
comprises two types of monomers: adipic acid monomer
Figure 7-11 (HOOC(CH2)4COOH). The second monomer is hexameth-
Nylon when prepared from ylene diamine (H2N(CH2)6NH2) and the resulting poly-
hexagonal methylene Di- mer is called (6,6 nylon), which is to contain each carbon
amine and adipic acid and monomer (6) carbon atoms.This is one of the most manu-
pull it in the form of fila- factured polymers frequently used. Nylon is characterized
ments. by its low water absorption, which makes it limited to use
in body contact clothes, but is also used in the manufacture
Do you know
of moisture resistant fabrics such as coats and umbrellas.
One type of polyamide is called
Figure (7-12) shows the most important uses of nylon.
fiber (Kevlar) is used in armor
bulletproof worn by military
personnel.

Figure 7-12
Nylon uses.

2. Proteins
When we mention polyamides we have to look at pro-
teins. They are polymers that contain thousands of (H2N-
X-COOH) molecules that bind together and have the same
bonding groups in Nylon, it is an amide group. It removes
water from amino acids when it polymerized, so it is one
of the condensation polymerization reactions. The polym-
erization of these acids can be represented as follows:

184
3. Polyesters
An example of polyesters is polyethyleneteraphatlate
(PET). It is polyster is one of the most famous condensate
polymers where it is prepared from the reaction of ethyl-
ene glycol (HOCH2CH2OH) with with terephthalic acid Do you know
(COOH-C6H4-COOH) in the presence of catalysts accord- that the cartoons that withdraw
ing to the following reaction: cash and those used to mobilize
phones Mobile charge and credit
cards in banks and others are
all made of the plastic is light,
long-lasting, strong, cheap and
terephthalic acid Ethylene Glycol
easy to print and save magnetic
information (CD) on them.

polyethyleneteraphatlate water

A water molecule is removed during the polymerization

process and the bonding group between the repeated units
in the polysters is the ester group. Polyethylene terephthal-
ate is used in most bottles of carbonated beverages and wa-
ter because it is a synthetic polymer that does not pollute
these beverages.

Table 7-2 Structures of condensate polymers, their repeated units and some of their uses

Polymer name Repeated polymer unit form Uses

Foam filler for bedding, seats and com-

C NH R NH C O Rَ O
Polyurethane O] O n
[ position, spray insulators, spare parts for
cars and industry footwear and packaging
As example R,Rَ = CH2 CH2
materials waterproof

Polyethylene
Terephthalate
(Polyester)
] O CH2 CH2 O C
O
C
O [ n
Tire, tapes, clothes and bottles and soft
drinks

6,6 Nylon
(Polyamide) ] NH ( CH2)6 NH C ( CH2)4 C
O O n [
Home furniture, clothes, fiber carpets and
fishing nets Fish

185
 4. Natural rubber and synthetic rubber
Natural rubber is extracted from the rubber tree (avail-
able in Malaysia and Brazil). Natural rubber is composed
of a repeated unit called isoprene in its cis form (cis-iso-
prene): a 2-methyl 1,3-butadiene, which is polymerized in
the following form:

poly
add

Isoprene poly isoprene

Rubber produced from other monomers, in addition to

isoprene, such as butadiene and styrene.
Extraction of natural rubber
from the rubber tree Note that the applications of pure natural rubber and
syntheses relatively few because when it heated, individual
molecules slide easily into front and back and on each oth-
er, so that the rubber become soft and viscous. It cannot be
used to make needs, so sulfur is added to the rubber. It can-
not be used to make needs, so sulfur is added to the rubber
in its manufacturing processes turn it into a solid and strong
material. Adding sulfur to rubber is called vulcanization. it
is a process that leads to the entanglement of adjacent rub-
ber molecules through sulfur atoms. Vulcanization makes
the rubber usable in a wide field as rubber pipe industry,
automotive tires ...etc. Note Figure (7-13). Another exam-
ple is the rubber made from polymerization of Neoprene,
which is 2-Chlorobutadiene with the following structural
structure similar to isoprene. Note that 2-chlorobutadiene
Figure 7-13
is similar to natural rubber isoprene monomer, except that
Some uses of synthetic rubber.
the chlorine atom has replaced the methyl group in carbon
Do you know atom number (2). As describe below
That the material Bakelite, the
first plastic material, its was dis-
coverer by Bakeland 1908, so H CH3 H Cl
attributed to him and uses bake- C C H C C H
lite in making hard tools such as H C C H C C
Radio, television and telephone H H H H
instruments. Isoprene 2-chlorobutadiene

186
 7-4 The world of plastics
Since the Second World War, plastics have become in-
creasingly important. Plastics has many advantages com-
pared to other materials are clean, cheap and transparent
and easy to form and non-corrosive and insulating good
lightweight, easy to shape and stays long and can be ex-
tremely strong. But they have one serious flaw. They do
not degrade and cause environmental pollution because Plastic waste causes killing
they can remain in the earth indefinitely. many revival
Plastic waste
It is known that waste is of two types:
A- Organic waste from natural products around us that is
biodegradable (naturally occurring in the presence of the
sun), such as paper and cardboard, which can be decom-
posed by bacteria into their essential components during
the days and months or years.
Plastic waste, which does not degrade naturally even in
natural conditions (sun and temperature) except during
plastic waste
decades unlike organic waste. For example, plastic bags
affect marine and terrestrial organisms because they are
Do you know
impermeable to air, causing many of these organisms to
Research is still ongoing to ob-
suffocate as they enter the bags.
tain biodegradable (natural)
There are positive things to follow to reduce contamina-
plastics so they can be eliminated
tion of plastics waste as follows:
and thus eliminate the problem of
1. Reduce the use of plastic bags for everyday purchases
pollution plastic waste. Research
2. Reuse bags more than once instead of throwing them
is conducted on the development
directly
of biodegradable plastics called
3. Throw the plastic bags in the designated places.
(degradable biopolymer).
4. Collection of the plastic products that have been dis-
It is made from fermentation of
carded and delivered to the companies that recycle them
sugar by bacteria. It is a typical
from waste to new raw materials.
plastic and can be formed into
Recycled products can be found in the presence of the re-
molds or slices.
cycling mark Most natural materials, such as cartons and
It costs 15 times the cost of its
paper, are degraded by microorganisms such as bacteria in
manufacture and its chemical
the soil. Plastic containers do not degrade naturally and
name is polyhydroxybutyrate
they are said to be non-biodegradable. Therefore, plastic
(PHB).
waste is an environmental problem, even if disposed of
by burning. They produce toxic fumes such as hydrogen
chloride from burning polyvinyl chloride and also toxic
cyanogen gas from burning polyurethane.

187
 Basic concepts

Addition polymerization Catalyst

Chemical reaction in which too many A substance that accelerates the rate of
small unsaturated particles (monomers) chemical reaction and its composition
are added to be one huge and large mol- does not change at the end of the reaction.
ecule, a polymer. Like adding ethylene Dehydration
molecules to each other to be a polyethyl- Pull up water from the material.
ene molecule. Thermoplastic polymer
Condensation Polymerisation It is a polymer that melts more than once
Bonding of monomers to polymer com- so it can be reshaped many times.
position with extract a small molecule Plastics
like water. Such as the formation of ny- Polymers are easy to form for low melting
lon and terylene polymers. point, they have relative properties. So
Polymer that they are used for many purposes in
A macromolecule consists of the bond- contemporary life and sometimes called
ing of a large number of small molecules plastic materials.
(monomers) such as polyethylene. Plastic
Monomer A large molecule is often a polymer that
A small molecule can bond to a large can be cast and formed such as polyeth-
number of it (polymerize) as ethylene is a ylene PE and polyvinyl chloride PVC .....
monomer of polyethylene. etc.
Thermosetting polymer Vulcanization
A polymer that does not melt after its for- Is the process of overlap between contigu-
mation and cannot be reshaped. ous polyisoprene molecules that it occurs,
Natural polymer when these molecules are heated with sul-
A huge macromolecule like carbohydrates fur atoms.
Proteins and fats. Polymerization
Synthetic Polymer Is the process of combining the number of
Man-made polymer such as plastic, syn- n monomers with some of them to form a
thetic fiber.... etc. polymer molecule.
Biodegradable
Natural decomposition of substances me- Polymerization process
diated by bacteria and light the sun.
process
Pollution
Damage to the environment as a result of
the use of dangerous chemical materials
which are called pollutants.

