BASIC INTRODUCTORY CHEMISTRY, Summaries of Chemistry

Download BASIC INTRODUCTORY CHEMISTRY and more Summaries Chemistry in PDF only on Docsity!

MALIKI MUNIRATU AND HILARY IFIJEN

BASIC INTRODUCTORY

BASIC

INTRODUCTORY

CHEMISTRY 1

BY

MALIKI M.

MARCH 2019

DEDICATION

This book dedicated to God almighty for his infinite mercies and for making impossibilities

possible

ACKNOWLEDGMENT

We would like to give God all the glory and adoration for granting us the courage, strength and

for using many wonderful persons to make this possible.

The success of this book depended on materials and works of other authors. We therefore

gratefully acknowledge and appreciate these authors. Our deep and passionate apperception goes to Prof. V. U. Okojie for making out time to edit this book

We gratefully appreciate the Management and staff of Edo University Iyamho (EUI) and Rubber

Research Institute of Nigeria (RRIN). We are particularly grateful to the Vice Chancellor of Edo

University Iyamho, Engineer Prof. E.O. Aluyor, the Deputy Vice Chancellor, Prof. S. M. Omodia

and the Managing Director of Rubber Research Institute of Nigeria, Dr. A. I. Aigbodion. We also

want to acknowledge the support of members of our research group especially Dr. S.O. Omorogbe

and Dr. N. O. Ogbebor.

Our sincere and heart-felt gratitude also goes to our family who supported and encouraged us

throughout the writing and publication of this book. We are particularly grateful to Mr C.I. Maliki

and Mrs. Precious Olohi Ifijen

.

CONTENTS

Title

Page

Chapter one Atomic theory and nature of atoms

Chapter two The periodic table

Chapter three The wave mechanical model

Chapter four Chemical symbols formulae and equations

Chapter five Stoichiometry

Chapter six State of matter and the gas law

Chapter seven Chemistry of the group 1, 2 and 3 element

Chapter eight The general concept of acid, bases and salts

Chapter nine Acid-base Reaction

Chapter ten Electrochemistry

Chapter eleven Oxidation/reduction reaction

Chapter twelve Chemical equilibrium

Chapter thirteen Introduction to solution chemistry

Chapter fourteen

Nuclear Radioactive chemistry

CHAPTER ONE

ATOMIC THEORY AND NATURE OF ATOMS

1.0 INTRODUCTION

Chemistry studies the composition of matter and the changes it undergoes. Matter is defined as anything that has weight and can occupy space. All materials are made up of matter.

For centuries, philosopher, chemist and physicist tried to answer the question of what matter was made up of using a variety of experiments and observations.

1.1 OBJECTIVES

At the end of this chapter, students will be able to:

• Understand the atomic structure

i. The early ideas of the atom ii. The modern atomic models

• Know what matter is made up of the discoverer of the three main atomic particle
• Understand the limitations of all atomic models
• Calculate wavelength, frequency and energy of spectra lines of hydrogen atom

1.2 THE EARLY IDEAS OF THE ATOM

A Greek philosopher named Democritus once came up with a conceived idea that matter was composed of very tiny and indestructible particles. He stated that these particles were the smallest unit of matter which he called atomos – meaning indivisible in Greek. The theory was not generally accepted because it was a mere philosophy (guess). In 1803, John Dalton proposed the atomic theory as follows:

1. All elements are made up of atoms and atoms are indivisible and indestructible particles
2. Atoms of the same elements are exactly alike (identical) in size, shape, mass and chemistry. While atoms of different elements are different.
3. Atoms combine with atoms of other elements in simple whole-number ratios to form chemical compounds.

Dalton’s atomic theory became one of the foundations of chemistry and the dawn of quantitative analysis, because soon after the theory, scientific investigation became rapid. Curiosity and the urge to understand the world became increased.

1.3 JOSEPH JOHN THOMSON ATOMIC MODEL

Discovery of the electrons In 1897, J.J. Thomson discovered the electrons while investigating the electrical conductivity of gases at very low pressure. In his experiment, a very high potential difference (pd) was placed across a glass tube containing gas at a very low pressure. At that experimental condition, a glow coming from the negative terminal (cathode) was observed and it was attracted to the anode. A

Source : https://www.testandmeasurementtips.com/wp-content/uploads/2017/08/CRT-components.jpg FIGURE 1.1 THE CATHODE RAY TUBE

By the early 1900s, it became clear that cathode rays have identical properties regardless of the element used to produce them. Also, that all elements must contain negatively charged electrons and atoms are neutral. Therefore, there must be positive particles in the atom to balance the negative charge of the electrons. Thus J.J. Thomson later proposed a revised model for the atom called the plum pudding model. This is shown in figure 1.2. He describe an atom as a uniform positive shpere of matter ( the dough) in which negatively charged electrons (raisins) were stucked.

