Chapter One
1. Introduction to study of modern chemistry
1.1. The atom in modern chemistry
A. Introduction to Chemistry, Atoms and Elements
Atomic Theory is the central theme of chemistry and most important idea in science. Chemistry
is central to science; it deals with matter (stuff of the world) and transformations of matter. It is
important and essential in physics, biology, geology, industry, medicine, nursing, engineering,
and philosophy….etc.
Chemistry is not just beakers and test tubes in labs
Chemical processes are going on everywhere all around us and inside us, all the stuff you see
around you is made of atoms and you are also made of atoms.
Examples:
 Breathing O2 binds to Fe in Hemoglobin in red blood cells and thousands of other
chemical processes going on in our cells and bodies
 We get energy from food we eat: sugar + oxygen → carbon dioxide + water + energy
 We get energy for industry by combustion:
Butane + oxygen → carbon dioxide + water +energy
Generally-Chemistry is the study of matter and its properties, the changes that matter undergoes,
and the energy associated with those changes.
Main areas of Chemistry:
Organic – compounds of carbon and Hydrogen
Inorganic – compounds that do not include carbon
Analytical – composition of matter and mixtures (what is there and how much)
Physical – applies ideas of math and physics to chemistry
Biochemistry – chemistry of living things (from bacteria to humans)
1. Dalton’s atomic theory
Dalton’s atomic theory of matter: (based on knowledge at that time):
1. All matter is made of atoms. These indivisible and indestructible objects are the ultimate
chemical particles.
2. All the atoms of a given element are identical, in both weight and chemical properties.
However, atoms of different elements have different weights and different chemical
properties.
3. Compounds are formed by the combination of different atoms in the ratio of small whole
numbers.
4. A chemical reaction involves only the combination, separation, or rearrangement of atoms;
atoms are neither created nor destroyed in the course of ordinary chemical reactions.
Two modifications have been made to Dalton’s theory
1. Subatomic particles and (2) Isotopes were discovered.
2. Early experiments to characterize the atom
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Based on the work of Dalton, Gay-Lussac, Avogadro, & others, chemistry was beginning to
make sense and the concept of the atom was clearly a good idea!
1. The electron
J.J. Thomson, English (1898-1903)—found that when high voltage was applied to an evacuated
tube, a “ray” he called a cathode ray [since it emanated from the (-) electrode or cathode when
you apply a voltage across it] was produced.
When the power was turned on, a "ray" could be seen striking the phosphor-coated end of the
tube and emitting a glowing spot of light. The rays were called cathode rays because they
originated at the negative electrode (cathode) and moved to the positive electrode (anode).
Cathode rays typically travel in a straight line, but in a magnetic field the path is bent, indicating
that the particles are charged, and in an electric field the path bends toward the positive plate.
The ray is identical no matter what metal is used as the cathode. It was concluded that cathode
rays consist of negatively charged particles found in all matter. The rays appear when these
particles collide with the few remaining gas molecules in the evacuated tube. Cathode ray
particles were later named electrons.
He then measured the deflection of beams of e- to determine the charge-to-mass ratio
e/m = -1.76108 C/g
2. The Nucleus
Figure bellow is representation of Rutherford's experiment. Tiny, dense, positively charged alpha
particles emitted from radium were aimed, like minute projectiles, at thin gold foil.
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The data showed that very few particles were deflected at all, and that only 1 in 20,000 was
deflected by more than 900 ("coming backwards"). It seemed that these few particles were
being repelled by something small, dense, and positive within the gold atoms. From the data,
Rutherford calculated that an atom is mostly space occupied by electrons, but in the center of
that space is a tiny region, which he called the nucleus, that contains all the positive charge and
essentially all the mass of the atom. He proposed that positive particles lay within the nucleus
and called them protons.
The modern view of atomic structure: an introduction element
All matter composed of only one type of atom is an element. There are 92 naturally occurring
elements, all others are manmade.
Atoms
Atom--the smallest particle of an element that retains the chemical properties of that element.
Nucleus-contains the protons and the neutrons; the electrons are located outside the nucleus.
Diameter of the nucleus is 10-13 cm. The electrons are located 10-8cm from the nucleus. A mass
of nuclear material the size of a pea would weigh 250 million tons! Very dense! Proton--positive
charge, responsible for the identity of the element, defines atomic number. Neutron--no charge,
same size & mass as a proton, responsible for isotopes, alters atomic mass number. Electron-negative charge, same size as a proton or neutron, but 1/2,000 the mass of a proton or neutron,
responsible for bonding, hence reactions and ionizations, easily added or removed. Atomic
number (Z) .The number of p+ in an atom, all atoms of the same element have the same number
of p+. Mass number (A): the sum of the number of neutrons and p+ for an atom. A different
mass number does not mean a different element--just an isotope.
ISOTOPES
• Isotopes--atoms having the same atomic number (# of p+) but a different number of neutrons
• Most elements have at least two stable isotopes, there are very few with only one stable isotope
(Al, F, P)
Elements and Compounds
Atoms are building blocks of nature/matter
Elements – composed of one type of atom e.g. Na
Compounds - composed of two or more types of elements/atoms e.g. NaCl
1.2. Chemical formulas, chemical equation and reaction yields
1. The meaning of a chemical equation
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A chemical equation is a chemist’s shorthand expression for describing a chemical change. As an
example, - consider what takes place when iron rusts. The equation for this change is:
Fe + O2 → Fe2O3
In this expression, the symbols and formulas of the reacting substances, called the reactants, are
written on the left side of the arrow and the products of the reaction are written on the right side.
The arrow is read as “gives”, “yields”, or “forms” and the plus (+) sign is read as “and”. When
the plus (+) sign appears between the formulas for two reactants, it can be read as “reacts with”.
(The + sign does not imply mathematical addition.)
The equation, above, can be read as iron reacts with oxygen to yield (or form) iron (III) oxide.
 Balancing a chemical equation
As it is written, the equation indicates in a qualitative way what substances are consumed in the
reaction and what new substances are formed. In order to have quantitative information about the
reaction, the equation must be balanced so that it conforms to the Law of Conservation of Matter.
That is, there must be the same number of atoms of each element on the right hand side of the
equation as there are on the left hand side.
Half-Reaction Method for Balancing Redox Reactions
The following steps are used in balancing a redox reaction by the half-reaction method:
Step1. Divide the skeleton reaction into two half-reactions
Step2. Balance the atoms and charges in each half-reaction
Step3. If necessary, multiply one or both half-reactions by an integer to make the number
of e - gained in the reduction equal the number lost in the oxidation.
Step4. Add the balanced half-reactions, and include states of matter.
Step5. Check that the atoms and charges are balanced.
Example: Balance the redox reaction between dichromate ion and iodide ion to form chromium
(III) ion and solid iodine, which occurs in acidic solution
Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s) [acidic solution]
2. The meaning of a chemical formula
A chemical formula is a shorthand method of representing the elements in a compound. The
formula shows the formulas of the elements in the compound and the ratio of the elements to one
another. For example, the formula for sodium chloride: NaCl tells us that the compound is
composed of the elements sodium, Na, and chlorine, Cl, in a one-to-one ratio, That is, one atom
of sodium combines with one atom of chlorine.
When elements combine in different ratios, subscripts are added, following the element symbol,
to -indicate that the number of atoms of that element in the compound if it is greater than one.
The subscript refers only to the element it immediately follows. For example, the formula for
magnesium bromide: MgBr2 tells us that two bromine atoms combine with one magnesium atom.
Types of Chemical Formulas
In a chemical formula, element symbols and numerical subscripts show the type and number of
each atom present in the smallest unit of the substance. There are several types of chemical
formulas for a compound:
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1. The empirical formula shows the relative number of atoms of each element in the compound.
It is the simplest type of formula and is derived from the masses of the component elements.
For example, in hydrogen peroxide, there is 1 part by mass of hydrogen for every 16 parts by
mass of oxygen. Therefore, the empirical formula of hydrogen peroxide is HO: one H atom
for every O atom.
2. The molecular formula shows the actual number of atoms of each element in a molecule of the
compound. The molecular formula of hydrogen peroxide is H2O2; there are two H atoms and
two O atoms in each molecule.
3. A structural formula shows the number of atoms and the bonds between them; that is, the
relative placement and connections of atoms in the molecule. The structural formula of
hydrogen peroxide is H-O-O-H; each H is bonded to an O, and the O's are bonded to each
other.
Example: During excessive physical activity, lactic acid (M = 90.08 g/mol) forms in muscle
tissue and is responsible for muscle soreness. Elemental analysis shows that this compound
contains 40.0 mass % C, 6.71 mass % H, and 53.3 mass %O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
Exercise: One of the most widespread environmental carcinogens (cancer-causing agents) is
benzo[a]pyrene (M = 252.30 g/mol). It is found in coal dust, in cigarette smoke, and even in
charcoal-grilled meat. Analysis of this hydrocarbon shows 95.21 mass % C and 4.79 mass % H.
What is the molecular formula of benzo[a]pyrene?
3. Reaction yields
Percent Yield: is the actual yield expressed as a percentage of the theoretical yield. It is used is
cases where a chemical transformation occurs.
To calculate the % yield, you need the following information:
1. The molar ratio of product to starting material.
2. Molecular weights of product and starting material.
3. Limiting reagent.
Percent yield is calculated as follows:
% yield =
x 100
Calculating percent yield actually involves a series of short calculations. Yield calculations can
be broken up into a series of six steps. These are:
1. Write a balanced equation for the reaction
2. Calculate the molecular weight of each reactant and product
3. Convert all amounts of reactants and products into moles
4. Figure out the limiting reagent
5. Calculate the theoretical yield
6. Calculate the percentage yield
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
Actual yield is the amount of product you actually got while theoretical is the maximum
possible yield. Be sure that actual and theoretical yields are both in the same units so that
units cancel in the calculation.
 Limiting reagent; the reactant run out first is the limiting reagent.
Exercise1: When 5.00 g of KClO3 is heated it decomposes according to the equation:
2KClO3→2KCl + 3O2
a) Calculate the theoretical yield of oxygen.
b) Give the % yield if 1.78 g of O2 is produced.
c) How much O2 would be produced if the percentage yield was 78.5%?
Exercise2: How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g
of aluminum and 15.0 g of sulfur?
a. How much of the non limiting reactant is in excess?
b. Give the % yield if 22.3 g of Al2S3 is produced.
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Chapter Two
2. Chemical bonding and molecular structure
2.1. Chemical bonding: the classical description
By classical, we mean models that do not take into account the quantum behavior of small
particles, notably the electron. These models generally assume that electrons and ions behave as
point charges which attract and repel according to the laws of electrostatics. Although this
completely ignores what has been learned about the nature of the electron since the development
of quantum theory. In the 1920’s, these classical models have not only proven extremely useful,
but the major ones also serve as the basis for the chemist’s general classification of compounds
into “covalent” and “ionic” categories.
1. The Ionic Bond
A chemical compound in which the component atoms exist as ions is called an ionic compound.
Potassium chloride, KCl, is a common ionic compound. The electronic configurations of the
potassium and chloride ions obey the octet rule.
The structure of crystalline KCl is shown in fig. below. In the KCl structure, which is typical of
many ionic compounds, each positive ion is surrounded by negative ions, and each negative ion
is surrounded by positive ions. The crystal structure is stabilized by an interaction between ions
of opposite charge. Such a stabilizing interaction between opposite charges is called an
electrostatic attraction. An electrostatic attraction that holds ions together, as in crystalline
KCl, is called an ionic bond. Thus, the crystal structure of KCl is maintained by ionic bonds
between potassium ions and chloride ions. The ionic bond is the same in all directions; that is, a
positive ion has the same attraction for each of its neighboring negative ions, and a negative ion
has the same attraction for each of its neighboring positive ions.
When an ionic compound such as KCl dissolves in water, it dissociates into free ions (each
surrounded by water). Each potassium ion moves around in solution more or less independently
of each chloride ion. The conduction of electricity by KCl solutions shows that the ions are
present. Thus, the ionic bond is broken when KCl dissolves in water.
To summarize, the ionic bond
1. is an electrostatic attraction between oppositely charged ions;
2. is the same in all directions—that is, it has no preferred orientation in space; and
3. is broken when an ionic compound dissolves in water.
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2. The Covalent Bond
Many compounds contain bonds that are very different from the ionic bond in KCl. these
compounds will not conduct electricity in solutions. This observation indicates that these
compounds are not ionic. According to this model, the chemical bond in a nonionic compound is
a covalent bond, which consists of an electron pair that is shared between bonded atoms. Let’s
examine some of the ideas associated with the covalent bond.
The Polar Covalent Bond
In many covalent bonds the electrons are not shared equally between two bonded atoms.
Consider, for example, the covalent compound hydrogen chloride, HCl. (Although HCl dissolves
in water to form H3O+ and Cl- ions, in the gaseous state pure HCl is a covalent compound). The
electrons in the H-Cl covalent bond are unevenly distributed between the two atoms; they are
polarized, or “pulled,” toward the chlorine and away from the hydrogen. A bond in which
electrons are shared unevenly is called a polar bond. The H-Cl bond is an example of a polar
bond. The tendency of an atom to attract electrons to itself in a covalent bond is indicated by its
electronegativity. When the ligand (Lewis base) donates the electron pair and the metal ion
(Lewis acid) accepts it to form one of the covalent bonds of the complex ion. Such a bond, in
which one atom in the bond contributes both electrons, is called a coordinate covalent bond,
although, once formed, it is identical to any covalent single bond.
2.2. Introduction to quantum mechanics
This chapter introduces some of the basic principles of quantum mechanics. First, it reviews the
experimental results that overthrew the concepts of classical mechanics. These experiments led
to the conclusion that particles may not have an arbitrary energy and that the classical concepts
of 'particle' and 'wave' blend together. In quantum mechanics, all the properties of a system are
expressed in terms of a wave function that is obtained by solving the Schrodinger equation.
Quantum mechanics is the branch of physics that mathematically describes the wave properties
of submicroscopic particles. it is a collection of postulates based on a huge number of
experimental observations.
Classical physics (mechanics) (1) predicts a precise trajectory for particles, with precisely
specified locations and momenta at each instant, and (2) allows the translational, rotational, and
vibrational modes of motion to be excited to any energy simply by controlling the forces that are
applied. These conclusions agree with everyday experience. Everyday experience, however, does
not extend to individual atoms, and careful experiments of the type described below have shown
that classical mechanics fails when applied to the transfers of very small energies and to objects
of very small mass.
Classical mechanics (newton's mechanics) and maxwell's equations (electromagnetics theory)
can explain macroscopic phenomena such as motion of billiard balls or rockets while quantum
mechanics is used to explain microscopic phenomena such as photon-atom scattering and flow of
the electrons in a semiconductor. In newtonian mechanics, the laws are written in terms of
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particle trajectories. It state descriptor of newtonian physics. A particle is an indivisible mass
point object that has a variety of properties that can be measured, which we call observables.
Quantum particles can act as both particles and waves in nature (wave-particle duality.
Quantization of energy is yet another property of "microscopic" particles.
The electromagnetic field is characterized by a wavelength,
neighbouring peaks of the wave, and its frequency,
(lambda), the distance between the
(nu), and the number of cycles per second
at which its displacement at a fixed point returns to its original value (Fig. 2). The frequency is
measured in hertz, where 1Hz = 1 s-1. The wavelength and frequency of an electromagnetic wave
are related by
Fig.2: The wave length
of a wave is the peak-to-peak distance.
The shorter the wavelength leads the higher the frequency. The characteristics of the wave are
also reported by giving the wave number, (nu tilde), of the radiation, where
The first clue in the development of quantum theory came with the discovery of the
deBroglie relation. In 1923, Louis de Broglie reasoned that if light exhibits particle aspects,
perhaps particles of matter show characteristics of waves. He postulated that a particle with mass
m and a velocity v has an associated wavelength. Light has both wave & particle properties. He
proposed that all moving objects have wave properties. Thus
 For light characteristics: E = h = hc/
 For particles: E = mc2 (Einstein)
Therefore, mc = h / and for particles (mass) x (velocity) = h / .where wave length ( ) for
particles is called the de Broglie wavelength.
Example: Determine the de Broglie wavelength of a person with a mass of 90 kg who is running
10 m/s?
The description and classification of the electromagnetic field according to its frequency and
wavelength were vied in the electromagnetic spectrum.
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Fig.2.1: The electromagnetic spectrum and the classification of the spectral regions.
In 1900 Max Planck found that he could account for the permitted energies of an electromagnetic
oscillator of frequency are integer multiples of hv. Light is a wave arriving as stream of
particles called "photons". Each photon is considered as quantum of energy. An increase in the
frequency the cause for increase in the energy similarly, an increase in the wavelength gives a
decrease in the energy of the photon.
E = hv =
, Where h (Planck's constant) = 6.63x1034Js.
Example: Determine the energies of photons with wavelengths of 650 nm, 700 nm?
Solution:
a. Frequencies 4.50x1014 s1, 6.50x1014 s1, respectively.
b. The total energy of a stream of particles (photons) of that energy will be:
E = nhv
Where n = 1, 2 ... (only discrete energies).
c. The wave number?
2.3. Quantum mechanics and atomic structure
In this chapter we see how to use quantum mechanics to describe the electronic structure of an
atom, the arrangement of electrons around a nucleus. It provides universal description of the
electron distribution in atoms. The concepts we meet are of central importance for understanding
the structures and reactions of atoms and molecules, and hence have extensive chemical
applications. Bohr’s theory established the concept of atomic energy levels but did not
thoroughly explain the “wave-like” behavior of the electron. Current ideas about atomic structure
depend on the principles of quantum mechanics, a theory that applies to subatomic particles such
as electrons.
Atomic structure elucidated by interaction of matter with light. Light properties: characterized by
wavelength,, and frequency,. Light is an electromagnetic radiation, a wave of oscillating
electric and magnetic influences causes the so called fields. Frequency and wavelength have
inversely proportional to each other. The structure of an atom is composed of subatomic particles
(proton, neutron, and electron). When it interacts with quanta will causes accumulation of energy
like potential, kinetic, translational etc.
c =  where c = the speed of light = 3.00x108 m/s; units  = s1,  = m
Examples: Calculate the frequency of light with a wavelength of 500 nm?
Schrödinger applied: idea of an electron behaving as a wave to the problem of electrons in
atoms. Although we cannot precisely define an electron’s orbit, we can obtain the probability of
finding an electron at a given point around the nucleus. Erwin Schrodinger defined this
probability in a mathematical expression called a wave function, denoted (psi). It is a function
of distance and two angles.
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
For 1 electron,