188
 Chapter seven Questions 7
8. Plastic has pollution problem because
7-1 Circle the correct answer symbol in many plastics are plastic.
all of the following: A- Very flammable
1. Which of the following materials is B- burns to create toxic fumes
used for non-stick surfaces? C- decomposes into toxic products
A- Dichloro difluoromethane 9. The polymer that does not melt when
B- Polychlorofluoroethylene (Teflon). heated but it retains its original form.
C- Tetrachloromethane A-Thermally unstable B-Thermally stable
D- Polyvinyl chloride C- linear D- branching
2. During the condensation polymeriza- 10. Large molecules consist of a number
tion reaction. of units the small ones are related in or-
A-The output becomes saturated ganic reactions are.
B- Often forms of water A- Monomers
C- Didn’t form by- product B- Polymers
3. Which of the following molecules is C- Functional groups
heated with sulfur atoms during the vul- D- carboxylic acids
canization process. 11. Which of the following are small units
A- Isoprene B -Polyisoprene associated to each other in a polymer dur-
C- 2-methyl 1,3Butadiene D- Butadiene ing organic chemical reactions.
4. In the vulcanization process, what hap- A- Monomers B- Micropolymers
pens to the contiguous molecules when C- branched polymers D- linear polymers
heated with sulfur atoms. 12. Small units that are associated during
A- branching organic reactions to be large molecules.
B- interrelated cross-link A- Must be similar
C- slip over each other B- must be different
D- collide C- can be similar or different
5. Any polymers of the following is added 13. Linear polymers
polymer. A- They have freely moving molecules
A- Nylon B- Polystyrene B- Have molecules that slide easily when
C- protein D- terlene heated
6. Any of the following hydrocarbon mol- C- (a) and (b) together
ecules can polymerize it. 14. The polymers are associated in cross-
A- CH4 B- C2H6 link shape
C- C3H6 D- C4H10 A- Thermally unstable
7. Biodegradable plastic. B- retains its shape when it is heated
A- It burns easily and produces toxic C- have side chains.
fumes. 15 . In an cross-link polymer, contiguous
B- can be absorbed through the roots of molecules.
plants A- linked to by chains
C- break down by bacterial decomposition. B- slide forward and back on each other

189
when heated for a condensation reaction to form a
C- regular in layers that can slip apart polymer.
when heated 5. Branched polymer needs to be fused to
16. Either of the following forms the two quantity of heat is greater than the quan-
main types for polymers? tity it needs linear polymer.
A- Thermally unstable and thermally 6. Vulcanized rubber withstand higher
fixed plastic temperatures of synthetic rubber.
B- Linear and branching polymers 7-3 Classify the following polymers into
C- Adding polymers and condensation polymers natural or synthetic .
polymers 1. cellulose 2. nylon
17. A polymer is formed during a chain 3. proteins 4. Polyisoprene
addition reaction among the monomers 5. Polypropylene
that have a bilateral bond? 7-4 Draw the structural composition of
A - polymer addition vinyl chloride.
B- condensation polymer 7-5 Write the polymerization reaction of
C- Branched polymer vinyl chloride to form polyvinyl chloride.
D- Cross-link polymer 7-6 Teflon is proven from any other poly-
18. In the addition reaction, monomers meric material, whether synthetic or natu-
associated through reaction involved ral.
A- Identical functional groups 7-7 Define the following: vulcanization,
B- Different functional groups plastics, polymers, monomers, polymers,
C- Bilateral links addition polymer.
19. A polymer usually consists of two al- 7-8 In a chemical reaction, two small
ternating monomers molecules were bound, a water molecule
A - polymers added produced what type of reaction occurred.
B- Condensation polymers 7-9 Does the reaction of addition increase
C- Branched polymers the saturation of a molecule or reduces it.
20. Linear molecules are stacked together 7-10 What are the two reactions can be
A- Cross-link polyethylene produced polymers.
B - high density polyethylene 7-11 What are the difference in Structural
C - low density polyethylene compositions of the three types of poly-
D - Neoprene ethylene HDPE, LDPE, CPE
7-2 Explain why: 7-12 What does the molecular structure
1. Natural rubber is thermally unstable of neoprene differs structure of natural
polymer. rubber.
2. Alkane is not suitable as a polymer for 7-13 Can ethanol acid be used as a mono-
the added polymer. mer in a condensation polymer. Justify
3. Cooking handles are made of a fixed your answer.
thermally polymer.
4. No single functional group molecule

190
 8
Chapter Eight
Aromatic Hydrocarbons

After completing this chapter, the student is expected to:

• Understand what’s aromatic compounds and their characteristics and identify
benzene and its composition, Draw its shape and accommodates its own properties.
• Recognize the meaning of resonance in chemical compounds and draw the
resonant shape of some aromatic compounds.
• Named aromatic derivatives of benzene.
• Distinguish between additive and substitution reactions in aromatic compounds.
• Recognize electrophilic substitution reactions in benzene and some of its
derivatives.
• Recognize phenols and their properties and some uses.
• Understand what are nonhomogeneous heterocyclic compounds and some
examples.

191
 8-1 Introduction
A group of organic compounds have been discovered
Do you know with a great deal of unsaturation, but they are unique in
that the name of aromatic
their stability and since the derivatives of these compounds
compounds is derived from
have aromatic odors called aromatic compounds or aro-
the Latin word (aroma),which
matics. This label has lost its actual meaning and gained
means fragrance or smell be-
a theoretical significance to describe the properties of that
cause these compounds have
type of unsaturated compounds.
distinctive scents. Though that
The first compound in this group is benzene that en-
is a lot of these compounds are
ables the scientist Faraday to obtain it of the first time in
toxic and some of them are con-
1825 from lighting gas. Coal tar is an important source of
sidered carcinogens as some
gasoline and large quantities is produced from petroleum
compounds produced from to-
hydrocarbons. This type includes organic compounds are
bacco smoking. Some of these
benzene and its derivatives and similar compounds in
compounds has great impor-
terms of chemical composition and effectiveness, such as
tance in the pharmaceutical in-
naphthalene and anthracene.
dustry such as aspirin and tetra-
cycline.

anthracene naphthalene benzene

8-2 Benzene composition

Benzene is the first aromatic compounds and its mol-
ecule consists of six carbon atoms linked together in the
form of a regular hexagonal ring and each carbon atom is
connected to a hydrogen atom and double bonds and sin-
gle-carbon atoms alternate between carbon atoms (compo-
sition I). Or each angle in the hexagon symbolizes a carbon
atom with a sphere combines with hydrogen and rotates
the double and single bonds (II), and can compensate for
the rotation of the double and a single by a circle inside the
loop installation(III).
Schematic representation
of a benzene molecule.

