Quntum Theory powerpoint
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This document discusses quantum theory and the electronic structure of atoms. It begins by introducing properties of waves and electromagnetic radiation. It then covers early discoveries and models in atomic structure, including Planck's quantization of energy, Einstein's explanation of the photoelectric effect using photons, Bohr's model of electron orbits, de Broglie's proposal that electrons exhibit wave-particle duality, and Schrodinger's wave equation describing electron probability distributions. The document concludes by discussing how the Schrodinger equation is used to determine electron configurations and orbital diagrams for atoms.
![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] 1s22s22p6
What are the possible quantum numbers for the
last (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](https://image.slidesharecdn.com/quntumtheorypowerpoint-121112083623-phpapp02/85/Quntum-Theory-powerpoint-51-320.jpg)
Quntum Theory powerpoint
- 1. Quantum Theory and the Electronic Structure of Atoms
- 2. Properties of Waves Wavelength (l) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 7.1
- 3. Properties of Waves Frequency (n) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = l x n 7.1
- 4. Maxwell (1873), proposed that visible light consists of electromagnetic 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. z Electric field component y x Magnetic field component
- 6. x
- 7. z Electric field component x
- 8. z Electric field component x
- 9. x
- 10. y x Magnetic field component
- 11. y x Magnetic field component
- 12. z Electric field component y x Magnetic field component
- 13. 7.1
- 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? l lxn=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. Mystery #1, “Black Body Problem” Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E=hxn Planck’s constant (h) h = 6.63 x 10-34 J•s 7.1
- 16. Mystery #2, “Photoelectric Effect” hn Solved by Einstein in 1905 Light has both: KE e- 1. wave nature 2. particle nature Photon is a “particle” of light 7.2
- 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=hxn E=hxc/l E = 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. Line Emission Spectrum of Hydrogen Atoms 7.3
- 20. 7.3
- 21. Bohr’s Model of the Atom (1913) 1. e- can only have specific (quantized) energy values 2. light is emitted as e- moves from one energy level to a lower energy level 1 En = -RH ( ) n2 n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J 7.3
- 22. E = hn E = hn 7.3
- 23. Why is e- energy quantized? De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nl l = h/mu u = velocity of e- m = mass of e- 7.4
- 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. Chemistry in Action: Element from the Sun In 1868, Pierre Janssen detected a new dark line in the solar emission spectrum that did not match known emission lines Mystery element was named Helium In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay).
- 26. Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described 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 space Schrodinger’s equation can only be solved exactly for the hydrogen atom. Must approximate its solution for multi-electron systems. 7.5
- 27. Schrodinger Wave Equation Y = fn(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus n=1 n=2 n=3 7.6
- 28. Where 90% of the e- density is found for the 1s orbital e- density (1s orbital) falls off rapidly as distance from nucleus increases 7.6
- 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 orbital Shape of the “volume” of space that the e- occupies 7.6
- 30. l = 0 (s orbitals) l = 1 (p orbitals) 7.6
- 31. l = 2 (d orbitals) 7.6
- 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 1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or 2 orientation of the orbital in space 7.6
- 33. ml = -1 ml = 0 ml = 1 ml = -2 ml = -1 ml = 0 ml = 1 ml = 2 7.6
- 34. Schrodinger Wave Equation Y = fn(n, l, ml, ms) spin quantum number ms ms = +½ or -½ ms = +½ ms = -½ 7.6
- 35. Oven Detecting screen Magnet Slit screen
- 36. Oven Detecting screen Magnet Slit screen
- 37. Oven Atom beam Detecting screen Magnet Slit screen
- 38. Oven Atom beam ms = + 1 – 2 Detecting screen Magnet Slit screen
- 39. ms = – 1 – 2 Oven Atom beam Detecting screen Magnet Slit screen
- 40. ms = – 1 – 2 Oven Atom beam ms = + 1 – 2 Detecting screen Magnet Slit screen
- 41. Schrodinger Wave Equation Y = fn(n, l, ml, ms) Existence (and energy) of electron in atom is described by its unique wave function Y. Pauli exclusion principle - no two electrons in an atom can 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. 7.6
- 43. Schrodinger Wave Equation Y = fn(n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of n and l Orbital – 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. How many 2p orbitals are there in an atom? n=2 If l = 1, then ml = -1, 0, or +1 2p 3 orbitals l=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. Energy of orbitals in a single electron atom Energy only depends on principal quantum number n n=3 n=2 1 En = -RH ( ) n2 n=1 7.7
- 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. “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. The most stable arrangement of electrons in subshells is the one with the greatest number 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. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7
- 50. Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 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. 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] 1s22s22p6 What are the possible quantum numbers for the last (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. Outermost subshell being filled with electrons 7.8
- 54. 1s 2s 3s 4s 5s 6s 7s
- 56. 3d 4d 5d 6d
- 57. 4f 5f
- 58. 1s 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 4f 5f
- 59. 7.8
- 60. Paramagnetic Diamagnetic unpaired electrons all electrons paired 2p 2p 7.8