electron is found.
Does not describe the exact location of the electron.

2
corresponds to an orbital — the region of space within which an
are proportional to the probability of finding an electron at a given point.
Uncertainty Principle: Problem of defining nature of electrons in atoms solved by Werner
Heisenberg in 1927 cannot simultaneously define the position and momentum (p = mv) of an
electron. We can no longer think of an electron as having a precise orbit in an atom. To describe
such an orbit would require knowing its exact position and velocity. Heisenberg’s uncertainty
principle is a relation that states that the product of the uncertainty in position (x) and the
uncertainty in momentum (mvx) of a particle can be no larger than h/4π.
x)(m
)≥
When m is large (for example, a baseball) the uncertainties are small, but for electrons, high
uncertainties disallow defining an exact orbit. At best we can describe the position and velocity
of an electron by a probability distribution, which is given by 2. The square of probability ( 2)
is proportional to the probability of finding an electron at a given point.
Orbital Quantum Numbers: An atomic orbital is defined by 3 quantum numbers: n, l, and ml.
Electrons are arranged in shells and subshells of orbitals.
n  shell
l  subshell
ml  designates an orbital within a subshell
Generally, the wave function of a particle (atom) must be;
– Single valued, Continuous, Continuous first derivative, Quadratically integrable

And
, where is volume element (i. e.
)
*d must be finite


Examples: Give the name, magnetic quantum numbers, and number of orbitals for each sublevel
with the given quantum numbers:
(a) n = 3, l = 2 (b) n = 2, l = 0 (c) n = 5, l = 1 (d) n = 4, l = 3
Answer
n
l
Possible ml Values
sublevel name
No. of Orbitals
(a)
3
2
-2, -1, 0, +1, + 2
3d
5
(b)
2
0
0
2s
1
(c)
5
1
-1,0,+1
5p
3
(d)
4
3
-3, -2, -1, 0, +1, +2, +3
4f
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To name the sublevel (subshell), we combine the n value and l letter designation. We know l, so
we can find the possible ml values, whose total number equals the number of orbitals.
2.4. Quantum mechanics and molecular structure
Quantum mechanics allows the prediction of molecular structure such as: bond lengths, bond
angles, and dissociation energies. The concepts developed in the structure of atomic orbitals, can
be extended to a description of the electronic structures of molecules. There are two principal
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quantum mechanical theories of molecular electronic structure. These are molecular orbital
theory (MO theory) and valence-bond theory is the concept of the shared electron pair. We see
how to write the wave function for such a pair, and how it may be extended to account for the
structures of a wide variety of molecules. The theory introduces the concepts of σ and π bonds,
promotion, and hybridization that are used widely in chemistry.
The concept of atomic orbital
is extended to that of molecular orbital, which is a wave function that spreads over all the atoms
in a molecule.
There are two major approaches to the calculation of molecular structure, valence bond theory
(VB theory) and The Schrodinger equation for a hydrogen atom can be solved exactly; an exact
solution is not possible for any molecule because the simplest molecule consists of three particles
(two nuclei and one electron).
We therefore adopt the Born-Oppenheimer approximation in which it is supposed that the nuclei,
being so much heavier than an electron, move relatively slowly and may be treated as stationary
while the electrons move in their field. We can therefore think of the nuclei as being fixed at
arbitrary locations, and then solve the Schrodinger equation for the wave function of the
electrons alone.
Valence bond theory: In VB theory, a bond is regarded as forming when an electron in an
atomic orbital on one atom pairs its spin with that of an electron in an atomic orbital on another
atom. The basic principle of VB theory is that a covalent bond forms when orbitals of two atoms
overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons.
("Orbital overlap" is another way of saying that the two wave functions are in phase, so the
amplitude increases between the nuclei).
Example1: Use partial orbital diagrams to describe how mixing of the atomic orbitals of the
central atom(s) leads to the hybrid orbitals in each of the following:
(a) Methanol, CH3OH
(b) Sulfur tetrafluoride, SF4
a
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b
Examples: Describe the types of bonds and orbitals in acetone, (CH3HCO.
From the Lewis structure, we determine the number and arrangement of electron groups around
each central atom, along with the molecular shape. From that, we postulate the type of hybrid
orbitals involved. Then, we write the partial orbital diagram for each central atom before and
after the orbitals are hybridized.
(a).For CH3OH. The electron-group arrangement is tetrahedral around both C and O atoms.
Therefore, each central atom is Sp3 hybridized. The C atom has four half filled Sp3 orbitals:
Answer: the shape around each central atom to postulate the type of the hybrid orbitals, paying
attention to the multiple bonding of the C and O bond.
Molecular orbital theory:
In VB theory, a molecule is pictured as a group of atoms bound together through localized
overlap of valence-shell atomic orbitals. In MO theory, a molecule is pictured as a collection of
nuclei with the electron orbitals delocalized over the entire molecule. The MO model is a
quantum-mechanical treatment for molecules. Just as an atom has atomic orbitals (AOs) with a
given energy and shape that are occupied by the atom's electrons, a molecule has molecular
orbitals (MOs) with a given energy and shape that are occupied by the molecule’s electrons.
Despite the great usefulness of MO theory, it too has a drawback: MOs are more difficult to
visualize than the easily depicted shapes of VSEPR theory or the hybrid orbitals of VB theory.
The Central Themes of MO Theory
Several key ideas of MO theory appear in its description of the hydrogen molecule and other
simple species. These ideas include the formation of MOs, their energy and shape, and how they
fill with electrons.
Formation of molecular orbital’s: because electron motion is so complex, we use
approximations to solve the Schrodinger equation for an atom with more than one electron.
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Similar complications arise even with H2, the simplest molecule, so we use approximations to
solve for the properties of MOs. The most common approximation mathematically combines
(adds or subtracts) the atomic orbitals (atomic wave functions) of nearby atoms to form MOs
(molecular wave functions).
When two H nuclei lie near each other, as in H2, their AOs will overlap. The two ways of
combining the AOs are as follows:
a. Adding the wave functions together: This combination forms a bonding MO, which has a
region of high electron density between the nuclei. Additive overlap is analogous to light
waves reinforcing each other, making the resulting amplitude higher and the light brighter.
For electron waves, the overlap increases the probability that the electrons are between the
nuclei see fig. a below.
Figure. An analogy between light waves and atomic wave functions, when light waves undergo
interference, their amplitudes either add together or subtract.
(a), when the amplitudes of atomic wave functions (dashed lines) are added, a bonding molecular orbital
(MO) results, and electron density (red line) increases between the nuclei.
(b) Conversely, when the amplitudes of the wave functions are subtracted, an antibonding MO results,
which has a node (region of zero electron density) between the nuclei.
b. Subtracting the wave functions from each other. This combination forms an antibonding MO,
which has a region of zero electron density (a node) between the nuclei (Figure above b).
Subtractive overlap is analogous to light waves canceling each other, so that the light
disappears. With electron waves, the probability that the electrons lie between the nuclei
decreases to zero.
The two possible combinations for hydrogen atoms HA and HB are
AO of HA + AO of HB = bonding MO of H2 (more e - density between nuclei)
AO of HA - AO of HB = antibonding MO of H2 (less e- density between nuclei)
Notice that the number of AOs combined always equals the number of MOs formed: two H
atomic orbitals combine to form two H2 molecular orbitals.
Energy and Shape of H2 Molecular Orbitals The bonding MO is lower in energy and the
antibonding MO higher in energy than the AOs that combined to form them. Let's examine
Figure below to see why this is so.
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Fig. Contours and energies of the bonding and anti bonding molecular orbitals (MOs) in H2• When two
H1s atomic orbitals (AOs) combine, they form two H2MOs. The bonding MO ( 1s) forms from
addition of the AOs and is lower in energy because most of its electron density lies between the
nuclei (shown as dots). The antibonding MO ( 1s*) forms from subtraction of the AOs and is higher
in energy because there is a node between the nuclei and most of the electron density lies outside the
internuclear region.
The bonding MO in H2 is spread mostly between the nuclei, with the nuclei attracted to the
intervening electrons. An electron in this MO can delocalize its charge over a much larger
volume than is possible in an individual AO in either H. Because the electron-electron repulsions
are reduced, the bonding MO is lower in energy than the isolated AOs. Therefore, when
electrons occupy this orbital, the molecule is more stable than the separate atoms. In contrast, the
anti bonding MO has a node between the nuclei and most of its electron density outside the
internuclear region. The electrons do not shield one nucleus from the other, which increases the
nucleus-nucleus repulsion and makes the antibonding MO higher in energy than the isolated
AOs. Therefore, when the antibonding orbital is occupied, the molecule is less stable than when
this orbital is empty.
Both the bonding and antibonding MOs of H2 are sigma ( ) MOs because they are cylindrically
symmetrical about an imaginary line that runs through the two nuclei. The bonding MO is
denoted by 1s that is, a MO formed by combination of 1s AOs. Antibonding orbitals are
denoted with a superscript star, so the orbital derived from the 1s AOs is
1s*
(spoken "sigma,
one ess, star").
To interact effectively and form MOs, atomic orbitals must have similar energy and orientation.
These orbitals on two H atoms have identical energy and orientation, so they interact strongly.
Filling Molecular Orbital’s with electrons is just the same as they fill AOs:
• MOs are filled in order of increasing energy (aufbau principle).
• An MO has a maximum capacity of two electrons with opposite spins (exclusion prin.).
• MOs of equal energy are half-filled, with spins parallel, before any of them is
completely filled (Hund's rule).
2.5. Bonding in organic molecules
It begins with an introduction to the important classes of organic molecules followed by a
description of chemical bonding in those molecules. Most organic molecules have much more
complex structures than most inorganic molecules.
1. Electron configuration, electro negativity, and covalent bonding: Carbon's ground-state
electron configuration of [He] 2S22p2 -four electrons more than He. Its electro negativity are
15
midway between the most metallic and nonmetallic elements of Period 2. Therefore, carbon
shares electrons to attain a filled outer (valence) level, bonding covalently in all its elemental
forms and compounds.
2. Bond properties, catenation, and molecular shape: The number and strength of carbon’s
bonds lead to its outstanding ability to catenate (form chains of atoms), which allows it to
form a multitude of chemically and thermally stable chain, ring, and branched compounds.
Through the process of orbital hybridization carbon forms four bonds in virtually all its
compounds, and they point in as many as four different directions.
3. Molecular stability. Although silicon and several other elements also catenate, none can
compete with carbon.
The Chemical Diversity of Organic Molecules
1. Bonding to heteroatoms. Many organic compounds contain heteroatoms, atoms other than C
or H. The most common heteroatoms are N and O, but S, P, and the halogens often occur as
well.
2. Electron density and reactivity. Most reactions start-that is, a new bond begins to form-when
a region of high electron density on one molecule meets a region of low electron density on
another.
For example, consider four bonds commonly found in organic molecules:
 The C-C bond: When C is singly bonded to another C, as occurs in portions of nearly
every organic molecule, the elecronegativity (EN) values are equal and the bond is
nonpolar. Therefore, in general, C-C bonds are uncreative.
 The C-H bond: This bond, which also occurs in nearly every organic molecule, is very
nearly nonpolar because it is short (109 pm) and the EN values of H (2.1) and C (2.5) are
close. Thus, C-H bonds are largely uncreative as well.
 The C-O bond: This bond, which occurs in many types of organic molecules, is highly
polar (EN = 1.0), with the O end of the bond electron rich and the C end electron poor.
As a result of this imbalance in electron density, the C-O bond is reactive and, given
appropriate conditions, a reaction will occur there.
 Bonds to other heteroatoms: Even when a carbon-heteroatom bond has a small EN,
such as that for C-Br (EN = 0.3), or none at all, as for C – S (EN = 0), heteroatoms
like these are large, arid so their bonds to carbon are long, weak, and thus reactive.
3. Nature of functional groups. One of the most important ideas in organic chemistry is that of
the functional group, a specific combination of bonded atoms that reacts in a characteristic
way, no matter what molecule it occurs in. In nearly every case, the reaction of an organic
compound takes place at the functional group. Functional groups vary from carbon-carbon
multiple bonds to several combinations of carbon-heteroatom bonds, and each has its own
pattern of reactivity. A particular bond may be part of one or more functional groups. For
example, the C-O bond occurs in four functional groups.
16
2.6. Bonding in transition metals compounds and coordination complexes
The transition elements (transition metals) make up the d block (B groups) and f block (inner
transition elements). In this chapter, we cover the d-block elements only. We first discuss some
properties of the elements and then focus on the most distinctive feature of their chemistry, the
formation of coordination compounds-substances that contain complex ions.
The electron configuration of the transition metal atom that correlates with physical properties
of the element: such as density and magnetic behavior, whereas it is the electron configuration of
the ion that determines the properties of the compounds.
Sample Problem:
1. Write the Electron Configurations of Transition Metal Atoms and Ions?
a) Fe and Ni
b) Cd2+ and V3+
2. Discus the physical and chemical properties of transition metals?
Coordination compounds
The most distinctive aspect of transition metal chemistry is the formation of coordination
compounds (also called complexes). These are substances that contain at least one complex ion,
a species consisting of a central metal cation (either a transition metal or a main-group metal)
that is bonded to molecules and/or anions called ligands.
17
The coordination compound is [Co(NH3)6]Cl3, the complex ion (always enclosed in square
brackets) is [Co(NH3 )6]3+ , the six NH3 molecules bonded to the central Co3+ are ligands, and the
three Cl- ions are counter ions. A coordination compound behaves like an electrolyte in water.
A complex ion is described by the metal ion and the number and types of ligands attached to it.
Its structure has three key characteristics-coordination number, geometry, and number of donor
atoms per ligand:
 Coordination number: The coordination numbers is the number of ligand atoms bonded
directly to the central metal ion.
 Geometry: The geometry (shape) of a complex ion depends on the coordination number
and nature of the metal ion.
Example: A complex ion whose metal ion has a coordination number of 2 is linear. The
coordination number 4 gives rise to either of two geometries-square planar or tetrahedral.
Most d8 metal ions form square planar complex ions. The d10 ions are among those that
form tetrahedral complex ions. A coordination number of 6 results in an octahedral
geometry, as shown by [Co (NH3)6]3+
 Donor atoms per ligand. The ligands of complex ions are molecules or anions with one
or more donor atoms that each donates a lone pair of electrons to the metal ion to form a
covalent bond.
Complex Ion and Ligands
Definitions
Complex ion - a metal ion with Lewis bases attached to it through coordinate covalent bonds.
A Complex (Coordination compound) is a compound consisting either of complex ions with
other ions of opposite charge or a neutral complex species.
Ligand - is the Lewis bases attached to the metal ion in a complex. Lewis bases are electron pair
donors. They can be ions or neutral molecules. Anions are the most common ionic Lewis bases.
Cations only very rarely form ligands with metal ions because the electron pair is usually held in
place by the positive charge.
18
Coordination number- the total number of bonds metal forms with ligands. The most common
coordination number is 6, although 4 are somewhat common also. Complex ions with
coordination number of 2 through 8 are known.
Monodentate – is ligand that bonds through one of its atoms to the metal ion.
Example H2O, NH3, CO, CN-)
Bidentate – is ligand that bond through two of its atoms with the metal ion.
Example C2O42- , ethylenediamine (C2H8N2) represented as "en")
Polydentate – is ligand that bonds through two or more of its atoms to the metal ion.
Example EDTA)
Chelate - a complex formed from polydentate ligands
Bonding and Properties of Complexes
In this section, we consider two models that address, in different ways, several key features of
complexes: how metal-ligand bonds form, why certain geometries are preferred, and why these
complexes are brightly colored and often paramagnetic.
According to Valence bond (VB) theory, in the formation of a complex ion, the filled ligand
orbital overlaps the empty metal-ion orbital. The ligand (Lewis base) donates the electron pair,
and the metal ion (Lewis acid) accepts it to form one of the covalent bonds of the complex ion
(Lewis adduct). Such a bond, in which one atom in the bond contributes both electrons, is called
a coordinate covalent bond.
VB concept of hybridization proposes the mixing of particular combinations of s, p, and d
orbitals to give sets of hybrid orbitals, which have specific geometries.
Let's discuss the orbital combinations that lead to octahedral, square planar, and tetrahedral
geometries.
a
b
c
2 3
1. Octahedral Complexes: d sp hybrid orbital. Fig. a above