192
 8-2-1 Resonance in benzene

In 1865, the scientific Kekule proposed the two formu-

las (IV) or (V) for benzene as follows

or

It is clear that the carbon atoms in the formulas (IV and

V) occupy similar places but differ in the places assigned
to the double bonds. The exchange of these links sites is A diagram of how carbon
called resonance (Ringing). That benzene is actually not atoms bonded in a benzene
a reciprocal state between the two states of resonance IV molecule.
and V are not part of the molecules in the form of reso-
nance and another part of the molecules in the form of the
other resonance. But the real form of benzene is modified
Do you know
for the two shapes is called the resonance hybrid. Where
The C6H6 benzene, we are
we observe it’s in place there is a double bond in one of
studying is completely different
the shapes, it will be single in other shape. When the aver-
from gasoline used in internal
age is taken, we get a hybrid form of resonance containing
combustion engines. The latter
of six carbon -carbon bonds, which are identical and of
is a mixture of 5 to 10 carbon
equal length as the average case between the length of the
hydrocarbons.
single bond and the length of the double bond.
Therefore, the shape of benzene is usually painted in the
form of a hexagonal ring, it has an inner loop (VI and VII
composition) rather than in the form of successive double
and double aces (IV and V composition) because it de-
scribes the situation is more accurate. However, drawing
in the form of double bonds and individual successive is
better in the case of monitoring the movement of electrons
as in reaction mechanism state.

or

In brief, benzene is a molecular formula C6H6 contain-

ing six identical carbon atoms and six hydrogen atoms.
The C-C bonds have equal lengths and it is in a median
case between the length of the single bond and the length
of the double bond.
193
 8-2-2 Special characteristics of benzene

1. Stability
Orbitals
It is a stable compound as indicated by its reactive proper-
ties and relative resistance to chemical changes. That most
unsaturated compounds tend to engage in addition reac-
tions in which binary or triple bonds are saturated into in-
dividual bonds. Note that cyclohexane reacts easily with
the bromine dissolved in carbon tetrachloride, forming
1,2-Dibromocyclohexane, whereas benzene does not fully
react under the same conditions.

π bond

In order for benzene to react with bromine, a catalyst

(FeBr3) must be used to enter a substitution reaction (sub-
σ bond
stitution) and not an addition interaction where the loop
remains preserved in its shape and this is evidence
on the stability of benzene.
An electronic cloud for a
π bond in a benzene mol-
ecule.

2. Resonance Energy
To understand the energy of the resonance, we explain
the following:
When a chemical reaction occurs, a change in thermal
energy. For example, cyclohexane hydrogenation is a heat-
emitting reaction, releasing 120 kJ energy per mole of cy-
clohexane.

+Heat (120kJ/mole)
Cyclohexene cyclohexane
At the hydrogenation of 1,3 cyclohexane (cyclohexane
containing two π-bonds) we expect the amount of energy
to be released 240 kJ per mole, twice the liberated energy
at the hydrogenation of cyclohexene because there is twice
the number of double bonds in the cyclohexene.The real
energy measured for this reaction was 232 kJ Per mole, it
is very close to expected.
194
 According to the same logic, we expect that the energy
released by hydrogenation of gasoline is three times the
liberated energy at hydrogenation of cyclohexane, ie 360
kJ per mole. The actual energy of this has been found. The
reaction is 208 kJ per mole and is very different to our ex-
pectation.

The difference between the expected energy value and

the real energy value is 152 kJ per mole and this amount
of energy is called a resonance energy. This means that
benzene contains 152 kJ per mole of energy is less than
expected. So that gasoline is more stable than expected by
152 kJ per mole.
3. Carbon-Carbon Bonds Length
Physical measurements showed that the lengths of car-
bon-carbon bonds in benzene are equal and are of average
length between the length of a single C-C and double C = C.
C-C C=C C-C
Exercise 8-1
Single double for benzene
1.54A o
1.34A o
1.40Ao Type the following com-
8-2-3 Benzene Derivatives Names pounds names as derivatives
There are some systems in the label as follows: of benzene.
1.1 Monocompensation Components: These compounds
are called as derivatives of benzene such as:

Ethylbenzene bromobenzene chlorobenzene Nitrobenzene

There are some commonly accepted compounds such
as:

methylbenzene hydroxybenzene amine benzene

(Toluene) (Phenol) (Aniline)
195
 2. Dual-compensation Compounds:
When there are two groups on the benzene ring, it is
not enough to mention the names of the two groups only,
but they must be located on the benzene ring. Where the
carbon atoms are numbered for the benzene ring and the
Exercise 8-2 two groups take the smallest figures, for example, 1,2 Di-
bromobenzene is:
Write the name of
components.

1,2 dibromobenzene
The terms ortho may be used to denote the site 2 and
(meta) to denote site 3 or (para) to denote on site 4 for the
main compensator) X
ortho ortho

meta meta
para
If the two groups are different, they are mentioned in
the nomenclature according to the alphabets one after the
other and followed by the word benzene with its location
at the beginning of the name or called the compound as a
derivative

1,4 chloronitrobenzene
para nitrochlorobenzene
para chloronitobenzene

Where we notice in the second label we considered that

the molecule of chlorobenzene is a benzene monocompen-
sation by chlorine. In the third label it was considered that
the benzene mono-compensation in nitro group, so it is
the basis in the designation. Dual-compensation benzene
compounds are named in this way and as in the following
examples:

1,3 Nitro hydroxybenzene 1,2 Bromo iodide benzene

Meta-nitrophenol Ortho iodide bromo benzene
Meta-hydroxy nitrobenzene Ortho bromo iodide benzene
3

196
3. Multiple Compensation Compounds:
When there are more than two groups on the benzene
ring in which case it should be numbered the locations of
these groups, where the name Ortho, Barra and Meta are
unacceptable. These groups are named as a derivative of
benzene or as a derivative of common names if any. If the
associated group are large, the compound is named in the
general designation of hydrocarbons. It takes the longest
hydrocarbon chain as the base of the name and benzene is
named as a compensating group, where it is called Phenyl.
If benzene is replaced by a group, it is called Aryl.
We will be satisfied with this stage by naming mono and
dual-compensation compounds. We postpone the designa-
tion of multi-compensation compounds to advanced levels.

8-2-4 Preparation of benzene

A-In laboratory
Benzene is prepared in laboratory by heating sodium ben-
zoate with sodium hydroxide (NaOH) in a glass distilla-
tion device where, benzene is obtained from the distillate
o
at 80 C.

B- Industrially:
Benzene industrially is prepared in several ways, the most
important of which are:
1. From phenol: Phenol is heated with Zn-dust in a distilla-
o
tion apparatus and then benzene is obtained from an 80 C
distillate.

2. From benzene sulfonic acid: benzene sulfonic acid is

heated with dilute hydrochloric acid or dilute sulfuric acid
to boiling point and under high pressure according to the
following equation:

197
 8-2-5 Properties of benzene
1. Physical properties
A- Colorless liquid, flammable, has a specific aromatic
smell and it is toxic.
B - boiling point 80 °C and freezing point 5 °C.
C - Its density is less than the density of water and didn’t
mixed with it.
D - a good solvent for non-polar organic materials such as
grease, oils, resins and others.
2. Chemical properties
Benzene is a relatively stable chemical compound com-
pared to unsaturated compounds for the presence of reso-
nance phenomenon.
It is not affected by concentrated bases or concentrated
hydrochloric acid nor by strong oxidizing agents such as
potassium permanganate but suffers from a number of re-
actions, such as combustion and addition and substitution.
A- Combustion: Benzene is burned with a bright and flame
due to the ratio of its high carbon content 92.3% and gives
carbon dioxide and water with heat liberation.

B- Additive reactions: Chlorine is added to benzene in the

presence of light and there is a reaction accompanied by
popping the reactants leading to the formation Hexachlo-
rocyclohexane as described in equation A:

Burning of benzene with sun light

a bright and flame.