Source: https://askeyphysics.files.wordpress.com/2015/01/plum-pudding-model-thomson.png?w= FIGURE 1.2 THE PLUM – PUDDING MODEL

1.4 RUTHERFORD EXPERIMENT Discovery of the nucleus In 1910, Hans Gieger and Ernest Marsden under the supervision of Ernest (later called Lord) Rutherford discovered the nucleus. Streams of positively charged particles were fired at a thin sheet of gold (gold foil). Rutherford’s team observed that most of the alpha particles passed through the gold foil without any deflection at all, while a few of the particles (1:800) were highly deflected through quite large angle. Sometimes also, but less frequently, α-particles would bounce back in the direction from which it came. This observation conflicted with Thomson’s model, because it was anticipated that since the positive charge in the Thomson’s atom was evenly spread

out, the beam (α-particle) should have easily passed through the foil with little or no deflection. Based on this observation, Rutherford proposed that most of the particles passed through the foil without any deflection because the atom consisted of largely empty spaces and does not interact much with the α-particles. However, occasionally one of the α-particle interacted with the very tiny and dense positively charged center which he called the nucleus and the very large repulsion of this positive charged particle and the nucleus resulted in the large deviation in the path of the α-particles, even to the extent of deflecting the α-particle along the direction from which it came. This model was clearly different from Thompson’s.

Rutherford’s model though very useful, it could not adequately explain the arrangement of electrons round the nucleus nor did it answer the question of what prevented the electrons from falling into the nucleus. This is because unlike charges attract and according to classical physics a moving charged body continually loses energy while it spirals inwards and finally collapses into the middle. Rutherford model of an atom fail to explain why electron in a given atoms do not collapse into the nucleus when in motion and could not explain the position and arrangement of electrons around the nucleus.

Source:https://upload.wikimedia.org/wikipedia/commons/thumb/c/c3/Rutherford_gold_foil_experiment_res ults.svg/672px-Rutherford_gold_foil_experiment_results.svg.png FIGURE 1.3 RUTHERFORD EXPERIMENT (EXPECTED AND ACTUAL RESULT)

Source:https://lh3.googleusercontent.com/proxy/YO_iH5ssP5OOKMzIjUHPLvOtP- 6UG6zQdxsTlOCoqa68AVziwWpA4umre_5v4O4b11cmKrwUFB86nRpJMKMBYO4A8XyfZDN0vLNgPa P50FpRQ FIGURE 1.4 RUTHERFORD’S MODEL

1.5 CHADWICK EXPERIMENT

Discovery of the neutrons After a series of experiments were conducted it was suspected that another type of subatomic particle exists. For example, experimental observations show that the mass of an oxygen atom was 16 times larger than the proton and oxygen has only eight positive charges, therefore, accounting for the remaining mass of the atom became a puzzle.

element and thus can be used in the qualitative and quantitative identification of elements. Since the wavelength of the emission spectrum tells us what the sample is and the intensity of each wavelength can tell us how much of the element is present.

Since all electromagnetic radiation have the same speed (i.e. speed is constant), then frequency of radiation relating to spectral line of wavelength (λ) is giving by the equation;

𝐶 λ ……………………. (Eq.1.2)

ƒλ = c ……………………. (Eq.1.3)

As can be seen from the equation, the wavelength of light increases as the frequency decreases

ƒ (or v) = frequency of radiation express as hertz (Hz) is equivalent to cycle “per second”

c = speed of light (3 ˟ 108 m/s.) λ = wavelength

The work of Max Planck has also showed that electromagnetic radiation may be regarded as stream of particles called photon. The energy carried by a photon is related to its frequency by the expression;

E = hƒ ……………………. (Eq.1.4)

But ƒ = 𝐶 λ ……………………. (Eq.1.2)

Therefore, E = ℎ𝐶 λ …………………….(Eq. 1.5)

Where h = Planck’s constant with a value of 6.63 ˟ 10 -34^ j-sec. A single photon carries one quantum of energy.

Example 1.1 Calculate the frequency of radiation having a wave length of 6.5 ˟ 10 -7meters.

Solution: ƒ =

𝐶 λ……………………. (Eq.1.2)

c =3 ˟ 108 m/s λ= 6.5 ˟ 10 -7m

3 x 108 m/s 6.5 X 10−7m = 4.6^ ˟^10

(^14) hertz

Example1.2 Calculate the frequency and energy of a photon having a wave length of 1. 7μm.

