NAME PER DATE DUE "ALICE" CHAPTER 12 MODERN VIEW OF ATOMIC STRUCTURE (PART 1)

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1 NAME PER DATE DUE ACTIVE LEARNING IN CHEMISTRY EDUCATION "ALICE" CHAPTER 12 MODERN VIEW OF ATOMIC STRUCTURE (PART 1) Atomic Structure & Spectra Nuclear Notation Orbital Notation Configuration Notation , A.J. Girondi

2 NOTICE OF RIGHTS All rights reserved. No part of this document may be reproduced or transmitted in any form by any means, electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author. Copies of this document may be made free of charge for use in public or nonprofit private educational institutions provided that permission is obtained from the author. Please indicate the name and address of the institution where use is anticipated A.J. Girondi, Ph.D. 505 Latshmere Drive Harrisburg, PA Website: , A.J. Girondi

3 SECTION 12.1 Early Discoveries About the Atom You have already learned about the concept of atoms and how they can combine to form compounds. You have also seen that each element has its own characteristic set of properties which help to distinguish it from all other elements. In this chapter we will study the structure of atoms and the laws governing the behavior of the particles that make up atoms. This knowledge will lead to an explanation of the properties of the elements and of their tendencies to form compounds. John Dalton regarded the atom as a particle with no internal parts. He believed an atom to be the smallest possible particle. However, certain experiments were being performed which gave definite indications that Dalton's view was not correct and that there was some sort of internal structure to the atom. It became apparent that atoms consisted of particles that had electrical charges and that these particles interacted according to the laws of electromagnetism. Charged particles carry either a positive (+) or negative (-) charge. We call two negative charges or two positive charges "like" charges. The laws of electromagnetism state that two like charges {1} each other, while unlike charges, {2} each other. Experiments performed during the late 1800's and early 1900's by chemists and physicists made it clear that atoms could, indeed, be broken into smaller parts, contrary to the ideas of {3}. In 1897, J.J. Thomson discovered that atoms could be "taken apart" when he studied the effects of electrical discharge on atoms of various gases. In his experiments he concluded that atoms were coming apart by yielding a stream of negatively charged particles with very small masses (compared to the masses of the atoms). These small negative particles became known as electrons. Thomson is credited with the discovery of electrons which were present in the atoms of all of the different gases that he examined. Another scientist, Ernest Rutherford, and his students performed experiments in England during the first decade of the 20th century in an attempt to determine the size of atoms. In 1906, Rutherford had his students direct a beam of positively-charged subatomic "alpha" particles at a very thin sheet of gold metal. It was known as the alpha scattering experiment." Since they believed that matter was mostly empty space, they expected the particles to pass through the thin sheet unhindered. To their surprise, they found that a small fraction of the particles bounced right back! This led Rutherford to believe that in the center of the atom was a small but very dense "nucleus" with which some of the alpha particles must have collided. He concluded that most of the mass of the atom was contained in the {4}. He also concluded that the nucleus was {5} charged since it repelled the positivelycharged alpha particles. He later said that "It was quite the most incredible event that has ever happened to me in my life. It was almost as if you fired a 15- inch shell into a piece of tissue paper and it came back and hit you." Rutherford realized that electrons were located at a considerable distance from the nucleus. If this were an accurate description of an atom and we could inflate the nucleus of a hydrogen atom to the size of a basketball, the electron would orbit this "basketball" nucleus at a distance of more than 15 miles away! Visualizing the atom like this enables you to realize that most of the atom is, indeed, nothing more than empty space! So much for the athlete who thinks he is "solid muscle!" With regard to the nucleus itself, it became obvious to scientists that the nucleus was composed of small particles. One of the particles in the nucleus is the proton. In 1914, Rutherford was given credit for discovering protons. A proton carries a charge equal to the charge of an electron, but opposite in character. The electron carries a charge of -1 while the proton carries a charge of {6}. In the nucleus of a neutral atom (that is, an atom with no overall electric charge), there must be an equal number of protons to balance the charges carried by the electrons. Unlike the electron, the proton is a particle with a relatively large mass, in atomic terms. The proton has a mass equal to 1,836 times that of the electron! Using the common unit of the gram to measure masses, the electron has a mass equal to 9.1 X grams, and the proton has a mass equal to X grams , A.J. Girondi

4 A second component of the nucleus was discovered in 1932 by another Englishman, James Chadwick. This particle became known as the neutron. A neutron is an electrically neutral particle, meaning that it carries no electric charge. The neutron was very difficult to discover. Because it has no charge of its own, it is neither attracted to nor repelled by an electrical charge. A neutron has a mass slightly larger than that of the proton, equal to X grams. The presence of this particle accounted for the observed masses of atoms, which were found to be greater than that predicted if only protons were present in the nucleus. Figure 12.1 and the accompanying table present an overall summary of the locations and properties of the components of an atom. Each element differs from all others in that atoms of each element contain a specific number of electrons, protons, and {7}. Indeed, the number of protons in the nucleus determines the actual identity of an element. Determining the number of electrons, protons, and neutrons in any given element is a relatively simple process. The atomic number of an element is equal to the number of protons found in the nucleus. The element with atomic number 19, potassium, has 19 protons. For atoms to be neutral, they must have equal numbers of positive (protons) and negative (electrons) charges. This means that potassium must have 19 protons and {8} electrons. Dalton s Atomic Theory (1803) Thomson Discovers the Electron (1897) Rutherford Discovers the Nucleus (1906) Bohr Develops the Planetary Model of the Atom (1913) Rutherford Discovers the Proton (1914) Chadwick Discovers the Neutron (1932) Table 12.1 Location and Properties of Subatomic Particles Particle Charge Comparative Mass Location electron -1 1/1836 outside nucleus proton +1 1 inside nucleus neutron 0 1 inside nucleus If an element has 10 protons in its nucleus, how many positive charges does it have?{9}. If an atom with 10 protons is neutral, how many electrons must it have?{10} The atomic number of oxygen is 8. How many protons does it have?{11} How many electrons does oxygen have?{12} SECTION 12.2 Mass Number, Atomic Mass, and Isotopes As you learned in an earlier chapter, the atomic masses of the elements that appear on the periodic table are really average masses. We will use the element hydrogen as an example. Most hydrogen atoms (99.85% of them) have no neutrons. A few hydrogen atoms (0.15% of them) have one neutron. These two slightly different kinds of hydrogen are often referred to as light and heavy hydrogen , A.J. Girondi

