Unformatted text preview: Contents
81 82 83 84 85 86 87 88 89 89 810 811 812
CHM 25 SP07 Slide 2 of 68 Chapter 8: Electrons in Atoms
CHM 25 Spring 2007 Dr. Berk & Dr. Koel
Slide 1 of 68 Electromagnetic Radiation Atomic Spectra Quantum Theory The Bohr Atom Two Ideas Leading to a New Quantum Mechanics Wave Mechanics Quantum Numbers and Electron Orbitals Quantum Numbers Interpreting and Representing Orbitals of the Hydrogen Atom Electron Spin Multielectron Atoms Electron Configurations Electron Configurations and the Periodic Table
CHM 25 SP07 Electromagnetic Radiation
♦ Electric and magnetic fields propagate as waves through empty space or through a medium. ♦ A wave transmits energy. Electromagnetic Radiation
wavelength Visible light Amplitude wavelength Node Ultaviolet radiation
Slide 3 of 68 CHM 25 SP07 Slide 4 of 68 CHM 25 SP07 Electromagnetic Radiation Frequency, Wavelength and Velocity
♦ Frequency (ν) in Hertz—Hz or s1. ♦ Wavelength ( ) in meters—m.
cm µm nm Å pm (102 m) (106 m) (109 m) (1010 m) (1012 m) Low ν ♦ Velocity (c)—2.997925 × 108 m s1.
High ν c= ν
Slide 5 of 68 CHM 25 SP07 Slide 6 of 68 = c/ ν ν= c/
CHM 25 SP07 CHM 25 SP07 1 Electromagnetic Spectrum Electromagnetic Spectrum
Short wavelength > Short high frequency high high energy increasing increasing frequency frequency
Slide 7 of 68 CHM 25 SP07 Slide 8 of 68 increasing increasing wavelength wavelength Electromagnetic Radiation Constructive and Destructive Interference Red light has λ = 700 nm. Calculate the frequency ( ).
700 nm • 1 x 10 9 m = 7.00 x 10 7 m 1 nm Freq =
Slide 9 of 68 3.00 x 10 8 m/s 7.00 x 10 7 m = 4.29 x 10 14 sec 1
CHM 25 SP07 Slide 10 of 68 CHM 25 SP07 Examples of Interference Refraction of Light Slide 11 of 68 CHM 25 SP07 Slide 12 of 68 CHM 25 SP07 '¢¤#¢¢£¢(¦¢ '¢¤#¢¢¢(¢ '¢¤#¢¢£¢(¦¢ ¢¢¢¢(¢¡¡¡¡¡¡¡¡ '¢¤#¢¢££¢(¦¦¢ ¢¤#¢¢¢(¢ '¢#¢¢£¢(¦¢ ''¢¤#¢¢££¢(¦¦¢ ' ¤¤ # £ ¦ '''&¢£¥¢%¢$¢ !" ¢ '&¢£¥¢%¢$¢ !" ¢ ¢¥¢¢¢######## " ¢¨¨¨¨¨¨¨¨ ¢£¥¢%¢$¢ !" ¢ '''&&¢£¥¢%%¢$¢ !" ¢ '¢£¥¢¢$¢ !" ¢ &&¢£¥¢%¢$¢ !" ¢ &¢¥¢%¢¢ !" ¢ & ££ % $$ ! ¢¤¥£¢ ¢¢¨¨©¦§¤¥££¢¡¢ ¢¤¥£¢ ¢¢©¦§¤¥£¢¢ ¢¤¥£¢ ¢¢¨¨©¦§¤¥£¢¡¡¢ ¤ £ ¦ ¤ £ ¡ ¢¤¥£¢ ¢¢©¦§¤¥¢¡¢ ¢¤¥£¢ ¢¢¨¨©¦§¤¥£¢¡¡¢ ¢¤¥£¢ ¢¢©¦§¤¥£¢¡¢ ¢¥¢ ¢¢¨¨©§¥¢¢ ¢¤¥£¢ ¢¢©¦§¤¥£¢¢
CHM 25 SP07 CHM 25 SP07 2 Atomic Spectra (a) (b) (c) (d) (e) Slide 13 of 68 CHM 25 SP07 Slide 14 of 68 CHM 25 SP07 • Excited atoms emit light of only certain wavelengths • The wavelengths of emitted light depend on the element. Hydrogen
Slide 15 of 68 CHM 25 SP07 Slide 17 of 68 CHM 25 SP07 @ §7 C I 8X S1 0 AH ¥U V5 CT T1B S P PQI ¥H H14F GD E2 C 0 [email protected] ¥ 0 97 85 63 34 0 1 §CB B 8SW13 3 I I A¥VC1SR R Q¥134GECB B A¥@ @ 98642 2 1) ) @ 7 I X W 0HU5 B I 3F D 2 [email protected] 0 7 5 0
Line emission spectra of hydrogen, mercury and neon. Excited gaseous elements produce characteristic spectra that can be used to identify the elements as well as to determine how much of each element is present in a sample.