Solution: ƒ =

𝐶 λ ……………………. (Eq.1.2)

c =3 ˟ 108 m/s 1micrometer = 10-6m 1.7micrometer = 1 .7 ˟ 10 -6m

λ= 1.7 ˟ 10 -6m

3X 10^8 1.7 𝑋 10−6^ = 1.77^ ˟^10

(^14) hertz

E = hƒ ……………………. (Eq.1.4)

ƒ = 1.77 ˟ 1014 hertz

h = 6.63 ˟ 10 -34^ j-sec

E = 1.77 ˟ 1014 ˟ 6.63 X 10- E= 1.174 ˟ 10 -19j

1.6.3 BOHR’S MODEL OF THE HYDROGEN ATOM

Bohr based his theory of the hydrogen atom on the electromagnetic spectrum that hydrogen produces. At room temperature, hydrogen gas does not emit light but when electricity is passed through a discharge tube containing hydrogen gas at low pressure, the molecules break down into atoms and the tube glows with a reddish pink light. When this pink light is passed through a prism, individual patterns of lines will be produced called spectra. That is, the emission spectrum reveals that the atom emits radiation in the form of lines. Several series of discrete lines with each corresponding to a different wavelength in the electromagnetic spectrum are observed. Figure 1.6 shows the emission spectrum of atomic hydrogen.

FIGURE 1.6: THE ATOMIC HYDROGEN SPECTRUM, SHOWING THE FIRST FIVE SERIES OF THE SPECTRAL LINES.

It can be observed from the spectrum that the intensity and distance between the lines decreases as the frequency increases after which a continuum is observed. That is the lines in each series become more closely spaced at increase frequency (decreasing wavelength)

The wave length (λ) of the radiation is related to the frequency (ƒ) by the equation

c ƒ ……………………. (Eq.1.6)

ƒ (or v) = frequency of radiation express as hertz (Hz) is equivalent to cycle “per second”

c = speed of light (3 ˟ 108 m/s.) λ = wavelength (m) In spectroscopy, frequency is expressed as wave number (∇) and the wave number is the reciprocal of wave length. ∇ = ( 1 λ)^ m

-1 (^) ……………………. (Eq.1.7)

1 λ = 10967876(^

1 12 −^

1 32 ) 1 λ= 10967876(

1 1 −^

1 9 ) 1 λ= 10967876^ ˟^ 0. 1 λ = 9761409.64m

λ = ( 1 9761409.64)^ = 1.024^ ˟^10

-7m

Third line: ( 1 λ)^ = R^ (^

1 n1^2 −^

1 n2^2 )…………………….(Eq.1.9).

1 λ = 10967876(^

1 12 −^

1 42 ) 1 λ = 10967876(

1 1 −^

1 16 ) 1 λ = 10967876(

15 16 ) 1 λ = 10967876^ ˟^ 0. 1 λ = 10282383.75 m

λ = ( 1 10282383.75 )^ = 9.73^ ˟^10

-8m

Example1.4 Calculate wavelengths and the energy of transition of the second and third lines in the Balmer series of a hydrogen atom. h = Planck’s constant = 6.63 ˟ 10 -34j-sec, c = speed of light 3 ˟ 108 m/s

Solution: Frequency (∇) = ( 1 λ)^ = R^ (^

1 n1^2 −^

1 n2^2 )……………………. (Eq.1.8)

For the Balmer series, n 1 = 2 therefore, the second line has n 2 = 4 and the third line has n 2 = 5. Substituting these values into the Rydberg equation we have;

Second line: ( 1 λ)^ = R^ (^

1 n1^2 −^

1 n2^2 )……………………. (Eq.1.8)

1 λ = 10967876(^

1 22 −^

1 42 ) 1 λ = 10967876(

1 4 −^

1 16 ) 1 λ = 10967876(

4− 16 )

1 λ = 10967876(^

3 16 ) 1 λ = 10967876^ ˟^ 0. 1 λ = 2056476.75 m

λ = ( 1 2056476.75)^ = 4.863^ ˟^10

-7m

Energy of transition E = ℎ𝐶 λ

E = (

6.63 X 10−34𝑋 3𝑋10^8 4.863 X 10−7 )^ = 4.01^ ˟^10

-19j

Third line: ( 1 λ)^ = R^ (^

1 n1^2 −^

1 n2^2 )…………………….(Eq.1.8)

1 λ= 10967876(^

1 22 −^

1 52 ) 1 λ= 10967876(

1 4 −^

1 25 ) 1 λ= 10967876(

25− 100 ) 1 λ= 10967876(^

21 100 ) 1 λ = 10967876^ ˟^ 0. 1 λ = 2303253.96 m

λ = ( 1 2303253.96)^ = 4.34^ ˟^10

-7 (^) m

Energy of transition E = ℎ𝐶 λ

E = (

6.63 X 10−34𝑋 3𝑋10^8 4.34 X 10−7 )^ = 4.58^ ˟^10

-19j

1.6.4 BOHR’S POSTULATE

The summaries of Bohr’s postulates are as follow:-

1. The electron moves in circular orbits at different energies around the nucleus.
2. The radius of the orbit is quantized. That is orbits are at definite distances from the nucleus.
3. The electron revolve only in the orbit which have fixed value of energy, thus an electron in an atom can have only definite or discrete value of energy.
4. As long as the electron remains in a particular orbit, it neither loses nor gain energy. These orbits are known as stationary state or ground state.