5 This situation is fairly typical of most elements. Elements exist in nature as a mixture of isotopes. Isotopes are atoms that have the same number of electrons and protons, but different numbers of {13}. As a result, {14} of an element are atoms that have different masses. The atomic masses that appear on the periodic table are merely "weighted averages" of the masses of all the naturally-occurring isotopes. Weighted averages are determined by including the relative amounts of each isotope found in nature when calculating the average mass. On the periodic table the atomic mass of carbon is Actually, no carbon atoms have this mass. Some have more mass than this, and some have less. The is the {15} mass of carbon atoms in nature. The number of protons in the nuclei of atoms of a specific element is constant, but the number of neutrons can vary. Therefore, the mass numbers (the sum of the protons and neutrons) of atoms of an element can vary. Atoms of a particular isotope of an element are identified by writing the mass number of the isotope after the name of the atom. For example, the two naturallyoccurring isotopes of hydrogen are hydrogen-1 and hydrogen-2. The most common form of carbon is carbon-12, while less common isotopes include carbon-13 and carbon-14. The difference in these mass numbers is a result of differences in the number of {16} in these atoms. The sum of the {17} and {18} in the nucleus of a particular isotope of an atom is known as its {19}. While the atomic mass of an atom of a particular isotope of an element is not found on the periodic table, it is very close to the sum of the masses of the neutrons and protons in the atom. {20} have such small masses that they are not even considered when atomic masses are calculated. This is the case even if an atom contains 100 or more electrons! The mass of a proton is expressed as 1 amu (atomic mass unit). Since the proton and the neutron have almost the same mass, the approximate mass of a neutron must also be 1 amu. (Actually, it is a tiny bit more than 1.) So, what is the mass of an isotope such as oxygen-16? Oxygen-16 is an atom of a particular kind of oxygen containing a total of 16 protons and neutrons. Remember, the sum of the protons and neutrons in the nucleus of an atom is called its mass number. Thus, the mass number of this atom is 16. Since each proton and each neutron has a mass of about 1 amu, you would expect that an atom of oxygen-16 would have a mass of about {21} amu. However, the actual mass of oxygen-16 atoms is amu. Why this tiny difference? This difference is a result of the fact that when protons and neutrons combine to form a nucleus, some matter is converted into energy. This is a kind of nuclear reaction. Therefore, whenever a nucleus forms, a bit of matter will be lost. Thus, chlorine-37 has an atomic mass of amu. You can determine the mass number of a specific isotope of an element if you round off its atomic mass to the closest integer (whole number). So, to get the mass number of chlorine-37 just round to {22}. This gives you the sum of the protons and neutrons in the nucleus of this specific form (isotope) of chlorine. Since the atomic mass of an element given on the periodic table is an average, rounding it off will usually give you the mass number of the most common isotope of the element. For example, on the periodic table the average mass given for carbon is {23}. Rounding will yield 12, which is the mass number of the most common isotope of carbon, which is carbon-12. Using this method, what if the mass number of the most common isotope of element #12, magnesium? {24}. What is the name of the most common isotope of element #15, phosphorus?{25}. Since the mass number of an atom is equal to the total number of protons and neutrons it contains, you can determine the number of neutrons in an atom by subtracting the atomic number (number of protons) from the mass number (number of protons + neutrons). atomic number = number of protons mass number = atomic mass rounded to closest integer neutrons = (mass number atomic number) , A.J. Girondi

6 The "nuclear notation" of an atom includes both the atomic number and the mass number of the atom. The atom's symbol is written with its mass number expressed as a superscript at the upper-left location and the atomic number written as a subscript at the lower-left location. For example: Nuclear notation of sodium -23: Na means mass number = 23 and atomic number = 11 Here are a couple of others: hydrogen-1: 1 1 H means 1 proton and 0 neutrons in the nucleus (atomic number = 1, and mass number = 1) hydrogen-1: 1 2 H means 1 proton and 1 neutron in the nucleus (atomic number =1, mass number = 2) Problem 1. For practice, determine the number of electrons, protons, and neutrons in the atoms of aluminum and cesium listed below. a. aluminum 27; Al protons: electrons: neutrons : b. cesium 133; Cs protons: electrons: neutrons : Problem 2. The element uranium has three different isotopes. In the blanks below, write the number of electrons, protons, and neutrons for each isotope of uranium. 234 a. 92 U protons: electrons: neutrons : 235 b. 92 U protons: electrons: neutrons : 238 c. 92 U protons: electrons: neutrons : Problem 3. Supply the missing information in the table below. Nuclear Isotope Notation At.No. Mass No. No. e's No. p's No. n's a. aluminum-27 b. bismuth-209 c. calcium-40 d. copper-64 e. 2 4 f g. 8 8 h _ i _ , A.J. Girondi