2. Square Planar Complexes: Metal ions with a d8 configuration usually form square planar
complexes and form four dsp2 hybrid orbitals. most of the time in strong ligand. Fig. b above
3. Tetrahedral Complexes: having sp3 hybrid orbital. Fig. c above
Crystal Field Theory: The VB model is easy to picture and rationalizes bonding and shape, but
it treats the orbitals as little more than empty "slots" for accepting electron pairs. Consequently,
it gives no insight into the colors of coordination compounds and sometimes predicts their
19
magnetic properties incorrectly. In contrast to the VB approach, crystal field theory provides
little insight about metal-ligand bonding but explains color and magnetism clearly.
Splitting of d Orbitals in an Octahedral Field of Ligands
The crystal field model explains that the properties of complexes result from the splitting of dorbital energies, which arises from electrostatic interactions between metal ion and ligands.
An energy diagram of the orbitals shows that all five d orbitals are higher in energy in the
forming complex than in the free metal ion because of repulsions from the approaching ligands,
but the orbital energies split, with two d orbitals higher in energy than the other three. The two
higher energy orbitals are called eg orbitals, and the three lower energy ones are t2g orbitals.
(These designations refer to features of the orbitals that need not concern us here.) The splitting
of orbital energies is called the crystal field effect, and the difference in energy between the eg
and t2g sets of orbitals is the crystal field splitting energy (). Different ligands create crystal
fields of different strength and, thus, cause the d-orbital energies to split to different extents.
Strong-field ligands lead to a larger splitting energy (larger ); weak-field ligands lead to a
smaller splitting energy (smaller ). For instance, H2O is a weak-field ligand, and CN- is a
strong-field ligand.
I-< Cl- < F- < OH- < H2O < SCN- < NH3 < en < NO2- < CN- < CO
Weak Field
Stronger Field
Smaller 
Larger 
Longer
Shorter
The orbital occupancy is affected by the ligand in one of the two ways:
 Weak-field ligands and high-spin complexes.
 Strong-field ligands and low-spin complexes.
Example:
Rank the ions [Ti (H2O)6]3+, [Ti (NH3)6]3+, and [Ti (CN)6]3- in terms of the relative values of 
and of the energy of visible light absorbed.
Solution:
The ligand field strength is in the order CN- > NH3 > H2O, so the relative size of  and energy of
light absorbed is
20
[Ti (CN)6]3- > [Ti (NH3)6]3+ > [Ti (H2O)6]3+
21
Chapter Three
3. Kinetic molecular description of the state of mater
3.1. The gaseous state
Gases are the least dense and the most mobile of the three states of matter and its molecules
move at very high velocities and have high kinetic energy. For example, average velocity of H2
molecules at 0oC is over 1600 m/s. mixture of gases are uniformly distributed within the
container, the same quantity of a substance occupies a much greater volume as a gas than other
two states. For example, 1 mole of water, v = 18 mL at 4oC, V = 22400 mL in the gaseous state.
Generally gass:
(1) Molecules are relatively far apart
(2) Can be greatly compressed
(3) Volume occupied by a gas is most empty space
The kinetic-molecular theory
Kinetic-molecular theory (KMT) is based on the motion of particles, particularly gas
molecules. a gas behaves exactly as outlined by KMT is known as an ideal gas
The principal assumptions of KMT are:
1. Gases consist of tiny particles
2. The distance between particles is large compared with the size of particles the volume
occupied by a gas consists mostly of empty space
3. Gas particles have no attraction for one another
4. Gas particles move straight lines in all directions colliding frequently with one another and
with the wall of container
5. No energy is lost by collision of a gas particle with another gas particle or with the wall of
container all collisions are perfectly elastic
6. The average kinetic energy for particles is the same for all gases at the same temperature and
its value is directly proportional to the Kelvin temperature
K = ½ mv2
The lighter molecules will have a greater velocity than the heavier ones
ex. the velocity of H2 is four times the velocity of O2
Graham’s law of effusion – the rate of effusion depends on the density of a gas. The rates of
effusion of two gases at the same temperature and pressure is inversely proportional to the square
roots of their densities or molar masses
=
=
Application of Graham’s law: separation of two gasses
Measurement of pressure of gases; pressure is defined as force per unit area
Pressure results from the collisions of gas molecules with the walls of container. The amount of
pressure exerted by a gas depends on the number of gas molecules, the temperature, the volume
in which the gas is confined.
Pressure units
22
 1 atm = 760 torr = 760 mm Hg = 76 cm Hg = 101.325 kPa = 1013 mbar
 1013 mbar = 29.9 in.Hg = 14.7 lb/in.2
Example: 12.1 740 mm Hg =? torr = ? Atm
740 mm Hg = 740 torr = 740/760 atm = 0.974 atm
Dependence of pressure on number of molecules and temperature
When the temperature and pressure are kept constant, the pressure is directly proportional to the
number of moles or molecules of gas present
The pressure of a gas in a fixed volume also varied with the temperature
Temperature is increased → KE of the molecules increases → more frequent and more energetic
collisions occur → pressure increases
o Boyle’s law – the relationship of P and V at constant temperature (T), the volume (V) of a
fixed mass of a gas is inversely proportional to the pressure (P)
V∝
or P1V1 = P2V2, PV = constant
Example: A gas occupies 200 mL at 400 torr pressure, what pressure will be when the volume is
changed to 75.0 mL?
(200 mL) × (400 torr) = (75 mL) × P, P = 1067 torr
o Charles’ law
Charles’ law – the effect of temperature (T) on the volume (V) of a gas at constant pressure the
volume of a fixed mass of any gas is directly proportional to the absolute temperature
V ∝ T or V1T2 = V2T1
Absolute zero – oK = -273.15oC
Example: 3L of H2 at -20oC warmed to room temperature of 27oC, V =? If P = constant
v = 3.56 L
3. Gay-Lussac’s law – relationship involving pressure and temperature at constant volume the
pressure of a fixed mass of a gas, at constant volume, is directly proportional to the Kelvin
temperature
P = kT or P1T2 =T1P2
Example: pressure of a container of He is 650 torr at 25oC, the sealed container is cooled to 0oC,
what will the pressure be?
P = 595 torr
4. Combined gas law
Common reference points of temperature and pressure were selected and called standard
condition or standard temperature and pressure (STP)
T = 273.15 K (0oC), P = 1 atm (760 torr or 760 mm Hg or 101.325 kPa)
T and P change at the same time
=
(i) T is constant Boyle’s law
23
(ii) P is constant Charles’ law
(iii) V is constant Guy-Lussac’s law
Example. 20.0 L of NH3 at 5oC and 730 torr, what is volume at 50oC and 800 torr?
V2 = 21.2 L
5. Dalton’s law of partial pressure
The total pressure of a mixture of gases is the sum of the partial pressure exerted by each of the
gases in the mixture
Ptotal = PA + PB + PC
6. Avogadro’s law equal volumes of different gases at the same temperature and pressure
contain the same number of molecules
This law afforded a firm foundation for the kinetic-molecular theory different gases at the same
temperature have the same kinetic energy. if two different gases at the same temperature, occupy
the same volumes, and exhibit equal pressure, must contain the same number of molecules
7. Ideal gas equation
Four variables in calculations involving gases: volume (V), pressure (P), temperature (T) and the
number of molecules or moles (n) combining these variables into a single expression
Ideal gas equation PV = nRT
R: ideal gas constant 0.0821 L-atm/mol-K
ex. 12.15 what pressure will be exerted by 0.400 mol of a gas in 5.0-L container at 17oC?
P=
= 1.90 atm
Density of gases
The density of a gas
d=
The volume of a gas depends on temperature and pressure → increasing the temperature will
reduce the density.
At STP, the density of a gas
dSTP = molar mass
dSTP = (79.90 g/mol)
)
)= 3.17 g/L
3.2. Solid, liquid and gas phase transition
It is particularly instructive to assemble graphs and diagrams that contrast the properties of a
compound's phases. For example, the phase diagram that is most famous is a plot of pressure vs.
temperature in which the various states of matter are identified.
Presented below are three ways we can draw a phase diagram:
 Certainly the simplest is a triangle with the 3 phases at the vertices and the names of the
phase changes along the sides of the triangle.
A second phase transition tracks the temperature change over time energy is added or removed
from the system. Note that there are five different constants for five different possible
calculations within specific temperature regions.
24
Note bellow that in the heating of a solid until it becomes a vapor there are five distinct points at
which you can calculate an amount of heat change.
A third and most informative phase diagram: the Pressure-Temperature (PT) Phase Diagram
As described below, (a) the PT diagram is used to determine the phase of a substance at any
combination of pressures and temperatures. In addition, the lines that separate the phases define
equilibrium transitions pressures and temperatures. (b) PT Diagram for water: spend some time
staring at this diagram and looking at the numbers, especially around phase transitions. Your
familiarity with the quantitative aspects of water’s phase changes will help you appreciate the
kind of information realized from this kind of graph.
Things to notice on any PT diagram:
1. If you are given a P and T, then you can read the actual phase (s, l, or g) at that PT point.
2. In general, and sensibly,
As T↑ gas forms
As T↓ solid forms
As P↑ solid forms
As P↓ gas forms
(3) The solid lines on a PT diagram (a) represent equilibrium phase transitions (the six shown on
the triangle below.) So for example, at the line separating solid from liquid, both liquid and
solid exist simultaneously and define the melting or freezing point for the compound. For
example, on a water PT diagram (b), at 1 atm pressure you would expect to see crossing
points at 0oC and 100 oC.
(4) When you are on a line, you are at equilibrium. In other words, if you see both phases present
simultaneously, G = 0. We will use this fact to do a calculation shortly.
(5) There are two special regions on a PT diagram.
i)The triple point where all 3 phases exist at equilibrium simultaneously, (For water this
occurs just below 0oC at 4.6 torr) Were you to create this environment you would see ice,
water, and steam all present at the same time kind of gives you chills just thinking about
the excitement of it all.
25
a
b
ii) Supercritical region an odd thing happens if you crank up the pressure and temperature.
You reach a place at the end of the gas liquid line where a new fluid, termed a
supercritical fluid, is formed. It has different properties than either the liquid or vapor of a
substance. (For example, this is the region where supercritical carbon dioxide solubilizes
caffeine in coffee--there was a commercial a few years ago for decaf coffee that referred to
“the natural effervescence found in nature” as a way to describe using supercritical CO2
rather than a dry cleaning solvent to extract the caffeine from coffee beans.) What is
interesting to note about the supercritical region is above the critical temperature it is not
possible to turn vapor into a liquid by increasing the pressure. For water the critical
temperature is 374 oC at a pressure of 218 atm. For CO2 it is 31oC at 73 atm.
Water is weird and its PT Diagram (above b) proves it.
The slope of the line at the solid/liquid phase diagram is like no other common material it is
negative!
The unique feature of the graph is the negative slope of the ice-liquid transition. (Every other
phase diagram shows a positive slope for the solid-liquid interface. Why is this negative slope for
H2O so interesting? It means that
 Ice is less dense than water-it is why ice floats on water but in all other substances the solid
form sinks. It is also why when you put a glass of water in the freezer, the frozen ice
increases in size and can break the glass.
 The PT diagram also suggests that for water, as the pressure goes up, ice melts. Think about
it this way: when you bite an ice cube it liquefies. When you bite any other solid, it stays a
solid!!
By the way, the reason for this peculiar behavior of water is the enormous H-bonding capacity of
H2O that more tightly packs it in liquid form/
26
3.3. Solution
Solutions are homogeneous mixtures consisting of a solute dissolved in a solvent through the
action of intermolecular forces. The solubility of a solute in a given amount of solvent is the
maximum amount that can dissolve at a specified temperature. (For gaseous solutes, the pressure
must be specified also.) In addition to the intermolecular forces that exist in pure substances, iondipole, ion-induced dipole, and dipole-induced dipole forces occur in solution.
If similar intermolecular forces occur in solute and solvent, a solution will likely form solution,
"like dissolves like". When ionic compounds dissolve in water, the ions become surrounded by
hydration shells of H-bonded water molecules.
1. What is the molecular formula of each compound?
(a) Empirical formula CH2 (M = 42.08 g/mol)
(b) Empirical formula NH2 (M = 32.05 g/mol)
2. (a) Cortisol (M = 362.47 g/mol), one of the major steroid hormones, is a key factor in the
synthesis of protein. Its profound effect on the reduction of inflammation explains its use
in the treatment of rheumatoid arthritis. Cortisol contains 69.6% C, 8.34% H, and 22.1%
O by mass. What is its molecular formula?
(b) A compound is found to contain 29.08% Na, 40.58% S, and 30.34% O by mass. What is
the empirical formula for this compound?
4. (a) Many metals react with oxygen gas to form the metal oxide.
For example, calcium reacts as follows:
2Ca(s) + O2(g) → 2CaO(s)
You wish to calculate the mass of calcium oxide that can be prepared from 4.20 g of Ca
and 2.80 g of O2.
(i) Which is the limiting reactant?
(ii) How many grams of CaO can form?
(b) Iron pyrites (FeS2) reacts with oxygen according to the following equation:
4FeS2 + 11O2 → 2Fe2O3 + 8SO2
27
If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced.
What is the percent yield of ferric oxide?
7. How many orbitals in an atom can have each of the following designations?
(a) 1s; (b) 4d; (c) 3p; (d) n = 3?
8. Give all possible ml, values for orbitals that have each of the following:
(a) l = 2; (b) n = 1; (c) n = 4, 1 = 3.
17. Are these statements true or false? Correct any that is false.
(a) Two bonds comprise a double bond.
(b) A triple bond consists of one
bond and two
bonds.
(c) Bonds formed from atomic s orbitals are always
(d) A
bond restricts rotation about the -bond axis.
(e) A
bond consists of two pairs of electrons.
bonds.
(f) End-to-end overlap results in a bond with electron density above and below the bond
axis.
19. How do the bonding and antibonding MOs formed from a given pair of AOs compare to each
other with respect to (a) energy; (b) presence of nodes; (c) internuclear electron density?
23 How does the energy of attraction between particles compare with their energy of motion in a
gas and in a solid? As part of your answer, identify two macroscopic properties that differ
between a gas and a solid.
24. (a) Why is the heat of fusion ( Hfus) of a substance smaller than its heat of vaporization ( Hvap)?
(b) Why is the heat of sublimation ( Hsub) of a substance greater than its Hvap?
(c) At a given temperature and pressure, how does the magnitude of the heat of vaporization
of a substance compare with that of its heat of condensation?
31. Calculate the Molality, polarity and mole fraction of the following and give the way in which
we prepare these solutions:
(a) A solution containing 88.4 g of glycine (NH2CH2COOH) dissolved in 1.250 kg of H2O
33. A dental hygienist uses x-rays (λ = 1.00 ) to take a series of dental radiographs while the
patient listens to a radio station ( = 325 cm) and looks out the window at the blue sky ( =
473 nm). What is the frequency (in s-1) and energy of the electromagnetic radiation from each
source? (Assume that the radiation travels at the speed of light (3.00 x l08 m/s.)
35. Using the periodic table and give the full and condensed electron configurations, partial
orbital diagrams showing valence electrons, and number of inner electrons for the following
elements:
(a) Ni (Z = 28) (b) Sr (Z = 38) (c) Po (Z = 84)
37. Permanganate ion is a strong oxidizing agent, and its deep purple color makes it useful as an
indicator in redox titrations. It reacts in basic solution with the oxalate ion to form carbonate
ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction
between NaMn04 and Na2C2O4 in acidic and basic solution:
MnO4-(aq) + C2O42-(aq) MnO2(S) + CO32-(aq)
28
39. Consider the phase diagram shown for substance X.
(a) What phase(s) is (are) present at point A? E? F? H? B? C?
(b) Which point corresponds to the critical point? Which point corresponds to the triple
point?
(c) What curve corresponds to conditions at which the solid and gas are in equilibrium?
(d) Describe what happens when you start at point A and increase the temperature at
constant pressure.
(e) Describe what happens when you start at point H and decrease the pressure at
constant temperature.
(f) Is liquid X more or less dense than solid X?
40. A sample of argon gas occupies 105 mL at 0.871 atm. If the temperature remains constant,
what is the volume (in L) at 26.3 kPa?
41. An engineer pumps air at 0 into a newly designed piston-cylinder assembly. The volume
measures 6.83 cm3. At what temperature (in K) will the volume be 9.75 cm3?
1. (a) C3H6
2. C21H30O5
4.
(a) 0.105 mol CaO
(b) 0.175 mol CaO (c) calcium (d) 5.88 g CaO
7. (a) one (b) five (c) three (d) nine
8. (a) ml:- 2, - 1 , 0, + 1 , + 2 (b)ml = 0 (if n = l , then l = 0) (c) ml,: - 3, -2, - 1 , 0, +1, +2, + 3
17.
(a) False. A double bond is one and one bond.
(b) False. A triple bond consists of one and two bonds.
(c) True (d) True (e) False. A bond consists of a second pair of electrons after a bond
has been previously formed.
(f) False. End-to-end overlap results in a bond with electron density along the bond axis.
19.
(a) Bonding MOs have lower energy than anti bonding MOs. Lower energy = more stable
23. In a solid, the energy of attraction of the particles is greater than their energy of motion; in a
gas, it is less. Gases have high compressibility and the ability to flow, while solids have
neither.
24. (a) because the intermolecular forces are only partially overcome when fusion occurs but
need to be totally overcome in vaporization.
(b) Because solids have greater intermolecular forces than liquids do.
(c) Hvap = Hcond
31.
(a) 0.942 m glycine (b) 1.29 m glycerol
29
4. Equilibrium in chemical reaction
Thermodynamic process and Thermochemistry
Chemical Thermodynamics is the study of the energetic of a chemical reaction.