Hexachlorocyclohexane

Benzene is also reduced to hydrogen at high temperatures

and under high pressure in the presence of a catalyst such
as platinum to cyclohexane, as in the following equation:

cyclohexane

198
C-Substitution reactions (compensation): One of the most
important reactions of benzene and its derivatives is the
possibility of replacing (compensation) one of the hydro-
gen atoms by atom or different group (e.g. alkyl group R,
nitro group NO2, sulfonic group SO3H, halide X or ace-
O
tyl group -C-CH in the presence of an appropriate catalyst,
-

3
which is help for reaction. Examples of substitution reac-
tions is:
A-Halogenation
It is the process of replacing one of the hydrogen atoms
with a halogen atom (such as chlorine Cl2 or Br2) with a
catalyst such as ferric chloride FeCl3 or ferric bromide
FeBr3 :

Chlorbenzene

B- Sulfonation Bromobenzene
It is the process of replacing one of the hydrogen atoms
with the sulfonic group SO3H, for example benzene reacts
with the dark concentrated sulfuric acid at room tempera-
ture to compose benzene sulfonic acid.

dark concentrated
sulfuric acid benzene sulfonic acid

C- Nitration
Replace a hydrogen atom with a nitro group NO2 where
benzene reacts with a mixture of concentrated nitric and
sulfuric acid at a temperature of 45 °C to compose of ni-
trobenzene.

Nitrobenzene
199
 D- Friedle- Crafts reactions
1. Friedel-crafts Al-kylation: It is the process of replacing
one hydrogen atom in the alkyl group (R) with an appropri-
ate catalyst. When benzene reacts with alkyl halide (R-X)
in presence of catalyst like dry aluminum chloride (AlCl3)
lead to compose of alkyl benzene

dry

alkyl benzene
Example 8-1
Type the Friedel-crafts Al-kylation for benzene using
methyl chloride (CH3Cl) in the presence of dry alumi-
num chloride as a catalyst and named the produced com-
pound.
Solution Benzene reacts with CH3Cl in the presence of
dry AlCl3 according to the following reaction:

dry
toluene
2. Friedel-Kraft acetylation: is the process of replacing an
o
atom hydrogen in the acetyl group (-R-C). Benzene reacts
o
with acetylcholine (R-C-Cl) in the presence of dry aluminum
chloride to compose of acetyl benzene.

Example 8-2
Type the Friedel-kraft acetylation of benzene and name
the produced compound.
Solution Benzene reacts with in the presence of dry
AlCl3 according to the following reaction:

acetylene chloride
acetophonon
200
 8-3 Mechanism of electrophilic substitution reactions
The substitution reactions above are called electrophilic
substitution reactions. Where there are electrons detectors
is also called electrophilic reagent (E). The electron elec-
trophilic reagent is a reagent that needs electrons and can
be in the form of a positive charge that can form covalent
bonds with carbon atoms like X+, NO2+,R+, RC+O.
A double link is rich in electrons therefore, it is a source
of electrons for the reagent that electrophilic reagents need
it. Electron-rich reagent are called Neocluphilic reagents
(Nu)(see chemistry book for the fourth stage). By review-
ing these reactions we can symbolize all these reactions by
the following general reaction:
E

Catalyst
+ E-Nu + H-Nu

These include three steps:

A-In the first step, the catalyst converts the reactant into an
electron reagent (E+) according to the equation:

E-Nu + Catalyst E+ + Nu-Catalyst

B- In the second step, one of the double bonds of the re-

agent benzene ring attacks the electron, forming a positive
carbonium ion.
H E H
+

+E+

carbonium ion
C- In the third step, a proton (positive hydrogen atom H+)
is withdrawn by Nu-Catalyst. The output from the first step
forms the other output H-Nu while the benzene ring re-
turns to the aromatic state.
E H E

+
+ Nu-Catalyst + H-Nu + Catalyst

This mechanism, known as electrophilic compensation

mechanism is a complex subject which we believe is above

201
 the level of student understanding at this stage. So we were
limited to describe general mechanical only to explain how
Do you know the reaction occurs in a briefly method, these mechanics
That photography is black and are studied in detail in advanced stages.
white, it mainly depends on the
oxidation of a phenol. It is hy- 8-4 Phenols
droquinone by silver bromide Aromatic organic compounds derived from benzene, which
with a photosensitive reaction. are formed the hydroxyl group (OH) is directly linked to
Since the film contains small the benzene ring like :-
granules of silver bromide, the
OH
light-grained particles are active OH OH
CH3
and then the film is treated with
hydroquinone solution which
is called the looking solution. OH

Where the activated granules are 1,4 dihydroxylbenzene Ortho methylphenol Phenol
reduced from silver bromide, It is a material with wide application, it has wide pharma-
leaving black silver deposits in cological properties such as its ability to act as an anti-
light-exposed places. As a re- mildew, antibacterial and anesthetic or topical analgesic.
sult, we get the negative image That’s why they are used in spray products ointments,
with dark areas in places hit by antiseptics, soaps, first aid materials and gargle solutions.
light. They are also used as antioxidants as they are added to
most foods and cosmetics, because of the easy oxidation
of phenols. They are used in the developer in photography
black and white kind.
Some phenols are found in natural sources such as for
example, vaniline gives vanilla, which is used as a food
moisturizer. OH
OCH3

vaniline

C O
H
and Eugenol compound which is extracted from clove oil.
OH
OCH3

Eugenol

CH2CH=CH2
and Thymol which is extracted from peppermint plant
OH
CH CH3
Thymol CH3

CH3

202
Some plants contain phenols consisting of three rings,
these compounds are called flavanoids. They are antioxi-
dant and extracted Do you know
of green tea. That there are bacteria in the soil
8-4-1 Properties of phenols feed on aromatic compounds
Physical properties: - The hydroxyl OH group found in where you turn them first to
phenols has a significant effect in determining physical phenols and then it breaks the
properties, due to the formation of hydrogen bonds be- aromatic ring and breaks the
tween their molecules. The high melting point and boiling resulting compounds into wa-
point of these molecules, which are higher than the melt- ter and carbon dioxide. These
ing point and boiling point of alcohols (ROH), phenols are bacteria abound in the soil on
easily dissolved in water due to the formation of hydrogen the roadsides where they feed on
bonds with water molecules . compounds that fall from their
Hydrogen bond O H..................... O H
carrier vehicles and pedestrians
on these roads.

8-4-2 Naming of phenolic compounds

According to the International Union of Pure and Applied Exercise 8-3
Chemistry (IUPAC) system, whose abbreviation (IUPAC), Type the names of the vehicles
phenols are named by addition section (ol) to the name of below by the system of nam-
the mother hydrocarbons such as: ing IUPAC time and common
OH OH names one more time
OH OH

CH3
NO2

3-methyl benzeneol Benzeneol Benzene Cl

Phenols are rarely called in this way, but they are named in Br
OH OH

common names, which is accepted by IUPAC. Where the

name came from the old name of benzene is (phene) where CH2CH3

the syllable is added (first ol) so the main name is Phenol.