7 Here is an example which shows how atomic masses on the periodic table are calculated: Sample Problem: Oxygen exists in nature in three forms. Oxygen-16 (99.759%), oxygen-17 (0.037%), and oxygen-18 (0.204%). Oxygen-16 atoms have a mass of amu; oxygen-17 atoms have a mass of amu, and the mass of oxygen-18 atoms is amu. What is the average mass of oxygen atoms in nature? To solve this problem, assume that you have a 100 atom sample of oxygen. In that case, since percent means parts per hundred, atoms in the sample would be oxygen-16 (even though atoms cannot exist in fractional parts); atoms would be oxygen-17, and atoms would be oxygen-18. The total mass of the oxygen-16 atoms in the sample is: Atomic masses such as these are determined by experiment atoms X amu 1 atom = amu The total mass of the oxygen-17 atoms in the sample is: atom X amu 1 atom = 0.63 amu The total mass of the oxygen-18 atoms in the sample is: atom X amu 1 atom = 3.67 amu The total mass of all 100 atoms in the sample would be: amu amu amu = amu. To get the average mass, we divide by the number of atoms in the sample: amu 100 = amu Does this answer agree with the atomic mass of oxygen on the periodic table? In the sample problem above, the mass of an atom of oxygen-16 is listed as amu. Assuming that protons and neutrons weigh 1 amu each, why doesn't oxygen-16 weigh 16 amu? {26} Problem 4. In nature, copper is found to exist in two forms: copper-63 and copper-65. Copper-63 atoms have a mass of amu, while copper-65 atoms have a mass of amu. Naturally-occurring copper contains 69.40% copper-63. Calculate the atomic mass of naturally-occurring copper atoms. Show all of your work , A.J. Girondi

8 Does your answer to Problem 4 compare favorably to the atomic mass of copper appearing on the periodic table? SECTION 12.3 The Spectrum of Visible Light Now that you know something about the makeup of atoms in terms of electrons, protons, and neutrons, we will take time to examine some of the experimental evidence that has contributed to our present understanding of atomic structure. In particular, we will be studying the light emitted by atoms when we add energy to them. Such light is called the "emission spectrum" of an atom. To understand the significance of emission spectra from atoms, we need to learn more about light spectra in general. In Figure 12.1, is a box that represents a continuous light spectrum. The numbers above the names of the colors represent the wavelengths of light in units called angstroms. Angstroms are denoted by the symbol Å which is a capital A with a small circle above it. An angstrom is a unit of length equal to 1 X m (which is very small). The wavelengths in Figure 12.1 range from 3900 Å to 7700 Å and comprise the continuous spectrum of visible light. These are the wavelengths of light that our eyes can see. When all of the colors (wavelengths) are present, they appear to us as white light. The numbers below the spectrum in Table 12.1 represent nanometers (nm). A nanometer is 1 X 10-9 meter ; therefore, a nanometer is ten times larger than an Angstrom. Both units are commonly used to measure wavelengths of light. <----- shorter wavelengths (higher energy) (lower energy) longer wavelengths -----> 3900 Å 4400 Å 5100 Å 5600 Å 6300 Å 7700 Å violet blues greens yellow reds 390 nm 440 nm 510 nm 560 nm 630 nm 770 nm Figure 12.1 Components of a Continuous Spectrum of Visible White Light A spectroscope is a laboratory instrument that separates the continuous spectrum of white light into its component wavelengths (colors). The spectrum is said to be continuous because all of the wavelengths from violet to red are present in white light. Light is a form of energy. The violet end of the spectrum represents light of higher energy than light in the red region, which represents visible light of the lowest energy. With this in mind, looking again at Figure 12.1, you can conclude that as the wavelength of the light gets {27}, the energy content of the light gets smaller. What does all this have to do with atomic structure? We shall soon see. When atoms absorb energy we describe them as being "excited." Excited atoms are capable of emitting light. However, atoms do not just give off light all by themselves. Something has to be done to them, first. Fluorescent lights do not emit light until an electrical current is passed through them. The sun emits light through the process of nuclear fusion which causes hydrogen atoms to combine to form helium. Electricity and nuclear fusion both involve energy changes in a system. When energy is added to a system (that is, through fusion or passing an electrical current through it), the atoms absorb that added energy. This "overload" of energy causes the electrons in the atoms to become very unstable. By unstable we mean that at least some of the electrons in the atoms have too much energy and they "want" to change their position or their motion in some way in order to get rid of the excess energy and, thereby, become more stable. Because excited electrons are so unstable, they will give off this added energy at the first opportunity. This energy that is emitted by the excited electrons may at least partly be in the form of visible light , A.J. Girondi

