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### Quntum Theory powerpoint

1. 1. Quantum Theory and theElectronic Structure of Atoms
2. 2. Properties of WavesWavelength (l) is the distance between identical points onsuccessive waves.Amplitude is the vertical distance from the midline of awave to the peak or trough. 7.1
3. 3. Properties of WavesFrequency (n) is the number of waves that pass through aparticular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n 7.1
4. 4. Maxwell (1873), proposed that visible light consists ofelectromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation lxn=c 7.1
5. 5. z Electric field component y xMagnetic field component
6. 6. x
7. 7. z Electric field component x
8. 8. z Electric field component x
9. 9. x
10. 10. y xMagnetic field component
11. 11. y xMagnetic field component
12. 12. z Electric field component y xMagnetic field component
13. 13. 7.1
14. 14. A photon has a frequency of 6.0 x 104 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? llxn=c l = c/n n l = 3.00 x 108 m/s / 6.0 x 104 Hz l = 5.0 x 103 m l = 5.0 x 1012 nm Radio wave 7.1
15. 15. Mystery #1, “Black Body Problem”Solved by Planck in 1900Energy (light) is emitted orabsorbed in discrete units(quantum). E=hxn Planck’s constant (h) h = 6.63 x 10-34 J•s 7.1
16. 16. Mystery #2, “Photoelectric Effect” hnSolved by Einstein in 1905 Light has both: KE e- 1. wave nature 2. particle naturePhoton is a “particle” of light 7.2
17. 17. When copper is bombarded with high-energy electrons, X rays are emitted. Calculate the energy (in joules) associated with the photons if the wavelength of the X rays is 0.154 nm.E=hxnE=hxc/lE = 6.63 x 10-34 (J•s) x 3.00 x 10 8 (m/s) / 0.154 x 10-9 (m)E = 1.29 x 10 -15 J 7.2
18. 18. Line Emission Spectrum of Hydrogen Atoms 7.3
19. 19. Fig. 7.p267middle
20. 20. 7.3
21. 21. Bohr’s Model of the Atom (1913)1. e- can only have specific (quantized) energy values2. light is emitted as e- moves from one energy level to a lower energy level 1 En = -RH ( ) n2n (principal quantum number) = 1,2,3,…RH (Rydberg constant) = 2.18 x 10-18J 7.3
22. 22. E = hnE = hn 7.3
23. 23. Why is e- energy quantized?De Broglie (1924) reasonedthat e- is both particle andwave. 2pr = nl l = h/mu u = velocity of e- m = mass of e- 7.4
24. 24. What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s?l = h/mu h in J•s m in kg u in (m/s)l = 6.63 x 10-34 / (2.5 x 10-3 x 15.6)l = 1.7 x 10-32 m = 1.7 x 10-23 nm 7.4
25. 25. Chemistry in Action: Element from the SunIn 1868, Pierre Janssen detected a new dark line in the solaremission spectrum that did not match known emission lines Mystery element was named HeliumIn 1895, William Ramsey discovered helium in a mineral ofuranium (from alpha decay).
