G11 Chemistry Textbook PDF

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 Take Good Care of this Textbook This textbook is the property of your school. Take good care not to damage or lose it. Here are 10 ideas to help take care of the book: 1. Cover the book with protective material, such as plastic, old newspapers or magazines. 2. Always keep the book in a clean dry place. 3. Be sure your hands are clean when you use the book. 4. Do not write on the cover or inside pages. 5. Use a piece of paper or cardboard as a bookmark. 6. Never tear or cut out any pictures or pages. 7. Repair any torn pages with paste or tape. 8. Pack the book carefully when you place it in your school bag. 9. Handle the book with care when passing it to another person. 10.When using a new book for the first time, lay it on its back. Open only a few pages at a time. Press lightly along the bound edge as you turn the pages. This will keep the cover in good condition.  CHEMISTRY TEXTBOOK GRADE 11 Writers: Tsegaye Girma (PhD) Abera Gure (PhD) Editors: Chala Regasa (MSc) (Content Editor) Taye Hirpassa (BSc., MA) (Curriculum Editor) Meseret Getnet (PhD) (Language Editor) Illustrator: Asresahegn Kassaye (MSc) Designer: Daniel Tesfay (MSc) Evaluators: Tolessa Mergo Roro (BSc., MEd) Nega Gichile (BSc., MA) Sefiw Melesse (MSc.) FEDERAL DEMOCRATIC REPUBLIC OF ETHIOPIA HAWASSA UNIVERSITY MINISTRY OF EDUCATION  Foreword Education and development are closely related endeavors. This is the main reason why it is said that education is the key instrument in Ethiopia’s development and social transformation. The fast and globalized world we now live in requires new knowledge, skill and attitude on the part of each individual. It is with this objective in view that the curriculum, which is not only the Blueprint but also a reflection of a country’s education system, must be responsive to changing conditions. It has been almost three decades since Ethiopia launched and implemented new Education and Training Policy. Since the 1994 Education and Training Policy our country has recorded remarkable progress in terms of access, equity and relevance. Vigorous efforts also have been made, and continue to be made, to improve the quality of education. To continue this progress, the Ministry of Education has developed a new General Education Curriculum Framework in 2021. The Framework covers all pre-primary, primary, Middle level and secondary level grades and subjects. It aims to reinforce the basic tenets and principles outlined in the Education and Training Policy, and provides guidance on the preparation of all subsequent curriculum materials – including this Teacher Guide and the Student Textbook that come with it – to be based on active-learning methods and a competency-based approach. In the development of this new curriculum, recommendations of the education Road Map studies conducted in 2018 are used as milestones. The new curriculum materials balance the content with students’ age, incorporate indigenous knowledge where necessary, use technology for learning and teaching, integrate vocational contents, incorporate the moral education as a subject and incorporate career and technical education as a subject in order to accommodate the diverse needs of learners. Publication of a new framework, textbooks and teacher guides are by no means the sole solution to improving the quality of education in any country. Continued improvement calls for the efforts of all stakeholders. The teacher’s role must become more flexible ranging from lecturer to motivator, guider and facilitator. To assist this, teachers have been given, and will continue to receive, training on the strategies suggested in the Framework and in this teacher guide. Teachers are urged to read this Guide carefully and to support their students by putting into action the strategies and activities suggested in it. For systemic reform and continuous improvement in the quality of curriculum materials, the Ministry of Education welcomes comments and suggestions which will enable us to undertake further review and refinement.  Content CONTENT nit u 1 ATOMIC STRUCTURE AND PERIODIC PROPERTIES OF THE ELEMENTS 1.1 Introduction 2 1.2 Dalton’s Atomic Theory and the Modern Atomic Theory 3 1.2.1 Postulates of Dalton’s Atomic Theory................................... 3 1.2.2 Postulates of Modern Atomic Theory...................................... 5 1.3 Early Experiments to Characterize the Atom 6 1.3.1 The Discovery of the Electron.............................................. 6 1.3.2 Radioactivity and the Discovery of the Nucleus...................... 9 1.3.3 Discovery of the Neutron...................................................11 1.4 Make-up of the Nucleus 12 1.4.1 Subatomic Particles..........................................................12 1.4.2 Atomic Mass and Isotopes.................................................13 1.5 Electromagnetic Radiation and Atomic Spectra 15 1.5.1 Electromagnetic Radiation.................................................15 1.5.2 The Quantum Theory and Photon.........................................18 1.5.3 Atomic Spectra.................................................................23 1.5.4 The Bohr Model of the Hydrogen Atom..................................25 1.5.5 Limitations of the Bohr Model.............................................32 1.5.6 The Wave-Particle Duality of Matter and Energy.....................33 1.6 The Quantum Mechanical Model of the Atom 34 1.6.1 The Heisenberg’s Principle..................................................35 1.6.2 Quantum Numbers.............................................................36 1.6.3 Shapes of Atomic Orbitals...................................................39 1.7 Electronic Configurations and Orbital Diagrams 41 1.7.1 Ground State Electronic Configuration of the Elements............42 UNIT 1 I 1.8 Electronic Configurations and the Periodic Table of the Elements 47 1.8.1 The Modern Periodic Table..................................................47 1.8.2 Classification of the Elements.............................................48 1.8.3 Periodic Properties...........................................................49 1.8.4 Advantages of Periodic Classification of the Elements..............58 nit u 2 CHEMICAL BONDING 2.1 Introduction 67 2.1.1 The Octet Rule.................................................................68 2.1.2 Types of Chemical Bonding................................................68 2.2 Ionic Bonds 69 2.2.1 Lewis Electron-Dot Symbols...............................................71 2.2.2 Formation of Ionic Bonds..................................................72 2.2.3 Exceptions to the Octet Rule in Ionic Compounds..................76 2.2.4 Properties of Ionic Compounds............................................79 2.3 Covalent Bonds and Molecular Geometry 83 2.3.1 Molecular Geometry......................................................... 100 2.3.2 Intermolecular Forces in Covalent Compounds...................... 107 2.4 Metallic Bonding 111 2.4.1 Formation of Metallic Bonding........................................... 111 2.4.2 Electron-Sea Model......................................................... 112 2.4.3 Properties of Metals and Bonding..................................... 112 2.5 Chemical Bonding Theories 114 2.5.1 Valence Bond (VB) Theory............................................... 115 2.5.2 Molecular Orbital Theory (MOT)......................................... 131 2.6 Types of Crystal 135 II CHEMISTRY GRADE 11  Content nit u 3 PHYSICAL STATES OF MATTER 3.1 Introduction 145 3.2 Kinetic Theory and Properties of Matter 147 3.2.1 The Kinetic Theory of Matter............................................. 148 3.2.2 Properties of Matter......................................................... 148 3.3 The Gaseous State 150 3.3.1 The Kinetic Molecular Theory of Gases................................ 152 3.3.2 The Gas Laws.................................................................. 153 3.4 The Liquid State 173 3.4.1 Energy Changes in Liquids................................................ 174 3.5 The Solid State 180 nit u 4 CHEMICAL KINETICS 4.1 Introduction 193 4.2 The Rate of a Reaction 193 4.3 Factors Affecting the Rate of a Chemical Reaction 207 UNIT 1 III  nit u 5 CHEMICAL EQUILIBRIUM 5.1 Introduction 222 5.2 Chemical Equilibrium 222 5.2.1 Reversible and Irreversible Reactions................................. 224 5.2.2 Attainment and Characteristics of Chemical Equilibria........... 225 5.2.3 Conditions for Attainment of Chemical Equilibria................. 226 5.2.4 Equilibrium Expression and Equilibrium Constant.................. 228 5.2.5 Applications of Equilibrium Constant................................. 239 5.2.6 Changing Equilibrium Conditions: Le-Chatelier’s Principle...... 245 5.2.7 Equilibrium and Industry.................................................. 253 nit u 6 SOME IMPORTANT OXYGEN-CONTAINING ORGANIC COMPOUNDS 6.1 Introduction 264 6.2 Alcohols and Ethers 265 6.2.1 Classification of Alcohols.................................................. 267 6.2.2 Nomenclature of Alcohols................................................. 269 6.2.3 Physical Properties of Alcohols.......................................... 270 6.2.4 Preparation of Alcohols.................................................... 272 6.2.5 Chemical Properties of Alcohols......................................... 277 6.2.6 Structure and Nomenclature of Ethers................................. 283 6.2.7 Physical Properties of Ethers............................................. 284 6.2.8 Preparation of Ethers....................................................... 285 6.2.9 Reactions of Ethers.......................................................... 286 IV CHEMISTRY GRADE 11  Content 6.3 Aldehydes and Ketones 287 6.3.1 Nomenclature................................................................. 288 6.3.2 Physical Properties of Aldehydes and Ketones...................... 289 6.4 Carboxylic Acids 292 6.4.1 Structure and Nomenclature of Carboxylic Acids................... 292 6.4.2 Physical Properties of Carboxylic Acids................................ 