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CHAPTER 4 The Structure of Atoms Chapter Outline Subatomic Particles 1. Fundamental Particles 2. The Discovery of Electrons 3. Canal Rays and Protons 4. Rutherford and the Nuclear Atom 5. Atomic Number 6. Neutrons 7. Mass Number and Isotopes 8. Mass spectrometry and Isotopic Abundance 9. The Atomic Weight Scale and Atomic Weights 10. The Periodic Table: Metals, Nonmetals, and Metalloids The Electronic Structures of Atoms 11. Electromagnetic radiation 12. The Photoelectric Effect 13. Atomic Spectra and the Bohr Atom 14. The Wave Nature of the Electron 15. The Quantum Mechanical Picture of the Atom 16. Quantum Numbers 17. Atomic Orbitals 18. Electron Configurations 19. Paramagnetism and Diamagnetism 20. The Periodic Table and 2 Electron Configurations Fundamental Particles Three fundamental particles make up atoms. The following table lists these particles together with their masses and their charges. Particle Mass (amu) Charge Electron (e-) 0.00054858 -1 Proton (p,p+) 1.0073 +1 Neutron(n,n0) 1.0087 0 3 The Discovery of Electrons • Humphrey Davy in the early 1800’s passed electricity through compounds and noted and concluded that: – the compounds decomposed into elements. – compounds are held together by electrical forces. • Michael Faraday in 1832-1833 realized that the amount of reaction that occurs during electrolysis is proportional to the electrical current passed through the compounds. 4 The Discovery of Electrons • Cathode Ray Tubes experiments performed in the late 1800’s & early 1900’s. – Consist of two electrodes sealed in a glass tube containing a gas at very low pressure. – When a voltage is applied to the cathodes a glow discharge is emitted. 5 The Discovery of Electrons • These “rays” are emitted from cathode (end) and travel to anode (+ end). – Cathode Rays must be negatively charged! • J.J. Thomson modified the cathode ray tube experiments in 1897 by adding two adjustable voltage electrodes. – Studied the amount that the cathode ray beam was deflected by additional electric field. 6 The Discovery of Electrons 7 The Discovery of Electrons 8 The Discovery of Electrons 9 The Discovery of Electrons • Thomson used his modification to measure the charge to mass ratio of electrons. Charge to mass ratio e/m = -1.75882 x 108 coulomb/g • Thomson named the cathode rays electrons. • Thomson is considered to be the “discoverer of electrons”. • TV sets and computer screens are cathode 10 ray tubes. The Discovery of Electrons • Robert A. Millikan won the Nobel Prize in 1923 for his famous oil-drop experiment. • In 1909 Millikan determined the charge and mass of the electron. 11 The Discovery of Electrons • Millikan determined that the charge on a single electron = -1.60218 x 10-19 coulomb. • Using Thomson’s charge to mass ratio we get that the mass of one electron is 9.11 x 10-28 g. – e/m = -1.75882 x 108 coulomb – e = -1.60218 x 10-19 coulomb – Thus m = 9.10940 x 10-28 g 12 Canal Rays and Protons • Eugene Goldstein noted streams of positively charged particles in cathode rays in 1886. – Particles move in opposite direction of cathode rays. – Called “Canal Rays” because they passed through holes (channels or canals) drilled through the negative electrode. • Canal rays must be positive. – Goldstein postulated the existence of a positive fundamental particle called the “proton”. 13 Rutherford and the Nuclear Atom • Ernest Rutherford directed Hans Geiger and Ernst Marsden’s experiment in 1910. – α- particle scattering from thin Au foils – Gave us the basic picture of the atom’s structure. 14 Rutherford and the Nuclear Atom 15 Rutherford and the Nuclear Atom Rutherford’s major conclusions from the αparticle scattering experiment 1. The atom is mostly empty space. 2. It contains a very small, dense center called the nucleus. 3. Nearly all of the atom’s mass is in the nucleus. 4. The nuclear diameter is 1/10,000 to 1/100,000 times less than atom’s radius. 16 Rutherford and the Nuclear Atom 17 Rutherford and the Nuclear Atom • Because the atom’s mass is contained in such a small volume: – The nuclear density is ~1015g/mL. – This is equivalent to ~3.72 x 109 tons/in3. – Density inside the nucleus is almost the same as a neutron star’s density. 18 Atomic Number • The atomic number is equal to the number of protons in the nucleus. – Sometimes given the symbol Z. – On the periodic table Z is the uppermost number in each element’s box. • In 1913 H.G.J. Moseley realized that the atomic number determines the element . – The elements differ from each other by the number of protons in the nucleus. – The number of electrons in a neutral atom is also equal to the atomic number. 19 Neutrons • James Chadwick in 1932 analyzed the results of α-particle scattering on thin Be films. • Chadwick recognized existence of massive neutral particles which he called neutrons. – Chadwick discovered the neutron. 