The rest of the compounds are called phenol derivatives
such as:
OH OH OH
Br

CH3

2- Bromo Phenol 3-Methylphenol Phenol

or ortho bromo phenol or meta methylphenol

203
 8-4-3 Phenolic acid
Phenols are organic acids and their acidic strength is com-
pared by measuring their ionization in water. Acids that
Do you know are fully ionized with water, such as HCl and HNO3, are
Phenol is a common name that called strong acids. The weak acids are partly ionized in
is carbolic acid and phenol more the aqueous solution where there is equilibrium between
acidic than water and causes irri- the ionizing state and the non ionizing state as in the fol-
tation or stirring the skin strong- lowing equation (see chapter four of this book):
↼
ly. HA + H2O ↼ H3O+ + A-
Most organic acids are weak acids with little ionization in
water,
phenol is one of them and is ionized in water according to
the following formula:
OH O-

↼ +
+ H2O ↼ + H3O

8-4-4 Phenol
Phenol was separated in the 19th century from coal tar.
Phenol has been used in most countries of the world as a
feedstock for the preparation of aspirin, dyes, writing ink,
glue, plastic industries and most commercial phenols can
be prepared in the laboratory or in the industry.
1-Phenol preparation
A- From Benzene sulfonic acid
This method is an old method used in the industry to pre-
pare phenol where benzene can be sulfonized first to ob-
tain benzene sulfonic treated with sodium hydroxide solu-
tion (NaOH) and then make acid solution We get phenol as
in the following equation:
SO3H OH

1) NaOH , 300°C
+ H2SO4 2) H+ , H2O

B) From heating chlorobenzene with sodium hydroxide

solution under high pressure and then making the solution
acidic we get phenol.
Cl O-Na+ OH

NaOH H+ , H2O
350°C , 150 atm

204
 8-5 Preparation of aspirin
Aspirin is prepared industrially in laboratory by the re-
action of salicyclic acid with organic acid, which is an-
hydrous aceticacid to produce the corresponding ester, ,
which is acetylsalicylic acid, which is known commercial-
ly as aspirin.
O
OH O O C CH3
C OH O O C OH
H+ Ejector beetle
C O C CH3 ↼
O
+ CH3 ↼ + CH3 COOH
Do you know
salicylic acid anhydrous acetic Aspirin
acid That there is a kind of ladybug,
which is called ejector beetle.
8-6 Detection of phenols When the insect is threatened,
For the detection of phenols, iron salts (III) are added to
the ladybug defends itself with
their solutions, the solution color is dark blue or dark
a hot, irritating solution of qui-
green depending on the chemical composition of phenol.
none (a phenol) on its enemies.
When iron potassium cyanide solution (III) is added to
The beetle mixes hydroquinone
their solutions, the solution is colored in red.
with hydrogen peroxide with
a special enzyme, during that
8-7 Heterocyclic compounds hydroquinone oxide to qui-
8-7-1 Introduction
none and this reaction emits
A heterocyclic compound is a compound in which one is
high heat lead to boiling solution
atoms of the ring or more are carbon, and most common
and strongly emitted from the
species contain on nitrogen, oxygen, or sulfur atoms in ad-
insect›s abdomen.
dition to carbon which makes up the high percentage. Het-
erocyclic compounds can be aliphatic or aromatic in nature
depending on electronic structure.
The heterocyclic aliphatic compounds are chemically sim-
ilar to a large degree with their corresponding compounds
open-chain aliphatic, as the cyclic heterogeneous aromatic
compounds have a similarities with their corresponding of Furan VIII
aromatic carbonate compounds.
Heterocyclic compounds are widespread in nature is es-
sential for life in multiple shapes. Most sugars and their
derivatives, including vitamin C (for example), are present
often in the form of heterocyclic compounds either pen-
tagonal like Furan (composition VIII) or hexagonal such Piran
as Piran (composition IX) where the ring contains one
atom of oxygen. Most alkaloids, nitrogenous bases found
in plants
205
 and many antibiotics, including penicillins, contain hetero-
geneous cyclic systems. It was found that Priden (compo-
sition X) a heterocyclic hexagonal compound containing
one nitrogen atom that is part of the nicotine structure. It
found pyrol (composition XI) is a pentagonal compound
X Priden
containing one nitrogen atom that is a unit of hemoglobin
and chlorophyll. There are also a large number of hetero-
cyclic compounds that it can be obtained through labora-
tory preparations having valuable qualities as therapeutic
chemical compounds and as drugs, dyes and copolymers.

8-7-2 Naming of heterocyclic compounds

XI Pyrol
There are many ways to number the atoms and compensat-
ed groups in the heterocyclic compounds. In monocyclic
compounds the atoms are numbered starting from hetero-
geneous atom which then carries the number -1- and then
revolves around the ring so that the compensating groups
in this ring give the lowest numbered and then arrange
them according to the order of the letters thealphabet for
XII 2-chloro - 3 methylpred-
example compound number(XII) named 2-chloer-3-meth-
ine
yl predine

8-7-3 Type of heterocyclic compounds

To facilitate the study of this type of compounds were di-
vided into the following:
1. Heterogeneous triple ring compounds:
XIII Ethylene Epoxide
They are the cyclic compounds corresponding to the cyclic
propane, but they contain one differ atom. Examples of
these compounds are epoxides such as composition for-
mula (XIII).
2. Heterogeneous quadrilateral compounds:
These are the heterocyclic compounds corresponding to
the cyclic butane. These compounds are more stable than
XIV oxctane
triple-ring compounds, such as the composition formula
oxctane (XIV).

206
3. Heterocyclic five-ring compounds:
They are the heterocyclic compounds corresponding to cy-
clopentadiene. The most common examples of these com-
pounds are thiophene (XV) and furan mentioned earlier
(VIII). Furan is a colorless liquid, boiling at 31 °C. It has a
chloroform-like odor and dissolves scarcely in water and
its completely soluble in organic solvents. In 1877, Bayer
proved furan structure and composition. It is best to look
at the furan molecule as a resonance hybrid because of the
not positional of the oxygen atom electron pair where the
a to d hypothetical structures form, where a and b are the XV Thiophene
most important.
- -
- -
O+ O+ O+ O+
)a) (b) (c) (d(

4. Heterocyclic hexagonal compounds:

Which are the heterocyclic compounds corresponding to VIII furan
benzene,
Predine as an examples with the composition formula (X)
mentioned earlier. These compounds also have aromatic
properties so it high stability characteristic.
Predine has heterogeneous system with hexagonal ring re-
sembles to benzene in its composition and stability. Its
like gasoline, it is resistant to oxidation and is often used
as a solvent for many chemical reactions. The stability of X predine
the installation is due to the following hybrids resonance:

+ +
::
::

::

N N N N N
- - -
:

It is a weak base because nitrogen contains two unshared

electrons that greatly affect its chemical properties, intro-
ducing more compensating reactions than additive reac-
tions.

207
 Basic concepts

Aromatic Hydrocarbons pounds case with multiple compensation

It is a group of organic compounds on should be used numbering method where
some degree of unsaturation is only unique the term ortho and meta parra is unaccept-
in its stability, therefore these compounds able.
tend to engage in substitution (compen- Electrophilic Substitution
sate) reactions where the aromatic ring It is the process of replacing one of the hy-
remains preserving their form and not ad- drogen atoms in the ring in atom or differ-
ditional reactions as in most unsaturated ent group such as halogenation, sulfona-
compounds. This type of organic com- tion, Nitration, alkylation , and acylation
pounds includes benzene and its deriva- , and is made by presence reagent looking
tives and similar compounds in terms of for electrons (electrophilic reagent), espe-
composition and chemical reactivity. cially in each process.
Resonance Phenols
It is a phenomenon of the lack of concen- Aromatic organic compounds derived
tration of electrons around nuclei of spe- from benzene it consists of the bonding
cific atoms in the molecule or ion making of the hydroxyl group to benzene ring
them more stable. In the case of benzene directly. The presence of the hydroxyl
and in order to justify strictly the fact the group has a significant effect in determin-
composition formula of the benzene mol- ing its physical properties due to the for-
ecule, so that the bonds of the -bonds mation of hydrogen bonds between their
appear in the double carbon atoms like molecules, which causes the high degree
they are in continuous motion, we draw of boiling point and melting point.
the hexagonal form of benzene and in- Heterocyclic Compounds
side it moving electrons in the form of A heterocyclic compound is a compound
sequential points or closed loop instead in which one or more atoms of the ring
of the double elements. That is, we make are not carbon may be an atom of nitrogen
the electrons of these bonds the property or oxygen or sulfur. It can be a heteroge-
of all six carbon atoms, this type of elec- neous ring triple, quadruple, pentagonal
tronic motion so-called resonance. or hexagonal.
Benzene Derivatives Names Benzene
Monocomponent compounds are called as Benzene is a molecular formula with six
derivatives of benzene or called by their identical carbon atoms and six identical
common names. In the case of derivatives hydrogen atoms, and the lengths of the
of dual compensation must be located on C-C are equal. It is an intermediate state
the ring either numbering where the two between the length of the single bond and
groups take the smallest a numbers and the double bond.
then mention the names of the two groups
by alphabets Ortho, Meta and Barra to in-
dicate the location of the groups. In com-