300 6.4.3 Chemical Properties of Carboxylic Acids............................... 303 6.4.4 Preparation of Carboxylic Acids.......................................... 306 6.4.5 Fatty Acids..................................................................... 310 6.4.6 Uses of Carboxylic Acids................................................... 312 6.5 Esters 313 6.5.1 Sources of Esters............................................................. 313 6.5.2 Nomenclature................................................................. 313 6.5.3 Physical Properties.......................................................... 315 6.5.4 Chemical Properties......................................................... 316 6.5.5 Preparation of Esters....................................................... 318 6.5.6 Uses of Esters................................................................. 318 6.6 Fats and Oils 319 6.6.1 Source of Fats and Oils.................................................... 319 6.6.2 Structure of Fats and Oils................................................. 320 6.6.3 Physical Properties of Fats and Oils.................................... 321 6.6.4 Hardening of Oils............................................................ 322 6.6.5 Rancidity....................................................................... 323 UNIT 1 V !"#$ 1 ATOMIC STRUCTURE AND PERIODIC PROPERTIES OF THE ELEMENTS Unit Outcomes At the end of this unit, you will be able to: ) discuss the historical development of atomic structure ) explain the experimental observations and inferences made by some famous scientists to characterize the atom ) list the subatomic particles ) identify atomic mass and isotope terms ) explain electromagnetic radiation, atomic spectra and Bohr models of the atom ) compute the calculations involving atomic structure ) describe the quantum mechanical model of the atom and the related postulates and principles ) demonstrate periodic law and how electronic configurations of atoms are related to the orbital diagrams and can explain periodic trends ) describe scientific enquiry skills along this unit: inferring, predicting, classifying, comparing and contrasting, communicating, asking questions and making generalizations. UNIT 1 1  CHEMISTRY GRADE 11 Start-up Activity Discuss the following questions in groups, and present your ideas to the whole class. 1. What are the basic building blocks of the following substances? a. water b. chalk c. sugar d. table salt Are the basic building blocks of these substances the same or different? 2. Why do different materials show different properties? For example, materials such as woods can burn; an iron nail can rust; table salt dissolves in water, etc. 1.1 !"#$%&'(#)%" At the end of this section, you will be able to: ) compare the views of different philosophers on the nature of matter. In Grade 9, you learned about the historical development of the atomic theories of matter. In this unit, we will briefly revise it and discuss in detail about early experiments to characterize an atom, the atomic spectra, different models of an atom, etc. Activity 1.1 Form a group and discuss the following questions, then present your views to the whole class. 1. How do the beliefs about the structure of matter evolve? 2. Describe the early developments leading to the modern concept of the atom. 3. Can we see atoms with our naked eyes? The philosophers of ancient Greece wondered about the composition of matters: is matter continuously divisible into ever smaller and smaller pieces, or is there an ultimate limit? Although most philosophers, including Plato and Aristotle, believed that matter is continuous, Democritus disagreed. The Greek philosopher Democritus (460–370 BC) suggested that if you divided matter into smaller and smaller pieces, you would eventually end up with tiny, indestructible particles called atomos, or “atoms”, meaning “indivisible”. 2 UNIT 1  Dalton’s Atomic Theory and the Modern Atomic Theory His ideas were based on philosophical speculation rather than experimental evidence. The ideas of Democritus were not widely accepted until 1808. After nearly 2000 years, John Dalton, developed an atomic theory that had gained broad acceptance. 1.2 !"#$%&'()*$%+,-)./0%12)"&3)$/0)4%301&)*$%+,-)./0%12) At the end of this section, you will be able to: ) state postulates of Dalton’s atomic theory ) state postulates of the modern atomic theory ) state the laws of conservation of mass, definite proportions, multiple proportions and the basis of each of these laws ) use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions ) evaluate postulates of Dalton’s and the modern atomic theories. !"#"!$ %&'()*+(,' &-.+*(&/0' 1(&234 56,&78$$ Activity 1.2 Recall your Grade 9 knowledge in order to answer the following questions in groups, and share your responses with the whole class. 1. Describe the five postulates of Dalton’s atomic theory. 2. How do the scientific ideas develop based on previous scientific findings? Scientific laws usually develop based on previous scientific findings. The laws that are the basis for Dalton’s atomic theory are the law of conservation of mass and the law of definite proportions. What is the law that Dalton formulated based on the law of conservation of mass and the law of definite proportions? Write its statement in your notebook. UNIT 1 3  CHEMISTRY GRADE 11 Activity 1.3 In your Grade 9 chemistry lesson, you learned about the law of conservation of mass and the law of definite proportion. Form a group and discuss the following questions, then present your responses to the whole class. 1. The mass of a piece of wood before and after it is burnt to ashes is not the same. Does this show that mass is created or destroyed? 2. What would be the mass of products if the burning of wood was carried out in a closed container? 3. Sugar consists of C, H, and O atoms. When a certain amount of sugar is burned in a crucible, it changes from white sugar to black carbon. Where has the hydrogen and oxygen gone? Example 1.1 1. What mass of hydrogen and oxygen can be obtained from a. 18.0 g of water b. 1.00 g of water Solutions: Water is always 11.2% hydrogen and 88. 8% oxygen by mass, so: a. mass of hydrogen in 18.0 g water = 0.112 × 18.0 g = 2.02 g mass of oxygen in 18.0 g water = 0.888 × 18.0 g = 15.98 g; or = 18.0 g – 2.02 g = 15.98 g b. mass of hydrogen in 1.00 g water = 0.112 × 1. 00 g = 0.112 g mass of oxygen in 1.0 g water = 0.888 × 1.0 g = 0.888 g; or = 1.0 g – 0.112 g = 0.888 g 2. The following data were collected for several compounds of nitrogen and oxygen: Compound A Compound B Compound C Mass of nitrogen that combines with 1 g of oxygen 1.750 g 0.8750 g 0.4375 g Show how these data illustrate the law of multiple proportions. 4 UNIT 1  Dalton’s Atomic Theory and the Modern Atomic Theory Solution: For the law of multiple proportions to hold, the ratios of the masses of nitrogen combining with 1 gram of oxygen in each pair of compounds should be small whole numbers. We therefore compute the ratios as follows: A 1.750 4 B 0.875 2 C 0.4375 1 = = = = = = C 0.4375 1 C 0.4375 1 C 0.4375 1 These results support the law of multiple proportions. Exercise 1.1 1. List the postulates of Dalton’s that continue to have significance (are retained in the modern atomic theory). 2. How does the atomic theory account for the fact that when 1.00 g of water is decomposed into its elements, 0.112 g of hydrogen and 0.888 g of oxygen are obtained regardless of the source of the water? !"#"# $%&'()*'+&,%-,.%/+01,2'%345,67+%08 Activity 1.4 Remember what you have learned in Grade 9 and discuss the following questions in a group of three or four, then present your responses to the whole class. 1. Which of Dalton’s postulates about atoms are inconsistent with later observations? Do these inconsistencies mean that Dalton was wrong? 2. Is Dalton’s model still useful? Most of the experiments conducted during the development of the modern atomic theory will be discussed in Sections 1.3-1.6. In this section, generalizations derived from the experiments are presented as postulates of the modern atomic theory. The modern atomic theory is generally said to begin with John Dalton. Dalton’s work was mainly about the chemistry of atoms how they combine to form new compounds rather than about the internal structure of atoms. The modern theories about the physical structure of atoms did not begin until J.J. Thomson discovered the electron in 1897. UNIT 1 5  CHEMISTRY GRADE 11 Exercise 1.2 1. Describe the limitations of Dalton’s atomic theory. 2. Explain the postulates of the modern atomic theory. 3. List the three fundamental laws of chemistry. 4. How does the modern atomic theory explain the three fundamental laws of chemistry? 1.3 !"#$%&!'()#*+),-.&-/&01"#"2-)#*3)&-1)&4-/+& At the end of this section, you will be able to: ) discuss the discovery of the electron ) describe the properties of cathode rays ) define the terms: radioactivity, radioactive decay and radio-isotope ) describe the common types of radioactive emissions ) illustrate the alpha scattering experiment, and summarize and interpret the major contribution of experiments of Thomson, Millikan and Rutherford concerning atomic structure. 56765 81)&9*.2/:)#%&/;&-1)&!