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Unformatted text preview: Chapter 7 Atomic Structure Chapter goals Describe the properties of electromagnetic radiation. Understand the origin of light from excited atoms and its relationship to atomic structure. Describe experimental evidence for waveparticle duality. Describe the basic ideas of quantum mechanics. Define the three quantum numbers (n, l , and ml ) and their relationship to atomic Electromagnetic Radiation Energy can be transferred between atoms & molecules in the form of light dual nature of light: wave and particle transverse wave: perpendicular oscillating electric and magnetic fields
wavelength, amplitude frequency, speed, c Electromagnetic Radiation
Wavelength, : distance traveled by wave in 1 complete oscillation; distance from the top (crest) of one wave to the top of the next wave; or successive troughs/nodes
wavelength Visible light
Amplitude Node wavelength All visible & invisible wavelengths are called electromagnetic radiation. Ultaviolet radiation Electromagnetic Spectrum
-the entire range of electromagnetic radiation- ROYGBIV, 380-750 nm short wavelength high frequency high ENERGY Long wavelength small frequency low ENERGY wavelength, measured in m, cm, nm, (angstrom) 1 = 1 1 0 -10 m = 1 1 0 -8 cm frequency, (nu), measured in s-1 (hertz) (Hz): number of complete oscillations or cycles passing a point per unit time (s) speed of propagation, c distance traveled by ray per unit time in vacuum; all electromagnetic radiation travel at same rate c = 2.998 x 1010 cm/s (speed of light) 2.998 108 m/s whichchas a frequency of 4.80 x 10 s ? eqn: = c 2.998 108 m s-1 = -- = ------------ = ? 4.80 1014 s-1 1 nm 6.25 10-7 m --------- = ? 1 10-9 m -can also calculate frequency: Names to remember
Max Planck: quantized energy E = h ~1900 Albert Einstein: photoelectric effect ~1905 Niels Bohr: 2-D version of atom En=(-RH)(1/n2) Balmer, 1885, then Bohr, 1913 Louis de Broglie: Wavelike properties of matter ~1915 Werner Heisenberg: Uncertainty Principle ~1923 Erwin Schrdinger: Schrdinger Equation ~1926 Planck's equation Planck studied black body radiation, such as that of a heated body, and realized that to explain the energy spectrum he had to assume that: 1. An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA of specific frequency ( ) 2. light has particle character (photons) c Planck's equation is E = h = h E = energy of one photon h = Planck's constant = 6.626 10-34 Compact disk players use lasers that emit red light with a wavelength of 685 nm. What is the energy of one photon of this light? What is the energy of one mole of photons of that red light?
Here's the plan: , nm , m , s-1 E, J/photon E, J/mole 10-9 m c Avogadro's ------- = -- E = h nm number Use the above for the various steps: 10-9 m 685 nm ------- = 6.85 10-7 m 1 nm c 2.998 108 m s-1 = -- = ------------ = 4.38 1014 s-1 6.85 10-7 m E = h = (6.626 10-34 J s/photon) 4.38 1014 s-1 =? = (2.90 10-19 J/photon) 6.022 1023
photons/mol =? The Photoelectric Effect Light can strike the surface of some metals causing electrons to be ejected. It demonstrates the particle nature of light. What are some practical uses of the photoelectric effect? Electronic door openers Light switches for street lights Exposure meters for cameras Albert Einstein explained the effect Explanation involved light having particlelike behavior. The minimum energy needed to eject the e- is The Photoelectric Effect E = h (Planck's equation) Einstein won the 1921 Nobel Prize in Physics for this work. From the value of energy we calculate the frequency ( ) and, with this we calculate lambda ( ). Firstly, we need to calculate the energy in J per atom; it is given in kJ per mol of atoms... kJ 1000 J 1 mol 2.0 102 x x = 3.3 10-19 Joule per atom mol kJ 6.022 1023 atoms E 3.3 10-19 Joule = = = ? s-1 h 6.626 10-34 J s c 2.998 108 m s-1 = = = ? 5.0 1014 s-1 (Visible light) E=h Now, speed of light, c = 1 nm 6.0 10-7 m = ? nm 1 10-9 m The energy of photon is calculated with Planck's Equation c E = h = h If calculated E 6.7 10-17 J, the switch will work. 1 10-9 m 540 nm = ? m nm 2.998 108 m s-1 E = 6.626 10-34 J s = ? J 5.40 10-7 m The switch won't open, because E < 6.7 10-17 J. has to be less than 540 nm. Atomic Line Spectra and the Bohr Atom (Niels Bohr, 1885-1962) 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. Atomic Line Spectra and the Bohr The Rydberg Atom