Thermodynamics deals with the absorption or release of energy (generally as heat) that
accompanies chemical reactions.

These dynamics generally give us an idea of whether a reaction will be spontaneous

Spontaneous describes a process that can occur without outside intervention (i.e.
changing the temperature, pressure or concentration)

Spontaneity does not imply that the process occurs quickly, but rather describes a
capability to proceed.

If a chemical reaction is found to be spontaneous in one direction, then under the same
conditions, the reverse reaction will be non-spontaneous

This does not mean that the reaction cannot occur, but that it will need some outside help.
4.1. Some commonly used words
The System and the Surroundings
The system is the specific part of the universe that is of interest to us (i.e. the reaction with
reactants & products) and the surroundings is the rest of the universe

When performing experiments, we measure changes to the surroundings then apply those
findings to the system.

Energy is transferred between the system and the surroundings
 Energy that is transferred from the system to the surroundings has a negative sign
 Energy that is transferred from the surroundings to the system has a positive sign
ΔE = Efinal - EInitial
State Functions
30
A State Function is a function or property whose value depends only on the present state
(condition) of the system not the path used to arrive at that condition.

The change in a state function is zero when the system returns to its original condition.

For non-state functions, the change is not zero if the path returns to the original condition
Here’s an example with a chemical reaction:

The two paths below give the same final state:
o N2H4(g) + H2(g) → 2NH3(g) + heat (188 kJ)
o N2(g) + 3H2(g) → 2NH3(g) + heat (92 kJ)

State properties include for the formation of NH3: temperature, total energy, pressure,
density, and [NH3]

Non-state properties include for the formation of NH3: heat and reactants used
Energy
Energy is the capacity to do work, or supply heat.
Energy = Work + Heat

Potential energy, PE, is stored energy; it results from position or composition.