9 In fluorescent bulbs, a stream of electrons flows between the metal ends of a glass tube which contains a mixture of argon gas and mercury vapor. The electron stream excites the {28} of the mercury atoms which then emit energy as invisible ultraviolet rays. These rays excite the atoms in the coating of the glass tube which then emit visible light. The light emitted by specific atoms (or more accurately, by excited electrons in specific atoms) does not compose a continuous spectrum (containing all wavelengths). The light emitted by excited electrons of a particular atom is composed of certain specific wavelengths and is called a bright-line spectrum, since only certain wavelengths of light appear as colored lines. Each element emits a different characteristic set of colored bright lines (wavelengths). No two elements will emit the same set of wavelengths of light. Atoms of different elements have {29} bright-line spectra. Figures12.2 through 12.6, show the visible portion of the emission spectra of 5 different elements. Each black line represents a wavelength of light which is emitted by atoms of that element. If this document were printed in color, the lines would have a different colors. ACTIVITY 12.4 The Emission Spectra of Elements (This activity may be done by all lab groups in your class at the same time. Ask your instructor.) Procedure - Part A. Now you are going to actually look at the light emitted by excited electrons in atoms of several metallic elements. A dark or dim room should be available for this activity. 1. Obtain a burner and a platinum or nichrome wire with a small loop at the end. Pour some of each of the six solutions provided into separate wells of a dropping plate. Each solution contains a different metal ion including Sr 2+, Li 1+, Ba 2+, Ca 2+, Cu, 2+ and Na 1+. Pour some dilute hydrochloric acid (HCl) into a seventh well on the plate. Make note of which solution is in each well. 2. Dip the loop of the wire into the lithium solution and then touch it to the edge of the flame of the burner. This should impart some color to the flame. The burst of color may last only a fraction of a second. For some solutions it may last longer. What color of light does the lithium produce? {30} Clean the wire well by running tap water over it and then dipping it into the hydrochloric acid. Follow this by rinsing the wire with tap water again. Touch the clean wire to the flame. If it is clean, the color will have disappeared. If not, repeat the cleaning procedure. 3. Repeat the procedure above with the other solutions, noting the color emitted by each solution (test the sodium solution last). The metal solutions which may include copper, strontium, calcium, barium, or others. Note your observations below. Since sodium is not easy to remove from the wire, clean the loop well several times before returning it to the materials shelf. The color which you saw emitted by each metallic element in the flame was actually a blend of various wavelengths of light. These wavelengths of light compose what is known as the bright-line spectrum or emission spectrum of each element. Metal in Solution Color Emitted sodium 4. Discard the solutions and rinse the dropping plate. Return the loop and all materials to the proper place. Scientists have made quantitative observations of bright-line spectra by measuring the wavelengths of the light waves emitted by elements. This is how we got the information revealed in Figures 12.2 through Other references are also available that can provide you with the spectra of , A.J. Girondi

10 other elements. There may be a "spectrum chart" available in your classroom. Use the chart in your classroom or Figures 12.2 through 12.6 to complete problem 5 below. Problem 5. Study the emission spectrum of sodium vapor in Figure 12.6, and provide a rough estimate of the seven wavelengths (in nm) emitted in the spaces provided below. Each line in Figure 12.6 represents a different wavelength of light being emitted. a. b. c. d. e. f. g. If you were to actually observe the emission spectrum of sodium vapor through a spectroscope, you would probably see only one or two bright lines instead of seven. The reason for this is that the yellow lines located near 570 nm are so bright, that they "wash out" the other lines. They are so bright, in fact, that sodium vapor lamps are commonly used as street lights in areas where strong illumination is needed such as high-crime neighborhoods. Procedure - Part B. Every element emits characteristic wavelengths, and no two elements emit the same set of wavelengths. In this way, spectra are like fingerprints. By studying atomic spectra, we can identify elements. This is how scientists know which elements are present in the sun. Wavelengths of the sun's light have been measured, separated, and matched with spectra of known elements to identify the contents of the sun. Your teacher may have equipment which is designed to "excite" atoms of gases, so that their bright-line spectra can be observed using small hand-held spectroscopes. Before going on, ask your teacher if this equipment is available in your lab. If so, your teacher will assist you in its use. Caution: high voltage equipment will be used. (See Figure 12.7) Most light is a blend of many wavelengths. The spectroscope you will use is a device that contains a plastic "grating" which separates the wavelengths emitted by a light source so that you can observe them separately. List below the gases (or vapors) that you observed using this high-voltage equipment. Figure 12.7 Spectra Observed: Ask your teacher if any kinds of "vapor" lights are used to illuminate any of the rooms in your school. If so, use a spectroscope to examine the light emitted by them, and see if you can identify the element in the lights by comparing the spectra they emit to those in Figures 12.2 through 12.6 or those found on a spectrum chart in your classroom. If you do have such lights in your school, go to the proper location and observe the light using your hand-held spectroscope. Supply the information requested below. Location of lights in school:. Symbol of element revealed by emission spectrum:. Check with your instructor to see if you identified the unknown element(s) correctly , A.J. Girondi

11 4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å 400 nm 450 nm 500 nm 550 nm 600 nm 650 nm Figure 12.2 Emission Spectrum of Hydrogen 4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å 400 nm 450 nm 500 nm 550 nm 600 nm 650 nm Figure 12.3 Emission Spectrum of Mercury 4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å 400 nm 450 nm 500 nm 550 nm 600 nm 650 nm Figure 12.4 Emission Spectrum of Calcium 4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å 400 nm 450 nm 500 nm 550 nm 600 nm 650 nm Figure 12.5 Emission Spectrum of Neon 4000 Å 4500 Å 5000 Å 5500 Å 6000 Å 6500 Å 400 nm 450 nm 500 nm 550 nm 600 nm 650 nm Figure 12.6 Emission Spectrum of Sodium SECTION 12.5 The Bohr Model of the Atom , A.J. Girondi