208
 Chapter eight Questions 8
8-1 How did the concept of aromatic 8-11 Draw the formulas for the following
changed? compounds:
8-2 What is the source of aromatic hy- 1. Para-dichlorobenzene
drocarbons? 2. 1, 3 - difluorobenzene
8-3 Why is the composition of benzene a 3. 1, 2 - iodochlorobenzene
resonance hybrid? 4. Para bromochlorobenzene
8-4 When benzene behaves as a non-sat- 5. Meta-bromo sulfonic acid
urated hydrocarbon? 6. Para nitrophenol
8-5 How can you distinguish of benzene 8-12 Complete the following equations?
in its reactions from alkanes and alkenes?
What caused the relative stability of its + H2SO4
molecule?
8-6 What are the possible isomers of ni-
trochlorobenzene? AlCl3
8-7 How do you distinguish between cy- + CH3CH2COCl
clohexane and benzene?
8-8 Write the chemical equations for the
reaction of benzene with : + CH3Cl
AlCl3

A- Cl2/FeCl3
B- CH3CH2Cl/AlCl3 SO3H
C- CH3COCl/AlCl3 1) NaOH , 300°C
8-9 Draw Resonance shape for the fol- 2) H+ , H O
2

lowing ::
:

O
Cl

NaOH
+ + 350°C , 150 atm

8-10 Named the following components O-Na+

Br
Br Cl I H+,H2O
Cl

OH
Br
Cl ↼
↼
F OH OH CH3 + H2O
CH3 Cl

Br
H2SO4
OH
OH + HNO3
NO2

CH2CH3

209
8-13 What are the Industrially methods ing an Hydrogen atom in benzene by sul-
for preparing benzene? Write them in de- fonic group .SO3H
tail? 7. Reagents rich in electrons are called re-
8-14 Write the composition formulas of agents looking for electrons.
the following compounds 8. The Furan molecule is made of five
1. 1, 3, 5 Tri-Bromo Benzene. carbon atoms bonded to form a pentago-
2. Meta - Chloro Toluene. nal ring.
3. Ethyl-benzene. 9. Benzene Molecular formula C6H6 con-
4. Para - dichlorobenzene. tains of six bonds three of which are sin-
8-15 Expressed in composition formulas gle and the other three being double.
the reaction of benzene with ethyl chlo- 10. The ester resulting from the reaction
ride? of salicylic acid and acetic acid is com-
8-16 By chemical equation and composi- mercially called aspirin.
tion formulas, write the product reduc- 11. Phenols are detected by mercury salt
tion process of benzene by hydrogen and solutions.
with the presence of platinum? 12. Predine is a heterocyclic compound
8-17 Write the reaction of chlorine addi- that contains one oxygen.
tion to benzene in presence of light ? 13. In nonhomogeneous heterocyclic
8-18 Draw of the following compounds: compounds the numbering begins starting
A- Meta-fluoro-chlorobenzene with the heterocyclic atom which is given
B- 2 - Bromo - 4 - Chloropridine the number1.
C- Ortho Ethylphenol 14. The heterocyclic triple-ring com-
8-19 Place a ( ) in front of the correct pounds are the corresponding cyclic com-
statement and ( ) in front of the wrong pounds of Butane.
statement: 8-20 What are nonhomogeneous hetero-
1. Aromatic compounds are organic com- cyclic compounds and how are they di-
pounds with high nonsaturation. vided?
2. The compensation reactions introduced 8-21 What are the physical properties of
by benzene are evidence on its high sta- furan and what are resonance shaps for it?
bility. 8-22 What is Predine and what are its res-
3. The terms Ortho, Meta and Para are onance shaps?
used in the designation of binary and mul- 8-23 When did phenol first isolated and
tiple benzene derivatives. what are its uses.
4. Benzene burns with a bright and smoky 8-24 write down the general electrophilic
flame due high carbon content. substitution reaction and what are its three
5. Benzene is easily reduced by hydrogen steps.
to cyclohexane.
6. Halogenation is the process of replac-

210
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

‫‪Cathod ray‬‬ ‫ا�شعة الكاثود‬
‫‪Electromaganetic ray‬‬ ‫ا�شعة كهرومغناطي�سية‬
‫‪Metalloids‬‬ ‫ا�شباه الفلزات‬
‫‪Ionic bond‬‬ ‫�آ�صرة ايونية‬
‫‪Covalent bond‬‬ ‫�آ�صرة ت�ساهمية‬
‫‪Polar covalent bond‬‬ ‫�آ�صرة ت�ساهمية م�ستقطبة‬
‫‪Single bond‬‬ ‫�آ�صرة منفردة (احادية)‬
‫‪Double bond‬‬ ‫�آ�صرة مزدوجة‬
‫‪Triple bond‬‬ ‫�آ�صرة ثالثية‬
‫‪Coordinate bond‬‬ ‫�آ�صرة تنا�سقية‬
‫‪Metalic bond‬‬ ‫�آ�صرة فلزية‬
‫‪Hydrogen bond‬‬ ‫�آ�صرة هيدروجينية‬
‫‪Quntaum numbers‬‬ ‫اعداد الكم‬
‫‪Electron‬‬ ‫الكترون‬
‫‪Valence electrons‬‬ ‫الكترونات التكاف�ؤ‬
‫‪Resonance‬‬ ‫الرنين(ظاهرة الروزنان�س )‬
‫‪Electron affimty‬‬ ‫الفة الكترونية‬
‫‪Nonmetal‬‬ ‫الالفلزات‬
‫‪Soulblity‬‬ ‫الذوبانية‬
‫‪Overall rate of reaction‬‬ ‫ال�سرعة العامة للتفاعل‬
‫‪s - Orbital‬‬ ‫اوربيتال ‪s‬‬
‫‪p - Orbital‬‬ ‫اوربيتال ‪p‬‬
‫‪d - Orbital‬‬ ‫اوربيتال ‪d‬‬
‫‪f - Orbital‬‬ ‫اوربيتال ‪f‬‬
‫‪Ion‬‬ ‫ايون‬
‫‪Hydronium ion‬‬ ‫�آيون الهيدرونيوم‬
‫‪Paramagnatic‬‬ ‫بارا مغناطي�سية‬
‫‪Proton‬‬ ‫بروتون‬
‫‪Polymer‬‬ ‫بوليمر‬

‫‪211‬‬
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

Plastics ) ‫بال�ستيكات (لدائن‬
Addition Polymers ‫بوليمرات الأ�ضافة‬
Condensation polymers ‫بوليمرات التكاثف‬
Thermoplastic polymers ‫بوليمرات مطاوعة للحرارة‬
Thermoset polymors ‫بوليمرات غير مطاوعة للحرارة‬
Synthetic polymers ‫بوليمرات م�صنعة‬
Chemical bonding ‫ت�آ�صر كيميائي‬
Hydrolysis ‫تحلل مائي‬
Concentration ‫تركيز‬
Molar concentration M ‫تركيز موالري‬
Molal concentration m ‫تركيز مواللي‬
Effective collision ‫ت�صادم فعال‬
Noneffective collision ‫ت�صادم غير فعال‬
Nutrlization ‫تعادل‬
Hyperdiazation ‫تهجين‬
Orbital hyperdiazation ‫تهجين اوربتالي‬
Frequency ‫تردد‬
Electron configuration ‫ترتيب الكتروني‬
Titration ‫ت�سحيح‬
Chemical reaction ‫تفاعل كيميائي‬
Pollution ‫تلوث‬
Rate constant ‫ثابت �سرعة التفاعل‬
Plank constant ‫ثابت بالنك‬
Perodic Table ‫جدول دوري‬
Acid ‫حام�ض‬
Conjugated acid ‫حام�ض قرين‬
Excited state ‫حالة م�ستثارة‬
oxidiation state ‫حالة الت�أك�سد‬
Atomic volume ‫حجم ذري‬
212
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