Particle Symbol Mass (g) Relative electrical charge

Approximate relative mass (a m u)

Electron e-^ 9.091^ ˟^10

• 28 1 -

Proton p+^ 1.67252 ˟ 10 -^24 1+ 1

Neutron n^0 1.67495 ˟ 10 -^24 0

Note : amu is atomic mass unit

1.8 ATOMIC NUMBER AND MASS NUMBER The proton number is referred to as the atomic number. The atomic number is the whole number that increases across each row of the periodic table from left to right. The sum total of the number of protons and neutrons in the nucleus of an atom is the mass number. If an element X is represented thus: AZX^ (example is helium 24 He^ )

A = Mass number Z = Atomic number. Therefore A-Z = Neutron number

1.9 THE RELATIVE ATOMIC MASS The mass number of an atom cannot be weighed directly because it is very small; however experimental methods are used to determine the mass of one atom relative to another. The internationally agreed standard is an atom of carbon-12. This is an isotope of carbon with six protons and six neutrons, having a mass of 12 atomic mass unit (amu). Based on this standard,

1amu is exactly equal to one- twelfth the mass of one carbon-12 atom ( 1 12th, the mass of carbon 12). It is because the actual masses of the proton and neutron are very small that for practical purpose, the atomic mass unit (amu) is used.

Mass of one carbon-12 atom = 12 atomic mass unit therefore, 1 atomic mass unit = mass of one carbon−12 atom 12

The atomic mass unit is related to the number of protons and neutrons. It has nearly the same value as the mass number.

1.10 THE MASS SPECTROMETER The process of producing and analysing spectra is called spectroscopy and the instrument for analysing the spectra is called a spectrometer. The mass spectrometer is an instrument designed to separate atoms of slightly different masses. First the sample to be separated is vapourized so as to allow it to move through the machine. The vapourized sample is then hit with high energy electrons, knocking off one or more electrons thus converting the atoms into positively charged ions. The ions are accelerated to a higher speed by an electric and magnetic field; also because ions are charged particles they can be deflected by the electric and magnetic field. The degree of deflection is thus dependent on the mass to charge ratio. The masses of each particle are determined by measuring the exact strength of the field and the degree of deflection of the particles in the mass spectrometer. Different elements have different mass which corresponds to a different deflection. Figure 1.7 shows the mass spectrometer

Source: https://d3fhk62vq6ub16.cloudfront.net/uploads/2018/07/1-55.jpg FIGURE 1.7: THE MASS SPECTROMETER

1.11 MASS SPECTRA AND ISOTOPES The proton number (atomic number) tells us what an element is. Every element has its specific number of proton; however the number of neutrons and electrons may vary. For example carbon 12 has a proton number of 6 and a neutron number of 6 but carbon 14 has a proton number of 6 but a neutron number of 8 with a mass number of 14. This phenomenon is known as isotopy. Carbon 12 and carbon 14 are known as isotopes of carbon. Thus isotopy is defined as a phenomenon in which an element has the same number of proton but different number of neutrons. In other words isotopy is the existence of an element with the same atomic number but different mass number. If an element X is represented thus: AZX

A = the mass number (values are different for isotopes) Z = the atomic number (values are same for isotopes) Therefore, A-Z = Neutron number (will also be different for isotopes)

Majority of elements found in nature are a mixture of isotopes. For example hydrogen has three isotopes namely: hydrogen 11 H^ , deuterium 12 H^ and tritium 13 H^ , which is unstable and disintegrates. Almost all hydrogen exists as hydrogen with no neutron. Also, Bromine exists in nature as 3579 Br and bromine 3581 Br^. A mass spectrometer is used to find out the masses of these isotopes using the carbon -12 scale and the proportion of each isotope. Any naturally occurring sample of bromine contains approximately 50.52% of 3579 Br^ , and 49.48% of 3581 Br

1.12 CALCULATING RELATIVE ATOMIC MASSES FROM SPECTRA The mass spectrum of an element will show the presence of isotopes. From the relative heights of the peaks, we can work out the relative atomic mass of the element.

Example 1.3 : Calculate the relative atomic mass of bromine from the mass spectra of bromine

Solution : Relative abundance of 3579 Br^ = 50.52% Relative abundance of 3581 Br^ = 49.48%