20 Mass Number and Isotopes • Mass number is given the symbol A. • A is the sum of the number of protons and neutrons. – Z = proton number N = neutron number – A=Z+N • A common symbolism used to show mass and proton numbers is A 12 48 Z 6 20 Can be shortened to this symbolism. E for example 14 63 N, Cu, C, Ca, 107 Ag, etc. 197 79 Au 21 Mass Number and Isotopes • Isotopes are atoms of the same element but with different neutron numbers. – Isotopes have different masses and A values but are the same element. • One example of an isotopic series is the hydrogen isotopes. 1H or protium is the most common hydrogen isotope. • one proton and no neutrons 2H or deuterium is the second most abundant hydrogen isotope. • one proton and one neutron 3H or tritium is a radioactive hydrogen isotope. • one proton and two neutrons 22 Mass Number and Isotopes • The stable oxygen isotopes provide another example. • 16O is the most abundant stable O isotope. • How many protons and neutrons are in 16O? 8 protons and 8 neutrons 17O is the least abundant stable O isotope. How many protons and neutrons are in 17O? 8 protons and 9 neutrons 18O is the second most abundant stable O isotope. How many protons and neutrons in 18O? 8 protons and 10 neutrons 23 Mass Spectrometry and Isotopic Abundances • Francis Aston devised the first mass spectrometer. – Device generates ions that pass down an evacuated path inside a magnet. – Ions are separated based on their mass. 24 Mass Spectrometry and Isotopic Abundances There are four factors which determine a particle’s path in the mass spectrometer. 1 accelerating voltage 2 magnetic field strength 3 masses of particles 4 charge on particles 25 Mass Spectrometry and Isotopic Abundances • Mass spectrum of Ne+ ions shown below. – How scientists determine the masses and abundances of the isotopes of an element. 26 The Atomic Weight Scale and Atomic Weights • If we define the mass of 12C as exactly 12 atomic mass units (amu), then it is possible to establish a relative weight scale for atoms. – 1 amu = (1/12) mass of 12C by definition – What is the mass of an amu in grams? 27 The Atomic Weight Scale and Atomic Weights Example 4-1: Calculate the number of atomic mass units in one gram. – The mass of one 31P atom has been experimentally determined to be 30.99376 amu. – 1 mol of 31P atoms has a mass of 30.99376 g. 28 The Atomic Weight Scale and Atomic Weights Example 4-1: Calculate the number of atomic mass units in one gram. – The mass of one 31P atom has been experimentally determined to be 30.99376 amu. – 1 mol of 31P atoms has a mass of 30.99376 g. 1 mole 31 P atoms ? amu 1.000 g 31 30.99376 g P 6.022 1023 31 P atoms 30.99376 amu 31 31 P atom 1 mole P atoms 23 31 6.022 10 amu P 29 The Atomic Weight Scale and Atomic Weights • The atomic weight of an element is the weighted average of the masses of its stable isotopes 30 Atomic Weight Scale and Atomic Weights Example 4-2: Naturally occurring Cu consists of 2 isotopes. It is 69.1% 63Cu with a mass of 62.9 amu, and 30.9% 65Cu, which has a mass of 64.9 amu. Calculate the atomic weight of Cu to one decimal place. atomic weight (0.691)(62 .9 amu) (0.309)(64 .9 amu) 63 Cu isotope 65 Cu isotope atomic weight 63.5 amu for copper 31 The Atomic Weight Scale and Atomic Weights Example 4-3: Naturally occurring chromium consists of four isotopes. It is 4.31% 2450Cr, mass = 49.946 amu, 83.76% 2452Cr, mass = 51.941 amu, 9.55% 2453Cr, mass = 52.941 amu, and 2.38% 54Cr, mass = 53.939 amu. Calculate the atomic 24 weight of chromium. You do it! 32 The Atomic Weight Scale and Atomic Weights Example 4-3: Naturally occurring chromium consists of four isotopes. It is 4.31% 2450Cr, mass = 49.946 amu, 83.76% 2452Cr, mass = 51.941 amu, 9.55% 2453Cr, mass = 52.941 amu, and 2.38% 54Cr, mass = 53.939 amu. Calculate the atomic 24 weight of chromium. atomic weight (0.0431 49.946 amu) (0.8376 51.941 amu) (0.0955 52.941 amu) (0.0238 53.939 amu) 2.153 43.506 5.056 1.284 amu 51.998 amu 33 The Atomic Weight Scale and Atomic Weights Example 4-4: The atomic weight of boron is 10.811 amu. The masses of the two naturally occurring isotopes 510B and 511B, are 10.013 and 11.009 amu, respectively. Calculate the fraction and percentage of each isotope. You do it! • This problem requires a little algebra. – A hint for this problem is x + (1-x) = 1 34 The Atomic Weight Scale and Atomic Weights Example 4-4: The atomic weight of boron is 10.811 amu. The masses of the two naturally occurring isotopes 510B and 511B, are 10.013 and 11.009 amu, respectively. Calculate the fraction and percentage of each isotope. 