$)2-#/,& Activity 1.5 Think back to Grade 9 and reflect on the following questions in a group of four. Then, present your responses to the whole class. a. How were electrons discovered? b. Are cathode rays visible to the naked eye? 6 UNIT 1  Early Experiments to Characterize the Atom Historical note: J.J. Thomson (1856-1940) was a British physicist and Nobel laureate. Sir Joseph John Thomson was born near Manchester, England, and educated at Owens College (now part of Victoria University of Manchester) and Trinity College, University of Cambridge. At Cambridge he taught mathematics and physics, served as Cavendish Professor of Experimental Sir Joseph Thomson Physics, and was (1918-40) master of Trinity College. Thomson’s work on cathode rays led to the discovery of the electron in 1897. His later work with positive ion beams led to a method of separating atoms and molecules by mass and the discovery of neon. Thomson was awarded the Nobel Prize in Physics in 1906 and was knighted in 1908. Cathode Rays One of the first experiments on subatomic particles was carried out by the English physicist J.J. Thomson in 1897. Figure 1.1 shows an experimental apparatus similar to the one used by Thomson. In this apparatus, two electrodes from a high-voltage source are sealed into a glass tube from which the air has been evacuated. The negative electrode is called the cathode; the positive one, the anode. When the high-voltage current is turned on, the glass tube emits a greenish light. Experiments showed that this greenish light is caused by the interaction of the glass with cathode rays, which are rays that originate from the cathode. Figure 1.1: Formation of cathode rays UNIT 1 7  CHEMISTRY GRADE 11 After the cathode rays leave the negative electrode, they move toward the anode, where some rays pass through a hole to form a beam (Figure 1.1). This beam bends away from the negatively charged plate and toward the positively charged plate. What did Thomson conclude from this observation? Figure 1.2 shows a similar experiment, in which cathode rays are seen to bend when a magnet is brought toward them. Thomson showed that the characteristics of cathode rays are independent of the material making up the cathode. From such evidence, he concluded that a cathode ray consists of a beam of negatively charged particles (or electrons) and that electrons are constituents of all matter. Figure 1.2: Bending cathode rays using a magnet By measuring the amount of deflection of a cathode ray beam in electric and magnetic fields of known strengths, Thomson was able to calculate the ratio of the mass of an electron, me, to its charge, e. The number he came up with is – 5.686 × 10–12 kg C−1 (kilograms per coulomb). However, he could not obtain either the mass or the charge separately. In 1909, Robert A. Millikan, an American physicist, measured the charge of the electron by measuring the effect of an electrical field on the rate at which charged oil drops fell under the influence of gravity. Based on careful experiments, Millikan established the charge on an electron as e = –1.602 × 10–19 C. He used this value and Thomson’s mass/charge ratio to calculate an electron’s mass to be 9.109 × 10–31 kg. m me = e × e = −5.686 × 10−12 kg C−1 × − 1.602 × 10−19 C e = 9.109 × 10−31 kg 8 UNIT 1  Early Experiments to Characterize the Atom !"#"$ %&'()&*+(,(+-.&/'.+01.2(3*),14-.)5.+01.67*8173. Activity 1.6 Answer the following questions individually, then share your answers with the whole class. 1. Describe the properties of cathode rays. 2. Which rays are used to see whether bones are broken or not? Radioactivity Radioactivity or radioactive decay is the spontaneous emission of particles and/or radiation from the unstable nuclei of certain atoms such as uranium, radium, etc. Does radioactivity support Dalton’s idea of atoms? Shortly after the discovery of radioactivity, three types of rays were identified in the emissions from radioactive substances. Two are deflected by oppositely charged metal plates (Figure 1.3). Alpha (α) rays consist of positively charged particles, called α particles. They have a mass of about four times that of a hydrogen atom and a charge twice the magnitude of an electron; they are identical to helium nuclei. Beta (β) rays, or β particles, are electrons coming from inside the nucleus and are deflected by the negatively charged plate. The third type of radioactive radiation consists of high-energy rays called gamma (γ) rays. They have no charge and are not affected by an external electric or magnetic field. Figure 1.3: Three types of ray emitted by radioactive elements UNIT 1 9  CHEMISTRY GRADE 11 The Discovery of the Nucleus Thomson proposed a “plum-pudding” model (Figure 1.4) for the atom in which the electrons and protons were randomly distributed in a positively charged cloud like plums in a pudding. Figure 1.4: Thomson’s “plum-pudding” model of an atom In1911, Ernest Rutherford worked with Thomson to test this model. In Rutherford’s experiment, positively charged particles were aimed at a thin sheet of gold foil (Figure 1.5). If the Thomson model were correct, the particles would travel in straight paths through the gold foil. Rutherford was greatly surprised to find that some of the particles were deflected as they passed through the gold foil, and a few particles were deflected so much that they went back in the opposite direction. Figure 1.5: (a) α-particles aimed at a piece of gold foil, (b) Magnified view of α-particles passing through and being deflected by nuclei 10 UNIT 1  Early Experiments to Characterize the Atom Activity 1.7 Form a group and discuss Rutherford’s experiment as shown in Figure 1.5 (a) and (b) and answer the following questions: 1. Why did most of the α-particles pass through the foil undeflected? 2. Why did only a small fraction of the α-particles show a slight deflection? 3. Why didn't all α-particles bounce at an angle of 180°? 4. Based on the findings of Rutherford’s experiment, what is your conclusion about an atom? 5. When Rutherford’s co-workers bombarded gold foil with α-particles, they obtained results that overturned the existing (Thomson) model of the atom. Explain. Report your responses to the whole class. !"#"# $%&'()*+,-(.-/0*-1*2/+(3- The neutron was also discovered by alpha-particle scattering experiments. When beryllium metal is irradiated with alpha rays, a strongly penetrating radiation is obtained from the metal. In 1932 the British physicist James Chadwick (1891– 1974) showed that this penetrating radiation consists of neutral particles, called neutrons. The neutron is a nuclear particle having a mass almost identical to that of the proton but with no electric charge. The mass of a neutron, mn = 1.67493 × 10–27 kg, which is about 1840 times the mass of an electron. Activity 1.8 Discuss the following questions in groups, and present your responses to the whole class. 1. A sample of a radioactive element is found to be losing mass gradually. Explain what is happening to the sample. 2. Describe the experimental basis for believing that the nucleus occupies a very small fraction of the volume of the atom. UNIT 1 11  CHEMISTRY GRADE 11 1.4 !"#$%&'()*(+,$(-&./$&0 At the end of this section, you will be able to: ) describe the make-up of the nucleus ) define atomic mass ) define isotope, and calculate the relative atomic mass (atomic mass) of naturally occurring isotopic elements. 12321 4&5"+)67.(8"9+7./$0( In 1919, Rutherford discovered that hydrogen nuclei, or what we now call protons, form when alpha particles strike some of the lighter elements, such as nitrogen. A proton is a nuclear particle having a positive charge equal in magnitude to that of the electron. A proton has a mass of mp = 1.67262 × 10–27 kg, which is about 1840 times the mass of electrons. The protons in a nucleus give the nucleus its positive charge. Table 1.1 compares the relative masses and charges of the three subatomic particles (note that "amu" stands for "atomic mass unit", which is equal to 1 the mass of an 12 atom of carbon-12). Table 1.1: Properties of subatomic particles Particle Actual mass (kg) Relative mass Actual charge Relative (amu) (C) charge Proton (p) 1.672622 × 10-27 1.007276 1.602 × 10-19 +1 Neutron (n) 1.674927 × 10 1.008665 0 0 -27 Electron (e−) 9.109383 × 10 5.485799 × 10-4 -1.602 × 0-19 -1 -31 The atomic number (Z) of an element equals the number of protons in the nucleus of each of its atoms. All atoms of a particular element have the same atomic number, and each element has a different atomic number from that of any other element. The total number of protons and neutrons in the nucleus of an atom is its mass number (A). The mass number and atomic number of an element X are often written with the symbol, 12 UNIT 1  Make-up of the Nucleus Activity 1.9 Based on your previous knowledge, write the similarities and differences between the pairs of atomic notations and present your responses to the whole class. a. 29 30 14 Si and 14 Si b. 14 6 C and 147 N c. 79 35 81 Br and 35 Br d. 79 35 Br − and 79 36 Kr !"#"$ %&'()*+,-..+-/0+1.'&'23.+ All atoms of an element are identical in atomic number but not in mass number. Isotopes of an element are atoms that have different numbers of neutrons and different mass numbers. For example, all carbon atoms have six protons in the nucleus (Z = 6) but only 98.89 % of naturally occurring carbon atoms have six neutrons in the nucleus (A = 12). A small percentage (1.11 %) have seven neutrons in the nucleus (A = 13), and even fewer (less than 0.01 %) have eight (A = 14). Hence, carbon has three naturally occurring isotopes: 12C, l3C, and 14C. Most elements found in nature are mixtures of isotopes. The average mass for the atoms in an element is called the atomic mass of the element and can be obtained as averages over the relative masses of the isotopes of each element, weighted by their observed fractional abundances. If an element consists of n isotopes, of relative masses A1, A2…An and fractional abundances of f1, f2… fn, then the average relative atomic mass (A) of the element is: A = A1 f1 + A2 f2 + … + An fn. UNIT 1 13  CHEMISTRY GRADE 11 Example 1.2 There are two naturally occurring isotopes of silver. Isotope 107Ag (106.90509 amu) accounts for 51.84% of the total abundance, and isotope 109Ag (108.90476) accounts for the remaining 48.16%. Calculate the atomic mass of silver? Solution: Find the portion of the atomic mass from each isotope: Portion of atomic mass from 107Ag: = isotopic mass × fractional abundance = 106.90509 amu × 0.5184 = 55.42 amu Portion of atomic mass from 109Ag: = 108.90476 amu × 0.4816 = 52.45 amu Find the atomic mass of silver: Atomic mass of Ag = 55.42 amu + 52.45 amu = 107.87 amu Exercise 1.3 1. How many protons and neutrons are in the nucleus of each of the following atoms? a. 27 13 Al b. 32 16 S 64 207 Pb c. 30 Zn d. 82 2. Element X is toxic to humans in high concentration but essential to life at low concentrations. It has four naturally occurring isotopes that contain 24 protons. Identify element X and give the atomic symbol for the isotopes with 26, 28, 29, and 30 neutrons. 