Kinetic energy, KE, is the energy matter has as a result of motion.
KE = 1/2 mv2
 Units for KE: 1 Joule = 1 kg.m2/s2 or 1 calorie = 4.184 J

Units of Energy the SI
unit for energy is the Joule (J)
 The J is a small amount of energy so scientists generally refer to Kilojoules (kJ)
 The calorie (cal) is another unit of energy, It is defined as the amount of energy
needed to raise the temperature of 1.00 g of water by one degree Celsius,
KE, Temperature & State
All substances have kinetic energy no matter what physical state they are in.
• Solids have the lowest kinetic energy, and gases have the greatest kinetic energy.
• As you increase the temperature of a substance, its kinetic energy increases.
Forms of Energy
There are six forms of energy: Heat, Radiant, Electrical, Mechanical, Chemical and Nuclear. We
can convert among these types of energy
31
Thermal Energy
Thermal Energy is the kinetic energy of molecular motion (translational, rotation, and vibration).
We measure this energy by finding the temperature of an object

Thermal energy is proportional to the temperature in degrees Kelvin


Ethermal α T(K)
Chemical Energy is the PE of a chemical compound where the chemical bonds act as
“storage” containers for the energy
Work
Work (w) is the application of a force (F) through a distance (d)
 This work produces an object’s movement
 If the system performs work on the surroundings, the sign of w is negative.
 If the surroundings perform work on the system, the sign of w is positive.
Work = Force • Distance
Units for W: Newton-meters (N-m) or Joules (J), 1J = 1N-m
Expansion Work
A system that contains one or more gases performs a specific type of work called “PV Work” or
expansion work. It perform work as the gas expands against an opposing pressure (P) exerted by
the surroundings.
w = - PΔV, ΔV = VFinal - VInitial, 1L • atm = 101 J

When a gas expands, w is negative because the system transfers energy to the surroundings

When a gas contracts, w is positive because the surroundings transfer energy to the system
Heat
Heat (q) is the energy that flows from a warmer object (higher temperature) to a cooler object
(lower temperature)

Heat is associated with the movement of particles

The faster the particles are moving, the more heat you generate (and vice versa)

If the system gives heat to the surroundings, the sign of q is negative.

If the surroundings give heat to the system, the sign of q is positive.
Zero Law of Thermodynamics
If substance ‘’A’’ is in thermal equilibrium with substance ‘’B’’ and if substance ‘’B’’ is in
thermal Equilibrium with ‘’C’’ always substance ‘’A’’ is in thermal equilibrium with ‘’C’’.
32
A
B
C
4.2. The First Law of Thermodynamics
It states that the internal energy (E) of a system, the sum of the kinetic and potential energy of all
its particles, changes when heat (q) and/or work (w) are added or removed, In a reaction, energy
is conserved. It is neither created nor destroyed.
Heat and/or work gained by the system is lost by the surroundings, and vice versa:
(q + w)sys = - (q + w)surr

It can be converted into a different type of energy though!

The energy change in a system equals the work done on the system + the heat added.
Total Energy of Reactants = Total Energy of Products
ΔE = Efinal – Einitial = q + w
4.3. Heat of reaction and thermochemistry
Most chemical reactions are performed in containers that are open to the atmosphere.

The volume of the system is allowed to change but the pressure (atmospheric pressure)
remains constant.

If the work performed is limited to PV work, a state function called Enthalpy (H), it can
be defined.
H = E + PV

The change in enthalpy (ΔH) is equal to the heat absorbed by the system under conditions
of constant pressure.
ΔH = ΔE + PΔV

There are two types of enthalpy changes:
1. The Enthalpy of Physical Change is the amount of heat required to change the state of matter
2. The Enthalpy of chemical change deals with the energy requirements of chemical reactions.
• The Heat of the Reaction (ΔHrxn) gives the amount of heat absorbed or released due to a
chemical reaction at constant pressure.
ΔHrxn = Hproducts – Hreactants
33
 In an Exothermic reaction, the reaction releases energy in the form of heat to the
surroundings (you feel the reaction get hot!)
o The energy of the reactants is greater than that of the products.

Eg. C(s) + O2 (g) → CO2 + Energy
 In an Endothermic reaction, the reaction absorbs energy in the form of heat from the
surroundings (you feel the reaction get cold!)
o The energy of the products is greater than that of the reactants.
Eg. 2H2O (l) + Energy → 2H2 (g) + O2 (g)
Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy
changes for the individual steps in the reaction.
Eg, 3H2(g) + N2(g) → 2NH3(g) ΔH° = –92.2 kJ
Reactants and products in individual steps can be added and subtracted to determine the overall
equation.
(a) 2H2(g) + N2(g) → N2H4(g)
ΔH°1 = ?
(b) N2H4(g) + H2(g) → 2 NH3(g)
ΔH°2 = –187.6 kJ
(c) 3 H2(g) + N2(g) → 2 NH3(g)
ΔH°3 = –92.2 kJ
ΔH°1 = ΔH°3 – ΔH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
Manipulating Chemical Enthalpy Changes
1. Reversing a reaction changes the sign of ΔH for a reaction:
Eg. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
3CO2(g) + 4H2O(l) → C3H8(g) + 5O2(g)
ΔH = –2219 kJ
ΔH = +2219 kJ
2. Multiplying a reaction increases ΔH by the same factor because ΔH is also
stoichiometrically related to the coefficients of the balanced chemical equation:
34
Eg. 3(C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
3C3H8(g) + 15O2(g) → 3CO2(g) + 12H2O(l)
ΔH = –2219 kJ)
ΔH = –6657 kJ
Standard Heats of Formation

Standard Heats of Formation (ΔH°f): The enthalpy change for the formation of 1 mole of
substance in its standard state from its constituent elements in their standard states.


C(s) + 2H2 (g) → CH4 (g) ΔHf° = -74.5 kJ
ΔH°f for any element in its standard state is defined as being zero.
ΔH°f = 0 for an element in its standard state
 The standard enthalpy change for any chemical reaction is found by subtracting the sum of the
heats of formation of all reactants from the sum of the heats of formation of all products
ΔH° = ΣΔHf° (Products) – ΣΔHf° (Reactants)
For a balanced equation, each heat of formation must be multiplied by the stoichiometric
coefficient.
aA + bB → cC + dD
ΔH° = [cΔH°f (C) + dΔH°f (D)] – [aΔH°f (A) + bΔH°f (B)]
Bond Dissociation Energy
There are over 18 million chemical compounds known at this time. Who do you think is making
all of the ΔH°f determinations for those compounds?
• It is possible to get an approximate ΔH°f for a compound based on the bonds present.
• The Bond Dissociation Energy can be used to determine an approximate value for ΔH°f.
ΔH°f = Reactant Bonds (Bonds Broken) – Product Bonds (Bonds Formed)
Introduction to Entropy

We have stated that chemical and physical processes occur spontaneously only if they go
“downhill” energetically (give off energy) so that the final state is more stable and lower in
energy than the initial state.

However, we know that reactions that require energy (endothermic reactions) will also occur;
this is possible because energy is more than just heat. There are other forms of energy present
in reactions that allow it to be spontaneous despite absorbing heat.

Another factor that affects spontaneity of a reaction is the molecular disorder of the reaction.
 The amount of molecular disorder in a system is called the system’s entropy; it has units
of J/K (Joules per Kelvin).
35
ΔS = Sfinal – Sinitial
 Positive value of ΔS indicates increased disorder and Negative value of ΔS indicates
decreased disorder.
Limitations of the 1st law of thermodynamics, it tells nothing about the direction of a
spontaneous change
Second Law of Thermodynamics:
Reactions proceed in the direction that increases the entropy of the system plus surroundings.
A quantitative statement of the second law is that, for any real spontaneous process,
ΔS = ΔSsys + ΔSsurr > 0
Entropy and Enthalpy
To decide whether a process is spontaneous, sometimes both enthalpy and entropy changes must
be considered:
Spontaneous process: Decrease in enthalpy (–ΔH) and Increase in entropy (+ΔS).
Non-spontaneous process: Increase in enthalpy (+ΔH) and Decrease in entropy (–ΔS).
Third law of thermodynamics
Both entropy and enthalpy are state functions, but the nature of their values differs in a
fundamental way. Recall that we cannot determine absolute enthalpies because we have no easily
measurable starting point, no baseline value for the enthalpy of a substance. Therefore, we
measure only enthalpy changes.
Third law of thermodynamics, states that a perfect crystal has zero entropy at a temperature of
absolute zero: Ssys = 0 at 0 K. "Perfect" means that all the particles are aligned flawlessly in the
crystal structure, with no defects of any kind. At absolute zero, all particles in the crystal have
the minimum energy, and there is only one way it can be dispersed
Entropy, Free Energy, and Work

The free energy change (ΔG) is a measure of the spontaneity of a process and of the useful
energy available from it.

free energy (G), is a function that combines the system's enthalpy and entropy:
G = H – TS
4.4. Spontaneous process and thermodynamic equilibrium
Spontaneous processes

Spontaneous process: takes place naturally, without continuous outside intervention
36
E.g. a drop of food coloring will spread in a glass of water or once ignited H2 gas
burns in O2 gas.

Nonspontaneous process: only takes place as a result of continuous intervention
E.g., electricity is required to convert water to H2 gas and O2 gas
Keep in mind that spontaneous processes may be fast or slow.
a. An explosion is fast and spontaneous.
b. The process of rusting is slow and spontaneous.
Consider the following spontaneous processes at room temperature:
 A drop of food coloring will spread in a glass of water.
 Methane (CH4) burns in O2 gas.
 Ice melts in your hand.
 Ammonium chloride dissolves in a test tube with water, making the test tube colder.
Example: What do these processes or reactions have in common? Are they all exothermic?
Are they all endothermic?
4.5. Chemical equilibrium
The concept of equilibrium and the equilibrium constant
Many chemical reactions do not go to completion but instead attain a state of chemical
equilibrium.
Chemical equilibrium: A state in which the rates of the forward and reverse reactions are equal
and the concentrations of the reactants and products remain constant. It is a dynamic process.
The conversions of reactants to products and products to reactants are still going on, although
there is no net change in the number of reactant and product molecules.
E.g. For the reaction: N2O4(g) ⇋ 2NO2(g)
1. The Equilibrium Constant
For a reaction: aA + bB ⇋ cC + dD
Equilibrium constant: Kc =
37
The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the
equilibrium concentrations of reactants each raised to the power of their stoichiometric
coefficients.
E.g. Write the equilibrium constant, Kc, for N2O4(g) ⇋2NO2(g)
Law of mass action - The value of the equilibrium constant expression, Kc, is constant for a
given reaction at equilibrium and at a constant temperature. The equilibrium concentrations of
reactants and products may vary, but the value for Kc remains the same.
Other Characteristics of Kc
1) Equilibrium can be approached from either direction.
2) Kc does not depend on the initial concentrations of reactants and products.
3) Kc does depend on temperature.
Magnitude of Kc

If Kc >> 1, the equilibrium lies to the right and the reaction mixture contains mostly products

If Kc <<1, the equilibrium lies to the left and the reaction mixture contains mostly reactants.

If 0.10 < Kc < 10, the mixture contains appreciable amounts of both reactants and products.
Writing equilibrium constant expressions
Calculating Equilibrium Constants, Kc
Kc values are listed without units
If equilibrium concentrations are known, simply substitute the concentrations into the
equilibrium constant expression:
Example- For the reaction, CO + 3H2 ⇋ CH4 + H2O,
Calculate Kc from the following equilibrium concentrations: [CO] = 0.0613 M; [H2] = 0.1839
M; [CH4] = 0.0387 M; [H2O] = 0.0387 M.
Heterogeneous Equilibria and Homogeneous equilibria
Homogeneous equilibria: reactants and products exist in a single phase.
For the gas phase reaction: N2O4(g) ⇋ 2NO2(g)
The equilibrium constant with the concentrations of reactants and products expressed in terms of
molarity, Kc, is:
Kc =
Gas Phase Expressions can also be expressed by Kp
38
⇒ The Kp expression is written using equilibrium partial pressures of reactants & products.
For the reaction given above, the Kp expression is:
Kp =
Kp is related to Kc
Since pressure and molarity are related by the Ideal Gas Law (PV = nRT), the following equation
relates Kp and Kc:
Kp = Kc(RT)Δn
Where R = 0.0821 L.atm/ K mol, T = temperature in Kelvin
Δn = moles of gaseous products, moles of gaseous reactants
Note that Kc = Kp when the number of gas molecules are the same on both sides.
Example- Does Kc = Kp for (a) H2(g) + F2(g) ⇋ 2HF(g)? (b) 2SO2(g) + O2(g) 2SO3(g)?
Example- For the reaction, 2SO2(g) + O2(g) ⇋ 2SO3(g) (a) write the equilibrium constant
expression, Kp. (b) What is the value for Kp if Kc = 2.8x102 at 1000 K?
Heterogeneous Equilibria and Solvents in Homogeneous Equilibria
Heterogeneous equilibria: reactants and products are present in more than one phase.
Pure solids and liquids: concentrations of pure solids and liquids are fixed by their density and
molar mass (both constants) and do not vary with the amount.
M=
=
=
x
Thus, the concentrations of solids and liquids are incorporated in the Kc value; they are not part
of the variable Kc expression:
Example: Write the Kc expression for CaCO3(s) ⇋ CaO(s) + CO2(g)

Omit concentration terms for solids and liquids from Kc and Kp expressions; only include
terms for gases (g) and aqueous substances (aq).
Example-Write the Kc expression for the following reaction:
3Cu(s) + 2NO3-(aq) + 8H+(aq) ⇋ 3Cu2+(aq) + 2NO(g) + 4H2O(l)
The relationship between chemical kinetics and chemical equilibrium
For the reaction N2O4(g)⇋ 2NO2(g)
Rate of forward reaction = Rate of reverse reaction
kf[N2O4] = kr[NO2]2
39
Rearrange:
=
= Kc
Thus, the equilibrium constant is simply the ratio of the forward and reverse rate constants which
are both constant values at a given temperature.
What does the equilibrium constant tell us?
 Predicting the Direction of Reaction
The reaction quotient, Q, is the resulting value when we substitute reactant and product
concentrations into the equilibrium expression.
1. If Q > K, the reaction will go to the left.
• The ratio of products over reactants is too large & the reaction will move toward
equilibrium by forming more reactants.
2. If Q < K, the reaction will go to the right.
• The ratio of products over reactants is too small & the reaction will move toward
equilibrium by forming more products.
3. If Q = K, the reaction mixture is already at equilibrium, so no shift occurs.
Example: For the reaction, B
2A, Kc = 2. Suppose 3.0 moles of A and 3.0 moles of B are
introduced into a 2.00 L flask.
(a) In which direction will the reaction proceed to attain equilibrium?
(b) Will the concentration of B increase, decrease or remain the same as the system
moves towards equilibrium?
Factors that affect chemical equilibrium
Le Chatelier's Principle: If a system at equilibrium is disturbed by an external stress, the system
adjusts to partially offset the stress as the system attains a new equilibrium position.
1. Changes in Concentration

Adding a reactant or product, the equilibria shifts away from the increase in order to
consume part of the added substance.