26. 26. Schrodinger Wave EquationIn 1926 Schrodinger wrote an equation thatdescribed both the particle and wave nature of the e-Wave function (Y) describes: 1. energy of e- with a given Y 2. probability of finding e- in a volume of spaceSchrodinger’s equation can only be solved exactlyfor the hydrogen atom. Must approximate itssolution for multi-electron systems. 7.5
27. 27. Schrodinger Wave Equation Y = fn(n, l, ml, ms)principal quantum number nn = 1, 2, 3, 4, ….distance of e- from the nucleus n=1 n=2 n=3 7.6
28. 28. Where 90% of thee- density is foundfor the 1s orbital e- density (1s orbital) falls off rapidly as distance from nucleus increases 7.6
29. 29. Schrodinger Wave Equation Y = fn(n, l, ml, ms) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 l=0 s orbital n = 1, l = 0 l=1 p orbital n = 2, l = 0 or 1 l=2 d orbital n = 3, l = 0, 1, or 2 l=3 f orbitalShape of the “volume” of space that the e- occupies 7.6
30. 30. l = 0 (s orbitals) l = 1 (p orbitals) 7.6
31. 31. l = 2 (d orbitals) 7.6
32. 32. Schrodinger Wave Equation Y = fn(n, l, ml, ms) magnetic quantum number ml for a given value of l ml = -l, …., 0, …. +l if l = 1 (p orbital), ml = -1, 0, or 1if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 orientation of the orbital in space 7.6
33. 33. ml = -1 ml = 0 ml = 1ml = -2 ml = -1 ml = 0 ml = 1 ml = 2 7.6
34. 34. Schrodinger Wave EquationY = fn(n, l, ml, ms)spin quantum number msms = +½ or -½ ms = +½ ms = -½ 7.6
35. 35. OvenDetecting screen Magnet Slit screen
36. 36. OvenDetecting screen Magnet Slit screen
37. 37. Oven Atom beamDetecting screen Magnet Slit screen
38. 38. Oven Atom beam ms = + 1 – 2Detecting screen Magnet Slit screen
39. 39. ms = – 1 – 2 Oven Atom beamDetecting screen Magnet Slit screen
40. 40. ms = – 1 – 2 Oven Atom beam ms = + 1 – 2Detecting screen Magnet Slit screen
41. 41. Schrodinger Wave Equation Y = fn(n, l, ml, ms)Existence (and energy) of electron in atom is describedby its unique wave function Y.Pauli exclusion principle - no two electrons in an atomcan have the same four quantum numbers. Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6
42. 42. 7.6
43. 43. Schrodinger Wave Equation Y = fn(n, l, ml, ms)Shell – electrons with the same value of nSubshell – electrons with the same values of n and lOrbital – electrons with the same values of n, l, and ml How many electrons can an orbital hold? If n, l, and ml are fixed, then ms = ½ or - ½ Y = (n, l, ml, ½) or Y = (n, l, ml, -½) An orbital can hold 2 electrons 7.6
44. 44. How many 2p orbitals are there in an atom?n=2 If l = 1, then ml = -1, 0, or +1 2p 3 orbitalsl=1 How many electrons can be placed in the 3d subshell? n=3 If l = 2, then ml = -2, -1, 0, +1, or +2 3d 5 orbitals which can hold a total of 10 e- l=2 7.6
45. 45. Energy of orbitals in a single electron atomEnergy only depends on principal quantum number n n=3 n=2 1 En = -RH ( ) n2 n=1 7.7
46. 46. Energy of orbitals in a multi-electron atom Energy depends on n and l n=3 l = 2 n=3 l = 1 n=3 l = 0 n=2 l = 1 n=2 l = 0 n=1 l = 0 7.7
47. 47. “Fill up” electrons in lowest energy orbitals (Aufbau principle) ?? Be3 electrons Li 54 electrons C B6 BBe 1s222s2 1 Li 1s 2s1 1s22s22p H 12 electrons He electron He 1s12 H 1s 7.7
48. 48. The most stable arrangement of electronsin subshells is the one with the greatestnumber of parallel spins (Hund’s rule). Ne F 8 O 7 electrons N 6 electrons C 9 10 electrons Ne 1s22s22p56 C 1s22s22p N O F 2 2 3 2 4 7.7
49. 49. Order of orbitals (filling) in multi-electron atom1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7
50. 50. Electron configuration is how the electrons aredistributed among the various atomic orbitals in anatom. number of electrons in the orbital or subshell 1s1 principal quantum angular momentum number n quantum number l Orbital diagram H 1s1 7.8
51. 51. What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6What are the possible quantum numbers for thelast (outermost) electron in Cl? Cl 17 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s23p5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n=3 l=1 ml = -1, 0, or +1 ms = ½ or -½ 7.8
52. 52. Outermost subshell being filled with electrons 7.8
53. 53. 1s 2s 3s 4s 5s 6s 7s
54. 54. 1s2p3p4p5p6p
55. 55. 3d 4d 5d6d
56. 56. 4f5f
57. 57. 1s 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 4f 5f
58. 58. 7.8
59. 59. Paramagnetic Diamagneticunpaired electrons all electrons paired 2p 2p 7.8