3. Naturally occurring boron consists of two isotopes, 10B and 11B, with the isotopic masses 10.013 amu and 11.009 amu, respectively. The observed atomic mass of boron is 10.811 amu. Calculate the abundance of each isotope 4. The two naturally occurring isotopes of lithium, lithium-6 and lithium-7, have masses of 6.01512 amu and 7.01600 amu, respectively. Which of these two occurs in greater abundance? 14 UNIT 1  Electromagnetic Radiation and Atomic Spectra 1.5 !"#$%&'()*+#%,$-.)/,)%,'+-)+/-0%'(,$-12#$%&) At the end of this section, you will be able to: ) characterize electromagnetic radiation (EMR) in terms of wavelength, frequency and speed ) calculate the wavelength and frequency of EMR ) explain the dual nature of light ) describe emission spectra of atoms as consisting a series of lines ) define a photon as a unit of light energy ) distinguish how the photon theory explains the photoelectric effect ) identify the relationship between a photon absorbed and an electron released ) state Bohr’s assumption of energy of the electron in an hydrogen atom ) calculate the radius of electron orbit, the electron velocity and the energy of an electron using Bohr's model ) explain that the line spectrum of hydrogen atom demonstrates the quantized nature of the energy of its electron ) explain that atoms emit or absorb energy when they undergo transitions from one state to another ) compose the limitations of Bohr’s theory. 34543 !"#$%&'()*+#%,$-.)/,)%,'+- Activity 1.10 Form a group and discuss the following questions, then present your responses to the whole class. 1. Explain how energy travels in space. 2. What is the importance of electromagnetic radiation (EMR) in chemistry? 3. What are the common features of different energy sources? UNIT 1 15  CHEMISTRY GRADE 11 In 1873, James Clerk Maxwell proposed that light consists of electromagnetic waves. According to his theory, an electromagnetic wave has an electric field and magnetic field components. These two components vibrate in two mutually perpendicular planes (Figure 1.6). EMR is the emission and transmission of energy in the form of electromagnetic waves. Figure 1.6: The electric field and magnetic field components of an electromagnetic waves Electromagnetic waves have three primary characteristics: wavelength, frequency and speed. Wavelength (λ, Greek lambda), is the distance the wave travels during one cycle (Figure 1.7). It is expressed in meters (m) and often, for very short wavelengths, in nanometers (nm), picometers (pm), or angstrom (Å). Frequency (ν, Greek letter nu) is the number of cycles the wave undergoes per second and is expressed in units of l/second (1/s; also called hertz, Hz). Figure 1.7: Frequency of waves 16 UNIT 1  Electromagnetic Radiation and Atomic Spectra Wave has a speed which depends on the type of wave and the nature of the medium through which it is traveling (for example, air, water, or a vacuum). The speed (c) of a wave is the product of its wavelength and its frequency: c = vλ (1.1) In a vacuum, electromagnetic waves travel at 3 ´ 108 m/s, which is a physical constant called the speed of light. EMR comes in a broad range of frequencies called the electromagnetic spectrum (Figure 1.8). A rainbow is an example of a continuous spectrum. Different wavelengths in visible light have different colors from red (λ = 750 nm) to violet (λ = 380 nm). Radiation provides an important means of energy transfer. For instance, the energy from the Sun reaches the Earth mainly in the form of visible and ultraviolet radiation. The glowing coals of a fireplace transmit heat energy by infrared radiation. In microwave ovens, microwave radiation is used to heat water in foods, causig the food to cook quickly. Figure 1.8: The electromagnetic spectrum UNIT 1 17  CHEMISTRY GRADE 11 Example 1.3 Ethiopian National Radio, Addis Ababa station broadcasts its AM signal at a frequency of 2400 kHz. What is the wavelength of the radio wave expressed in meters? Solution: We obtain the wavelength of the radio wave by rearranging Equation 1.1 c 3.00 × 108 m / s so, λ = = = 125.0 m v 2.4 ×106 s Exercise 1. 4 1. The most intense radiation emitted by the Earth has a wavelength of about 10.0 µm. What is the frequency of this radiation in hertz? 2. Addis Ababa Fana FM radio station, broadcasts electromagnetic radiation at a frequency of 98.1 MHz. What is the wavelength of the radio waves, expressed in meters? !"#"$ %&'()*+,-*.(%&'/01(+,2(3&/-/, Why are waves treated as particles? In 1900, Max Planck, the German physicist, came to an entirely new view of matter and energy. He made a revolutionary proposal, energy like matter is discontinuous. According to Planck, atoms and molecules could emit or absorb energy only in discrete quantities, like small packages or bundles. Each of these small “packets” of energy is called a quantum. The energy of a quantum is proportional to the frequency of the radiation. The energy E of a single quantum is given by: E = hν (1.2) Where h is called Planck’s constant and ν is the frequency of radiation. The value of Planck’s constant is 6.63 ´ 10-34 J. s. 18 UNIT 1  Electromagnetic Radiation and Atomic Spectra Since ν = c/λ, Equation 1.2 can also be expressed as: hc E= (1.3) λ According to quantum theory, energy is always emitted or absorbed in integral multiples of hν; for example, hν, 2hν, 3hν, etc. A system can transfer energy only in whole quanta. Thus, energy seems to have particulate properties. Example 1.4 The blue color in fireworks is often achieved by heating copper (I) chloride (CuCl) to about 1200 °C. Then the compound emits blue light having a wavelength of 600 nm. What is the increment of energy (the quantum) that is emitted at 600 nm by CuCl? Solution: The quantum of energy can be calculated from the Equation 1.2: E = hv The frequency (ν) for this case can be calculated as follows: c v= λ 3.0 × 108 ms -1 = 6.0 × 10-7 m = 0.5 × 1015 s -1 So, E = hv = 6.63 × 10-34 J. s × 0.50 × 1015s-1 = 3.315 × 10-19 J = 3.32 × 10−19 J The Photoelectric Effect In 1905, Albert Einstein used the quantum theory to explain the photoelectric effect. The photoelectric effect is a phenomenon in which electrons are ejected from the surface of certain metals exposed to light of at least a certain minimum frequency, called the threshold frequency, νo. UNIT 1 19  CHEMISTRY GRADE 11 These observations can be explained by assuming that EMR is quantized (consists of photons), and the threshold frequency represents the minimum energy required to remove the electron from the metal’s surface. Photons are particles of light or energy packet. The minimum energy required to remove an electron is: Eo = hνo (1.4) Where Eo is the minimum energy (of the photon), and νo, the threshold frequency. A photon with energy less than Eo (ν < νo) cannot remove an electron, or a light with a frequency less than the νo produces no electrons. On the other hand, if a light has ν > νo, the energy in excess of that required to remove the electron is given to the electron as kinetic energy (KE): KEe = ½ mv2 = hν - hνo (1.5) Where KEe is the kinetic energy of an electron, m is mass of an electron, v is the velocity of an electron, hν, is the energy of an incident photon, and hνo is the energy required to remove an electron from the metal’s surface. Intensity of light is a measure of the number of photons present in a given part of the beam: a greater intensity means that more photons are available to release electrons (as long as ν > νo for the radiation). In his theory of relativity in 1905, Einstein derived the famous equation: E = mc2 (1.6) Rearranging this equation, we have E m = 2 (1.7) c Where E is energy, m is mass, and c is speed of light. The main significance of Equation 1.7 is that energy has mass. Using this equation, we can calculate the mass associated with a given quantity of energy or the apparent mass of a photon. For electromagnetic radiation of wavelength, λ, the energy of each photon is given by the expression: hc Ephoton = (1.8) λ 20 UNIT 1  Electromagnetic Radiation and Atomic Spectra Then, the apparent mass of a photon of light with wavelength is given by: E m= 2 = ( ) hc λ = h (1.9) c c2 cλ We can summarize the important conclusions from the work of Planck and Einstein as follows: Energy is quantized. It can occur only in discrete units called photon or quanta. EMR, which was previously thought to exhibit only wave properties, seems to show certain characteristics of particulate matter as well. This phenomenon is sometimes referred to as the dual nature of light and is illustrated Figure 1.9. Figure 1.9: The dual nature of light Example 1.5 1. Compare the wavelength for an electron (mass = 9.11 × 10-31 kg) traveling at a speed of 1.00 × 107 m/s with that for a ball (mass = 0.10 kg) traveling at 35 m/s. Solution: We use the equation m = h/cλ, Where h = 6.63 × 10-34 J.s = 6.63×10-34 kg. m2.s Since 1 J = 1 kg. m2/s2 For the electron: 6.63 × 10−34 kg.m 2.s λ= 9.11× 10−31 kg × 1.00 × 107 m s = 7.28 × 10−11 m = 7.28 nm For the ball: 6.63 × 10−34 kg.m 2.s λ = = 1.89 × 10−34 m = 189 × 10−25 nm 0.10 kg × 35 m / s UNIT 1 21  CHEMISTRY GRADE 11 2. The maximum kinetic energy of the photoelectrons emitted from a give metal is 1.5 × 10–20 J when a light that has a 750 nm wavelength shines on the surface. Determine the threshold frequency, νo , for this metal. Calculate the corresponding wavelength, λo. Solution: a) Determination of the threshold frequency, νo Solve for v from c = v ´ λ Thus, v = c/λ = (3.0 ´ 108 m/s)/7.5 ´ 10-7 m) = 0.4 ´ 1015 s-1 = 4.0 ´ 1014 s-1 Similarly, rearrange Equation 1.5 and solve vo hv − KE vo = h 6.63 ×10−34 J.s × 4.0 × 1014 s −1 − 1.5 × 10 –20 J = 6.63 ×10−34 J. s 26.52 ×10−20 J − 1.5 × 10 –20 J 24.02 × 10−20 J = = 6.63 ×10−34 J. s 6.63 × 10−34 J.s = 3.77 × 1014 s −1 Therefore, a frequency of 3.77 ´ 1014 Hz is the minimum (threshold) required to cause the photoelectric effect for this metal. b) Calculate the corresponding wavelength λo : c λo = v° 3.0 × 108 m s = 3.77 × 1014 s -1 = 0.796 × 10-6 m = 796 nm 22 UNIT 1  Electromagnetic Radiation and Atomic Spectra Exercise 1.5 1. The following are representative wavelengths in the infrared, ultraviolet, and !