equation is an empirical equation that relates the wavelengths of the lines in the hydrogen spectrum. Lines are due to transitions n2 upper level n1 lower level 1 1 1 = R 2 - 2 n n2 1 R is the Rydberg constant 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 Example 5-8. What is the wavelength in angstroms of light emitted when the hydrogen atom's energy changes from n = 4 to n = 2? 1 1 = R n = n = 4 to 2 - 2 2 n 1 n2 1 1 1 7 -1 1 = 1.097 10 m 2 - 2 4 2 1 = 1.097 107 m -1 ( 0.250 - 0.0625) 1 =1.097 10 7 m -1 ( 0.1875) 1 = 2.057 10 6 m -1 Therefore, = ? This is the green line in the H spectrum Atomic Line 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. Every element has a unique spectrum. Thus we can use spectra to identify elements. This can be done in the lab, stars, fireworks, etc. Bohr Model of the Atom planetary model considers only the particle nature of the electron p+ & n packed tightly in nucleus electrons traveling in circular paths, orbits, in space surrounding nucleus size, energy, and e capacity of orbits increase as does distance from nucleus (orbital radius) orbits quantized e e e
P+ n e e e e Energy Levels E n=6 n=5 n=4 n=3 n=2 n=1 E n=6 n=5 n=4 n=3 n=2 Exciting the Atom from ground level (n = 1) to upper levels (n > 1) Energy is absorbed n=1 E n=6 n=5 n=4 n=3 n=2 Decay of the Atom from upper levels to lower levels: Energy is emitted Emission of Photons h n=1 E n=6 n=5 n=4 n=3 n=2 Balmer Series, n = 2, for hydrogen. There are other series. h n=1 Calculating E Difference Between two Levels A school teacher was the first to find this! Johann Balmer 1 1 E= |Efinal - Einitial| = RH( - ) nf2 ni2 RH= 2.18 x 10-18 J/atom = 1312 kJ/mol ni and nf = principal quantum numbers of the initial and final states: nf < ni 1,2,3,4.... Problem: Calculate E and of Balmer series of H. n for the violet line =6 n =2 1 1 E= RH( - ) RH= 2.18 x 10-18 J/atom nf2 ni2 1 1 E= 2.18 x 10-18 J( - ) = ? J 22 62 E = h = c/ Then, E = hc/ hc 6.626 10-34 J s 2.998x108 ms-1 = = E 4.84 x 10-19 J = ? nm Bohr Model of the Atom Bohr's theory correctly explains the H emission spectrum and those of hydrogenlike ions (He+, Li2+ ... 1e- species) The theory fails for all other elements because it is not an adequate theory. 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 The Wave-Particle Duality of the Electron 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 (not for macroscopic ones.) The Wave-Particle Duality of the Electron Example: Determine the wavelength, in m and , of an electron, with mass 9.11 x 10-31 kg, having a velocity of 5.65 x 107 m/s. h = 6.626 x 10-34 Js = 6.626 x 10-34 kg m2/s h 6.626 10-34 kg m2s-1 = = mv 9.11 10-31kg 5.65x107 ms-1
=?m =? The Wave-Particle Duality of the Electron Example: 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. h = 6.626 x 10-34 Js = 6.626 x 10-34 kg m2/s h 6.626 10-34 kg m2s-1 = = mv 3.89 10-3kg 395 ms-1 =? Quantum Mechanical Model of the Atom
considers both particle and wave nature of electrons Heisenberg and Born 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). Consequently, we must speak of the electrons' position about the atom in terms of probability functions, i.e., wave equation written for each electron. These probability functions are represented as orbitals in quantum mechanics. They are the wave equations squared and plotted in 3 dimensions. Schrdinger's Model 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). 1. 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 Schrdinger's Model of the Atom
1. 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 Schrdinger, Heisenberg & Dirac equations. Four quantum numbers are necessary to describe energy states of electrons in atoms.