10.811 amu x(10.013 amu) 1 x (11.009 amu) 10 B isotope 11 B isotope 10.013 x 11.009 - 11.009 x amu 10.811 -11.009 amu 10.013 x - 11.009 x amu - 0.198 -0.996 x 0.199 x 35 The Atomic Weight Scale and Atomic Weights • Note that because x is the multiplier for the 10B isotope, our solution gives us the fraction of natural B that is 10B. • Fraction of 10B = 0.199 and % abundance of 10B = 19.9%. • The multiplier for 11B is (1-x) thus the fraction of 11B is 1-0.199 = 0.801 and the % abundance of 11B is 80.1%. 36 The Periodic Table: Metals, Nonmetals, and Metalloids • 1869 - Mendeleev & Meyer – Discovered the periodic law • The properties of the elements are periodic functions of their atomic numbers. 37 The Periodic Table: Metals, Nonmetals, and Metalloids • Groups or families – Vertical group of elements on periodic table – Similar chemical and physical properties 38 The Periodic Table: Metals, Nonmetals, and Metalloids • Period – Horizontal group of elements on periodic table – Transition from metals to nonmetals 39 The Periodic Table: Metals, Nonmetals, and Metalloids • Some chemical properties of metals 1. 2. 3. 4. Outer shells contain few electrons Form cations by losing electrons Form ionic compounds with nonmetals Solid state characterized by metallic bonding 40 The Periodic Table: Metals, Nonmetals, and Metalloids • Group IA metals – Li, Na, K, Rb, Cs, Fr • One example of a periodic trend – The reactions with water of Li 41 The Periodic Table: Metals, Nonmetals, and Metalloids • Group IA metals – Li, Na, K, Rb, Cs, Fr • One example of a periodic trend – The reactions with water of Li, Na 42 The Periodic Table: Metals, Nonmetals, and Metalloids • Group IA metals – Li, Na, K, Rb, Cs, Fr • One example of a periodic trend – The reactions with water of Li, Na, & K 43 The Periodic Table: Metals, Nonmetals, and Metalloids • Group IIA metals – alkaline earth metals • Be, Mg, Ca, Sr, Ba, Ra 44 The Periodic Table: Metals, Nonmetals, and Metalloids • Some chemical properties of nonmetals 1. 2. 3. Outer shells contain four or more electrons Form anions by gaining electrons Form ionic compounds with metals and covalent compounds with other nonmetals Form covalently bonded molecules; noble gases are monatomic 4. 45 The Periodic Table: Metals, Nonmetals, and Metalloids • Group VIIA nonmetals – halogens – F, Cl, Br, I, At 46 The Periodic Table: Metals, Nonmetals, and Metalloids • Group VIA nonmetals – O, S, Se, Te 47 The Periodic Table: Metals, Nonmetals, and Metalloids • Group 0 nonmetals – noble, inert or rare gases – He, Ne, Ar, Kr, Xe, Rn 48 The Periodic Table: Metals, Nonmetals, and Metalloids • Stair step function on periodic table separates metals from nonmetals. • Metals are to the left of stair step. – Approximately 80% of the elements • Best metals are on the far left of the table. 49 The Periodic Table: Metals, Nonmetals, and Metalloids • Stair step function on periodic table separates metals from nonmetals. • Nonmetals are to the right of stair step. – Approximately 20% of the elements • Best nonmetals are on the far right of the table. 50 The Periodic Table: Metals, Nonmetals, and Metalloids • Stair step function on periodic table separates metals from nonmetals. • Metalloids have one side of the box on the stair step. 51 The Periodic Table: Metals, Nonmetals, and Metalloids • Periodic trends in metallic character 52 Electromagnetic Radiation • The wavelength of electromagnetic radiation has the symbol λ . • Wavelength is the distance from the top (crest) of one wave to the top of the next wave. – Measured in units of distance such as m,cm, Å. – 1 Å = 1 x 10-10 m = 1 x 10-8 cm • The frequency of electromagnetic radiation has the symbol . • Frequency is the number of crests or troughs that pass a given point per second. – Measured in units of 1/time - s-1 53 Electromagnetic Radiation • The relationship between wavelength and frequency for any wave is velocity = λ υ. • For electromagnetic radiation the velocity is 3.00 x 108 m/s and has the symbol c. • Thus c = λ υ for electromagnetic radiation. 54 Electromagnetic Radiation 55 Electromagnetic Radiation 56 Electromagnetic Radiation • Molecules interact with electromagnetic radiation. – Molecules can absorb and emit light. • Once a molecule has absorbed light (energy), the molecule can: 1. Rotate 2. Translate 3. Vibrate 4. Electronic transition 57 Electromagnetic Radiation • For water: – Rotations occur in the microwave portion of spectrum. – Vibrations occur in the infrared portion of spectrum. – Translation occurs across the spectrum. – Electronic transitions occur in the ultraviolet portion of spectrum. 58 Electromagnetic Radiation Example 4-5: What is the frequency of green light of wavelength 5200 Å? 59 Electromagnetic Radiation Example 4-5: What is the frequency of green light of wavelength 5200 Å? c c 1 x 10 -10 m 5.200 10 -7 m (5200 Å) 1Å 3.00 108 m/s 5.200 10-7 m 5.77 1014 s -1 60 Electromagnetic Radiation • In 1900 Max Planck studied black body radiation and realized that to explain the energy spectrum he had to assume that: 1. energy is quantized 2. light has particle character • Planck’s equation is E h or E hc h Planck’ s constant 6.626 x 10 -34 J s 61 Electromagnetic Radiation Example 4-6: What is the energy of a photon of green light with wavelength 5200 Å? What is the energy of 1.00 mol of these photons? 