12 Niels Bohr developed a theory to account for the location of electrons around the nucleus of an atom. He believed his theory would explain the bright-line spectra emitted by excited atoms. He proposed that electrons followed specific paths or orbits around the nucleus. These paths or energy levels, as they are also called, are numbered starting with the lowest one (closest to the nucleus) as 1, the next farther from the nucleus as 2, the next as 3, and so forth. Figure 12.8 illustrates 4 of the energy levels of a n atom. Bohr's model resembled a planetary system like our solar system in which he suggested that the electrons revolve around the nucleus. The first energy level, nearest the nucleus, is represented as number 1. Each level thereafter is increased by one. A total of 7 energy levels are needed to explain the structure of all of the elements Figure 12.8 Energy Levels The orbits around the nucleus are called energy levels because there are different and very specific energies associated with each level. Bohr knew that energy was being added to an atom when it was being heated or when an electrical current was passed through an element. This extra energy has to go somewhere. The added energy is absorbed by the electrons that are in the outermost orbit (farthest from the nucleus). Since there are specific amounts of energy associated with each energy level, the electron that absorbs all of this extra energy can no longer stay in the orbit (energy level) in which it normally belongs which is its "ground state." Instead, it will move to another energy level. It is now in an "excited state" and is what we earlier referred to as an "excited" electron. This electron is not doomed to spend the rest of its time in this higher energy level. Indeed, the excited electron is now very unstable. Because this is an unstable state, the electron will soon return to its ground state. This is where the atomic spectra enters into the picture. The energy that the electron emits when it returns to its ground state is in the form of light and heat. Figure 12.9 gives a general picture of this process. In "A", the electron is excited and jumps to a new energy level that is further from the nucleus. As the electron falls back to its ground state in "B", energy is given off in the form of light. LIGHT EMITTED A B Figure 12.9 Emission of Light from an Excited Electron As electrons move to lower energy levels, the energy they emit is given off in a "burst" or quantity of energy with a well-defined wavelength. The quantities of emitted energy are called quanta. Bohr's theory of the atom gave birth to the quantum theory, a name that reflects the notion that atoms must absorb and emit energy in specific amounts. Therefore, we say that the energy is "quantized." A quantum of energy can be defined as the amount of energy needed to move an electron from one energy level to the next higher one. Similarly, it can be defined as the amount of energy emitted when an , A.J. Girondi

13 electron moves from its present energy level to a lower one. In his theory, Bohr proposed that electrons were only "allowed" to exist at certain distances from the nucleus. These distances became his energy levels. He believed that electrons were not "allowed" to exist between these levels. Although the reasons for this behavior were unclear, the idea did explain the existence of bright-line spectra. The spectrum of hydrogen contains a specific set of bright lines. Of these, only 3 or 4 are clearly visible to the naked eye. (See Figure 12.2) So Bohr asked, why only this set of lines? Why does the excited hydrogen electron emit only this specific set of wavelengths? According to Bohr, the fact that there was always the same set of bright lines in the hydrogen spectrum was evidence that only certain energy changes were possible for the hydrogen electron. The electron could only make certain "jumps" and, therefore, could only emit certain wavelengths of light. The number of lines was limited because there were only a few "excited" energy levels to which the electron was "allowed" to move. He explained that the electron was - for whatever reason - not "permitted" to exist between these levels. Since every element has a specific number of electrons, different wavelengths of light will be emitted by excited electrons of atoms of different elements. Even without a spectroscope you can see that the glow of a neon sign is quite different from the fluorescent lights in your classroom. You probably did not realize that what you were seeing was the result of excited electrons at work. To better understand how an electron behaves as it returns to its stable ground state, consider how you would jump down a staircase. Just as an electron can only stop at certain energy levels, you can only stop at certain levels (steps). The electron cannot exist between energy levels, and you cannot stop between the steps. To get down to the bottom of the staircase, you could jump all the way down at once; or one step at a time; or one step followed by a two-step jump; or perhaps a two-step jump followed by one step, and then by a three-step jump. A variety of different jumps is possible, each yielding a different amount of energy Likewise, suppose an electron jumps out to level 4 in Figure There are ten different sizes of jumps possible as it returns to its ground state. So how many different wavelengths of light could be emitted by this electron? {31} A sample containing many excited atoms of this element would emit all possible wavelengths. 1 Ground State Figure Energy Level Diagram SECTION 12.6 The Theory of Quantum Mechanics Bohr's model worked well in explaining the emission spectra of hydrogen. But, it really did not work well for atoms with more electrons in them. It even failed to explain some of the complexities of the spectrum of hydrogen atoms. In addition, it could not explain why electrons are not pulled into the nucleus of the atom (which would cause the atom to collapse). We know, today, that electrons do not travel around the nucleus as planets do around the sun. The failure of Bohr's model led to the development of the theory of quantum mechanics. Quantum mechanics is a very complex theory. It involves describing the approximate location of any electron in any atom. The theory provides an explanation for the behavior of electrons in atoms. The details of quantum mechanics, while beyond the scope of this course, will be discussed briefly later in this chapter. This will , A.J. Girondi