-ray regions of the electromagnetic spectrum, respectively: 1.0 ´ 10-6 m, 1.0 ´ 10-8 m, and 1.0 ´10-10 m. a. What is the energy of a photon of each radiation? b. Which has the greatest amount of energy per photon? c. Which has the least? 2. A clean metal surface is irradiated with light of three different wavelengths λ1, λ 2, and λ 3. The kinetic energies of the ejected electrons are as follows: λ1: 7.2 ´ 10 -20 J; λ2: approximately zero; λ 3: 5.8 ´ 10 -19 J. Which light has the shortest wavelength and which has the longest wavelength? Determine the threshold frequency, ν o, for this metal. 3. The minimum energy required to cause the photoelectric effect in potassium metal is 3.69 ´ 10 -19 J. Will photoelectrons be produced when visible light, 520 nm and 620 nm, shines on the surface of potassium? What is or are the velocities of the ejected electron/s? !"#"$ %&'()*+,-.*&/0 Activity 1.11 Discuss the following ideas in pairs and share with the rest of the class. 1. Why do you observe different colors when you are watching fireworks? 2. As it is shown in the Figure 1.10, when compounds of the alkali metals: lithium, sodium, and potassium are excited in the gas flames they give different colored flames. Why do they emit different colors? Figure 1.10: Flame colors of lithium, sodium, and potassium compounds UNIT 1 23  CHEMISTRY GRADE 11 Atomic or line spectra are produced from the emission of photons of electromagnetic radiation (light). Different kinds of spectrum are observed when an electric discharge, or spark, passes through a gas such as hydrogen. The electric discharge is an electric current that excites, or energizes, the atoms of the gas. More specifically, the electric current transfers energy to the electrons in the atoms raising them to excited states. The atoms then emit the absorbed energy in the form of light as the electrons return to lower energy states. When a narrow beam of this light is passed through a prism, we do not see a continuous spectrum, or rainbow, as sunlight does. Rather, only a few colors are observed, displayed as a series of individual lines. This series of lines is called the element’s atomic spectrum or emission spectrum. The wavelengths of these spectral lines are characteristic of the element producing them, and used for their identification. For example, the emission (line) spectrum of hydrogen atom is show in Figure 1.11. Figure 1.11: The hydrogen line spectrum, containing only a few discrete wavelengths Changes in energy between discrete energy levels in hydrogen will produce only certain wavelengths of emitted light, as shown in Figure 1.12. For example, a given change in energy from a high to a lower level would give a wavelength of light which can be expressed using Planck’s equation: ∆E = hν = hc/λ 24 UNIT 1  Electromagnetic Radiation and Atomic Spectra Figure 1.12: A change between two discrete energy levels emits a photon of light !"#"$ %&'()*&+(,*-'.(*/(0&'(12-+*3'4(50*6 Activity 1.12 Form a group and discuss the following questions. Share your ideas with the rest of the class. The nature of the nucleus of the atom was explained by Rutherford. However, he was not able to explain the position and velocity of electrons in the atom. 1. Is it possible to know the exact location of an electron in the atom? Defend your suggestion. 2. Explain why an electron does not enter the nucleus, even though they are oppositely charged. In 1913, Niels Bohr (a Danish physicist) explained why the orbiting electron does not radiate energy as it moves around the nucleus. He introduced the fundamental idea that the absorption and emission of light by hydrogen atoms was due to energy changes of the electrons within the atoms. The fact that only certain frequencies are absorbed or emitted by an atom tells us that only certain energy changes are possible. Thus, energy changes in an atom are quantized. UNIT 1 25  CHEMISTRY GRADE 11 Bohr used Planck’s and Einstein’s ideas about quantized energy and proposed the following assumptions: 1. The electron in an hydrogen atom travels around the nucleus in a circular orbit. 2. The energy of the electron in an atom is proportional to its distance from the nucleus. The further an electron is from the nucleus, the more energy it has. 3. Only limited number of orbits with certain energies are allowed. This means, the orbits are quantized. 4. The only orbits that are allowed are those for which the angular momentum of the electron is an integral multiple of h/2π. 5. As long as an electron stays in a given orbital, it neither gains or losses energy. That means, the atom does not change its energy while the electron moves within an orbit. 6. The electron moves to a higher energy orbit only by absorbing energy in the form of light, and emitting light when it falls to a lower energy orbit. The energy (photon) of the light absorbed or emitted is exactly equal to the difference between the energies of the two orbits. Furthermore, Bohr showed that the radii, r, of the permitted orbits or energy levels for an atom of hydrogen atom are related to Planck’s constant, h, the electron’s charge, e, and its mass, m. Consider hydrogen atom with an electron with constant speed, v, circulating the nucleus in an orbit of radius, r. The total energy of the electron is the sum of the kinetic energy (energy of movement) and the potential energy (energy of position): 1 2 e2 E = mv + (1.10) 2 r For an electron to exist in a stable orbit of a constant radius, the centripetal force (attracting the electron to the nucleus), e2/r2, and the centrifugal force (pulling away the electron from the nucleus), mv2/r must be equal: e2 mv 2 = (1.11) r2 r Bohr then introduced an additional requirement that the angular momentum, mvr, of the electron can take only certain permitted values, that is, an integral multiple of h/2π. This requirement is called a quantum condition: 26 UNIT 1  Electromagnetic Radiation and Atomic Spectra nh mvr = (1.12) 2π Solving Equation 1.12 for v gives: nh v= (1.13) 2π mr Substituting for v in Equation 1.11 and solving for r gives: n2 h2 2 h2  (1.14) r= =n  2  4πme 2  4πme  Here, n is positive integer (n = 1, 2, 3...) and is called quantum number. It is known that h, π, m, and e are constants, thus Equation 1.14 can be simplified to: r = n 2 ao (1.15) According to Equation 1.15 the only permitted values of the radii of the electron path in the hydrogen atom are those proportional to the square of a whole number, n. The numerical values of ao is 0.53 Å. Thus, r = (0.53 Å)n2 For instance, for n = 1, r = 0.53 Å. This is the first Bohr radius. The larger the values of n, the further the electron from the nucleus. Figure 1.13: Bohr’s energy levels of a hydrogen atom UNIT 1 27  CHEMISTRY GRADE 11 Bohr showed that the energies that an electron in hydrogen atom can occupy are given by: R En = - H2 (1.16) n Where RH, the Rydberg constant for the hydrogen atom and has the value 2.18 x 10-18 J. Thus, Equation 1.15 can be written as: 2.18 × 10−18 J En = − n2 The negative sign in Equation 1.16 is an arbitrary convention, signifying that the energy of the electron in the atom is lower than the energy of a free electron, which is an electron that is infinitely far from the nucleus. The energy of a free electron is given a value of zero. As the electron gets closer to the nucleus (as n decreases), En becomes larger in absolute value, but also more negative. The most negative value, then, is reached when n = 1, which corresponds to the most stable energy state. We call this the ground state or ground level, which refers to the lowest energy state of an atom. The stability of the hydrogen electron diminishes for n = 2, 3.... Each of these levels is called an excited state, or excited level, which is higher in energy than the ground state. A hydrogen electron for which n is greater than 1 is said to be in an excited state. Using Bohr’s Equation 1.16, relating the energy (En ) and energy level (n) for an electron it is possible to calculate the energy of a single electron in a ground state or excited state, or the energy change when an electron moves between two energy levels. 28 UNIT 1  Electromagnetic Radiation and Atomic Spectra Example 1.6 Consider the n = 5 state of hydrogen atom. Using the Bohr model, calculate the radius of the electron orbit, the velocity and the energy of the electron. Solution: To determine the radius, Equation 1.15 is used: r = ( 0.53 Å ) × 52 = 15.24 Å = 1.524 nm Velocity of the electron is determined using Equation 1.13: nh v= 2πmr = ( 5 6.63 × 10−34 kg m 2 s −1 ) 2 × 3.14 × 9.11 × 10−31 kg × 1.524 × 10−9 m = 3.8 × 105 m / s Equation 1.16 is used to solve E5 2.18 × 10−18 J E5 = − 2 = 8.72 × 10−20 J 5 Calculate the energies of the hydrogen electron in n = 1, n = 2 and n = 3 Solution: Using Equation 1.16: 2.18 × 10−18 J For n = 1, E1 = − = − 2.18 × 10−18 J 12 2.18 × 10−18 J n = 2, E2 = − 2 = − 5.54 × 10−19 J 2 2.18 × 10−18 J n = 3 , E3 = − 2 = − 2.42 × 10−19 J 3 Bohr’s model also quantitatively explained the line spectra of hydrogen atom. He proposed that the absorptions and emissions in line spectra correspond to the transfer of the electron from one orbit to another. Energy must be absorbed for the electron to move from one orbit to another one having a bigger radius. Whereas, energy is emitted when an electron moves from the higher orbital energy level, ni, to the lower orbital, nf, energy level. Thus, the change in energy, E, is the difference in the energy between the final and the initial state electron: UNIT 1 29  CHEMISTRY GRADE 11 ∆E = Ef − Ei (1.17) This equation is similar to:  1 1  ∆E = − RH  2 − 2   nf ni  (1.18)  1 1  = −2.18 × 10−18 J  2− 2  nf ni  Where ni and nf represent quantum numbers for initial and final states respectively. But, ∆E = hν, thus we have:  1 1  ∆E = hv = − 2.18 × 10−18 J  2 − 2  (1.19)  nf ni  Notice that when nf > ni, ∆E is positive, indicating that the system has absorbed energy. But, ni > nf, ∆E is negative and this corresponds to emission of energy. Example 1.7 Calculate the energy emitted when an electron moves from the n = 3 to the n = 2 energy level. Determine