Slide 16 of 68 CHM 25 SP07 Black Body Radiation Quantum Theory
Blackbody Radiation: Max Planck, 1900 Energy, like matter, is discontinuous.
CHM 25 SP07 Slide 18 of 68 = hν
CHM 25 SP07 3 An object can gain or lose energy by absorbing or emitting radiant energy in QUANTA. Experiment demonstrates the particle nature of light. Experiment Energy of radiation is proportional to frequency E = h•ν E = h•ν h = Planck’s constant = 6.6262 x 1034 J•s Light with large λ (small ) has a small E. Light with a short λ (large ) has a large E.
Slide 19 of 68 CHM 25 SP07 Slide 20 of 68 CHM 25 SP07 The Photoelectric Effect
♦ Heinrich Hertz, 1888 • Light striking the surface of certain metals causes ejection of electrons. Using Planck’s Equation Understand experimental observations if light of consists of particles called discrete energy. PROBLEM: Calculate the energy of 1.00 mol of photons Calculate of red light. of ♦λ = 700. nm ♦ = 4.29 x 1014 sec1 4.29
Slide 22 of 68 CHM 25 SP07 • ν > νo
threshold frequency • #e ∝ I
Slide 21 of 68 Albert Einstein 1905
CHM 25 SP07 EXAMPLE
Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 s1 before it can emit an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to produce this effect? (b) What wavelength radiation will provide a photon of this energy? Equations so far…. Slide 23 of 68 CHM 25 SP07 Slide 24 of 68 CHM 25 SP07 CHM 25 SP07 4 Bohr’s greatest contribution to Bohr greatest science was in building a simple model of the atom. It was based on an understanding of the understanding
Niels Bohr (18851962)
Slide 25 of 68 One view of atomic structure in early 20th One century was that an electron (e) traveled century traveled about the nucleus in an orbit. about + Electron orbit SHARP LINE SHARP EMISSION SPECTRA of EMISSION of
excited atoms. excited
CHM 25 SP07 Slide 26 of 68 1. Any orbit should be possible and so is any energy. 2. But a charged particle moving in an electric field should emit energy. End result should be destruction!
CHM 25 SP07 The Bohr Atom
Bohr said classical view is wrong. Proposed a new theory — now called QUANTUM or WAVE MECHANICS. e can only exist in certain discrete orbits — called stationary states. e is restricted to QUANTIZED energy states.
E= RH n2 RH = 2.179 × 1018 J Energy of state =  RH/n2
Slide 27 of 68 rn = n2a0
a0 = 53 pm (0.53Å)
CHM 25 SP07 Slide 28 of 68 CHM 25 SP07 where n = quantum no. = 1, 2, 3, 4, .... EnergyLevel Diagram Ionization Energy of Hydrogen
E = RH ( 1 1 – ) = hν ni2 nf2 As nf goes to infinity for hydrogen starting in the ground state: E = Ef – Ei = = RH ( RH RH –2 nf2 ni hν = RH ( 1 ) = RH ni2 1 1 – ) = hν = hc/ ni2 nf2 This also works for hydrogenlike species such as He+ and Li2+. hν = Z2 RH
Slide 30 of 68 CHM 25 SP07 Slide 29 of 68 CHM 25 SP07 CHM 25 SP07 5 Atomic Spectra and Bohr Atomic Spectra and Bohr
Calculate E for e “falling” from high energy level (n = 2) to low energy level (n = 1). E = Efinal  Einitial = RH[(1/12)  (1/2)2] = (0.75)RH RH (called the Rydberg constant) = 2.179 x 1018 J so, E of emitted light = (0.75)RH = 1.63 x 1018 J From = E/h you get = 2.46 x 1015 sec1 and from λ = c/ν the λ = 121.6 nm
Slide 31 of 68 CHM 25 SP07 Emission and Absorption Spectroscopy Slide 32 of 68 CHM 25 SP07 Without the use of a calculator, indicate which of the following transitions in the hydrogen atom results in the emission of light of the greatest energy?
∆E ∝ 1 1 −2 n2 nf i Two Ideas Leading to a New Quantum Mechanics
♦ WaveParticle Duality
• Einstein suggested particlelike properties of light could explain the photoelectric effect. • Diffraction patterns suggest photons are wavelike. 1. 2. 3. 4. 5. n=4 to n=3 n=1 to n=2 n=3 to n=2 n=2 to n=1 n=1 to n=3
CHM 25 SP07 ♦ deBroglie, 1924
• Small particles of matter may at times display wavelike properties.