14 provide you with an appreciation and a general knowledge of the principles of the theory. In these next sections, you will see how each of the energy levels is divided into sublevels. We will concern ourselves with how the electrons distribute themselves among the energy levels and sublevels of atoms. Each energy level can hold a maximum number of electrons. The number of electrons in any given energy level is limited to 2n 2. This expression means 2(n 2 ), not (2n) 2! For example, the energy level 3 can be occupied by 2(3 )2, or 18, electrons at the most. Problem 6. Calculate the maximum number of electrons that could occupy the first four energy levels in Energy Level Maximum No. Electrons (2n 2 ) Energy levels are sometimes called "shells" in which case each one is designated by a letter, K through Q (see Figure 12.13). You may see this notation in your future studies in chemistry, so it is important that you know something about it. Each main energy level consists of one or more sublevels. The number of sublevels is equal to the number of the energy level. Therefore, the third energy level has three sublevels, and the fifth energy level has five sublevels. Which energy level has four sublevels? {32}. Which energy level has two sublevels? {33}. The energy level diagram can now be redrawn. The main energy levels and the sublevels within each energy level are labeled in Figure Each sublevel is designated by a small letter (s,p,d,f). By overlapping we mean that a higher energy level begins before a lower one ends. Notice on figure how the 4th energy level begins before the third is full. Notice the overlapping of the main energy levels that occurs as you go beyond the third energy level. This means that a higher energy level begins before a lower one ends. Clearly, atoms are more complicated than a staircase. Certainly, the third step on a staircase cannot begin before the second step ends! However, you must keep in mind that energy levels are not "physical things," whereas staircases are real physical things. Energy levels are conceptual things regions of space. The complexity of overlapping energy levels becomes more noticeable as you progress from the third energy level to the seventh. The first sublevel of each energy is designated as an "s" sublevel. This is followed by a "p" sublevel, the a "d", then an "f", depending upon how many electrons there are. Note Figure 12.13, in which the energy sublevels s, p, d, f are each shown with one or more circles. These circles represent orbitals. The number of circles represents the number of orbitals in a particular sublevel. At this point, you can think of an orbital as a region of space where up to two electrons can coexist. We think of energy levels as a staircase, and electrons can exist only in the steps, not between them, right? We think of orbitals as "houses" where electrons live. Each of these orbital "houses" can contain a maximum of two electrons, and as you can see there are different numbers of orbital "houses" on different steps , A.J. Girondi

15 7s 6d 6p 6s 5f 5d 5p 5s 4f 4d 4p 4s 3d 3p 7th Energy Level 6th Energy Level 5th Energy Level 4th Energy Level 3rd Energy Level 3s 2p 2s 2nd Energy Level 1s 1st Energy Level Nucleus Figure Overlapping of Energy Levels We will learn more about orbitals later in this chapter, and we will change the definition somewhat at that time. Looking at Figure 12.13, notice that every s sublevel has only one orbital. Look at the p sublevels in figure How many orbitals does the p sublevel have? {34}. How many orbitals does the d sublevel have? {35}. How many orbitals does the f sublevel have? {36} Since each orbital can hold a maximum of two electrons, what is the capacity of an "s" sublevel? {37} Of a "p" sublevel? {38} Of a "d" sublevel? {39} Of an "f" sublevel? {40}. To help you remember the order in which the energy levels and sublevels fill, you should become familiar with the diagonal rule. It is shown in Figure By simply following the arrows along the diagonals, you have the correct order of filling. Use this diagonal rule to see if the levels and sublevels are properly arranged in Figure The order of filling is: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d In theory there are other orbitals such as 5g, 6f, etc., but they are never used since there are no atoms that have enough electrons to fill them. 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 5d 5f 6s 6p 6d 7s Figure The Diagonal Rule , A.J. Girondi

16 ORBITALS Energy Level 7 (Q Shell) 5f 7s 6d Energy Level 6 (P Shell) 6p Energy Level 5 (O Shell) 5d 4f 6s 5p 5s 4d Energy Level 4 (N Shell) 4p 3d Energy Level 3 (M Shell) 3p 4s 3s 2p 2s Energy Level 2 (L Shell) Energy Level 1 (K Shell) 1s Figure Atomic Energy Levels and Sublevels SECTION 12.7 Orbital Notation of Electrons Now that we have the sublevels and orbitals identified, we can use this information to describe where the electrons are located in the various atoms represented on the periodic table. To do this, you must keep several basic rules in mind: 1. Electrons fill up the energy sublevels , A.J. Girondi

17 2. The lowest energy sublevel must be completely filled before the next higher sublevel can begin to be filled. For example, according to the diagonal rule, the 4s sublevel must be filled before the 3d sublevel can be used. Caution: note that this rule refers to sublevels, not energy levels. 3. Each orbital (represented by a circle in figure 12.13) can hold a maximum number of two electrons. We will represent electrons as slashed lines. The symbol (/) represents a single electron in an orbital. A full orbital would contain two electrons and would be represented like this: (X). This method of representing the electrons in an atom is called orbital notation. 4. Electrons repel each other and will not pair up in a single orbital in any given sublevel until all orbitals in that sublevel are half full (have one electron in them). The incorrect and correct way to represent a 2p orbital containing three electrons is shown below: Incorrect 2p Orbital Notation Correct 2p Orbital Notation 2p 2p (X)(/)( ) (/)(/)(/) Problem 7. You should now fill in the correct orbital notation to the right of the incorrect notation shown below for a 3d sublevel with 5 orbitals that contain 6 electrons: Incorrect 3d Orbital Notation Correct 3d Orbital Notation 3d 3d (X)(X)(X)( )( ) ( )( )( )( )( ) Orbital notation can be easily written for each element on the periodic table. For example, to write the orbital notation for an atom of iron, we must first look up the atomic number of iron on the periodic table. The atomic number of iron is {41}. How many electrons does iron have? {42}. The orbital diagram shown below illustrates the order in which the sublevels and orbitals are filled for iron according to the diagonal rule. Iron: 1s 2s 2p 3s 3p 4s 3d (X) (X) (X)(X)(X) (X) (X)(X)(X) (X) (X)(/)(/)(/)(/) Each "X" represents two slashed lines. Is the number of slashed lines shown equal to the number of electrons in iron? {43} In Table 12.2, the orbital notations for the first ten elements are illustrated using this set of rules. Notice that even empty orbitals are shown if a sublevel contains at least one electron (see boron and carbon). Table 12.2 Orbital Notation of the First Ten Elements Element 1s 2s 2p Hydrogen (/) Helium (X) Lithium (X) (/) Beryllium (X) (X) Boron (X) (X) (/)( )( ) Carbon (X) (X) (/)(/)( ) Nitrogen (X) (X) (/)(/)(/) Oxygen (X) (X) (X)(/)(/) Fluorine (X) (X) (X)(X)(/) Neon (X) (X) (X)(X)(X) , A.J. Girondi