62 Electromagnetic Radiation Example 4-6: What is the energy of a photon of green light with wavelength 5200 Å? What is the energy of 1.00 mol of these photons? From Example 4-5, we know that 5.77 x 1014 s-1 E h E (6.626 10-34 J s)(5.77 1014 s -1 ) E 3.83 10-19 J per photon For 1.00 mol of photons : (6.022 1023 photons)(3 .83 10-19 J per photon) 231 kJ/mol 63 The Photoelectric Effect Light can strike the surface of some metals causing an electron to be ejected. 64 The Photoelectric Effect • What are some practical uses of the photoelectric effect? You do it! • • • • Electronic door openers Light switches for street lights Exposure meters for cameras Albert Einstein explained the photoelectric effect – Explanation involved light having particle-like behavior. – Einstein won the 1921 Nobel Prize in Physics for this 65 work. Atomic Spectra and the Bohr Atom An emission spectrum is formed by an electric current passing through a gas in a vacuum tube (at very low pressure) which causes the gas to emit light. – Sometimes called a bright line spectrum. 66 Atomic Spectra and the Bohr Atom 67 Atomic Spectra and the Bohr Atom • An absorption spectrum is formed by shining a beam of white light through a sample of gas. – Absorption spectra indicate the wavelengths of light that have been absorbed. 68 Atomic Spectra and the Bohr Atom • Every element has a unique spectrum. • Thus we can use spectra to identify elements. – This can be done in the lab, stars, fireworks, etc. 69 Atomic Spectra and the Bohr Atom • Atomic and molecular spectra are important indicators of the underlying structure of the species. • In the early 20th century several eminent scientists began to understand this underlying structure. – Included in this list are: – Niels Bohr – Erwin Schrodinger – Werner Heisenberg 70 Atomic Spectra and the Bohr Atom Example 4-7: An orange line of wavelength 5890 Å is observed in the emission spectrum of sodium. What is the energy of one photon of this orange light? You do it! 71 Atomic Spectra and the Bohr Atom Example 4-7: An orange line of wavelength 5890 Å is observed in the emission spectrum of sodium. What is the energy of one photon of this orange light? 1 10 -10 m 5.890 10 7 m 5890 Å Å E h 6.626 10 34 hc J s 3.00 10 m/s 7 5.890 10 m 19 3.375 10 J 8 72 Atomic Spectra and the Bohr Atom • The Rydberg equation is an empirical equation that relates the wavelengths of the lines in the hydrogen spectrum. 1 1 R 2 2 n1 n 2 R is the Rydberg constant 1 R 1.097 107 m -1 n1 n 2 n’ s refer to the numbers of the energy levels in the emission spectrum of hydrogen 73 Atomic Spectra and the Bohr Atom Example 4-8: What is the wavelength of light emitted when the hydrogen atom’s energy changes from n = 4 to n = 2? 74 Atomic Spectra and the Bohr Atom Example 4-8. What is n 2 4 and n1 2 the wavelength of light 1 1 1 R 2 2 emitted when the n1 n 2 hydrogen atom’s energy 1 1 7 -1 1 1.097 10 m 2 2 changes from n = 4 to n 2 4 1 1 = 2? 7 -1 1 1.097 10 m 4 16 1 1 1 1.097 107 m -1 0.250 0.0625 1.097 107 m -1 0.1875 2.057 106 m -1 75 Atomic Spectra and the Bohr Atom Notice that the wavelength calculated from the Rydberg equation matches the wavelength of the green colored line in the H spectrum. 76 Atomic Spectra and the Bohr Atom • In 1913 Neils Bohr incorporated Planck’s quantum theory into the hydrogen spectrum explanation. • Here are the postulates of Bohr’s theory. 1. Atom has a number of definite and discrete energy levels (orbits) in which an electron may exist without emitting or absorbing electromagnetic radiation. As the orbital radius increases so does the energy 1<2<3<4<5...... 77 Atomic Spectra and the Bohr Atom 2. An electron may move from one discrete energy level (orbit) to another, but, in so doing, monochromatic radiation is emitted or absorbed in accordance with the following equation. E 2 - E1 E h hc E 2 E1 Energy is absorbed when electrons jump to higher orbits. n = 2 to n = 4 for example Energy is emitted when electrons fall to lower orbits. n = 4 to n = 1 for example 78 Atomic Spectra and the Bohr Atom 3. An electron moves in a circular orbit about the nucleus and it motion is governed by the ordinary laws of mechanics and electrostatics, with the restriction that the angular momentum of the electron is quantized (can only have certain discrete values). angular momentum = mvr = nh/2π h = Planck’s constant n = 1,2,3,4,...