Removing a reactant or product, the equilibria shifts toward the decrease to replace part
of the removed species.
E.g. for H2 + I2 ⇋ 2HI, does the equilibria shift left or right if we:
a) add H2? b) Remove I2?
Changes in Volume and Pressure
40
Because the pressure of gases is related directly to the concentration changing the pressure by
increasing/decreasing the volume of a container will disturb an equilibrium system.

If P increases (V decreases), the system shifts to the side with a smaller number of gas
molecules (this effectively reestablishes equilibrium by decreasing the pressure).

If P decreases (V increases), the system shifts to the side with a greater number of gas
molecules.
Example: For N2(g) + 3H2(g) ⇋ 2NH3(g), does the equilibrium shift left or right if the pressure is
increased?
Changes in Temperature
Heat can be considered a reactant in an endothermic rxn and a product in an exothermic rxn.
Endothermic (ΔH > 0) R + Heat⇋ Products
Exothermic (ΔH < 0) R ⇋ Products + Heat
Recall that both Kc and the position of the equilibrium system will vary with temperature:

Kc is larger when the reaction shifts right. This occurs if T is increased for an
Endothermic Reaction or T is decreased for an exothermic reaction.

Kc is smaller when the reaction shifts left. This occurs if T is decreased for an
Endothermic Reaction or T is increased for an exothermic reaction.
Example: If the temperature is decreased for the reaction: 2CO2 ⇋ 2CO + O2, ΔH = 566 kJ.
a) Will the equilibrium shift left or right? b) Does Kc become larger or smaller?
Effect of a Catalyst

Catalysts lower Ea for the reaction, so a catalyst decreases the amount of time taken to
reach equilibrium for both the forward and reverse reactions.

The catalyst does not affect the equilibrium concentrations of reactants and products in
the equilibrium mixture; thus, the Kc value does not change.
4.6. Acid base equilibrium
Acid-base reactions
We are going to be working with acid-base equilibria in aqueous solution, and we will use the
Brønsted-Lowry definitions that an acid is a source of H+ and a base is an acceptor of H+.
Acid ionization constant Ka
A convenient way to write the reaction of an acid HA in water is
HA(aq) + H2O(l) ⇋ H3O+(aq) + A-(aq)
41
Here water is acting as a base, accepting the H+; the result, H3O+, called the conjugate acid of
H2O, since H3O+ can donate H+ to reform H2O. In a similar way, A- is called the conjugate base
of the acid HA, since A- can accept H+ to reform HA. The equilibrium constant is known as the
acid ionization constant Ka,
HA(aq) + H2O(l) ⇋ H3O+(aq) + A-(aq)
ka =
With the understanding that […] stands for the numerical value—without units—of the
concentration in mol/L. As usual, Ka is unit-less. An example acid ionization is
CH3COOH(aq)
H+(aq) + CH3COO- (aq)
Kc =
If Ka is much greater than 1, the acid is mostly dissociated and so is said to be a strong acid. If
Ka is much less than 1, the acid is dissociated only to a small extent and so is said to be a weak
acid.
Base ionization constant Kb
Similarly, we can write the reaction of a base B in water as
H2O(l) + B(aq) ⇋ HB+ (aq) + OH-(aq).
Here water is acting as an acid, donating the H+; the result, OH-, is called the conjugate base of
H2O, since OH- can accept H+ to reform H2O. Analogously, HB+ is called the conjugate acid of
the base B, since HB+ can donate H+ to reform B. The equilibrium constant is known as the base
ionization constant Kb,
kb
Here is an example base ionization
NH3 + H2O ⇋ NH4+ + OH-
,
kb
If Kb is much greater than 1, the base reacts nearly completely with water and so is said to be a
strong base. If Kb is much less than 1, the base reacts hardly at all with water and so is said to be
a weak base.
Dissociation of water and pH of aqueous solutions
Fredrich Kohlrausch, around 1900, found that no matter how pure water is, it still conducts a
minute amount of electric current. This proves that water self-ionizes.
42
Since the water molecule is amphoteric, it may dissociate with itself to a slight extent.
Only about 2 in a billion water molecules are ionized at any instant!
H2O(l) + H2O(l) ⇋ H3O+(aq) + OH−(aq)
The equilibrium expression used here is referred to as the autoionization constant for water
In pure water or dilute aqueous solutions, the concentration of water can be considered to be a
constant (55.6 M), so we include that with the equilibrium constant and write the expression as:
Kw = [H3O+][OH−] = [H3O+]2 = [OH–]2= 1.008 × 10−14 at 25°C = Ka × Kb
Knowing this value allows us to calculate the OH− and H+ concentration for various situations.
[OH−] = [H+] solution is neutral (in pure water, each of these is 1.0×10−7)
[OH−] > [H+] solution is basic
[OH−] < [H+] solution is acidic
The pH Scale
The pH and pOH scales provide a convenient way to express the acidity and basicity of dilute
aqueous solutions. The pH scale ranges from 0 to 14. The pH and pOH of a solution are defined
as
pH = −log [H+] → pOH = −log[OH−]→ pH + pOH = 14
Calculate pH and concentration of OH- ion of a solution in which the concentration of H3O+ is
0.050mol/L.
Solution
pH = -log [H3O+] = -log [0.050]= 1.3 and pOH = 14 - pH = 13.7
[OH] = 10-pOH = 10-13.4 = ……
Exercise
1. Calculate either the [H+] or [OH−] from the information given for each of the following
solutions at 25°C, and state whether the solution is neutral, acidic, or basic.
1.0 × 10−5 M OH−
b. 1.0 × 10−7 M OH−
c. 10.0 M H+
Common Ion Effect
Addition of a substance having an ion in common with the equilibrium mixture of the common
ion causes the equilibrium to shift left; this suppresses the ionization of a weak acid or a weak
base. The source of the common ion is typically provided by adding a strong acid, a strong base
or a soluble salt to the equilibrium reaction mixture.
43
For example, ionization of weak acid, acetic acid is suppressed in the presence of sodium acetate
than when it is dissolved in pure aqueous solution.
Since acetate ion is common ion, according to Le-Chatelier’s Principle, the equilibrium of
CH3COOH shift to the left as CH3COO- combines with H3O+ to form non-ionized CH3COOH
and H2O. The result is a drastic decrease in [H3O+] in the solution and ionization of CH3COOH
also depressed.
What happens to the pH of the acetic acid solution if we add (a) HCl?
Buffer Solutions
Buffers are a solution which resists changes in pH upon addition of small amounts of an acid or
base on to it. It contains a conjugate acid–base pair with both the acid and base in reasonable
concentrations, the acidic component reacts with added strong bases and the basic component
reacts with added strong acids H+.
A buffers solution can be classified either as an acidic or basic buffers.

Acidic buffers (pH<7) are those composed of weak acid plus a soluble ionic salt of the
weak acid (conjugate salt of acid) e.g., CH3COOH plus NaCH3COO.

Basic buffers (pH>7) are composed of weak base plus a soluble ionic salt of the weak
base (conjugate salt of base) e.g., NH3 plus NH4Cl.
Acid-Base Titration
Acid-Base titration is based on the titration of bases by a standard acid (acidimetry) or titration of
acids by a standard base (alkalimetry). Acid-Base titration is also known as a neutralization
titration which is widely used to determine the amounts of acids and bases and it might also be
used to monitor the progress of reactions that produce or consume hydrogen ions.
The standard reagents used in acid/base titrations are always strong acids or strong bases, most
commonly HCl, HClO4, H2SO4, NaOH, and KOH. Weak acids and bases are never used as
standard reagents because they react incompletely with analytes. A titration curve in acid-base
titration is a plot of pH as ordinate versus the amount (usually volume) of acid or base added as
abscissa. It displays graphically the change in pH as acid or base is added to a solution and
shows how pH changes.
44
Titration curve is evaluated experimentally by determination of the pH at various stages during
the titration by a potentiometric method or it may be calculated from theoretical principles. In
order to construct the curve we have to calculate pH at three different stages: pre-equivalent
point, equivalence point, and post equivalent. In the titration of a:

strong acid and strong base – the salt is neutral, pH at equivalence point = 7

strong acid and weak base – the salt is acidic, pH at equivalence point < 7

strong base and weak acid – the salt is basic, pH at equivalence point > 7
Solubility Product Principle
Solubility, Solubility Equilibria and Solubility Product
Solubility-Solubility is the maximum concentration (in terms of the solid) of a substance that can
exist in solution before precipitation begins … if sufficiently seeded.
Suppose we add one gram of solid barium sulfate, BaSO4, to 1.0 liter of water at 25°C and stir
until the solution is saturated. Very little BaSO4 dissolves and careful measurements of
conductivity show that one liter of a saturated solution of barium sulfate contains only 0.0025
gram of BaSO4, no matter how much more BaSO4 is added. The BaSO4 that does dissolve is
completely dissociated into its constituent ions.
BaSO4(s) ⇋ Ba2+(aq) + SO42-(aq)
In equilibria that involve slightly soluble compounds in water, the equilibrium constant is called
a solubility product constant, Ksp. The activity of the solid BaSO4 is one. Hence, the
concentration of the solid is not included in the equilibrium constant expression. For a saturated
solution of BaSO4 in contact with solid BaSO4, we write
BaSO4(s) ⇋ Ba2+(aq) + SO42-(aq) and Ksp = [SO42-][Ba2+]
In general, the solubility product expression for a compound is the product of the
concentrations of its constituent ions, each raised to the power that corresponds to the number of
ions in one formula unit of the compound. The quantity is constant at constant temperature for a
saturated solution of the compound. This statement is the solubility product principle. The very
small amount of solid zinc phosphate, Zn3(PO4)2, that dissolves in water gives three zinc ions
and two phosphate ions per formula unit.
Zn3(PO4)2(s) ⇋ 3Zn2+ (aq) + 2PO43-(aq) Ksp = [Zn2+]3 [PO43- ]2 = 9.1 x 10-33
Generally, we may represent the dissolution of a slightly soluble compound and its Ksp
expression as
45
MyXz(s) ⇋ yMz+(aq) + zXy-(aq) and Ksp = [Mz+]y[Xy-]z
In some cases a compound contains more than two kinds of ions. Dissolution of the slightly
soluble compound magnesium ammonium phosphate, MgNH4PO4, in water and its solubility
product expression are represented as
MgNH4PO4(s) ⇋ Mg2+(aq) + NH4+(aq) + PO43-(aq) Ksp = [Mg2+][NH4+][PO43-] = 2.5 x10-12
The molar solubility of a compound is the number of moles that dissolve to give one liter of
saturated solution. If the molar solubility of a compound is known, the value of its solubility
product can be calculated.
We know the molar solubility of BaSO4. The dissolution equation shows that each formula unit
of BaSO4 that dissolves produces one Ba2+ ion and one SO42- ion.
BaSO4(s)
1.1x10-5mol/L
⇋
Ba2+(aq)
SO42-(aq)
+
1.1x10-5mol/L
1.1x10-5mol/L
In a saturated solution [Ba2+] = [SO42-] = 1.1x10-5 M.
Ksp = [SO42-][Ba2+] = (1.1x10-5mol/L) (1.1x10-5mol/L) = 1.2 x 10-10
46
5. Electrochemistry
Redox Reactions, Reducing and Oxidizing Agents
Redox reactions
The term ‘redox’ is used as an abbreviation for the processes of reduction and oxidation. These
two processes usually occur simultaneously. Redox reactions are electron transfer reactions. The
separate equations showing which substance gains electrons and which substance loses electrons
are known as half-equations. Oxidation is defined as the loss of electrons and reduction is
defined as the gain of electrons. Oxidizing agents are substances which accept electrons;
reducing agents are substances which donate electrons.
Example 1
When a piece of iron is immersed in copper (II) sulphate solution, it soon becomes coated with
copper. The reaction may be represented by the ionic equation:
Fe(s) + Cu2+(aq) ⇋ Fe2+(aq) + Cu(s)
This shows that iron is the reducing agent and it is oxidized to Fe2+(aq) :
Oxidation half-equation: Fe(s) ⇋ Fe2+(aq) + 2eThe oxidizing agent is Cu2+(aq) and it is reduced to copper :
Reduction half-equation: Cu2+(aq) + 2e- ⇋ Cu(s)
Redox Reactions in Electrochemical Cells and Electrode Potential
Electrochemical cell:
An electrochemical cell is a device which produces an electromotive force (e.m.f.) as a result of
chemical reactions taking place at the electrodes. An electrochemical cell is thus a device which
converts chemical energy into electrical energy. Each cell consists of two half-cells. In one halfcell an oxidation half-reaction takes place and in the other a reduction half-reaction takes place.