18 Table12.3 summarizes much of the important information about the location of electrons in the first four energy levels. Table 12.3 Data Summary for Energy Levels 1 Through 4 Energy Level 1st 2nd 3rd 4th Shell K L M N Shell Capacities No. of Sublevels Sublevel Types s s,p s,p,d s,p,d,f Sublevel Capacities 2 2,6 2,6,10 2,6,10,14 No. of orbitals per sublevel: s = 1; p = 3; d = 5; f = 7 Problem 8. Now it is time for you to use what you have learned. Fill in the electrons for the elements below. Follow the rules listed in section 12.7 as well as the diagonal rule. The finished product will be the orbital notation for an atom of each element. Element 1s 2s 2p 3s 3p 4s 3d a. beryllium ( ) ( ) b. magnesium ( ) ( ) ( )( )( ) ( ) c. nitrogen ( ) ( ) ( )( )( ) d. silicon ( ) ( ) ( )( )( ) ( ) ( )( )( ) e. calcium ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) f. potassium ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) g. argon ( ) ( ) ( )( )( ) ( ) ( )( )( ) h. sodium ( ) ( ) ( )( )( ) ( ) i. chlorine ( ) ( ) ( )( )( ) ( ) ( )( )( ) j. copper ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( ) k. phosphorus ( ) ( ) ( )( )( ) ( ) ( )( )( ) l. nickel ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( ) Note: The 3d sublevel is a "higher energy" level than the 4s because electrons in the 3d have more energy than those in the 4s. Electrons fill the "lower energy" levels first. However, the lower energy 4s sublevel is farther from the nucleus than the higher energy 3d. A "higher" energy sublevel is not necessarily farther from the nucleus than a "lower" energy sublevel. The distance of the energy sublevels from the nucleus is: Increasing Distance from Nucleus-----> 1s 2s,2p 3s,3p,3d 4s,4p,4d,4f 5s,5p,5d,5f 6s,6p,6d,etc. (closest) (farthest) SECTION 12.8 Electron Configuration Notation There is another form of notation which reveals which energy level and sublevel the electrons of an atom are in, but it does not show the individual orbitals. This method is known as electron configuration notation. A superscript is used to reveal the number of electrons in each sublevel. For example, the configuration notation for lithium, element number 3, is: 1s 2 2s 1. This means that there are 2 electrons in , A.J. Girondi

19 sublevel "s" of energy level 1, and there is one electron in sublevel "s" of energy level 2. Configuration notation does not reveal whether the electrons are alone or paired in their orbitals like orbital notation does. Note that the sum of the superscripts should equal the number of electrons in the atoms. (A lithium atom contains three electrons.) If an atom has a configuration notation of 1s 2 2s 2 2p 3, how many electrons does the atom contain? {44} How many of these electrons are found in the second energy level? {45} How many are found in the first energy level? {46} More examples of configuration notation are shown below. Element Configuration Notation hydrogen 1s 1 helium 1s 2 lithium 1s 2 2s 1 beryllium 1s 2 2s 2 boron 1s 2 2s 2 2p 1 manganese 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 5 selenium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 4 antimony 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 3 Problem 9. Give the electron configuration notation for the elements listed below. Element Atomic Number Configuration Notation a. C 6 b. N 7 c. O 8 d. F 9 1s 2 2s 2 2p 5 e. Ne 10 f. P 15 g. Ar 18 h. Ca 20 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 i. Mn 25 j. Zn 30 k. Kr , A.J. Girondi

20 SECTION 12.9 Review Problems Problem 10. Naturally-occurring iridium (Ir) consists of two isotopes. Iridium 191 (atomic mass = composes 37.58% of it, while iridium 193 (atomic mass = ) composes the remaining 62.7%. Calculate the average atomic mass of iridium atoms. Problem 11. Write the nuclear notation for the following isotopes: a. iridium 191 b. chlorine 37 c. chromium 50 Problem 12. Indicate the number of protons, electrons and neutrons in each of the following: protons electrons neutrons iridium-191 chlorine-37 chromium-50 Problem 13. Write the orbital notation for: a. sulfur: 1s 2s 2p 3s 3p 4s 3d ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( ) b. vanadium: 1s 2s 2p 3s 3p 4s 3d ( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( )( )( ) Problem 14. Write the configuration notation for: a. sulfur: b. vanadium: ACTIVITY An Optional Enrichment Activity - The Colors of Transition Metal Chemistry The elements with atomic numbers 21 through 30 exhibit many properties different from those of other elements in the same period (row). The elements directly below in the next two periods, however, have similar properties. Together these elements are known as the transition metals. The purpose of this activity is to compare the properties of some substances which contain transition metals with those which , A.J. Girondi