Heat of solution ‫حرارة المحلول‬
Chemical kinetics ‫حركيات كيميائية‬
Linear ‫خطي‬
Golligative properties ‫خوا�ص جمعية‬
Diamagnatic ‫دايا مغناطي�سية‬
Temperature ‫درجة الحرارة‬
Boiling point ‫درجة حراة الغليان‬
Freezing point ‫درجة حرارة االنجماد‬
Indicaters ‫دالئل‬
Lewis symbol ‫رمز لوي�س‬
Burret ‫�سحاحة‬
Rate of reaction ‫�سرعة التفاعل‬
Light speed ‫�سرعة ال�ضوء‬
Crystal lattice ‫�شبكية بلورية‬
Tetrahedral ‫�شكل رباعي االوجه منتظم‬
Trigonal ‫�شكل مثلثي منتظم‬
Pressure ‫�ضغط‬
Partial pressure ‫�ضغط جزئي‬
Total pressure ‫�ضغط كلي‬
Vapour pressure ‫�ضغط بخاري‬
Osmotis pressure ‫�ضغط ازموزي‬
Activation energy ‫طاقة التن�شيط‬
Resonance energy ‫طاقة الرنين‬
Ionization energy ‫طاقة التاين‬
Wave length ‫طول الموجة‬
Flame spectrum ‫طيف اللهب‬
Line emisson spectrum ‫طيف االنبعاث الخطي‬
Nucleuphilic reagent ‫عامل باحث عن النواة‬
Electrolphilic reagent ‫عامل باحث عن الألكترونات‬
213
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

d - Block d ‫مجموعة‬
s - Block s ‫مجموعة‬
f -Block f ‫مجموعة‬
Solutions ‫محاليل‬
Aqueous solution ‫محاليل مائية‬
Nonaqueous solution ‫محاليل غير مائية‬
Colliodal solution ‫محاليل غروية‬
Saturated solution ‫محلول م�شبع‬
Unsaturated solution ‫محلول غير م�شبع‬
Supesaturated solution ‫محلول فوق الم�شبع‬
Ideal solution ‫محلول مثالي‬
Standard solution ‫محلول قيا�سي‬
Solute ‫مذاب‬
Solvent ‫مذيب‬
Reaction order ‫مرتبة التفاعل‬
Intermediate ‫مركب و�سطي‬
Hetrocyclic compouds ‫مركبات حلقية غير متجان�سة‬
Ground state )‫م�ستوى ار�ضي (م�ستقر‬
Rubber ‫مطاط‬
Coordination complexes ‫معقدات تنا�سقية‬
Activated-Complex ‫معقد من�شط‬
Salt ‫ملح‬
Amphoteric substances ‫مواد �أمفوتيرية‬
Molarity M ‫موالرية‬
Molality m ‫مواللية‬
Monomer ‫مونمر‬
Quntaum mechanics ‫مكانيكيا كمية‬
Reaction Mechanism ‫ميكانيكية التفاعل‬
Ionic radius ‫ن�صف قطر االيون‬
214
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

Catalyst ‫عامل م�ساعد‬
Atomic number ‫عدد ذري‬
Atomic mass ‫عدد الكتلة‬
Principal quantum number ‫عدد الكم الرئي�س‬
Magnetic quantum number ‫عدد الكم المغناطي�سي‬
Spin quantum number ‫عدد الكم المغزلي‬
Momentum quantum number )‫عدد الكم الثانوي(الزخمي‬
Polymerization ‫عملية البلمرة‬
Element ‫عن�صر‬
Transition element ‫عنا�صر انتقالية‬
Inner Transition element ‫عنا�صر انتقالية داخلية‬
Rare -earth elements ‫عنا�صر االتربه النادرة‬
Coordinated number ‫عدد تنا�سقي‬
Alkalin -earth elements ‫عنا�صر االتربة القلوية‬
Metal ‫فلز‬
Vulcanization ‫فلكنة‬
Photon ‫فوتون‬
Ferromagnatic ‫فيرومغناطي�سية‬
Base ‫قاعدة‬
Hund،s rule ‫قاعدة هوند‬
Pauli- excluded principle ‫قاعدة اال�ستثناء الباوي‬
Octate rule ‫قاعدة الثمان‬
Congugated base ‫قاعدة قرينة‬
Rate law ‫قانون �سرعة التفاعل‬
Raoult,s law ‫قانون را�ؤلت‬
Electron mass ‫كتلة االلكترون‬
Atomic mass ‫كتلة ذرية‬
Mole fraction x ‫ك�سر مولي‬
Electronegatvity ‫كهر�سلبية‬
215
 ‫امل�صطلحات‬

‫امل�صطلح باللغة االنكليزية‬ ‫امل�صطلح باللغة العربية‬

‫‪Collision Theory‬‬ ‫نظرية الت�صادم‬
‫‪Transitsition- state theory‬‬ ‫نظرية الحالة االنتقالية‬
‫‪Activated- complex theory‬‬ ‫نظرية المعقد المن�شط‬
‫‪Arrhenius Theory‬‬ ‫نظرية ارينو�س‬
‫‪Bronsted lowry theory‬‬ ‫نظرية برون�شتد ‪ -‬لوري‬
‫‪Lewis theory‬‬ ‫نظرية لوي�س‬
‫‪Quantum theory‬‬ ‫نظرية الكم‬
‫‪Valence - bond theory‬‬ ‫نظرية ا�صرة التكاف�ؤ‬
‫‪Valence shell electron pair repulsion theory‬‬ ‫نظرية تنافر وازدواج الكترونات غالف التكاف�ؤ‬
‫‪Equivalent point‬‬ ‫نقطة التكاف ؤ�‬
‫‪End point‬‬ ‫نقطة الأنتهاء‬
‫‪Neutron‬‬ ‫نيوترون‬
‫‪Armatic hydrocarbon‬‬ ‫هيدروكاربونات اروماتية‬
‫‪Rate-determing step‬‬ ‫الخطوة المحددة لل�سرعة‬
‫‪Electron charge‬‬ ‫�شحنة االلكترون‬
‫‪Dilution Law‬‬ ‫‪ ( V1 M1 = V2 M2‬قانون التخفيف )‬

