Problems with the atomic model?

Transcription

1 Modern Atomic Theory- Electronic Structure of Atoms DR HNIMIR-CH7 Where should (-) electrons be found? Problems with the atomic model?

2 First, a Little About Electromagnetic Radiation- Waves

3 Another Look

4 Relationship λ ν = c

5 Planck s Relationship black body radiation experiment E = h ν

6 Problem For green visible light of wavelength 486 nm a. what is the frequency of this radiation? λ ν = c b. Calculate E E = h ν

7 Meaning of results? green light of wavelength nm comes "packaged" in packets (called quanta in general and photons for EM) of 4.086x10-19 J 4.086x10-19 J is the minmum amount of energy to be gotten from this green light. If you want more, you have to take 2 x 4.086x10-19 J (or 3x, 4x, etc) 4.086x10-19 J

8 Two More Things For red light of wavelength nm E is 2.740x10-19 J 2.740x10-19 J When you calculate E the units are actually J/photon

9 The EM Spectrum

10 Einstein s Involvement "particle / wave dualism"

11 The Photoelectric Effect

12 Frauhoffer Lines With good optics it was noticed that the continuous spectrum obtained from the sun had very fine black lines in it

13 Line Spectra The Bohr Atom

14 The Hydrogen Spectrum

15 The Bohr Model Niels Bohr interprets the lines as quantized energy emitted by electrons between allowed energy levels an electron absorbs energy and is promoted to a higher energy level the excited electron returns to a lower energy level and emits a specific amount of energy seen as a sharp line of light E

16 Bohr Orbits Bohr calculates the energy of the n th orbit by the formula E n = -B n 2 where B= 2.180x10-18 J for the hydrogen atom

17 Problem Calculate the energy of the 2nd and 4th energy levels E n = x10-18 J n 2 Calculate E for the transition ( E= E 2 -E 4 )

18 Calculate the Wavelength (in nm) Associated with the Answer from the Last Slide E= x10-19 J λ ν = c E = h ν

19 Interpretation of the Last Slide s Results This spectral line is due to a 4 2 electron transition 400 nm 500 nm 600 nm 700 nm λ= nm

20 A Useful Derived Equation E= E final - E initial = E inner - E outer x10-18 J E= - n i x10-18 J n o 2

21 Redo the previous problem using this equation 1 E= 2.180x10-18 J - 2 n o 1 n i 2

22 Regions of Spectral Lines UV Lyman series VIS Balmer series IR Paschen series

23 DeBroglie Assumes that if light (waves) could have particle properties then particles could have wave properties E = h ν E = mc 2 h ν= mc 2 h c = mc 2 λ λ = h mc

24 Interpretation The wavelength of an electron must be an integer multiple of the Bohr orbits (quantized)

25 Evidence for Wave Nature of the Electron

26 Interference Patterns Observed

27 Heisenberg Uncertainty Principle

28 Schroedinger s Equation He combines the mathematics of waves and probability He constructs 3D probability maps for electrons, called "orbitals"

29 Orbitals "s" "p"

30 s,p and d orbitals

31 Energy Levels, Sub-levels and Orbitals energy level sub-level orbitals 4 4f 4d 4p 4s 3 3d 3p 3s ENERGY 2 2p 2s 1 1s

32 O 8 8 electrons Electron Configurationsputting it all together start with the lowest energy level 1 s sublevel is only one in 1st energy level the one orbital can hold 2 electrons this fills the first energy level next electrons goes in the 2nd energy level the s sub-level is lower than p two electrons fill the s sublevel there is a p sublevel in the 2nd energy level the remaining electrons can go into the p sub-level of the 2nd energy level

33 Order of electron fillingfirst 36 elements (through krypton) 1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p

34 Out of Order 4th energy level 3rd energy level 3d 3p 3s 4f 4d 4p 4s 3rd energy level 4f 4d 4p 3d 4s 3p 3s

35 Maximum number of electrons: 1 s 2 s 2 p 3 s 3 p 4 s 3 d 4 p

36 Simplifying Electron Configurations "inert gas core" notation 19K

37 Electron Spin- Stern Gerlach Experiment N Ag furnace beam of gaseous silver atoms S powerful magnet

38 Electron Spin Pairing representing paired electrons in an energy diagram

39 Energy Diagrams- visualizing electron configurations 1s 2s 2p 2s 2p 1s

40 Examplecarbon 6C 1s 2 2s 2 2p 2 2s 2p 1s

41 Q5 Write electron configurations for Si and Ge

42 Q5 cont. Construct energy diagrams for P and F

43 Q5 cont.

44 Transition Metals 4s 3d Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Why the difference?

45 Half-filled Sub-levels are More Stable 4s 3d

46 Problem do the same for Cu 4s 3d

47 Magnetic Properties of Atoms

48 Quantum Numbers

49 Principle Q.N. n n gives the energy level of the electron n = 1,2,3,... (any integer)

50 Angular Momentum Q.N. l the values of l are: 0, 1,..., n-1 gives the number of sublevels in any energy level (if n =2, then l= 0 and 1) meaning: in the second energy level (n=2) there are two sub-levels (s and p) l sub-level 0 s 1 p 2 d 3 f

51 Orientation Quantum Number m l gives the number of orbitals in a sublevel m l = -l...,0,... +l if l = 1 (p sub-level) m l = -1,0,+1 (three p orbitals in the p sub-level)

52 Spin Quantum Number m s tells whether an electron is unpaired (the first into that orbital) or paired m l = +1/2 for the first electron in an orbital m l = -1/2 for the paired electron

53 How it Works 8O 1s 2 2s 2 2p 4 2s 2p 1s

54 Q6 Provide the four QN's for the characteristic electron of phosphorous Which element would have the following characteristicqn's? n = 2 l = 1 m l = 0 m s = -1/2