21 do not contain non-transition metals. We will then try to relate any unique properties of the transition metals to the electrons which they contain. Sc Ti V Cr Mn Fe Co Ni Cu Zn Figure The Transition Metals of Row Four Materials: 96-well microplate; plastic micropipets; toothpicks; 0.1M solutions of compounds containing period 4 metals: KNO3, Ca(NO3)2, NH4VO3, Cr(NO3)2, Mn(NO3)2, Co(NO3)2, Fe(NO3)3, Ni(NO3)2, Cu(NO3)2, Zn(NO3)2; 6.0M ammonia (NH3); 1M potassium thiocyanate, KSCN; 6M hydrochloric acid (HCl); one sheet of white paper. Procedure: 1. Place a 96-well microplate on a sheet of white paper. The numbered columns should be at the top with the lettered rows to your left. 2. Place 5 drops of KNO3 in each of wells A1, B1, C1, and D1. 1 = KNO3 4 = Cr(NO3)3 7 = Fe(NO3)3 10 = Zn(NO3)2 2 = Ca(NO3)2 5 = Mn(NO3)2 8 = Ni(NO3)2 3 = NH4VO3 6 = Co(NO3)2 9 = Cu(NO3)2 NH3(aq) -----> A KSCN > B HCl > C Control > D Figure Microplate 3. Place 5 drops of Ca(NO3)2 in each of wells A2, B2, C2, and D2. 4. Continue to repeat this process of placing 5 drops of solutions of metal compounds in subsequent , A.J. Girondi

22 columns of the microplate. Use chemicals in the order in which they are given in the materials list above. The last drops will be in column Add 5 drops of 6M NH3 to each well in row A. Mix well with a toothpick. Rinse the toothpick with water before moving from well to well. CAUTION: ammonia solution is caustic and has strong vapors. 6. Add 5 drops of KSCN solution to each well in row B. Mix well with a toothpick, rinsing the toothpick between uses as before. 7. Finally, add 5 drops of HCl to each well in row C. CAUTION: Do not mix HCl and KSCN. Clean up all spills immediately. Mix well. Use row D as a control for comparing with the other rows. Data and Observations: 1. Observe row D of your microplate. Record your observations - particularly colors, if any - in Table Compare the solutions in the wells in rows A, B, and C in each column to the solution in row D in that column. Where there was a change, record what you observe. A1 A2 A3 A4 A5 A6 A7 A8 A9 A10 B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 D1 D2 D3 D4 D5 D6 D7 D8 D9 D10 TABLE 12.4 Transition Metal Chemistry Analysis and Conclusions: Comment on the order of the metal ions in the wells from left to right in the rows of your microplate when compared to the positions of the metals in the periodic table: {47} Did you observe any colors in columns 1 and 2 of your microplate? {48} The metals used in columns 1 and 2 of your microplate are not transition metals. With respect to sublevel , A.J. Girondi

23 (s,p,d,f), what kind of electrons do the transition metals have that these metals do not have? {49} Review what you observed in columns 3 through 9 of your microplate. What do the results in these columns have in common? {50} Did the metal in column 1 and 2 produce any colors? {51} Did the metal in column 10 produce any colors? {52} Is the "d" sublevel of electrons completely filled in the metal atoms found in columns 3 through 9 of your microplate? {53} What is true about the distribution of electrons in the transition metals in columns 3 through 9 that is responsible for the fact that they produce colored products? {54} With respect to color, the results in column 10 should have been different from the results in columns 3 through 10. Why? {55} What property would help to identify a compound as one which contains a transition metal ion? { 56} SECTION Learning Outcomes This is the end of chapter 12. Check off each of the learning outcomes on the next page which you have completed. When you have mastered them all, arrange to take the chapter 12 exam, and go to chapter 13 which is part 2 of your study of atomic structure. 1. List the subatomic particles and describe their location and properties. 2. Determine the number of electrons, protons, and neutrons in an atom of any element given information such as its atomic number, mass number, or atomic mass. 3. Define isotope, give an example of one, and calculate the atomic mass of an element given the masses of its isotopes and their abundance in nature. 4. Describe the Bohr model of the atom. 5. Explain how atomic spectra help to explain the behavior of electrons. 6. Explain what is meant by energy levels, sublevels, orbitals, and electron configurations. 7. Give the capacities of energy levels and sublevels. 8. Write the electron configuration notation of atoms. 9. Write the orbital notation for atoms , A.J. Girondi

24 SECTION Answers to Questions and Problems Questions: {1} repel; {2} attract; {3} Dalton; {4} nucleus; {5} positively; {6} +1; {7} neutrons; {8} 19; {9} 10 {10} 10; {11} 8; {12} 8; {13} neutrons; {14} isotopes; {15} average; {16} neutrons; {17} protons; {18} neutrons; {19} mass number; {20} electrons; {21} 16; {22} 37; {23} 12.01; {24} 24; {25} phosphorus-31; {26} Some matter is converted into energy when an atom of oxygen forms; {27} longer; {28} electrons; {29} different (or unique); {30} reddish; {31} 10; {32} fourth; {33} second; {34} 3; {35} 5; {36} 7; {37} 2; {38} 6; {39} 10; {40} 14; {41} 26; {42} 26; {43} yes; {44} 7; {45} 5; {46} 2; {47} the same; {48} no; {49} d; {50} they have color; {51} no; {52} no; {53} no; {54} They have an incomplete (unfilled) "d" sublevel; {55} Elements in column 10 on the microplate have a filled "d" sublevel; {56} color Problems: 1. a. 13,13,14 b. 55, 55, a. 92, 92, 142 b. 92, 92, 143 c. 92, 92, Nuclear Isotope Notation At.No. Mass No. No. e's No. p's No. n's 27 a. aluminum-27 Al _ b. bismuth-209 Bi _ _ c. calcium-40 20Ca _ d. copper-64 Cu _ e. _helium-4 2 He f. _lead Pb 82 _ _ g. _oxygen O h. _tin Sn 50 _ _ i. _zinc Zn _ _ These are estimates: 490 nm, 495 nm, 568 nm, 572 nm, 592 nm, 595 nm, 615 nm 6. 2, 8, 18, d (X)(/)(/)(/)(/) , A.J. Girondi