‫‪216‬‬
 Element Eiectronic Configurations
sublevels
Elements
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p
1-Hydrogen 1
2-Helium 2
3-Lithium 2 1
4-Berylium 2 2
5-Boron 2 2 1
6-Carbon 2 2 2
7-Nitrogen 2 2 3
8-Oxygen 2 2 4
9-Fluorine 2 2 5
10-Neon 2 2 6
11-Sodium 2 2 6 1
12-Magnesium 2 2 6 2
13-Aluminum 2 2 6 2 1
14-Silicon 2 2 6 2 2
15-Phosphorus 2 2 6 2 3
16-Sulfur 2 2 6 2 4
17-Chlorine 2 2 6 2 5
18-Argon 2 2 6 2 6
19-Potassium 2 2 6 2 6 1
20-Calcium 2 2 6 2 6 2
21-Scandium 2 2 6 2 6 1 2
22-Titanium 2 2 6 2 6 2 2
23-Vanadium 2 2 6 2 6 3 2
24-Chromium 2 2 6 2 6 5 1
25-Manganese 2 2 6 2 6 5 2
26-Iron 2 2 6 2 6 6 2
27-Cobalt 2 2 6 2 6 7 2
28-Nickle 2 2 6 2 6 8 2
29-Copper 2 2 6 2 6 10 1
30-Zinc 2 2 6 2 6 10 2
31-Gallium 2 2 6 2 6 10 2 1
32-Germanium 2 2 6 2 6 10 2 2
33-Arsenic 2 2 6 2 6 10 2 3
34-Selenium 2 2 6 2 6 10 2 4
35-Bromium 2 2 6 2 6 10 2 5
36-Krypton 2 2 6 2 6 10 2 6
37-Rubidium 2 2 6 2 6 10 2 6 1
38-Strontium 2 2 6 2 6 10 2 6 2
39-Yttrium 2 2 6 2 6 10 2 6 1 2
40-Zirconium 2 2 6 2 6 10 2 6 2 2
217
 Element Eiectronic Configurations
sublevels
Elements
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p
41-Niobium 2 2 6 2 6 10 2 6 4 1
42-Molybdenum 2 2 6 2 6 10 2 6 5 1
43-Technetium 2 2 6 2 6 10 2 6 5 2
44-Ruthenum 2 2 6 2 6 10 2 6 7 1
45-Rhodium 2 2 6 2 6 10 2 6 8 1
46-Palladium 2 2 6 2 6 10 2 6 10
47-Silver 2 2 6 2 6 10 2 6 10 1
48-Cadmium 2 2 6 2 6 10 2 6 10 2
49-Indium 2 2 6 2 6 10 2 6 10 2 1
50-Tin 2 2 6 2 6 10 2 6 10 2 2
51-Antimony 2 2 6 2 6 10 2 6 10 2 3
52-Tellurium 2 2 6 2 6 10 2 6 10 2 4
53-Iodine 2 2 6 2 6 10 2 6 10 2 5
54-Xenon 2 2 6 2 6 10 2 6 10 2 6
55-Cesium 2 2 6 2 6 10 2 6 10 2 6 1
56-Barium 2 2 6 2 6 10 2 6 10 2 6 2
57-Lanthanum 2 2 6 2 6 10 2 6 10 2 6 1 2
58-Cerium 2 2 6 2 6 10 2 6 10 1 2 6 1 2
59-Prasedoymium 2 2 6 2 6 10 2 6 10 3 2 6 2
60-Neodymium 2 2 6 2 6 10 2 6 10 4 2 6 2
61-Promethium 2 2 6 2 6 10 2 6 10 5 2 6 2
62-Samarium 2 2 6 2 6 10 2 6 10 6 2 6 2
63-Europium 2 2 6 2 6 10 2 6 10 7 2 6 2
64-Gadolinum 2 2 6 2 6 10 2 6 10 7 2 6 1 2
65-Terbium 2 2 6 2 6 10 2 6 10 9 2 6 2
66-Dysprosium 2 2 6 2 6 10 2 6 10 10 2 6 2
67-Holmium 2 2 6 2 6 10 2 6 10 11 2 6 2
68-Erbium 2 2 6 2 6 10 2 6 10 12 2 6 2
69-Thulium 2 2 6 2 6 10 2 6 10 13 2 6 2
70-Yetterbium 2 2 6 2 6 10 2 6 10 14 2 6 2
71-Luteium 2 2 6 2 6 10 2 6 10 14 2 6 1 2
72-Hafanium 2 2 6 2 6 10 2 6 10 14 2 6 2 2
73-Tantalium 2 2 6 2 6 10 2 6 10 14 2 6 3 2
74-Tungesten 2 2 6 2 6 10 2 6 10 14 2 6 4 2
75-Rhenium 2 2 6 2 6 10 2 6 10 14 2 6 5 2
76-Osmium 2 2 6 2 6 10 2 6 10 14 2 6 6 2
77-Iridium 2 2 6 2 6 10 2 6 10 14 2 6 7 2
78-Platinum 2 2 6 2 6 10 2 6 10 14 2 6 9 1
79-Gold 2 2 6 2 6 10 2 6 10 14 2 6 10 1
80-Mercury 2 2 6 2 6 10 2 6 10 14 2 6 10 2
218
 Element Eiectronic Configurations
sublevels
Elements
1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s 7p
81-Thalium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 1
82-Lead 2 2 6 2 6 10 2 6 10 14 2 6 10 2 2
83-Bismuth 2 2 6 2 6 10 2 6 10 14 2 6 10 2 3
84-Polonium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 4
85-Astatine 2 2 6 2 6 10 2 6 10 14 2 6 10 2 5
86-Radon 2 2 6 2 6 10 2 6 10 14 2 6 10 2 6
87-Francium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 6 1
88-Radium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 6 2
89-Actinium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 6 1 2
90-Thorium 2 2 6 2 6 10 2 6 10 14 2 6 10 2 6 2 2
91-Protactinum 2 2 6 2 6 10 2 6 10 14 2 6 10 2 2 6 1 2
92-Uranium 2 2 6 2 6 10 2 6 10 14 2 6 10 3 2 6 1 2
93-Neptonium 2 2 6 2 6 10 2 6 10 14 2 6 10 4 2 6 1 2
94-Plutonium 2 2 6 2 6 10 2 6 10 14 2 6 10 6 2 6 2
95-Amerecium 2 2 6 2 6 10 2 6 10 14 2 6 10 7 2 6 2
96-Curium 2 2 6 2 6 10 2 6 10 14 2 6 10 7 2 6 1 2
97-Berkelium 2 2 6 2 6 10 2 6 10 14 2 6 10 9 2 6 2
98-Californium 2 2 6 2 6 10 2 6 10 14 2 6 10 10 2 6 2
99-Einestanium 2 2 6 2 6 10 2 6 10 14 2 6 10 11 2 6 2
100-Fermium 2 2 6 2 6 10 2 6 10 14 2 6 10 12 2 6 2
101-Mendelevium 2 2 6 2 6 10 2 6 10 14 2 6 10 13 2 6 2
102-Nobelium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 2
103-Lawrencium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 1 2
104-Rutherium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 2 2
105-Dubinum 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 3 2
106-Seaborgium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 4 2
107-Bohrium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 5 2
108-Hassium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 6 2
109-Meitnerium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 7 2
110-Darmstadium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 9 1
111-Unununium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 10 1
112-Unubium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2
113- Ununquadium 2 2 6 2 6 10 2 6 10 14 2 6 10 14 2 6 10 2 2

219
 ‫امل�صادر باللغة االنكليزية‬

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‫امل�صادر باللغة العربية‬

‫ القاهرة‬، ‫ البرادعي للن�شر التربوي‬، ‫ نور المعرفة‬، ‫ الكيمياء لل�صف الثالث الثانوي‬- 1
.)2007(
‫ االردن‬، ‫ �إدارة المناهج والكتب المدر�سية‬، ‫ وزارة التربيةوالتعليم‬، ‫ الكيمياء ال�صف العا�شر‬- 2
.)2006(
‫ �شركة المطبوعات‬، ‫ �شركة جيوبروجكت�س‬، ‫ جزء ثاني‬- ‫ الكيمياء لل�صف الثاني ع�شر‬- 3
. ) 2008-2009( ‫ لبنان‬،‫للتوزيع والن�شر‬
،‫ ادارة المناهج والكتب المدر�سية‬، ‫ الكيمياء للمرحلة الثانوية للم�ستويان االول والثاني‬- 4
.)2006( ‫االردن‬

220