Louis de Broglie Nobel Prize 1918
CHM 25 SP07 Slide 33 of 68 Slide 34 of 68 de Broglie and Matter Waves
E = mc2 hν = mc2 hν/c = mc = p p = h/ = h/p = h/mu
Slide 35 of 68 CHM 25 SP07 XRay Diffraction If the distance between the objects that the waves scatter from is about the same as the wavelength of the radiation, diffraction occurs and an interference pattern is observed.
Slide 36 of 68 CHM 25 SP07 CHM 25 SP07 6 !"
Calculate the wavelength (in nm) associated with an electron of mass (m) = 9.109 x 1028g that travels at 40.0% of the speed of light. Uncertainty Principle
Problem of defining nature of Problem of defining nature of electrons in atoms solved by electrons in atoms solved by W. Heisenberg. W. Heisenberg. Cannot simultaneously define Cannot simultaneously define the position and momentum the position and momentum (= m•v) of an electron. (= m•v) of an electron. We define e energy exactly We define e energy exactly but accept limitation that we but accept limitation that we do not know exact position. do not know exact position.
CHM 25 SP07 Compare this to a Baseball (115 g) at 100 mph = 1.3 x 1025 nm
Slide 37 of 68 CHM 25 SP07 W. Heisenberg 19011976 Slide 38 of 68 The Uncertainty Principle
♦ Werner Heisenberg
xp h 4 !"
Schrodinger applied idea of e behaving behaving as a wave to the problem of electrons in atoms. in He developed the WAVE EQUATION He W AVE Solution gives set of math expressions Solution called W AVE FUNCTIONS, Ψ Each describes an allowed energy state Each of an eof Quantization introduced naturally.
CHM 25 SP07 E. Schrodinger 18871961
Heisenberg and Bohr
Slide 39 of 68 CHM 25 SP07 Slide 40 of 68 # $% & ' (Ψ Wave Functions
The standing waves of a particle in a 'onedimensional' box. • Ψ is a function of distance and two angles. angles. • Each Ψ corresponds to an ORBITAL — Each corresponds ORBITAL the region of space within which an electron the is found. is • Ψ does NOT describe the exact location of does the electron. the • Ψ2 iis proportional to the probability of s finding an e at a given point. finding
Slide 41 of 68 CHM 25 SP07 Slide 42 of 68 ♦ The wave function changes sign at the nodes. ♦ Remember, there is little physical meaning for . It is the 2, which represents the probability of an electron existing in a given position that has importance.
= 2 nπ x sin L L
CHM 25 SP07 CHM 25 SP07 7 Probability of Finding an Electron
The probabilities of a particle in a 'onedimensional' box. •There is no chance of finding the particle at the points where the curve hits zero  these are nodes. •For a given energy level n, there are n1 nodes. •This solution to the Schrödinger equation shows Bohr' quantized s energy levels are the necessary conclusion to how electrons behave around atoms. s orbitals & p orbitals Slide 43 of 68 CHM 25 SP07 Slide 44 of 68 CHM 25 SP07 p orbitals & d orbitals % %" %" ) The shape, size, and energy of each orbital is a shape, of function of 3 quantum numbers: function n (Principle electronic shell) > shell (Angular momentum) > subshell  subshell m (Magnetic) > designates an orbital  designates within a subshell subshell
Slide 45 of 68 CHM 25 SP07 Slide 46 of 68 CHM 25 SP07 *
n=1 n=2 n=3 n=4 +
Symbol n (major) QUANTUM NUMBERS Values 1, 2, 3, .. Description Orbital size Orbital and energy where E = RH(1/n2) Orbital shape or type Orbital (subshell) (angular) (angular) 0, 1, 2, .. n1 0, m (magnetic) (magnetic)  … 1, 0, 1…+ Orbital orientation orientation # of orbitals in subshell = 2 + 1 of orbitals in subshell
CHM 25 SP07 Slide 47 of 68 CHM 25 SP07 Slide 48 of 68 CHM 25 SP07 8 * *
When n = 1, then = 0 and m = 0 and Therefore, in n = 1, there is 1 type of Therefore, subshell subshell and that subshell has a single orbital and subshell (m has a single value > 1 orbital) This subshell iis labeled s This subshell s Each shell has 1 orbital labeled s, and it Each is ) '& in shape. in