(energy levels) v = velocity of electron m = mass of electron r = radius of orbit 79 Atomic Spectra and the Bohr Atom • Light of a characteristic wavelength (and frequency) is emitted when electrons move from higher E (orbit, n = 4) to lower E (orbit, n = 1). – This is the origin of emission spectra. • Light of a characteristic wavelength (and frequency) is absorbed when electrons jump from lower E (orbit, n = 2) to higher E (orbit, n= 4) – This is the origin of absorption spectra. 80 Atomic Spectra and the Bohr Atom 81 Atomic Spectra and the Bohr Atom • Bohr’s theory correctly explains the H emission spectrum. • The theory fails for all other elements because it is not an adequate theory. 82 The Wave Nature of the Electron In 1925 Louis de Broglie published his Ph.D. dissertation. – A crucial element of his dissertation is that electrons have wave-like properties. – The electron wavelengths are described by the de Broglie relationship. h mv h Planck’ s constant m mass of particle v velocity of particle 83 The Wave Nature of the Electron 84 The Wave Nature of the Electron • De Broglie’s assertion was verified by Davisson & Germer within two years. • Consequently, we now know that electrons (in fact - all particles) have both a particle and a wave like character. – This wave-particle duality is a fundamental property of submicroscopic particles. 85 The Wave Nature of the Electron Example 4-9: Determine the wavelength, in m, of an electron, with mass 9.11 x 10-31 kg, having a velocity of 5.65 x 107 m/s. – Remember Planck’s constant is 6.626 x 10-34 Js which is also equal to 6.626 x 10-34 kg m2/s. 86 The Wave Nature of the Electron Example 4-9. Determine the wavelength, in m, of an electron, with mass 9.11 x 10-31 kg, having a velocity of 5.65 x 107 m/s. – Remember Planck’s constant is 6.626 x 10-34 Js which is also equal to 6.626 x 10-34 kg m2/s. h mv 6.626 10 34 kg m 2 / s 2 s -31 7 9.1110 kg 5.65 10 m/s 1.29 10 11 m 87 The Wave Nature of the Electron Example 4-10: Determine the wavelength, in m, of a 0.22 caliber bullet, with mass 3.89 x 10-3 kg, having a velocity of 395 m/s, ~ 1300 ft/s. You do it! 88 The Wave Nature of the Electron Example 4-10. Determine the wavelength, in m, of a 0.22 caliber bullet, with mass 3.89 x 10-3 kg, having a velocity of 395 m/s, ~ 1300 ft/s. You do it! h mv 34 2 2 6.626 10 kg m / s s -3 3.89 10 kg 395 m/s 4.3110 34 m 89 The Quantum Mechanical Picture of the Atom • Werner Heisenberg in 1927 developed the concept of the Uncertainty Principle. • It is impossible to determine simultaneously both the position and momentum of an electron (or any other small particle). – Detecting an electron requires the use of electromagnetic radiation which displaces the electron! • Electron microscopes use this phenomenon 90 The Quantum Mechanical Picture of the Atom • Consequently, we must speak of the electrons’ position about the atom in terms of probability functions. • These probability functions are represented as orbitals in quantum mechanics. 91 The Quantum Mechanical Picture of the Atom Basic Postulates of Quantum Theory 1. Atoms and molecules can exist only in certain energy states. In each energy state, the atom or molecule has a definite energy. When an atom or molecule changes its energy state, it must emit or absorb just enough energy to bring it to the new energy state (the quantum condition). 92 The Quantum Mechanical Picture of the Atom 2. Atoms or molecules emit or absorb radiation (light) as they change their energies. The frequency of the light emitted or absorbed is related to the energy change by a simple equation. E h hc 93 The Quantum Mechanical Picture of the Atom 3. The allowed energy states of atoms and molecules can be described by sets of numbers called quantum numbers. • • Quantum numbers are the solutions of the Schrodinger, Heisenberg & Dirac equations. Four quantum numbers are necessary to describe energy states of electrons in atoms. .. Schr o dinger equation b2 2 2 2 2 2 2 2 V E 8 m x y z 94 Quantum Numbers • The principal quantum number has the symbol n. n = 1, 2, 3, 4, ...... “shells” n = K, L, M, N, ...... The electron’s energy depends principally on n . 95 Quantum Numbers • The angular momentum quantum number has the symbol . = 0, 1, 2, 3, 4, 5, .......(n-1) = s, p, d, f, g, h, .......(n-1) • tells us the shape of the orbitals. • These orbitals are the volume around the atom that the electrons occupy 90-95% of the time. This is one of the places where Heisenberg’s Uncertainty principle comes into play. 96 Quantum Numbers • The symbol for the magnetic quantum number is m. m = - , (- + 1), (- +2), .....0, ......., ( -2), ( -1), • If = 0 (or an s orbital), then m = 0. – Notice that there is only 1 value of m. This implies that there is one s orbital per n value. n 1 • If = 1 (or a p orbital), then m = -1,0,+1. – There are 3 values of m. Thus there are three p orbitals per n value. n 2 97 Quantum Numbers • If ℓ = 2 (or a d orbital), then mℓ = -2,-1,0,+1,+2. – There are 5 values of m. Thus there are five d orbitals per n value. n 3 • If ℓ = 3 (or an f orbital), then mℓ = -3,-2,-1,0,+1,+2, +3. – There are 7 values of mℓ. Thus there are seven f orbitals per n value, n ≥ 4 • Theoretically, this series continues on to g,h,i, etc. orbitals. – Atoms that have been discovered or made up to this point in time only have electrons in s, p, d, or f orbitals in their ground state configurations. 