Cathode is the electrode at which reduction occurs. The

Anode is the electrode at which an oxidation takes place.
Galvanic, or voltaic, cells store electrical energy. The reactions at the two electrodes in such
cells tend to proceed spontaneously and produce a flow of electrons from the anode to the
cathode via an external conductor. An electrolytic cell, in contrast to a voltaic cell, requires an
external source of electrical energy for operation. Note that in the electrolytic cell, the direction
of the current is the reverse of that in the galvanic cell and the reactions at the electrodes are
reversed as well.
47
Examples
This combination of half-cells is an electrochemical cell (galvanic cells or voltaic cells).
Fig. Movement of charge in a galvanic cell
Salt bridge:
Simple electrochemical cells used in a laboratory often consist of two half-cells separate by a salt
bridge. It may consist of an inverted U-tube containing salt solution. It is plugged at both ends by
cotton wool. Alternatively it may consist of strip of filter paper soaked in salt solution. Salts used
for this purpose include NH4NO3, KNO3 and KCl. The salt used is chosen so that it does not
react with the ions in either half-cell.
Salt bridge thus serves two main functions:

It completes the circuit by allowing ions carrying charge to move towards one half-cell
from the other.

It provides cations and anions to replace those consumed at the electrodes.
Difference between galvanic and Electrolytic Cells
Electrochemical cell (Galvanic Cell)
Electrolytic cell
A Galvanic cell converts chemical energy into An electrolytic cell converts electrical energy into
electrical energy.
chemical energy.
Here, the redox reaction is spontaneous and is
responsible for the production of electrical energy.
The redox reaction is not spontaneous and
electrical energy has to be supplied to initiate the
reaction.
The two half-cells are set up in different containers,
being connected through the salt bridge or porous
partition.
Both the electrodes are placed in a same container
in the solution of molten electrolyte.
Here the anode is negative and cathode is the positive Here, the anode is positive and cathode is the
48
Electrochemical cell (Galvanic Cell)
Electrolytic cell
electrode. The reaction at the anode is oxidation and negative electrode. The reaction at the anode is
that at the cathode is reduction.
oxidation and that at the cathode is reduction.
The electrons are supplied by the species getting The external battery supplies the electrons. They
oxidized. They move from anode to the cathode in the enter through the cathode and come out through
external circuit.
the anode.
Measurement of electromotive force (e.m.f.) in electrochemical cells:
By using a high resistance voltmeter, the current in the external circuit is virtually zero and the
cell registers its maximum potential difference. This maximum p.d. is called the e.m.f. This
e.m.f. gives a quantitative measure of the likelihood of the redox reaction taking place in the cell.
Ecell is the e.m.f. of the cell in volts. Changes that occur spontaneously have positive e.m.f., while
changes that do not occur spontaneously have negative e.m.f. We can set up electrochemical
cells with the Cu (s)|Cu2+(aq) half-cell as the oxidation half-cell another metal-metal ion
(Ag+(aq)/Ag(s)) half-cells as the reduction half-cell. E.m.f.
Short-Hand Cell notation: Convention: Anode on Left
Cu (s) /Cu2+(aq) (0.1M) ‖ Ag+(aq)(0.1M) / Ag(s) Liquid-liquid interface
Electrode Potential
Standard Reduction Potentials
• Each half-reaction has a cell potential
• Each potential is measured against a standard which is the standard hydrogen electrode
[consists of a piece of inert Platinum that is bathed by hydrogen gas at 1 atm].
The hydrogen electrode is assigned a value of zero volts.
• Standard conditions—1 atm for gases, 1.0M for solutions and 25°C for all (298 K)
• Naught, °-we use the naught to symbolize standard conditions [Experiencing a thermo
flashback?] That means
Ecell, Emf, or εcell become Ecello, Emfo , or εcello when measurements are taken at standard
conditions. You’ll soon learn how these change when the conditions are nonstandard!
The diagram below (c) illustrates what really happens when a Galvanic cell is constructed from
zinc sulfate and copper (II) sulfate using the respective metals as electrodes.
49
(c)

Notice that 1.0 M solutions of each salt are used

Notice an overall voltage of 1.10 V for the process

Reading the reduction potential chart
Elements that have the most positive reduction potentials are easily reduced (non-metals)
Elements that have the least positive reduction potentials are easily oxidized (metals)
 The table below used to tell the strength of various oxidizing and reducing agents.
The electrochemical series
When redox systems are arranged in order of their standard electrode potentials, the
electrochemical series is obtained. The following table shows standard electrode potentials of
some common redox systems:
Uses of the standard electrode potential values
1. To compare the strength of oxidizing/reducing agents
The standard electrode potential of a half-cell is a measure of the oxidizing or reducing power of
the species in it, i.e. their ability to compete electrons. In general the stronger an oxidizing agent,
the more positive its electrode potential and a strong reducing agent has a large negative
electrode potential.
Examples
Ozone gas is a more powerful oxidizing agent than chlorine gas:
O3(g) + 2H+(aq) + 2e- ⇋ O2(g) + H2O(l) EO = + 2.08 V
Cl2(g) + 2e- ⇋ 2 Cl- (aq) EO = + 1.36 V
Calcium metal is a more powerful reducing agent than lead metal:
Pb2+(aq) + 2e- ⇋ Pb(s) EO = - 0.13 V
50
Ca2+(aq) + 2e- ⇋ Ca(s) EO = - 2.87 V
2. To calculate the e.m.f. of cells:
Calculating Standard Cell Potential Symbolized by E°cell or Emf° ORεcell
1. Decide which element is oxidized or reduced using the table of reduction potentials.
Remember: The more positive reduction potential gets to be reduced.
2. Write both equations AS IS from the chart with their voltages.
3. Reverse the equation that will be oxidized and change the sign of the voltage [this is now
E° oxidation]
4. Balance the two half reactions “do not multiply voltage values”
5. Add the two half reactions and the voltages together.
6. E°cell = E°oxidation + E°reduction (E°cathode - E°anode) °means standard conditions Devices that come
in handy when constructing spontaneous cell--one that can act as a battery:
Oxidation occurs at the anode (may show mass decrease)
51
Reduction occurs at the cathode (may show mass increase)
The electrons in a voltaic or galvanic cell always flow from the Anode to the cathode the cathode
is positive in galvanic (voltaic) cells
Example 1
Zn|ZnSO4 (aZn2+ = 1.00)||CuSO4(aCu2+ =1.00)|Cu
Anode
cathode
Zn2+ + 2e- ⇋ Zn
E0 = -0.763 V
Cu2+ + 2e- ⇋Cu
E0 = +0.337V
Zn reaction spontaneously backward - forms negative electrode - place of oxidation - anode
If a=1.00 M, E = E0:
Ecell = Ecathode - Eanode
= +0.337 - (-0.763) = +1.100 V
Applications of Galvanic Cells
Batteries: cells connected in series; potentials add together to give a total voltage.
Examples:

Lead-storage batteries (car)--Pb anode, PbO2 cathode, H2SO4 electrolyte

Dry cell batteries
 Acid versions: Zn anode, C cathode; MnO2 and NH4Cl paste
 Alkaline versions: some type of basic paste, ex. KOH
 Nickel-cadmium – anode and cathode can be recharged

Fuel cells
 Reactants continuously supplied (spacecraft –hydrogen and oxygen)
Electrolysis and Electrolytic Cells
Electrolysis- the use of electricity to bring about chemical change, Literal translation “split with
electricity”
Electrolytic cells [NON spontaneous cells]:
 Used to separate ores or plate out metals.
 Important differences between a voltaic/galvanic cell and an electrolytic cell:
1) Voltaic cells are spontaneous and electrolytic cells are forced to occur by using an electron
pump or battery or any DC source.
52
2) A voltaic cell is separated into two half cells to generate electricity; an electrolytic cell
occurs in a single container.
3) A voltaic [or galvanic] cell is a battery; an electrolytic cell NEEDS a battery
4) AN OX and RED CAT still apply BUT the polarity of the electrodes is reversed. The
cathode is Negative and the anode is Positive (remember E.P.A – electrolytic positive
anode). Electrons still flow FATCAT.
5) Usually use inert electrodes
Predicting the Products of Electrolysis:
If there is no water present and you have a pure molten ionic compound, then:
 the cation will be reduced (gain electrons/go down in charge)
 the anion will be oxidized (lose electrons/go up in charge)
If water is present and you have an aqueous solution of the ionic compound, then:
 You’ll need to figure out if the ions are reacting or the water is reacting.
 you can always look at a reduction potential table to figure it out but, as a rule of thumb:
 no group IA or IIA metal will be reduced in an aqueous solution
–
Water will be reduced instead.
 no polyatomic will be oxidized in an aqueous solution
–
Water will be oxidized instead.
*Since water has the more positive potential, we would expect to see oxygen gas produced at the
anode because it is easier to oxidize than water or chloride ion. Actually, chloride ion is the first
to be oxidized.
The voltage required in excess of the expected value (called the overvoltage) is much greater for
the production of oxygen than chlorine, which explains why chlorine is produced first. Causes of
overvoltage are very complex. Basically, it is caused by difficulties in transferring electrons from
53
the species in the solution to the atoms on the electrode across the electrode-solution interface.
Therefore, E values must be used cautiously in predicting the actual order of oxidation or
reduction of species in an electrolytic cell.
Half Reactions for the electrolysis of water
If Oxidized: 2H2O → O2 + 4H+ + 4e−
If Reduced: 2H2O + 2e− → H2 + 2OHApplications of electrolytic cells:
 production of pure forms of elements from mined ores
 Aluminum from Hall-Heroult process
 Separation of sodium and chlorine (Down's cell)
 Purify copper for wiring
 electroplating—applying a thin layer of an expensive metal to a less expensive one
 Jewelry --- 14 K gold plated
 Charging a battery --- i.e. your car battery when the alternator functions
 Corrosion—process of returning metals to their natural state, the ores.
 Involves oxidation of the metal which causes it to lose its structural integrity and
attractiveness.
 The main component of steel is iron.
 20% of the iron and steel produced annually is used to replace rusted metal!
 Most metals develop a thin oxide coating to protect them, patinas, tarnish, rust,
etc.
Rates of chemical and physical process
Chemical kinetics
I. Chemical Kinetics: The study of reaction rate or speed of a chemical reaction.
Factors that affect speeds of reactions and in how reactions can be controlled are
1. Chemical nature of the reaction- the ease with which the bonds can be broken and formed
affects the rate of the reaction. Depending the on the nature of the atoms, the speed at which
they are broken and formed will differ.
2. Concentration of the reactants- directly proportional to the rate of the reaction (if
Concentration increases → colliding molecules increase → effective collision increases → the
rate of reactions increase)
54
3. Temperature of the system- if temperature increases → KE of colliding molecules increase →
effective collision increases → the rate of reactions increase.
4. Catalysts- substances that increase the rates of chemical reactions without being used up.
It causes lower activation energy by providing a different mechanism for the reaction, and
thus a new, lower energy pathway.
II. Rates
A rate is the ratio of change concentration per unit of time.
Rate with respect to x =
, The units are mol L-1s-1.
III. Rate laws
Rate law is proportional to the product of the molar concentrations of the reactants. Rate law is
an expression that is used to calculate the rate of the reaction at any set of known values of
concentrations. You cannot predict the rate law from the overall reaction instead you must have
experimental data in order to predict the number of reactants in the rate law, and their exponents.
Order of the reaction- an exponent in the rate law corresponding to the reactant.

Occasionally there may be is a negative or fractional exponent; Negative exponents
indicate a term that should be in the denominator, which means that as the concentration
of the species increases, the reaction rate decreases.

Overall order of a reaction- the sum of the orders with respect to each reactant in the rate
law. Just add the individual orders to get the overall,

Zero order reactions- reactions that have rates that are independent of concentration of
any reactant.
55
The reaction order must be determined experimentally: by looking at the rates associated at the
differing concentrations of reactants the rate constant and the orders of the reactants can be
determined.
IV. Integrated rate laws give concentration as a function of timeThe relationship between the concentration of a reactant and time can be derived from the rate
law by integrating the rate law between zero and some specified time.
1. First order- the integral simplifies to
ln

] = [Ao]e-kt
= kt or can be rearranged to yield
If a linear plot is desired for the relationship between the concentration and time,
rearrange the equation above, and simplify to obtain:
ln[At] = -kt + ln[Ao] which is in the form of y = mx + b
2. Second order- starting with the rate = k [B]2 the equations can be simplified to:
which is then rearranged to give
=
-
= kt
+ kt in linear form.
B. Half LifeA way to describe how fast it reacts and it is the length of time required for the concentration of
the reactant to be decreased to half of its initial value

First order- t1/2 =
=
*Note-the half life of first order reactions are independent of concentration.