Slide 49 of 68 CHM 25 SP07 * Typical p orbital Typical p orbital When n = 2, then = 0 and 1 When n = 2, then = 0 and 1 When and Therefore, in n = 2 shell there Therefore, in n = 2 shell there Therefore, are 2 types of orbitals — 2 are 2 types of orbitals — 2 orbitals planar node planar node subshells subshells subshells When = 1, there is For = 0 m = 0 When For = 0 m = 0 For a tthis is a s subshell his this is a s subshell subshell thru the For = 1 m = 1, 0, +1 For = 1 m = 1, 0, +1 For nucleus. nucleus. tthis is a p subshell his this is a p subshell subshell with 3 orbitals with 3 orbitals with orbitals
Slide 50 of 68 CHM 25 SP07 d Orbitals d Orbitals Orbitals
When n = 3, what are the values of ? When * *
s orbitals have no planar orbitals have node ( = 0) and so are node spherical. spherical. p orbitals have = 1, and orbitals have have 1 planar node, have and so are “dumbbell” and shaped. * ,. /0 ! /
CHM 25 SP07 Slide 52 of 68 typical d orbital planar node = 0, 1, 2
and so there are 3 subshells in the shell. subshells For = 0, m = 0 For 0, > s subshell with single orbital subshell For = 1, m = 1, 0, +1 For 1, 1, > p subshell with 3 orbitals subshell with orbitals For = 2, m = 2, 1, 0, +1, +2 For 2, 2, >
Slide 51 of 68 planar node d subshell with 5 orbitals subshell with orbitals CHM 25 SP07 Orbital Energies Which of the following sets of quantum numbers is not matched with the orbital designation. 1. 2. 3. 4. 5. n=1, n=2, n=3, n=3, n=3, =0: a 1s orbital =0: a 2p orbital =0: a 3s orbital =1: a 3p orbital =2: a 3d orbital Slide 53 of 68 CHM 25 SP07 Slide 54 of 68 CHM 25 SP07 CHM 25 SP07 9 Electron Spin: A Fourth Quantum Number
Electrons in atoms are arranged as Electrons SHELLS (n) •Each orbital can be assigned no more than 2 electrons! existence of a 4th quantum number, the electron spin quantum number, ms.
CHM 25 SP07 SUBSHELLS ( ) •This is tied to the ORBITALS (m )
Slide 55 of 68 Can be proved experimentally that electron has a spin. Two Can spin directions are given by ms where ms = +1/2 and 1/2. spin +1/2
Slide 56 of 68 CHM 25 SP07 Electron Configurations
♦ Aufbau process.
• Build up and minimize energy. Orbital Energies ♦ Pauli exclusion principle.
• No two electrons can have all four quantum numbers alike. ♦ Hund’s rule.
• Degenerate orbitals are occupied singly first.
Slide 57 of 68 CHM 25 SP07 Slide 58 of 68 CHM 25 SP07 ! ! & & (1 (1 Orbital Filling ♦ Zeff iis the nuclear charge experienced by s the outermost electrons. the
♦ Explains why E(2s) < E(2p) ♦ Zeff iincreases across a period owing to ncreases incomplete shielding by inner electrons. incomplete ♦ Estimate Zeff by > [ Z  (no. inner electrons) ] Estimate by (no. ♦ Charge felt by 2s e iin Li n Z* = 3  2 = 1 ♦ Be Zeff = 4  2 = 2 Be ♦B Zeff = 5  2 = 3 and so on!
Slide 59 of 68 CHM 25 SP07 Slide 60 of 68 CHM 25 SP07 CHM 25 SP07 10 & &
Two ways of Two ways of writing configs. writing configs. One is called One is called the spdf the spdf & &
Two ways of Two ways of writing writing configs. Other configs. Other is called the is called the orbital box orbital box notation. notation.
ORBITAL BOX NOTATION for He, atomic number = 2 spdf notation for H, atomic number = 1 notation. notation. 1s 1 no. of electrons value of l 1s 2
1s Arrows depict electron spin value of n
Slide 61 of 68 One electron has n = 1, l = 0, ml = 0, ms = + 1/2 Other electron has n = 1, l = 0, ml = 0, ms =  1/2
Slide 62 of 68 CHM 25 SP07 CHM 25 SP07 Aufbau Process and Hunds Rule Filling p Orbitals Slide 63 of 68 CHM 25 SP07 Slide 64 of 68 CHM 25 SP07 Filling the d Orbitals Slide 65 of 68 CHM 25 SP07 Slide 66 of 68 CHM 25 SP07 CHM 25 SP07 11 Electron Configurations and the Periodic Table Which of the following is the condensed electron configuration for copper? [Xe] accounts for 54 electrons. 1. 2. 3. 4. 5. [Ar]3s23d9 [Ar]4s22d9 [Ar]3s14d10 [Ar]4s14d10 [Ar]4s13d10 Slide 67 of 68 CHM 25 SP07 Slide 68 of 68 CHM 25 SP07 CHM 25 SP07 12 ...
View
Full Document
 Spring '06
 x
 Atom, Electron, Atomic orbital

Click to edit the document details