98 Quantum Numbers • The last quantum number is the spin quantum number, ms. • The spin quantum number only has two possible values. ms = +1/2 or -1/2 • This quantum number tells us the spin and orientation of the magnetic field of the electrons. • Wolfgang Pauli in 1925 discovered the Exclusion Principle. – No two electrons in an atom can have the same set of 4 quantum numbers. 99 Atomic Orbitals • Atomic orbitals are regions of space where the probability of finding an electron about an atom is highest. • s orbital properties: – There is one s orbital per n level. –=0 1 value of m 100 Atomic Orbitals • s orbitals are spherically symmetric. 101 Atomic Orbitals • p orbital properties: – The first p orbitals appear in the n = 2 shell. • p orbitals are peanut or dumbbell shaped volumes. – They are directed along the axes of a Cartesian coordinate system. • There are 3 p orbitals per n level. – The three orbitals are named px, py, pz. – They have an = 1. – m = -1,0,+1 3 values of m 102 Atomic Orbitals • p orbitals are peanut or dumbbell shaped. 103 Atomic Orbitals • d orbital properties: – The first d orbitals appear in the n = 3 shell. • The five d orbitals have two different shapes: – 4 are clover leaf shaped. – 1 is peanut shaped with a doughnut around it. – The orbitals lie directly on the Cartesian axes or are rotated 45o from the axes. • There are 5 d orbitals per n level. – The five orbitals are named – d xy , d yz , d xz , d x 2 - y 2 , d z 2 – They have an = 2. – m = -2,-1,0,+1,+2 5 values of m 104 Atomic Orbitals • d orbital shapes 105 Atomic Orbitals • f orbital properties: – The first f orbitals appear in the n = 4 shell. • The f orbitals have the most complex shapes. • There are seven f orbitals per n level. – The f orbitals have complicated names. – They have an ℓ = 3 – mℓ = -3,-2,-1,0,+1,+2, +3 7 values of mℓ – The f orbitals have important effects in the lanthanide and actinide elements. 106 Atomic Orbitals • f orbital shapes 107 Atomic Orbitals • Spin quantum number effects: – Every orbital can hold up to two electrons. • Consequence of the Pauli Exclusion Principle. – The two electrons are designated as having – one spin up and one spin down • Spin describes the direction of the electron’s magnetic fields. 108 Atomic Orbitals 109 Paramagnetism and Diamagnetism • Unpaired electrons have their spins aligned or • This increases the magnetic field of the atom. • Atoms with unpaired electrons are called paramagnetic . – Paramagnetic atoms are attracted to a magnet. 110 Paramagnetism and Diamagnetism • Paired electrons have their spins unaligned . – Paired electrons have no net magnetic field. • Atoms with paired electrons are called diamagnetic. – Diamagnetic atoms are repelled by a magnet. 111 Paramagnetism and Diamagnetism 112 Paramagnetism and Diamagnetism • Because two electrons in the same orbital must be paired, it is possible to calculate the number of orbitals and the number of electrons in each n shell. • The number of orbitals per n level is given by n2. • The maximum number of electrons per n level is 2n2. – The value is 2n2 because of the two paired electrons. 113 Paramagnetism and Diamagnetism Energy Level n # of Orbitals n2 Max # of e2n2 1 1 2 2 4 8 3 9 18 4 16 32 114 The Periodic Table and Electron Configurations • The principle that describes how the periodic chart is a function of electronic configurations is the Aufbau Principle. • The electron that distinguishes an element from the previous element enters the lowest energy atomic orbital available. 115 The Periodic Table and Electron Configurations • The Aufbau Principle describes the electron filling order in atoms. 116 The Periodic Table and Electron Configurations 117 The Periodic Table and Electron Configurations • There are two ways to remember the correct filling order for electrons in atoms. 1. You can use this mnemonic. 118 The Periodic Table and Electron Configurations 2. Or you can use the periodic chart . 119 The Periodic Table and Electron Configurations • Now we will use the Aufbau Principle to determine the electronic configurations of the elements on the periodic chart. • 1st row elements. 1s 1 H 2 He Configurat ion 1 1s 1s 2 120 The Periodic Table and Electron Configurations • 2nd row elements. •Hund’s rule tells us that the electrons will fill the p orbitals by placing electrons in each orbital singly and with same spin until half-filled. Then the electrons will pair to finish the p orbitals. 