Second order- Dependent on concentration of the reactants.
t1/2 =
V. Reaction rate theories
A. Collision theory:
1. A postulate that the rate of a reactions proportional to the number of effective collisions per
second among the reactant molecules.
Rate
An effective collision is a collision that actually yields a product, dependent on concentration
of reactants, kinetic energy and orientation of collisions.
B. Transition state theory:
56
1. Head on collisions result in the slowing down of the colliding molecules. As this happens the
kinetic energy is being transferred into potential energy.
a. Molecules usually fly apart unchanged,
b. Some undergo a reaction and then products are formed.
2. Potential energy diagram- a way of visualizing the relationship between the activation energy
and the development of total potential energy.
a. The vertical axis is the change in potential energy.
b. Reaction coordinate (reaction progress) - the horizontal axis that represents the extent to
which the reactants have changed to products.
c. the activation energy is represented as a “hill” or barrier for the reactants to overcome to
transition into products.
3. Exothermic reaction- because the products have lower potential energy than that of the
reactants, there is a release of heat during the reaction, therefore a negative enthalpy. The
potential energy is decreasing during these reactions b/c transferred to kinetic energy of
products.
4. Endothermic reaction- the potential energy of the products is higher than that of the reactants.
Heat must be put into the system in order for the reactants to attain the necessary activation
energy to transition into the products. This results in a negative enthalpy.
a. Endothermic reaction
b. exothermic reaction
5. Transition state- the unstable chemical species that results upon a successful collision that
momentarily exists with partially formed and partially broken bonds, Sometimes referred to as
an activated complex.
a. Corresponds to the highest point on the potential energy diagram.
VI. Activation Energy equations
57

Arrhenius equation is an equation that relates the rate constant to the activation energy,
k=
In which A- is proportionality constant called the frequency factor, R is Gas constant 8.314 J
mol-1K-1,T= temp in Kelvin, k = the rate constant

Activation energy can be determined graphically. If you manipulate the above equation
with natural logs and a little algebra, the following equation results:
lnk = lnA – (EA/R)1/T which the independent x is 1/T; the dependent y is ln k.

If two rate constants can be obtained at two differing temperatures, activation energy can
be calculated using the following equation:
ln ( ) = -
(
-
)
VII. How catalysts affect rates
58
6. Materials
6.1. Structures and bonding in solids
Bonding in solids, relation between bonding, structure and properties of materials
It forms between a metal and a non metal.
There is electron transfer from the less electronegative atom to the more electronegative .
Bonding forces ⇒ F electrostatic attraction between opposite charged ions.
• Pure ionic bond: ideal.
⇒ Always exists covalent participation
Ionic compounds are crystalline solids, It is a non directional bond formed by strong electrostatic
interactions
Born-Haber cycle for LiF.
lattice energy:
Energy released when a mole of ionic solid is formed from its ions in the gas state. Many
properties are dependent on the lattice energy (melting point, hardness, thermal expansion
coefficient)
59
General properties of ionic compounds
Strong electrostatic attraction → High melting and evaporation points
Hard and brittle solids at room temperature
They do not conduct electricity (except in molten state or when dissolved in water) Water
soluble.
Covalent bond
Generally it forms between the non metallic elements of the periodic table
It forms by electron sharing
Polar covalent bond
60
Properties of the compounds with covalent bonds
Covalent Solids
Formed by a system of continuous covalent bonds, Non conductive lattices both in the solid and
in the molten state, Diamond, boron nitride, quartz (SiO2), silicon carbide (SiC)
Metallic bonds
Model of a sea of electrons
– Atomic nucleus surrounded from a sea of e-
61
Intermolecular Forces
Primary bonds: (strong)
- Ionic bond (> 150 kcal/mol): Transfer of e- ⇒ cations and anions ⇒ Fcoulomb (non directional)
- Covalent bond (> 50-150 kcal/mol): Sharing e- (directional)
- Metallic bond (> 20-120 kcal/mol): e- shared externally and very little tightened by the nucleus
(non directional)
Secondary bonds - weak (Van der Waals)
Permanent dipole (fig. a) 1-10 kcal/mol It forms between molecules:
• That presents constant dipole moment μ
• made of atoms with different electronegativity that are united with covalent bonding.
a
b
c
Fluctuating dipole (fig. b): < 2 kcal/mol. It forms in non polar molecules in crystalline lattice
• Instantaneous fluctuations of electron charge distribution ⇒ fluctuating dipoles
Hydrogen bond 7 kcal/mol in H2O (permanent dipole) (fig. c):
• In molecules with H and electronegative atoms (Polar covalent bond: O-H, N-H, F-H).
• Asymmetric distribution of charge ⇒ Permanent dipole
Bonding and Properties
62
General Properties of Materials
6.2. Inorganic materials
6.3. Polymeric materials and soft condensed matter
6.3.1. Polymeric materials
POLYMER
“Organic compound, natural or synthetic, with high molecular weight made of repetitive
structural units ” Large size chains formed from the covalent union of various monomer units
(macromolecule)
PLASTIC
1. Polymer whose fundamental property is plasticity (thermoplastic). It is deformed plastically
under the action of pressure and or heat.
2. Mixture (of a polymer with additives) that can be transformed by flowing or moulding in
liquid or molten state.
1. A substance that changes the rate of a chemical reaction without itself being used up.
a. Can speed up a reaction, positive catalysis,
b. Can slow down a reaction, negative catalysis.
2. Reduces the activation energy of the reaction, making it easier for the reactants to change into
products, allowing a larger fraction of molecules to attain activation energy.
63
3. Homogeneous catalysts- exist in the same phase as the reactants.
4. Heterogeneous catalysts- the catalyst is in a separate phase than the reactants.
1. How does an increase in pressure affect the rate of a gas phase reaction? Explain.

Reaction rate is proportional to concentration. An increase in pressure will increase the
concentration, resulting in an increased reaction rate.
2. How does an increase in temperature affect the rate of a reaction? Explain the two factors
involved.

An increase in temperature affects the rate of a reaction by increasing the number of
collisions between particles, but more importantly, the energy of collisions increases.
Both these factors increase the rate of reaction.
3. for the reaction
4A(g) + 3B(g) → 2C(g)
The following data were obtained at constant temperature:
Initial [A]
Initial [B]
Initial Rate
Exp.
(mol/L)
(mol/L)
(mol/L'min)
1
0.1 00
0.1 00
5.00
2
0.300
0.1 00
45.0
3
0.1 00
0.200
10.0
4
0.300
0.200
90.0
(a) What is the order with respect to each reactant? (b) Write the rate law. (c) Calculate k (using
the data from experiment 1).

second order - A; first order - B (b) rate = k[A]2[B] (c) 5.00 x 103 L2/mol2 ·min
4. The rate constant of a reaction is 4.7 x 10-3 s-1 at 25°C, and the activation energy is 33.6
kJ/mol. What is k at 75°C?

0.033 s-1
5. Does a catalyst increase reaction rate by the same means as a rise in temperature does?
Explain.

No. A catalyst changes the mechanism of a reaction to one with lower activation energy.
Lower activation energy means a faster reaction. An increase in temperature does not
influence the activation energy, but increases fraction of collisions with sufficient energy
to equal or exceed the activation energy.
64
6. Change in reaction conditions increases the rate of a certain forward reaction more than that of
the reverse reaction. What is the effect on the equilibrium constant and the concentrations of
reactants and products at equilibrium?
(a) Is K very large or very small for a reaction that goes essentially to completion? Explain.

If the change is one of concentrations, it results temporarily in more products and less
reactants. After equilibrium is reestablished, the Kc remains unchanged because the ratio
of products and reactants remains the same. If the change is one of temperature, [product]
and Kc increase and [reactant] decreases.
7. A certain reaction at equilibrium has more moles of gaseous products than of gaseous
reactants.
(a) Is Kc larger or smaller than Kp?
(b)Write a general statement about the relative sizes of Kc and Kp for any gaseous equilibrium.

(a) Smaller (b) Assuming that RT > 1 (T >12.2K), Kp > Kc if there are more moles of
products than reactants at equilibrium, and Kp < Kc if there are more moles of reactants
than products.
8. How would you adjust the volume of the reaction vessel and concentration of reactant in order
to maximize product yield in each of the following reactions?
(a) Fe3O4(s) + 4H2(g) ⇋ 3Fe(s) + 4H2O(g)
(b) 2C(s) + O2(g) ⇋ 2CO(g)

(a) no change (b) increase volume
9. Predict the effect of increasing the temperature on the amounts of products in the following
reactions:
(a) CO(g) + 2H2(g) ⇋ CH3OH(g) Hrxn = - 90.7 kJ
(b) 2NO2(g) ⇋: 2NO(g) + O2(g) (endothermic)

(a) amount decreases (b) amount increases
10. When a small amount of H3O+ is added to a buffer, does the pH remain constant? Explain.

The pH of a buffer decreases only slightly with added H3O +
13. Define oxidation and reduction in terms of electron transfer and change in oxidation number.

Oxidation is the loss of electrons and results in a higher oxidation number; reduction is
the gain of electrons and results in a lower oxidation number.
14. Consider the following balanced redox reaction
65
MnO4-(aq) + 10Cl-(aq) → 2Mn2+(aq) + Cl2(g)
(a) Which species is being oxidized?
(b) Which species is being reduced?
(c) Which species is the oxidizing agent?
(d) Which species is the reducing agent?
(e) From which species to which does electron transfer occur?

(a) Cl- (b) MnO4- (c) MnO4- (d) Cl- (e) from Cl- to MnO4(f) 8H2SO4(aq) + 2KMnO4(aq) + 10KCl(aq) → 2MnSO4(aq) + 5Cl(g) + 8H2O(l) + 6K2SO4(aq)
15. A voltaic cell is constructed with an Sn/Sn2+ half-cell and a Zn/Zn2+ half-cell. The zinc
electrode is negative.
(a) Write balanced half-reactions and the overall reaction.
(b) Draw a diagram of the cell, labeling electrodes with their charges and showing the
directions of electron flow in the circuit and of cation and anion flow in the salt bridge.

a) Oxidation: Zn(s) → Zn2+ (aq) + 2e-
Reduction: Sn2+ (aq) + 2e - →Sn(s)
Overall: Zn(s) + Sn2+ (aq) → Zn2+ (aq) + Sn(s)
16. Consider the following voltaic cell:
(a) In which direction do electrons flow in the external circuit?
(b) In which half-cell does oxidation occur?
66
(c) In which half-cell do electrons enter the cell?
(d) At which electrode are electrons consumed?
(e) Which electrode is negatively charged?
(f) Which electrode decreases in mass during cell operation?
(g) Suggest a solution for the cathode electrolyte.
(h) Suggest a pair of ions for the salt bridge.

(a) left to right (b) left (c) right (d) Ni (e) Fe (f) Fe (g) 1M NiSO4 (h) K+ and NO3 - (i) Fe
(j) From right to left
(k) Ox.: Fe(s) → Fe2+ (aq) + 2e-
Red.: N i2+ (aq) + 2e- → Ni(s)
Overall: Fe(s) + Ni2+ (aq) →Fe2+ (aq) + Ni(s)
17. In basic solution, Se2- and SO32- ions react spontaneously:
2Se2-(aq) + 2SO32- (aq) + 3H2O(l) → 2Se(s) + 6OH-(aq) + S2O32- (aq)
= 0.35 V
(a) Write balanced half-reactions for the process.
(b) If Eo is sulfite is -0.57V, calculate Eo for selenium

(a) Oxidation: Se2-(aq) → Se(s) + 2eReduction: 2SO32-(aq) + 3H2O(l) + 4e- → S2O32- (aq) + 6OH-(aq)
(b)
=
= -0.57 V - 0.35 V = -0.92 V
18. What product forms at each electrode in the aqueous electrolysis of the salts bellow?
CuSO4?

Anode: 2H2O(l) → O2(g) + 4H+ (aq) + 4e-
Cathode: Sn2+(aq) + 2e- → Sn(s)
19. A system receives 425 J of heat and delivers 425 J of work to its surroundings. What is the
change in internal energy of the system (in J)?
0J
20. Classify the following processes as exothermic or endothermic: (a) freezing of water; (b)
boiling of water; (c) a person is running; (d) a person growing; (e) wood being chopped; (f)
heating with a furnace.

(a)Exothermic (b) endothermic (c) exothermic (d) endothermic (e) endothermic (f)
exothermic
21. Calculate q when 12.0 g of water is heated from 20oC to 100oC.
4.0 x 103J
22. What is the difference between the standard heat of formation and the standard heat of
reaction?
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The standard heat of reaction
is the enthalpy change for any reaction where all
substances are in their standard states. The standard heat of formation
is the enthalpy
change that accompanies the formation of one mole of a compound in its standard state from
elements in their standard states.
21. Distinguish between the terms spontaneous and nonspontaneous. Can a nonspontaneous
process occur? Explain.

A spontaneous process occurs by itself, whereas a nonspontaneous process requires a
continuous input of energy to make it happen. It is possible to cause a nonspontaneous
process to occur, but the process stops once the energy source is removed. A reaction that
is nonspontaneous under one set of conditions may be spontaneous under a different set
of conditions.
22. How does the entropy of the surroundings change during an exothermic and endothermic
reaction? Other than the examples cited in text, describe a spontaneous endothermic process.

In an exothermic reaction, Ssurr > O, in an endothermic reaction, Ssurr < O, A chemical
cold pack for injuries is an example of an application using a spontaneous endothermic
process.
23. Which of these processes are spontaneous: (a) water evaporating from a puddle in summer;
(b) a lion chasing an antelope; (c) an unstable isotope undergoing radioactive disintegration?

(a), (b), and (c)
24. Predict the sign of
Ssys for each process: (a) a piece of wax melting, Positive; (b) silver
chloride precipitating from solution; negative (c) dew forming, negative
25. Predict the sign of SO for
(a) CaCO3(s) + 2HCl(aq)
(b) 2NO(g) + O2(g)
(c) 2KC1O3(s)
26. If
CaCl2(aq) + H2O(l) + CO2 (g) positive
2NO2(g)
2KCl(s) + 3O2(g)
negative
positive
< < 0 for a reaction, what do you know about the magnitude of K? Of Q?
K >> 1 . Q depends on initial conditions, not equilibrium conditions.
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