Schrdinger's eqn: b2 2 2 2 - 2 2 + 2 + 2 + V = E 8 m x y z Orbital region of space within which one can expect to find an electron no solid boundaries electron capacity of 2 per orbital space surrounding nucleus divided up into large volumes called shells shells subdivided into smaller volumes called subshells orbitals located in subshells as shells get further from nucleus, energy, size, and electron capacity increase shells, subshells, and orbitals described by quantum numbers Quantum Numbers The principal quantum number has the symbol n n = 1, 2, 3, 4, ... indicates shell K, L, M, N, ... shells as n increases, so does size, energy, and electron capacity The electron's energy depends principally on n . Quantum Numbers The angular momentum (azimuthal) quantum number has the symbol . It indicates subshell. = 0, 1, 2, 3, 4, 5, .......(n-1) = s, p, d, f, g, h, ....... Subshells h 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. 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 Magnetic Quantum Number, m 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 If = 2 (or a d orbital), then m = -2, -1, 0, +1, +2. There are 5 values of m . Thus there are five d orbitals per n value. n 3 If = 3 (or an f orbital), then m = -3, -2, -1, 0, +1, +2, +3. There are 7 values of m . Thus there are seven f orbitals per n value, n 4 Magnetic Quantum Number, m Theoretically, this series continues on to g, h, i, etc. orbitals. Practically speaking 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. Electrons Indicated by Shell and Subshell
Symbolism #electrons nl #
principal number 4s1 3s2 4d12 letter: s, p, d,.. orbital 3p7 4f5 4f14 5p4 there are 4 electrons in the 5p orbitals ml n shell l subshell s 0 1 K 0 1 s 0 2 L 0 1 1,0,1 1 p 3 0 3 M 0 s 1 1,0,1 1 p 3 2 d 5 -2,-1,0,1,2 0 4 N 0 s 1 1,0,1 1 p 3 2 d 5 -2,-1,0,1,2 3 f -3,-2,-1,0,1,2,3 7 #orbitals Max n2 #e 1 2 2 8 4 6 2 6 18 9 10 2 6 16 10 32 14 Maximum two electrons per orbital The Shape of Atomic Orbitals s orbitals are spherically symmetric. A plot of the surface density as a function of the distance from the nucleus for an s orbital of a hydrogen atom It gives the probability of finding the electron at a given distance from the nucleus p 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 p Orbitals
y z x y x z y z px x py pz 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 d orbitals y z x d Orbitals
z y x x y z y z dx2y2 dxy dz2
x y z x dxz dyz f 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 = 3 m = -3, -2, -1,0, +1,+2, +3 7 values of m The f orbitals have important effects in the lanthanide and actinide elements. f orbitals f orbital shapes n = 5, l=2 five possible ml : -2, -1, 0, +1, +2 Then, there are five sets of (n, l , ml ) (5, 2, -2) (5, 2, -1) (5, 2, 0) (5, 2, +1) (5, 2, +2) Prob. 7-34, textbook: Which of the following represent valid sets of quantum numbers? For a set that is invalid, explain briefly why it is not correct.
a) n = 3, l = 3, ml : 0 b) n = 2, l = 1, ml : 0 c) n = 6, l = 5, ml : -1 d) n = 4, l = 3, ml : -4 No: No: Prob. 7-36, textbook: What is the maximum number of orbitals that can be identified by each of the following sets of quantum numbers? When "none" is the correct answer, explain the reason.
Answer a) n = 4, l = 3 b) n = 5, c) n = 2, l = 2 d) n = 3, l = 1, ml : -1 Why? ml : -3, -2, -1, 0, 1, 2, 3 n2 maximum l = n - 1 Prob. 7-38, textbook: State which of the following are incorrect designations for orbitals according to the quantum theory: 3p, 4s, 2f, and 1p
Answer a) 3p b) 4s c) 2f d) 1p maximum l = 1 (l = 3 for f) maximum l = 0 (l = 1 for p) Why? n = 3, l = 1 maximum l = 3-1=2 ...
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This note was uploaded on 08/24/2008 for the course CHM 2045 taught by Professor Geiger during the Summer '08 term at University of Central Florida.
- Summer '08