121 The Periodic Table and Electron Configurations • 3rd row elements 3s 3p Configurat ion Na Ne 11 12 Mg Ne 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar Ne Ne Ne Ne Ne Ne Ne 3s1 Ne 3s22 Ne 3s 2 3p1 Ne 3s2 3p22 Ne 3s 2 3p3 Ne 3s 2 3p4 Ne 3s2 3p5 Ne 3s2 3p6 122 The Periodic Table and Electron Configurations • 4th row elements 3d 19 K Ar 4s 4p Configurat ion Ar 4s1 123 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar 4p Configurat ion Ar 4s1 Ar 4s2 124 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar 21 Sc You do it! 4p Configurat ion Ar 4s1 Ar 4s2 125 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar Sc Ar 21 4p Configurat ion Ar 4s1 Ar 4s2 Ar 4s2 3d1 126 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar Sc Ar 21 4p Configurat ion Ar 4s1 Ar 4s2 Ar 4s2 3d1 Ti You do it! 22 127 The Periodic Table and Electron Configurations 3d 19 K Ar 20 Ca Ar 21Sc Ar 22Ti Ar 4s 4p Configurat ion Ar 4s1 Ar 4s2 Ar 4s2 3d1 Ar 4s2 3d 2 128 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar Sc Ar 22 Ti Ar 23 V Ar 21 4p Configurat ion Ar 4s1 Ar 4s2 Ar 4s2 3d1 Ar 4s2 3d 2 Ar 4s2 3d 3 129 The Periodic Table and Electron Configurations 3d 4s 19 K Ar 20 Ca Ar Sc Ar 22 Ti Ar 23 V Ar Cr Ar 21 24 4p Configurat ion Ar 4s1 Ar 4s2 Ar 4s2 3d1 Ar 4s2 3d 2 Ar 4s2 3d3 Ar 4s1 3d5 There is an extra measure of stability associated with half - filled and completely filled orbitals. 130 The Periodic Table and Electron Configurations 3d 25 Mn Ar 4s 4p Configurat ion Ar 4s2 3d5 131 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 4s 4p Configurat ion Ar 4s2 3d5 Fe You do it! 132 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 Fe Ar 4s 4p Configurat ion Ar 4s2 3d5 Ar 4s2 3d 6 133 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 27 4s Fe Ar Co Ar 4p Configurat ion Ar 4s2 3d5 Ar 4s2 3d6 Ar 4s2 3d7 134 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 27 28 4s Fe Ar Co Ar Ni Ar 4p Configurat ion Ar 4s2 3d5 Ar 4s2 3d6 Ar 4s2 3d7 Ar 4s2 3d8 135 The Periodic Table and Electron Configurations 3d 25 Mn Ar 4s Fe Ar Co Ar 28 Ni Ar 29 Cu You do it! 26 27 4p Configurat ion Ar 4s2 3d5 Ar 4s2 3d6 Ar 4s2 3d7 Ar 4s2 3d8 136 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 27 28 29 4s 4p Fe Ar Co Ar Ni Ar Cu Ar Configurat ion Ar 4s2 3d5 Ar 4s2 3d 6 Ar 4s2 3d 7 Ar 4s2 3d8 Ar 4s1 3d10 Another exception like Cr and for essentiall y the same reason. 137 The Periodic Table and Electron Configurations 3d 25 Mn Ar 26 27 28 29 30 4s Fe Ar Co Ar Ni Ar Cu Ar Zn Ar 4p Configurat ion Ar 4s2 3d5 Ar 4s2 3d 6 Ar 4s2 3d 7 Ar 4s2 3d8 Ar 4s1 3d10 Ar 4s2 3d10 138 The Periodic Table and Electron Configurations 3d 4s 31 Ga Ar 4p Configurat ion Ar 4s2 3d10 4p1 139 The Periodic Table and Electron Configurations 3d 4s 31 Ga Ar 32 4p Configurat ion Ar 4s2 3d10 4p1 Ge You do it! 140 The Periodic Table and Electron Configurations 3d 4s 31 Ga Ar 32 4p Ge Ar Configurat ion Ar 4s2 3d10 4p1 Ar 4s2 3d10 4p2 141 The Periodic Table and Electron Configurations 3d 4s 4p 31 Ga Ar 32 33 Ge Ar As Ar Configurat ion Ar 4s2 3d10 4p1 Ar 4s2 3d10 4p2 Ar 4s2 3d10 4p3 142 The Periodic Table and Electron Configurations 3d 4s 4p 31 Ga Ar 32 Ge Ar 33 As Ar 34 Se You do it! Configurat ion Ar 4s2 3d10 4p1 Ar 4s2 3d10 4p2 Ar 4s2 3d10 4p3 143 The Periodic Table and Electron Configurations 3d 4s 4p 31 Ga Ar 32 33 34 Ge Ar As Ar Se Ar Configurat ion Ar 4s2 3d10 4p1 Ar 4s2 3d10 4p2 Ar 4s2 3d10 4p3 Ar 4s2 3d10 4p4 144 The Periodic Table and Electron Configurations 3d 4s 4p Configurat ion 31 Ga Ar 32 33 34 35 Ge Ar As Ar Se Ar Br Ar Ar 4s2 3d10 4p1 Ar 4s2 3d10 4p2 Ar 4s2 3d10 4p3 Ar 4s2 3d10 4p4 Ar 4s2 3d10 4p5 145 The Periodic Table and Electron Configurations 3d 4s 4p Configurat ion 31 Ga Ar Ar 4s 2 3d10 4p1 2 10 2 Ge Ar Ar 4s 3d 4p 32 2 10 3 As Ar Ar 4s 3d 4p 33 2 10 4 Se Ar Ar 4s 3d 4p 34 2 10 5 Br Ar Ar 4s 3d 4p 35 2 10 6 Kr Ar Ar 4s 3d 4p 36 146 The Periodic Table and Electron Configurations • Now we can write a complete set of quantum numbers for all of the electrons in these three elements as examples. – Na – Ca – Fe • First for 11Na. – When completed there must be one set of 4 quantum numbers for each of the 11 electrons in (remember Ne has 10 electrons) 3s 11 Na Ne 3p Configurat ion Ne 3s 1 147 The Periodic Table and Electron Configurations 1st e- n m 1 0 0 ms 1/2 148 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e- 1 0 0 1st e- ms 1/2 1 s electrons 1/2 149 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e - 1 0 0 3rd e - 0 0 1st e - 2 ms 1/2 1 s electrons 1/2 1/2 150 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e- 1 0 0 3rd e- 2 0 0 4 th e- 2 0 0 1st e- ms 1/2 1 s electrons 1/2 1/2 2 s electrons 1/2 151 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e - 1 0 0 3rd e - 