1. Chapter 8: The Quantum Mechanical Atom Chemistry: The Molecular Nature of Matter, 6E Jespersen/Brady/Hyslop

2. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electromagnetic Energy Electromagnetic Radiation  Light energy or wave  Travels through space at speed of light in vacuum  c = speed of light = 2.9979 × 10 8 m/s  Successive series of these waves or oscillations Waves or Oscillations  Systematic fluctuations in intensities of electrical and magnetic forces  Varies regularly with time  Exhibit wide range of energy 2

3. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Properties of Waves Wavelength ( )  Distance between two successive peaks or troughs  Units are in meters, centimeters, nanometers Frequency ( )  Number of waves per second that pass a given point in space  Units are in Hertz (Hz = cycles/sec = 1/sec = s –1 ) Related by  = c 3

4. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Properties of Waves Amplitude  Maximum and minimum height  Intensity of wave, or brightness  Varies with time as travels through space Nodes  Points of zero amplitude  Place where wave goes though axis  Distance between nodes is constant 4 nodes

5. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check: Converting from Wavelength to Frequency The bright red color in fireworks is due to emission of light when Sr(NO 3 ) 2 is heated. If the wavelength is ~650 nm, what is the frequency of this light? 5 = 4.61 × 10 14 s –1 = 4.6 × 10 14 Hz

6. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! WCBS broadcasts at a frequency of 880 kHz. What is the wavelength of their signal? A. 341 m B. 293 m C. 293 mm D. 341 km E. 293 mm 6

7. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electromagnetic Spectrum 7 high energy, short waveslow energy, long waves  Comprised of all frequencies of light  Divided into regions according to wavelengths of radiation

8. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electromagnetic Spectrum Visible light  Band of wavelengths that human eyes can see  400 to 700 nm  Make up spectrum of colors 8 White light  Combination of all these colors  Can separate white light into the colors with a prism

9. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Important Experiments in Atomic Theory Late 1800’s:  Matter and energy believed to be distinct  Matter: made up of particles  Energy: light waves Beginning of 1900’s:  Several experiments proved this idea incorrect  Experiments showed that electrons acted like:  Tiny charged particles in some experiments  Waves in other experiments 9

10. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Particle Theory of Light  Max Planck and Albert Einstein (1905)  Electromagnetic radiation is stream of small packets of energy  Quanta of energy or photons  Each photon travels with velocity = c  Waves with frequency =  Energy of photon of electromagnetic radiation is proportional to its frequency  Energy of photon E = h  h = Planck’s constant = 6.626 × 10 –34 J s 10

11. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check What is the frequency, in sec –1, of radiation which has an energy of 3.371 × 10 –19 joules per photon? 11 = 5.087 × 10 14 s –1

12. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! A microwave oven uses radiation with a frequency of 2450 MHz (megahertz, 10 6 s –1 ) to warm up food. What is the energy of such photons in joules? A. 1.62 × 10 –30 J B. 3.70 × 10 42 J C. 3.70 × 10 36 J D. 1.62 × 10 44 J E. 1.62 × 10 –24 J 12

13. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Photoelectric Effect  Shine light on metal surface  Below certain frequency ( )  Nothing happens  Even with very intense light (high amplitude)  Above certain frequency ( )  Number of electrons ejected increases as intensity increases  Kinetic energy (KE) of ejected electrons increases as frequency increases  KE = h – BE  h = energy of light shining on surface  BE = binding energy of electron 13

14. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Means that Energy is Quantized  Can occur only in discrete units of size h  1 photon = 1 quantum of energy  Energy gained or lost in whole number multiples of h E = nh  If n = N A, then one mole of photons gained or lost E = 6.02 × 10 23 h If light is required to start reaction  Must have light above certain frequency to start reaction  Below minimum threshold energy, intensity is NOT important 14

15. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check How much energy is contained in one mole of photons, each with frequency 2.00 × 10 13 ? E = 6.02 × 10 23 h 15 E = (6.02×10 23 mol –1 )(6.626×10 –34 J∙s)(2.00×10 13 s –1 ) E = 7.98 × 10 3 J/mol

16. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! If a mole of photons has an energy of 1.60 × 10 –3 J/mol, what is the frequency of each photon? Assume all photons have the same frequency. A. 8.03 × 10 28 Hz B. 2.12 × 10 –14 Hz C. 3.20 × 10 19 Hz D. 5.85 × 10 –62 Hz E. 4.01 × 10 6 Hz 16

17. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E For Example: Photosynthesis  If you irradiate plants with infrared and microwave radiation  No photosynthesis  Regardless of light intensity  If you irradiate plants with visible light  Photosynthesis occurs  More intense light now means more photosynthesis 17

18. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electronic Structure of Atom Clues come from: 1. Study of light absorption  Electron absorbs energy  Moves to higher energy “excited state” 2. Study of light emission  Electron loses photon of light  Drops back down to lower energy “ground state” 18

19. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Continuous Spectrum  Continuous unbroken spectrum of all colors  i.e., visible light through a prism  Sunlight  Incandescent light bulb  Very hot metal rod 19

20. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Discontinuous or Line Spectrum  Consider light given off when spark passes through gas under vacuum  Spark (electrical discharge) excites gas molecules or atoms  Spectrum that has only a few discrete lines  Also called atomic spectrum or emission spectrum  Each element has unique emission spectrum 20

21. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Line Spectrum 21

22. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Atomic Spectra  Atomic line spectra are rather complicated  Line spectrum of hydrogen is simplest  Single electron  First success in explaining quantized line spectra  First studied extensively  J.J. Balmer  Found empirical equation to fit lines in visible region of spectrum  J. Rydberg  More general equation explains all emission lines in H atom spectrum (infrared, visible, and UV) 22

23. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Rydberg Equation  Can be used to calculate all spectral lines of hydrogen  The values for n correspond to allowed energy levels for atom R H = 109,678 cm –1 = Rydberg constant = wavelength of light emitted n 1 and n 2 = whole numbers (integers) from 1 to  where n 2 > n 1 If n 1 = 1, then n 2 = 2, 3, 4, … 23

24. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check: Using Rydberg Equation Consider the Balmer series where n 1 = 2 Calculate (in nm) for the transition from n 2 = 6 down to n 1 = 2. 24 = 410.3 nmViolet line in spectrum = 24,373 cm –1

25. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check A photon undergoes a transition from n higher down to n = 2 and the emitted light has a wavelength of 650.5 nm? 25 n 2 = 3

26. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is the wavelength of light (in nm) that is emitted when an excited electron in the hydrogen atom falls from n = 5 to n = 3? A. 1.28 × 10 3 nm B. 1.462 × 10 4 nm C. 7.80 × 10 2 nm D. 7.80 × 10 –4 nm E. 3.65 × 10 –7 nm 26

27. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Significance of Atomic Spectra  Atomic line spectra tells us  When excited atom loses energy  Only fixed amounts of energy can be lost  Only certain energy photons are emitted  Electron restricted to certain fixed energy levels in atoms  Energy of electron is quantized  Simple extension of Planck's Theory  Any theory of atomic structure must account for  Atomic spectra  Quantization of energy levels in atom 27

28. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E What Does Quantized Mean?  Energy is quantized if only certain discrete values are allowed  Presence of discontinuities makes atomic emission quantized 28 Potential Energy of Rabbit

29. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Bohr Model of Atom  First theoretical model of atom to successfully account for Rydberg equation  Quantization of energy in hydrogen atom  Correctly explained atomic line spectra  Proposed that electrons moved around nucleus like planets move around sun  Move in fixed paths or orbits  Each orbit has fixed energy 29

30. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Energy for Bohr Model of H  Equation for energy of electron in H atom  Ultimately b relates to R H by b = R H hc  OR  Where b = R H hc = 2.1788 × 10 –18 J/atom  Allowed values of n = 1, 2, 3, 4, …   n = quantum number  Used to identify orbit 30

31. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Energy Level Diagram for H Atom  Absorption of photon  Electron raised to higher energy level  Emission of photon  Electron falls to lower energy level 31  Energy levels are quantized  Every time an electron drops from one energy level to a lower energy level  Same frequency photon is emitted  Yields line spectra

32. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Bohr Model of Hydrogen Atom  n = 1First Bohr orbit  Most stable energy state equals the ground state which is the lowest energy state  Electron remains in lowest energy state unless disturbed How to change the energy of the atom?  Add energy in the form of light:E = h  Electron raised to higher n orbit n = 2, 3, 4, …   Higher n orbits = excited states = less stable  So electron quickly drops to lower energy orbit and emits photon of energy equal to  E between levels  E = E h – E l h = higher l = lower 32

33. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Bohr’s Model Fails  Theory could not explain spectra of multi-electron atoms  Theory doesn’t explain collapsing atom paradox  If electron doesn’t move, atom collapses  Positive nucleus should easily capture electron  Vibrating charge should radiate and lose energy 33

34. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! In Bohr's atomic theory, when an electron moves from one energy level to another energy level more distant from the nucleus, A.energy is emitted B.energy is absorbed C.no change in energy occurs D.light is emitted E.none of these 34

35. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Light Exhibits Interference Constructive interference  Waves “ in-phase ” lead to greater amplitude  They add together Destructive interference  Waves “ out-of-phase ” lead to lower amplitude  They cancel out 35

36. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Diffraction and Electrons  Light  Exhibits interference  Has particle-like nature  Electrons  Known to be particles  Also demonstrate interference 36

37. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Standing vs. Traveling Waves Traveling wave  Produced by wind on surfaces of lakes and oceans Standing wave  Produced when guitar string is plucked  Center of string vibrates  Ends remain fixed 37

38. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Standing Wave on a Wire  Integer number (n) of peaks and troughs is required  Wavelength is quantized:  L is the length of the string 38

39. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E How Do We Describe an Electron?  Has both wave-like and particle-like properties  Energy of moving electron on a wire is E =½ mv 2  Wavelength is related to the quantum number, n, and the wire length: 39

40. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron on Wire—Theories Standing wave  Half-wavelength must occur integer number of times along wire’s length de Broglie’s equation relates the mass and speed of the particle to its wavelength  m = mass of particle  v = velocity of particle 40

41. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron on Wire—Theories Starting with the equation of the standing wave and the de Broglie equation Combining with E = ½mv 2, substituting for v and then λ, we get Combining gives: 41

42. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E de Broglie Explains Quantized Energy  Electron energy quantized  Depends on integer n  Energy level spacing changes when positive charge in nucleus changes  Line spectra different for each element  Lowest energy allowed is for n =1  Energy cannot be zero, hence atom cannot collapse 42

43. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check: Calculate Wavelength for an Electron What is the de Broglie wavelength associated with an electron of mass 9.11 × 10 –31 kg traveling at a velocity of 1.0 × 10 7 m/s? 43 = 7.27 × 10 –11 m

44. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Calculate the de Broglie wavelength of a baseball with a mass of 0.10 kg and traveling at a velocity of 35 m/s. A. 1.9 × 10 –35 m B. 6.6 × 10 –33 m C. 1.9 × 10 –34 m D. 2.3 × 10 –33 m E. 2.3 × 10 –31 m 44

45. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Wave Functions Schrödinger’s equation  Solutions give wave functions and energy levels of electrons Wave function  Wave that corresponds to electron  Called orbitals for electrons in atoms Amplitude of wave function squared  Can be related to probability of finding electron at that given point Nodes  Regions where electrons will not be found 45

46. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Orbitals Characterized by Three Quantum Numbers: Quantum Numbers:  Shorthand  Describes characteristics of electron’s position  Predicts its behavior n = principal quantum number  All orbitals with same n are in same shell ℓ = secondary quantum number  Divides shells into smaller groups called subshells m ℓ = magnetic quantum number  Divides subshells into individual orbitals 46

47. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E n = Principal Quantum Number  Allowed values: positive integers from 1 to   n = 1, 2, 3, 4, 5, …   Determines:  Size of orbital  Total energy of orbital  R H hc = 2.18 × 10 –18 J/atom  For given atom,  Lower n = Lower (more negative) E = More stable 47

48. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E ℓ = Orbital Angular Momentum Quantum Number  Allowed values: 0, 1, 2, 3, 4, 5…(n – 1)  Letters: s, p, d, f, g, h Orbital designation number n ℓ letter  Possible values of ℓ depend on n  n different values of ℓ for given n  Determines  Shape of orbital 48

49. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E m ℓ = Magnetic Quantum Number  Allowed values: from – ℓ to 0 to + ℓ  Ex. when ℓ =2 then m ℓ can be  –2, –1, 0, +1, +2  Possible values of m ℓ depend on ℓ  There are 2 ℓ +1 different values of m ℓ for given ℓ  Determines orientation of orbital in space  To designate specific orbital, you need three quantum numbers  n, ℓ, m ℓ 49

50. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Table 8.1 Summary of Relationships Among the Quantum Numbers n, ℓ, and m ℓ 50

51. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Orbitals of Many Electrons 51 Orbital Designation  Based on first two quantum numbers  Number for n and letter for ℓ  How many electrons can go in each orbital?  Two electrons  Need another quantum number

52. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Spin Quantum Number, m s  Arises out of behavior of electron in magnetic field  electron acts like a top  Spinning charge is like a magnet  Electron behave like tiny magnets  Leads to two possible directions of electron spin  Up and down  North and south 52 Possible Values: +½  ½  

53. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Pauli Exclusion Principle  No two electrons in same atom can have same set of all four quantum numbers (n, ℓ, m ℓ, m s )  Can only have two electrons per orbital  Two electrons in same orbital must have opposite spin  Electrons are said to be paired 53

54. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Number of Orbitals and Electrons in the Orbitals 54

55. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Know from Magnetic Properties  Two electrons in same orbital have different spins  Spins paired—diamagnetic  Sample not attracted to magnetic field  Magnetic effects tend to cancel each other  Two electrons in different orbital with same spin  Spins unpaired—paramagnetic  Sample attracted to a magnetic field  Magnetic effects add  Measure extent of attraction  Gives number of unpaired spins 55

56. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following is a valid set of four quantum numbers (n, ℓ, m ℓ, m s )? A. 3, 2, 3, +½ B. 3, 2, 1, 0 C. 3, 0, 0, –½ D. 3, 3, 0, +½ E. 0, –1, 0, –½ 56

57. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is the maximum number of electrons allowed in a set of 4p orbitals? A. 14 B. 6 C. 0 D. 2 E. 10 57

58. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Ground State Electron Arrangements Electron Configurations  Distribution of electrons among orbitals of atom 1.List subshells that contain electrons 2.Indicate their electron population with superscript e.g. N is 1s 2 2s 2 2p 3 Orbital Diagrams  Way to represent electrons in orbitals 1.Represent each orbital with circle (or line) 2.Use arrows to indicate spin of each electron e.g. N is 58 1s1s2s2s2p2p

59. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Energy Level Diagram for Multi Electron Atom/Ion  How to put electrons into a diagram?  Need some rules 59

60. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Aufbau Principle  Building-up principle Pauli Exclusion Principle  Two electrons per orbital  Fill following the order suggested by the periodic table  Spins must be paired 60

61. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Hund’s Rule  If you have more than one orbital all at the same energy  Put one electron into each orbital with spins parallel (all up) until all are half filled  After orbitals are half full, pair up electrons Why?  Repulsion of electrons in same region of space  Empirical observation based on magnetic properties 61

62. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Orbital Diagram and Electron Configurations: e.g. N, Z = 7 Each arrow represents electron 1s 2 2s 2 2p 31s 2 2s 2 2p 3 62

63. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Orbital Diagram and Electron Configurations: e.g. V, Z = 23 Each arrow represents an electron 1s 2 2s 2 2p 2 3s 2 3p 2 4s 2 3d 31s 2 2s 2 2p 2 3s 2 3p 2 4s 2 3d 3 63

64. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check Give electron configurations and orbital diagrams for Na and As Na Z = 11 As Z = 33 64 1s 2 2s 2 2p 2 3s 11s 2 2s 2 2p 2 3s 1 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3

65. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What is the correct ground state electron configuration for Si? A. 1s 2 2s 2 2p 6 3s 2 3p 6 B. 1s 2 2s 2 2p 6 3s 2 3p 4 C. 1s 2 2s 2 2p 6 2d 4 D. 1s 2 2s 2 2p 6 3s 2 3p 2 E. 1s 2 2s 2 2p 6 3s 1 3p 3 65

66. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Periodic Table  Divided into regions of 2, 6, 10, and 14 columns  This equals maximum number of electrons in s, p, d, and f sublevels 66

67. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E  Each row (period) represents different energy level  Each region of chart represents different type of sublevel 67 Sublevels and the Periodic Table

68. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Now Ready to Put Electrons into Atoms Electron configurations must be consistent with: Pauli Exclusion principle  Two electrons per orbital, spins opposite Aufbau principle  Start at lowest energy orbital  Fill, then move up Hund’s rule  One electron in each orbital of same energy, spins parallel  Only pair up if have to 68

69. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Where Are The Electrons? n= 11H1H 2 He n= 23 Li 4 Be 5B5B 6C6C 7N7N 8O8O 9F9F 10 Ne n= 311 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar n= 419 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr n= 537 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe n= 655 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn n= 787 Fr 88 Ra 89 Ac 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 69  Each box represents room for electron.  Read from left to right “ns” orbital being filled “np” orbital being filled “(n – 1)d” orbital being filled “( n – 2)f” orbital being filled

70. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Read Periodic Table to Determine Electron Configuration – He  Read from left to right  First electron goes into period 1  First type of sublevel to fill = “ 1s ”  He has 2 two electrons  electron configuration for He is: 1s 2 70 n= 11H1H 2 He n= 23 Li 4 Be n= 311 Na 12 Mg n= 419 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni n= 537 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd n= 655 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt n= 787 Fr 88 Ra 89 Ac 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds “ns” orbital being filled “np” orbital being filled “(n – 1)d” orbital being filled “( n – 2)f” orbital being filled

71. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron Configuration of Boron (B) 71 n= 11H1H 2 He n= 23 Li 4 Be 5B5B 6C6C 7N7N 8O8O 9F9F 10 Ne n= 311 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar n= 419 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr n= 537 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe n= 655 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn n= 787 Fr 88 Ra 89 Ac 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg  B has 5 electrons  Fill first shell…  Fill two subshells in second shell, in order of increasing energy  Electron Configuration B = 1s 2 2 s 2 2p 1

72. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning Check Write the correct ground state electron configuration for each of the following elements. List in order of increasing n and within each shell, increasing ℓ. 1. K Z = 19 = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 2.Ni Z = 28 = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8 = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 8 4s 2 3.PbZ = 82 = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 14 5s 2 5p 6 5d 10 6s 2 6p 2 72

73. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Chemical Reactivity  Periodic table arranged by chemical reactivity  Depends on outer shell electrons (highest n)  Each row is different n  Core electrons  Inner electrons are those with n < n max  Buried deep in atom 73

74. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Abbreviated Electron Configurations - Noble Gas Notation  [noble gas of previous row] and electrons filled in next row  Represents core + outer shell electrons  Use to emphasize that only outer shell electrons react e.g. Ba = [Xe] 6s 2 Ru = [Kr] 4d 6 5s 2 S = [Ne] 3s 2 3p 4 74

75. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Noble Gas Core Notation for Mn n= 11H1H 2 He n= 23 Li 4 Be 5B5B 6C6C 7N7N 8O8O 9F9F 10 Ne n= 311 Na 12 Mg 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar n= 419 K 20 Ca 21 Sc 22 Ti 23 V 24 Cr 25 Mn 26 Fe 27 Co 28 Ni 29 Cu 30 Zn 31 Ga 32 Ge 33 As 34 Se 35 Br 36 Kr n= 537 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Tc 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe n= 655 Cs 56 Ba 57 La 72 Hf 73 Ta 74 W 75 Re 76 Os 77 Ir 78 Pt 79 Au 80 Hg 81 Tl 82 Pb 83 Bi 84 Po 85 At 86 Rn n= 787 Fr 88 Ra 89 Ac 104 Rf 105 Db 106 Sg 107 Bh 108 Hs 109 Mt 110 Ds 111 Rg 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 75 “ns” orbital being filled “np” orbital being filled “(n – 1)d” orbital being filled “( n – 2)f” orbital being filled  Find last noble gas that is filled before Mn  Next fill sublevels that follow [Ar] 4s4s3d3d 25

76. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! The ground state electron configuration for Ca is: A. [Ar] 3s 1 B. 1s 2 2s 2 2p 6 3s 2 3p 5 4s 2 C. [Ar] 4s 2 D. [Kr] 4s 1 E. [Kr] 4s 2 76

77. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Look at Group 2A ZElectron ConfigurationAbbrev Be 41s 22s 21s 22s 2 [He] 2s 2 Mg121s 22s 22p 63s 21s 22s 22p 63s 2 [Ne] 3s 2 Ca201s 22s 22p 63s 23p 64s 21s 22s 22p 63s 23p 64s 2 [Ar] 4s 2 Sr381s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 5s 2 [Kr] 5s 2 Ba561s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 5s 2 5p 6 6s 2 [Xe] 6s 2 Ra881s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6 4d 10 4f 14 5s 2 5p 6 5d 10 6s 2 6p 6 7s 2 [Rn] 7s 2 77  All have ns 2 outer shell electrons  Only difference is value of n

78. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! An element with the electron configuration [Xe]4f 14 5d 7 6s 2 would belong to which class on the periodic table? A. Transition elements B. Alkaline earth elements C. Halogens D. Lanthanide elements E. Alkali metals 78

79. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Shorthand Orbital Diagrams 79 S[Ne]     3s 3p  Write out lines for orbital beyond Noble gas  Higher energy orbital to right  Fill from left to right Abbreviated Orbital Diagrams Ru[Kr]      4d4d5s5s

80. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Valence Shell Electron Configurations  One last type of electron configuration  Use with representative elements (s and p block elements) – longer columns  Here only electrons in outer shell important for bonding  Only electrons in s and p subshells  Valence shell = outer shell = occupied shell with highest n  Result – use even more abbreviated notation for electron configurations  Sn = 5s 2 5p 2 80

81. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electronic Configurations  A few exceptions to rules ElementExpectedExperimental Cr Cu Ag Au [Ar] 3d 4 4s 2 [Ar] 3d 9 4s 2 [Kr] 4d 9 5s 2 [Xe] 5d 9 6s 2 81 [Ar] 3d 5 4s 1 [Ar] 3d 10 4s 1 [Kr] 4d 10 5s 1 [Xe] 5d 10 6s 1  Exactly filled and exactly half-filled subshells have extra stability  Promote one electron into ns orbital to gain this extra stability

82. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! The orbital diagram corresponding to the ground state electron configuration for nitrogen is: A. B. C. D. E. 82

83. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following choices is the correct electron configuration for a cobalt atom? 4s 3d A. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ B. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑↓ C. [Ar] ↑ ↑↓ ↑↓ ↑↓ ↑ ↑ D. [Ar] ↑↓ ↑↓ ↑↓ ↑↓ ↑ E. [Ar] ↑↓ ↑↓ ↑↓ ↑ ↑ ↑ 83

84. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Heisenberg’s Uncertainty Principle  Can’t know both exact position and exact speed of subatomic particle simultaneously  Such measurements always have certain minimum uncertainty associated with them 84 x = particle position mv = particle momentum = mass × velocity of particle h = Planck’s constant = 6.626 × 10 –34 J s

85. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Heisenberg’s Uncertainty Principle Macroscopic scale  Errors in measurements much smaller than measured value Subatomic scale  Errors in measurements equal to or greater than measured value  If you know position exactly, know nothing about velocity  If you know velocity exactly, know nothing about position 85

86. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Consequence of Heisenberg’s Uncertainty Principle  Can’t talk about absolute position  Can only talk about electron probabilities  Where is e – likely to be?  ψ = wavefunction  Amplitude of electron wave  ψ 2 = probability of finding electron at given location  Probability of finding an electron in given region of space equals the square of the amplitude of wave at that point 86

87. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron Cloud Electron dot picture = snapshots  Lots of dots shown by large amplitude of wave function  High probability of finding electrons Electron density  How much of electrons charge packed into given volume  High Probability  High electron density or Large electron density  Low Probability  Low electron density or Small electron density 87

88. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 1s Orbital Representations a.Dot-density diagram b.Probability of finding electron around given point, ψ 2, with respect to distance from nucleus c.Radial probability distribution = probability of finding electron between r and r + x from nucleus  r max = Bohr radius 88

89. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron Density Distribution  Determined by  Electron density  No sharp boundary  Gradually fades away  “Shape”  Imaginary surface enclosing 90% of electron density of orbital  Probability of finding electrons is same everywhere on surface Shape Sizen Orientationm 89

90. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E  In any given direction probability of finding electron same  All s orbitals are spherically shaped  Size increases as n increases 90 Effect of n on s Orbital

91. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Spherical Nodes  At higher n, now have spherical nodes  Spherical regions of zero probability, inside orbital  Node for electron wave  Imaginary surface where electron density = 0  2s, one spherical node, size larger  3s, two spherical nodes, size larger yet In general:  Number of spherical nodes = n – 1 91

92. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E  Possess one nodal plane through nucleus  Electron density only on two sides of nucleus  Two lobes of electron density  All p orbitals have same overall shape  Size increases as n increases  For 3p have one spherical node 92 p Orbitals

93. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Representations of p Orbitals  Constant probability surface for 2p orbital  Simplified p orbital emphasizing directional nature of orbital  All 2p orbitals in p sub shell  One points along each axis 93 2px2px 2py2py 2pz2pz

94. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E There Are Five Different d Orbitals  Four with four lobes of electron density  One with two lobes and ring of electron density  Result of two nodal planes though nucleus  Number of nodal planes through nucleus = 94

95. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which sketch represents a p z orbital? 95 A. B. D. E. C.

96. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Periodic Properties: Consequences of Electron Configuration  Chemical and physical properties of elements  Vary systematically with position in periodic table  i.e. with element's electron configuration  To explain, must first consider amount of positive charge felt by outer electrons (valence electrons)  Core electrons spend most of their time closer to nucleus than valence (outer shell) electrons  Shield or cancel out (screen out, neutralize) some of positive charge of nucleus 96

97. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Learning check: Li 1s 2 2s 1  Three protons in nucleus  Two core electrons in close (1s)  Net positive charge felt by outer electron  Approximately one proton Effective Nuclear Charge (Z eff )  Net positive charge outer electron feels  Core electrons shield valence electrons from full nuclear charge 97

98. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Shielding  Electrons in same subshell don't shield each other  Same average distance from nucleus  Trying to stay away from each other  Spend very little time one below another  Effective nuclear charge determined primarily by  Difference between charge on nucleus (Z ) and charge on core (number of inner electrons) 98

99. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! What value is the closest estimate of Z eff for a valence electron of the calcium atom? A.1 B.2 C.6 D.20 E.40 99

100. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Atomic Size  Theory suggests sizes of atoms and ions indistinct  Experiment shows atoms/ions behave as if they have definite size  C and H have ~ same distance between nuclei in large number of compounds Atomic Radius (r)  Half of distance between two like atoms  H—H C—C etc.  Usually use units of picometer  1 pm = 1 × 10 –12 m  Range 37 – 270 pm for atoms 100

101. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Trends in Atomic Radius (r) Increases down Column (group)  Z eff essentially constant  n increases, outer electrons farther away from nucleus and radius increase Decreases across row (period)  n constant  Z eff decreases, outer electrons feel larger Z eff and radius decreases Transition Metals and Inner Transition Metals  Size variations less pronounced as filling core  n same (outer electrons) across row  Decrease in Z eff and r more gradually 101

102. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Atomic and Ionic Radii (in pm) 102

103. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Ionic Radii  Increases down column (group)  Decreases across row (period ) Anions larger than parent atom  Same Z eff, more electrons  Radius expands Cations smaller than parent atom  Same Z eff, less electrons,  Radius contracts 103

104. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following has the smallest radius? A. Ar B. K + C. Cl – D. Ca 2+ E. S 2– 104

105. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Ionization Energy  Energy required to remove electron from gas phase atom  Corresponds to taking electron from n to n =   First ionization energyM (g)  M + (g) + e –  IE =  E Trends:  Ionization energy decreases down column (group) as n increases  Ionization energy increases across row (period) as Z eff increases 105

106. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Comparing First Ionization Energies 106  Largest first ionization energies are in upper right  Smallest first ionization energies are in lower left

107. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Successive Ionization Energies  Increases slowly as remove each successive electron  See big increase in ionization energy  When break into exactly filled or half filled subshell 107

108. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E 108 Table 8.2: Successive Ionization Energies in kJ/mol for H through Mg

109. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Place the elements C, N, and O in order of increasing ionization energy. A. C, N, O B. O, N, C C. C, O, N D. N, O, C E. N, C, O 109

110. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Electron Affinity (EA)  Potential energy change associated with addition of one electron to gas phase atom or ion in the ground state X (g) + e –  X – (g)  O and F very favorable to add electrons  Comparing first electron affinities usually negative (exothermic)  Larger negative value means more favorable to add electron 110

111. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Table 8.3 Electron Affinities of Representative Elements 111

112. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Trends in Electron Affinity (EA)  Electron affinity becomes less exothermic down column (group) as n increases  Electron harder to add as orbital farther from nucleus and feels less positive charge  Electron affinity becomes more exothermic across row (period) as Z eff increases  Easier to attract electrons as positive charge increases 112

113. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Successive Electron Affinities  Addition of first electron – often exothermic  Addition of more than one electron requires energy  Consider addition of electrons to oxygen: 113 Change:EA(kJ/mol) O (g) + e –  O – (g) –141 O – (g) + e –  O 2– (g) +844 Net: O (g) + 2 e –  O 2– (g) +703

114. Jespersen/Brady/Hyslop Chemistry: The Molecular Nature of Matter, 6E Your Turn! Which of the following has the largest electron affinity? A. O B. F C. As D. Cs E. Ba 114