the wavelength of the emitted energy. Solution: Equation 1.19 is used, to determine the emitted energy:  1 1 ∆E = −2.18 × 10−18 J  2 − 2  = 3.03 × 10−19 J 2 3  To obtain λ, Equation 1.19: hc ∆E = hv = λ m 6.63 × 10−34 J s × 3.0 × 108 hc s Thus, = = ∆E 3.03 × 10−19 J = 6.56 × 10−7 m = 656 nm 30 UNIT 1  Electromagnetic Radiation and Atomic Spectra Exercise 1.6 1. What is the wavelength of a photon (in nanometers) emitted during a transition from the n = 5 state to the n = 2 state in the hydrogen atom? 2. Calculate the frequency of the green line arising from the electron moving from n = 4 to n = 20 in the visible spectrum of the hydrogen atom using Bohr’s theory. Each spectral line in the emission spectrum corresponds to a particular transition in a hydrogen atom. When we study a large number of hydrogen atoms, we observe all possible transitions and hence the corresponding spectral lines. For instance, Figure 1.14 illustrates line spectra of a hydrogen atom when its electron moves from n = 4 to n = 1; n = 3 to n = 1 and n = 2 to n = 1. Figure 1.14: The emission spectrum of a hydrogen atom Each horizontal line represents an allowed energy level for the electron in a hydrogen atom. The energy levels are labeled with their principal quantum numbers. The emission spectrum of hydrogen includes a wide range of wavelengths from the infrared to the ultraviolet. Table 1.2 shows the series of transitions in the hydrogen spectrum; they are named after their discoverers. The Balmer series was particularly easy to study because some its lines fall in the visible range. UNIT 1 31  CHEMISTRY GRADE 11 Table 1.2: The various series in atomic hydrogen emission spectrum Series nf ni Spectrum region Lyman 1 2, 3, 4, … Ultraviolet Balmer 2 3, 4, 5, … Visible and ultraviolet Paschen 3 4, 5, 6,... Infrared Brackett 4 5, 6, 7,... Infrared For a larger orbit radius (i.e. a higher atomic energy level), the further the electron drops, the greater is the energy (higher v, shorter λ) of the emitted photon. Exercise 1.7 1. Calculate the energies of the states of the hydrogen atom with n = 2 and n = 3, and calculate the wavelength of the photon emitted by the atom when an electron makes a transition between these two states. 2. What is the wavelength of a photon emitted during a transition from the ni = 10 state to the n f = 2 state in the hydrogen atom? !"#"# $%&%'('%)*+,)-,'./,0).1,2)3/4 The Bohr Model was an important step in the development of atomic theory. He introduced the idea of quantized energy states for the electron in a hydrogen atom. The model explains atoms and ions containing only one electron such as H, He+ and Li2+. However, the model has several limitations: It doesn't explain the atomic spectra of more complicated atoms and ions, even that of helium, the next simplest element. It doesn't explain about further splitting of spectral lines in the hydrogen spectra on application of a magnetic field. It considers electrons to have both known radius and orbit, which is impossible according to Heisenberg's uncertainly principle. 32 UNIT 1  Electromagnetic Radiation and Atomic Spectra !"#"$ %&'()*+',-*./012'(34*20/5(67(8*//'.(*9:(;9'. 0 52 UNIT 1  Electronic Configurations and the Periodic Table of the Elements The second ionization energy (IE2) removes the second electron. Since the electron is being pulled away from a positively charged ion, IE2 is always larger than IE1: Ion+ (g) → Ion2+ (g) + e– ∆E = IE2 (always > IE1) The first ionization energy is a key factor in an element’s chemical reactivity because, atoms with a low IE1 tend to form cations during reactions, whereas those with a high IE1, (except the noble gases) often form anions. Ionization energies display a periodic variation when plotted against atomic number, as Figure 1.20 shows. Within any period, values tend to increase with atomic number. Thus, the lowest values in a period are found for the Group IA elements (alkali metals). It is characteristic of reactive metals such as these to lose electrons easily. The largest ionization energies in any period occur for the noble-gas elements. This general trend increasing ionization energy with atomic number in a given period is due to the fact that as we move across a period, Zeff generally increases so atomic radii become smaller. As a result, the attraction between the nucleus and the outer electrons increases, so an electron becomes more difficult to remove. Figure 1.20: Ionization energy versus atomic number Small deviations from this general trend occur. A Group IIIA element (ns2np1) has a smaller ionization energy than the preceding Group IIA element (ns2). UNIT 1 53  CHEMISTRY GRADE 11 Apparently, the np electron of the Group IIIA element is more easily removed than one of the ns electrons of the preceding Group IIA element. Also note that a Group VIA element (ns2np4) has a smaller ionization energy than the preceding Group VA element. As a result of electron repulsion, it is easier to remove an electron from the doubly occupied np orbital of the Group VIA element than from a singly occupied orbital of the preceding Group VA element. Group Assignment 1.1 Form a group and discuss on the following questions. Write your reflections and submit it to your teacher. 1. The first ionization energy in general increases across a period, however, large drops occur when a new period begins. Why? 2. Explain the irregularities in the trends across periods: a. Boron has a smaller first ionization energy than beryllium. b. The first ionization energy of nitrogen is higher than oxygen. Exercise 1.13 1. Choose the element with the higher ionization energy from each pair: a. As or Bi b. As or Br c. Al or In d. K or Ge 2. The first and second ionization energies of potassium, K, are 419 kJ mol–1 and 3052 kJ mol–1 and those of calcium, Ca, are 590 kJ mol–1 and 1145 kJ mol –1, respectively. Compare their values and comment on the differences. 3. Based on their positions in the periodic table, predict which atom of the following pairs will have the larger first ionization energy: a. Ga or Ge b. Br or Sb c. K or Cr d. Mg or Sr e. O or Ne 4. Name the period 3 element with the following ionization energies (in kJ/mol), and write its electron configuration: IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,330 54 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Electron Affinity (EA) The electron affinity is the energy change for the process of adding an electron to a neutral atom in the gaseous state to form a negative ion: Atom (g) + e−  → Ion− (g) ∆! = EA1 An electron approaching a neutral atom experiences an attraction for the positively charged nucleus. Repulsion of the incoming electron by electrons already present in the atom tends to offset this attraction. Still, in many cases the incoming electron is absorbed by the atom and energy is evolved in the process. Thus, first electron affinity (EA1) is usually negative. The second electron affinity (EA2), however, is always positive because energy must be absorbed to overcome electrostatic repulsions and add another electron to a negative ion. For example, the EA1 and EA2 of an oxygen atom can be written as: O(g) + e– → O– (g) EA1 = –141 kJ mol–1 O–(g) + 1e– → O2–(g) EA2 = +744 kJ mol–1 Figure 1.21 gives the electron affinities of the main-group elements. Negative values indicate that energy is released when the anion forms. Positive values, which occur in Group VIIIA (18), indicate that energy is absorbed to form the anion; in fact, these anions are unstable and the values are estimated. Figure 1. 21: Electron affinities of the main-group elements (in kJ/mol) UNIT 1 55  CHEMISTRY GRADE 11 Activity 1.24 Discuss the following questions in groups, then present your answers to the whole class. 1. Which group in the periodic table has elements with high (endothermic) IE1 and very negative (exothermic) first electron affinities (EA1)? Give the charge on the ions these atoms form. 2. Silicon has an electron affinity of –134 kJ/mol. The electron affinity of phosphorus is –72 kJ/mol. Give a reason for this difference. 3. Why are the electron affinities of the alkaline earth metals either negative or small positive values? Electron affinities, EA, have a periodic variation, just as atomic radii and ionization energies do. Broadly speaking, the general trend is toward more negative electron affinities from left to right in any period. What could be the reason the IIA element and the VA element tend to have smaller electron affinities (Figure 1.21) than the preceding element? Note that the Group VIA and Group VIIA elements have the largest negative electron affinities of any of the main-group elements. The trend in going across a period is rather clear but there is no simple trend in going down a column of elements. In most cases, the added electron goes into an energy sublevel that is already partly filled. For Group IIA and VIIIA atoms, however, the added electron would be required to enter a significantly higher energy level, the np level for the Group IIA atoms and the s level for the next principal level for the Group VIIIA atoms. In these cases, a stable anion does not form. Electronegativity Electronegativity indicates the extent of attraction by which the electrons of the bond pair are attracted by an atom linked by this bond. Linus Pauling, an American scientist, in 1922 assigned arbitrarily a value of 4.0 to fluorine, the element considered to have the greatest ability to attract electrons. This is known as the Pauling scale. Approximate values for the electronegativity of a few elements are given in Table 1.4. 56 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Table 1.4: (a) Electronegativity values (on Pauling scale) across the periods Atom (Period II) Li Be B C N O F Electronegativity 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Atom (Period III) Na Mg Al Si P S Cl Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Table 1.4: (b) Electronegativity values (on Pauling scale) down a group Atom Atom Electronegativity Electronegativity (Group IA) (Group VIIA) value value Li 1.0 F 4.0 Na 0.9 Cl 3.0 K 0.8 Br 2.8 Rb 0.8 I 2.5 Cs 0.7 At 2.2 Electronegativity generally increases across a period from left to right (say from lithium to fluorine) and decreases down a group (say from fluorine to astatine) in the periodic table. The attraction between the valence electrons and the nucleus increases as the atomic radius decreases in a period. The electronegativity also increases. On the same account electronegativity values decrease with the increase in atomic radii down a group. The trend is similar to that of ionization energy. Activity 1.25 Discuss the following questions in groups, and present your answers to the whole class. 