2 0 0 4 th e - 2 0 0 5th e - 2 1 -1 1st e - ms 1/2 1 s electrons 1/2 1/2 2 s electrons 1/2 1/2 152 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e- 1 0 0 3rd e- 2 0 0 4 th e- 2 0 0 5th e- 2 1 -1 1/2 2 s electrons 1/2 1/2 6 th e- 2 1 0 1/2 1st e- ms 1/2 1 s electrons 1/2 153 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e - 1 0 0 3rd e - 2 0 0 4 th e - 2 0 0 5th e - 2 1 -1 1/2 2 s electrons 1/2 1/2 6 th e - 2 1 0 1/2 7 th e - 2 1 1 1/2 1st e - ms 1/2 1 s electrons - 1/2 154 The Periodic Table and Electron Configuration n m 1 0 0 2 nd e - 1 0 0 3rd e - 2 0 0 4 th e - 2 0 0 5th e - 2 1 -1 1/2 2 s electrons 1/2 1/2 6 th e - 2 1 0 1/2 7 th e - 2 1 1 1/2 8th e - 2 1 1 1/2 1st e - ms 1/2 1 s electrons 1/2 155 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e- 1 0 0 3rd e- 2 0 0 4 th e- 2 0 0 5th e- 2 1 -1 1/2 2 s electrons 1/2 1/2 6 th e- 2 1 0 1/2 7 th e- 2 1 1 1/2 8th e- 2 1 1 1/2 1 0 1/2 1st e- 9 th e- 2 ms 1/2 1 s electrons 1/2 156 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e - 1 0 0 3rd e - 2 0 0 4 th e - 2 0 0 5th e - 2 1 6 th e - 2 1 7 th e - 2 1 8th e - 2 1 9 th e - 2 1 10 th e - 2 1 1/2 0 1/2 1 1/2 2 p electrons 1 1/2 0 1/2 1 1/2 1st e - ms 1/2 1 s electrons 1/2 1/2 2 s electrons 1/2 -1 157 The Periodic Table and Electron Configurations n m 1 0 0 2 nd e - 1 0 0 3rd e - 2 0 0 4 th e - 2 0 0 5th e - 2 1 6 th e - 2 1 7 th e - 2 1 8th e - 2 1 9 th e - 2 1 10 th e - 2 1 11th e - 3 0 1st e - ms 1/2 1 s electrons 1/2 1/2 2 s electrons 1/2 1/2 0 1/2 1 1/2 2 p electrons 1 1/2 0 1/2 1 1/2 0 1/23 s electron -1 158 The Periodic Table and Electron Configurations • Next we will do the same exercise for 20Ca. – Again, when finished we must have one set of 4 quantum numbers for each of the 20 electrons in Ca. • We represent the first 18 electrons in Ca with the symbol [Ar]. 3d 20 Ca [Ar] 4s 4p Configurat ion Ar 4s 2 159 The Periodic Table and Electron Configurations n m ms [Ar ] 19 th e- 4 0 0 1/2 160 The Periodic Table and Electron Configurations n m [Ar ]19 th e - 4 0 0 20 th e - 4 0 0 ms 1/2 4 s electrons 1/2 161 The Periodic Table and Electron Configurations • Finally, we do the same exercise for 26Fe. – We should have one set of 4 quantum numbers for each of the 26 electrons in Fe. • To save time and space, we use the symbol [Ar] to represent the first 18 electrons in Fe 3d 26 Fe Ar 4s 4p Configurat ion Ar 4s2 3d6 162 The Periodic Table and Electron Configurations n m ms [Ar ] 19 th e - 4 0 0 1/2 163 The Periodic Table and Electron Configurations n m [Ar ]19 th e - 4 0 0 20 th e - 4 0 0 ms 1/2 4 s electrons 1/2 164 The Periodic Table and Electron Configurations n [Ar ] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 m ms 1/2 4 s electrons 1/2 1/2 165 The Periodic Table and Electron Configurations n m [Ar ] 19 th e - 4 0 0 20 th e - 4 0 21st e - 3 2 22 nd e - You do it! ms 1/2 4 s electrons 0 1/2 - 2 1/2 166 The Periodic Table and Electron Configurations n m ms [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 1/2 4 s electrons 1/2 1/2 22 nd e - 3 2 -1 1/2 167 The Periodic Table and Electron Configurations n m ms [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 1/2 4 s electrons 1/2 1/2 22 nd e - 3 2 -1 1/2 23rd e - 3 2 0 1/2 168 The Periodic Table and Electron Configurations n m [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 1/2 4 s electrons 1/2 1/2 22 nd e - 3 2 -1 1/2 23rd e - 3 2 0 1/2 24 th e - 3 2 1 1/2 ms 169 The Periodic Table and Electron Configurations n m [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 22 nd e - 3 2 -1 23rd e - 3 2 0 24 th e - 3 2 1 25 th e - 3 2 2 ms 1/2 4 s electrons 1/2 1/2 1/2 1/2 half - filled d shell 1/2 1/2 170 The Periodic Table and Electron Configurations n m [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 1/2 4 s electrons 1/2 1/2 22 nd e - 3 2 -1 1/2 23rd e - 3 2 0 1/2 24 th e - 3 2 1 1/2 25 th e - 3 2 2 1/2 26 th e - You do it! ms 171 The Periodic Table and Electron Configurations n m [Ar] 19 th e - 4 0 0 20 th e - 4 0 0 21st e - 3 2 -2 1/2 4 s electrons 1/2 1/2 22 nd e - 3 2 -1 1/2 23rd e - 3 2 0 1/2 24 th e - 3 2 1 1/2 25 th e - 3 2 2 1/2 26 th e - 3 2 -2 1/2 ms 172 Synthesis Question • What is the atomic number of the element that should theoretically be the noble gas below Rn? • The 6 d’s are completed with element 112 and the 7p’s are completed with element 118. Thus the next noble gas (or perhaps it will be a noble liquid) should be element 118. 173 Group Question • In a universe different from ours, the laws of quantum mechanics are the same as ours with one small change. Electrons in this universe have three spin states, -1, 0, and +1, rather than the two, +1/2 and -1/2, that we have. What two elements in this universe would be the first and second noble gases? (Assume that the elements in this different universe have the same 174 symbols as in ours.)