1. What is the electronegativity of an atom? How is it different from electron affinity? 2. In a given family of the periodic table, what is the general relationship between electronegativity and size? UNIT 1 57  CHEMISTRY GRADE 11 Metallic Character Activity 1.26 Form a group and summarize the trend in metallic character in the periodic table. Is it the same as the trend observed for atomic size and ionization energy? Present your responses to the whole the class. Metallic character refers to the chemical properties associated with elements classified as metals. These properties arise from the elements, ability to lose electrons. Metallic character decreases as you move across a period in the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell. Metallic character increases as you move down an element group in the periodic table. This is because electrons become easier to lose as the atomic radius increases, where there is less attraction between the nucleus and the valence electrons because of the increased distance between them. !"#"$ %&'()*(+,-./0.1,23/&34.56(--374(*3/)./0.*8,.96,:,)*- Activity 1.27 Form a group and discuss the following questions, then share your answers with the whole class. 1. In which region of the periodic table are metals located? 2. In which region are the elements with general electronic configuration of ns2p5 located? Give the group number. 3. Write the group number and period of an element with atomic number 26. 4. Is an element with the electronic configuration 1s2 2s2 2p 6 3s1 a metal or a non- metal? Will it form a cation or an anion readily? Explain your answer. 58 UNIT 1  Electronic Configurations and the Periodic Table of the Elements Some of the advantages of periodic classification of elements are: 1. The classification of elements is based on the atomic number, which is a fundamental property of an element. 2. Isotopes are averaged in one place, as the classification is on the basis of atomic number (For example, chlorine (atomic number 17 ) is given the atomic mass 35.5.) 3. It explains the periodicity of the properties of the elements and relates them to their electronic configurations. 4. The position of the elements that were misfits on the basis of mass number (anomalous pairs like argon and potassium) could be justified on the basis of atomic number. 5. The lanthanides and actinides are placed separately at the bottom of the periodic table. 6. The table is a simple, systematic and an easy way of remembering the properties of various elements, as it is based on the electronic configuration. UNIT 1 59  CHEMISTRY GRADE 11 UNIT SUMMARY ~ Modern chemistry began with eighteenth century discoveries leading to the formulation of two basic laws of chemical combination: the law of conservation of mass and the law of constant composition (definite proportions). Dalton proposed another law of chemical combination, the law of multiple proportions. ~ The first clues to the structure of atoms came through the discovery and characterization of cathode rays (electrons). Key experiments were those that established the mass-to-charge ratio and then the charge on an electron. ~ The principal types of radiation emitted by radioactive substances are alpha particles, beta particles, and gamma rays. ~ Studies on the scattering of particles by thin metal foils (Rutherford’s atomic model) led to the concept of the nuclear atom − a tiny, but massive, positively charged nucleus surrounded by lightweight, negatively charged electrons. ~ Electromagnetic radiation is characterized by its wavelength (λ), frequency (ν), and speed (c = 3.0 × 108 m/s), which are related by the formula: c = λν ~ Electromagnetic radiation can be viewed as a stream of “particles” called photons, each with energy hν, where h is Planck’s constant (6.63 × 1034 J.s). ~ When light strikes a metal surface, electrons are emitted. ~ Analysis of the kinetic energy and numbers of the emitted electrons led Einstein to suggest that electromagnetic radiation can be viewed as a stream of photons. ~ The photoelectric effect is the emission of an electron from the surface of a metal, caused by electromagnetic radiation of a certain minimum energy; the resulting current increases with increasing intensity of radiation. ~ Bohr’s theory requires the electron in a hydrogen atom to be in one of a discrete set of energy levels. The fall of an electron from a higher to a lower energy level releases a discrete amount of energy as a photon of light with a characteristic frequency. ~ Bohr’s theory accounts for the observed atomic spectrum of hydrogen atom. 60 UNIT 1  Electronic Configurations and the Periodic Table of the Elements ~ An emission spectrum is the spectrum associated with the emission of electromagnetic radiation by atoms (or other species) resulting from electron transitions from higher to lower energy states. ~ The electron in a hydrogen atom can be viewed as a matter-wave enveloping the nucleus. ~ The matter-wave is represented by a wave equation, and solutions of the wave equation are wave functions. ~ Each wave function is characterized by the value of four quantum numbers: the principal quantum number, n; the angular momentum quantum number ℓ; the magnetic quantum number, mℓ; and the spin quantum number, ms. ~ According to Heisenberg's Uncertainty Principle, it is impossible to determine both the momentum and position of an electron. ~ An orbital describes a region in an atom that has a high probability of containing an electron or a high electron change density. ~ Orbitals with the same value of n are in the same principal energy level. ~ Orbitals with the same value of n and of ℓ are in the same sublevel. ~ The shape of orbitals depend on the values of ℓ. Thus, the s orbital (ℓ = 0) is spherical and the p orbital (ℓ = 1) is dumbbell-shaped. ~ The n, ℓ and mℓ quantum numbers define an orbital, but a fourth quantum number, ms, is also required to characterize an electron in an orbital. ~ The Aufbau (“building up”) Principle is a guide for predicting the order in which electrons fill subshells and shells in atoms. ~ According to Pauli's Exclusion Principle, no two electrons in the same atom may have identical sets of four quantum numbers. ~ Hund’s Rule states that each orbital of a given subshell is occupied by a single electron before pairing begins. ~ Electron configuration refers to the distribution of electrons among orbitals in an atom. Introduced here are the subshell notations (or s, p, d, and f ) and the orbital diagram. UNIT 1 61  CHEMISTRY GRADE 11 ~ Elements with similar valence-shell electron configurations fall in the same group of the periodic table. The period number is the same as the highest number of principal shell containing electrons (the outer shell). ~ Certain atomic properties vary periodically, when atoms are arranged in terms of increasing atomic number. These include atomic size, metallic character, ionization energy, and electron affinity. CHECKLIST KEY TERMS  Amplitude  Mass number  Atomic mass  Non-metal  Atomic mass unit (amu)  Pauli’s Exclusion Principle  Atomic number  Period  Auf bau Principle  Periodic law  Cathode  Periodic table  Cathode rays  Photon  Charge/mass ratio  Proton  Dalton’s atomic theory  Quantum numbers  Effective nuclear charge  Representative element  Electromagnetic radiation  s-block elements  Electronic configuration  Schrödinger equation  Exited state  Transition metal  Frequency  Uncertainty principle  Ground state  Wavelength  Hund’s Rule  Inner-transition metal  Isotope 62 UNIT 1  Electronic Configurations and the Periodic Table of the Elements REVIEW EXERCISE Part I: Multiple Choice Questions: Choose the correct answer from the given alternatives. 1. Which of the following scientists did not contribute to determining the structure of the atom? a. Thomson c. Becquerel b. Rutherford d. Dalton 2. What is the frequency in Hz of the gamma radiation from a radioactive cobalt-60 source if its wave length is 1.0 × 10-9 nm? a. 3.3 × 10-27 Hz c. 3.0 × 1018 Hz b. 3.3 × 10-8 Hz d. 3.0 × 10-26 Hz 3. What is the energy of the electronic transition associated with the sodium-D line having a wavelength of 589 nm? a. 6.63 × 10-34 J c. 3.38 × 10-19 J b. 1.13 × 10-27 J d. 5.82 × 102 J 4. Among the following, which colour corresponds to light of the highest frequency? a. Green b. Red c. Yellow d. Blue 5. Which of the following orbital designations does not exist? a. 1s b. 2d c. 3p d. 4f 6. Which of the following sets of quantum numbers are not allowed in the hydrogen atom? a. n = 3, ℓ = 2, mℓ = 2 c. n = 1, ℓ = 0, mℓ = 0 b. n = 4, ℓ = 3, mℓ = 4 d. n = 2, ℓ = 1, mℓ = -1 7. The species that has the same number of electrons as 32S is: a. 32Cl b. 33P+ c. 34S+ d. 28Si2- 8. Which of the following has the lowest electronegativity? a. Carbon c. Beryllium b. Magnesium d. Boron 9. How many unpaired electrons are there in the Cr3+ ion? a. 0 b. 1 c. 2 d. 3 10. Which of the following species would require the highest energy for the removal of one electron? a. Mg2+ b. Na+ c. Ne d. F– UNIT 1 63  CHEMISTRY GRADE 11 !"#$ %%& '()*+,$, $-,.(++(/012 3$"$,),1$34 11. The atomic number gives the number of _______ in the nucleus. 12. In an atom, the number of electrons is equal to the number of _______. 13. When n = 2, the values of ℓ could be _________ and _____ 14. When ℓ = 1, the values of mℓ could be _______, ________ and _______ as well as the subshell has the letter label _____. 15. When ℓ = 1, the subshell is called a _____ subshell with _______ total orbitals. 16. When the subshell is labeled p, there are ______ values of mℓ and _______ orbitals occur within the subshell. 17. When the subshell is labeled f, there are ______ values of mℓ and _______ orbitals occur within the subshell. !"#$ %%%& 5-(#$ "13/,# 67,3$0(13 "18 *#(9+,)34 18. For the following atoms, determine the number of protons, neutrons, and electrons: a. 114 48 Cd c. 199 79 Au 222 b. 98 43Tc d. 86 Rn 19. Iridium has two naturally occurring isotopes. 1911r and 1931r have atomic masses of 1 90.9609 amu and 1 92.9633 amu, respectively. The average atomic mass for iridium is 192.22 amu. What is the percent natural abundance for each isotope? 20. The following are representative wavelengths in the IR, UV and x-ray regions of the electromagnetic spectrum, 1.0 × 10-6 m, 1.0 × 10-8 m and 1.0 × 10-10 m, respectively. a. What is the energy of a photon form each transition? b. Which of them has the highest energy per photon? 21. What is the wavelength of a beam of protons having a velocity of 1.38 × 107 cm/s? The mass of a proton is 1.76 × 10-24 kg. 22. The photon emitted by a cyclotron has a velocity of 1.50 × 103 m/s. What is the wavelength of this photon? The mass of photon = 1.676 × 10–27 kg and Planck’s constant = 6.62 × 10–34 J.s.s 64 UNIT 1  Electronic Configurations and the Periodic Table of the Elements 23. Calculate the wavelength of the light emitted when an electron falls from n = 3 to the n = 1 state in hydrogen atom. 24. Write the number and the letter for the orbital that corresponds to the following pairs of n and l quantum numbers: a. n = 3, l = 1 c. n = 3, l = 2 b. n = 4, l = 0 d. n = 5, l = 3 25. What type of orbital (i.e. 3s, 4p, …) is designated by these quantum numbers? a. n = 5, ℓ = 1, mℓ = 0 c. n = 2, ℓ = 0, mℓ = 0 b. n = 4, ℓ = 2, mℓ = -2 d. n = 4, ℓ = 3, mℓ = -3 26. Write the ground-state electron configurations for the following elements: a. Br (Z = 35) c. W (Z = 74) b. Mo (Z = 42) 27. Write the period number for each of the following and state if they are metal, non-metal or metalloid: a. P c. F b. Ge d. W UNIT 1 65  2 CHEMISTRY GRADE 11 !"#$ CHEMICAL BONDING Unit Outcomes At the end of this unit, you will be able to: ) explain that a chemical bond is an attractive force between particles ) demonstrate an understanding of the formation and general properties of substances containing ionic, covalent and metallic bonds ) draw Lewis structure for simple ionic and covalent compounds ) identify the origin of polarity within molecules ) describe the formation and nature of hydrogen bonds, dipole-dipole forces and London forces ) explain the bonding models (Lewis model, valence bond model and molecular orbital model) and show the usefulness of the bonding theories in explaining and predicting molecular properties (bond angle, bond length, bond energy, etc…) ) explain how the properties of a substance (solid or liquid) depends on the nature of the particles present and the type of intermolecular forces ) discuss the importance of intermolecular forces in plant and animal life ) explain how the Valence Shell Electron Pair Repulsion (VSEPR) model can be used to predict molecular shape ) conduct experiments to observe and analyze the physical properties of different substance to determine the type of bonding present ) describe scientific enquiry skills along this unit: observing, inferring, predicting, classifying, comparing and contrasting, making models, communicating, asking questions, applying concepts, relating cause and effect and making generalization. 66 UNIT 2  Introduction Start-up Activity Discuss the following issues in a group and present your ideas to the class. Why do the substances around us behave differently? For instance, table salt is a hard, brittle, high-melting solid that conducts electricity only when molten or dissolved in water. Candle wax melts at a low temprature, is soft, and non-conducting. Copper (and most other metallic substances) is shiny, malleable, and able to conduct electricity whether it is molten or solid. 2.1 !"#$%&'(#)%" At the end of this section, you will be able to: ) define chemical bonding ) explain why atoms form chemical bonds ) illustrate chemical bonding using the octet rule ) describe the types of chemical bonding and the mechanisms of the bonding process. In Grade 9, you learned about chemical bonding and its types, such as ionic, covalent and metallic bonding. In this unit, we will discuss some new concepts about chemical bonding, like intermolecular forces, molecular geometry, theories of chemical bonding, etc. Activity 2.1 Discuss the following issues in groups, then present your ideas to the whole class. 1. Why do atoms combine? 2. Why are molecules more stable than free atoms? 3. What forces act on two atoms as they come together? 4. Explain how the potential energy changes when two hydrogen atoms form a bond. The attractive force which holds atoms, ions, and molecules together is called a chemical bond. Since these forces of attracstion are intramolecular forces, they have an effect on the chemical properties as well as the physical properties of the chemical. UNIT 2 67  CHEMISTRY GRADE 11 !"#"# $%&'()*&*'+,-&' Activity 2.2 From your knowledge of Grade 9 chemistry discuss the following questions in a group of four, then present your responses to the whole class. 1. Why do some atoms combine while others do not? Give some examples in your answer. 2. Why do different atoms form different types of bonding? All noble gases except helium (1s2) have ns2np6 electron configurations (where n indicates the highest occupied shell). The noble gases are quite unreactive because they have very stable electron configurations, as reflected by their high ionization energies and low electron affinities. Because all the noble gases (except helium) have outer shells with eight electrons, many atoms undergoing reactions also attain eight valence electrons (ns2np6). This rule has become known as the octet rule, as follows: Atoms tend to gain or lose electrons until they have achieved an outer shell that contains an octet of electrons (eight electrons). Do you know any compound whose central atom does not obey the octet rule? !"#"! $./&0'12'3%&45)6-'718958:' Activity 2.3 Discuss the following questions in pairs and share your ideas with the whole class. 1. Why do metals tend to form cations and why do non-metals tend to form anions? 2. Why do two non-metals combine to form compounds while no similar process exists for two metals to form a compound? In general, there is a gradual change from metallic to non-metallic character as you move from left to right across a period and from bottom to top within most groups in the periodic table. Accordingly, atoms can combine to form three types of bond: metal with non-metal (ionic bond), non-metal with non-metal (covalent bond), and metal with metal (metallic bond). 68 UNIT 2  Ionic Bonds 2.2 !"#$%&'"#() At the end of this section, you will be able to: ) define ionic bonding ) use Lewis electron dot symbols for main group elements ) describe ionic bonding using Lewis electron dot symbols ) list the favourable conditions for the formation of ionic bonds ) explain the formation of ionic bonding ) give examples of ionic compounds ) define lattice energy ) calculate lattice energy of ionic crystals from given data using the Born- Haber cycle ) discuss the exceptions to the octet rule ) describe the properties of ionic bonding ) carry out an activity to demonstrate the effect of electricity on ionic compounds (PbI2 and NaCl) ) carry out an investigation into the melting point and solubility of some ionic compounds (NaCl and CuCl2). Activity 2.4 Form a group of four and discuss the following questions, then present your responses to the whole class. 1. Why do elements form ions in certain chemical reactions? 2. Which of the following pairs of elements are likely to form an ionic compound? a. sodium and chlorine b. nitrogen and fluorine c. lithium and oxygen An ionic bond is formed by the electrostatic attraction between positive and negative ions. The bond forms between two atoms when one or more electrons are transferred from the valence shell of one atom to the valence shell of the other. The atom that loses electrons becomes a cation (positive ion), and the atom that gains electrons becomes an anion (negative ion). Any given ion tends to attract as many neighboring ions of opposite charge as possible. UNIT 2 69  CHEMISTRY GRADE 11 When large numbers of ions gather together, they form an ionic solid. The solid normally has a regular, crystalline structure that allows for the maximum attraction of ions, given their particular sizes. Example 2.1 The formation of NaCl from sodium and chlorine can be explained as: Na → Na+ + e− [Ne]3s1 [Ne] Cl + e− → Cl− [Ne]3s23p5 [Ne]3s23p6 or [Ar] Na+ + Cl− → NaCl or Na+Cl− The bond formed, as a result of the electrostatic attraction between the positive and negative ions is called the electrovalent bond or ionic bond. Note: Ionic compounds are usually formed when metal cations bond with non-metal anions. What about an ionic compound containing an ammonium ion? Exercise 2.1 1. Explain the formation of bonds in the following pairs of elements: a. potassium and chlorine c. sodium and oxygen b. magnesium and oxygen 2. Which of the following elements will form an ionic bond with chlorine, why? a. calcium b. carbon c. oxygen d. silicon 3. Identify the species found in the following ionic compounds: CaCl2, MgO, and Al 2O 3. From this what can you conclude about the formation of these compounds? 4. Why do elements located on the opposite sides of the periodic table tend to form ionic bonds? 70 UNIT 2  Ionic Bonds !"!"# $%&'()*+%,-./012/-)3456/+() The American Chemist Gilbert N. Lewis (1875–1946) developed a special set of symbols for his theory. A Lewis symbol consists of a chemical symbol to represent the nucleus and core (inner-shell) electrons of an atom, together with dots placed around the symbol to represent the valence (outer-shell) electrons. For example, the Lewis symbol for chlorine, which has the electron configuration, [Ne]3s23p5, is Cl A Lewis structure is a combination of Lewis symbols that represents either the transfer or the sharing of electrons in a chemical bond. For example, the formation of sodiu