1. Quantum Theory and the Electronic Structure of Atoms Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Max Planck (1858 - 1947) => new era of physics ~ 1900 scientists tried to understand the atom using classical mechanics => How are atoms held together? => Why is radiation emitted at certain temperatures? Planck suggested the revolutionary idea that atoms & molecules are not governed by classical mechanics => Planck suggested that energy comes in discrete quantities or quanta => Need to understand wave to look at Planck’s theory

3. Wave can be thought of as a vibrating disturbance by which energy is transmitted.

4. Properties of Waves Wavelength ( ) 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

5. Properties of Waves Frequency ( ) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = x 7.1

6. Calculate the wavelength of a wave that has a frequency of 87.4 Hz and a speed of 1.52 x 10 3 cm/s. The speed (u) of the wave = x u = 1.52 x 10 3 cm/s  87.4 Hz  u /  1.52 x 10 3 cm/s = 17.4 cm 87.4 1/s => Let the units help you remember the equation. There are many types of waves ………. water, sound, light, ……...

7. 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 10 8 m/s All electromagnetic radiation x  c Two components with the same wavelength and frequency. Perpendicular planes Usually in nm (10 -9 m)

8. High Energy Low Energy

9. x = c = c/ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz = 5.0 x 10 3 m Radio wave A photon has a frequency of 6.0 x 10 4 Hz. Convert this frequency into wavelength (nm). Does this frequency fall in the visible region? = 5.0 x 10 12 nm 7.1

10. Mystery #1, “Black Body Problem” Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x Planck’s constant (h) h = 6.63 x 10 -34 J s Amount of radiation emitted by an object at a certain temperture depends on its wavelength. Smallest quantity of energy that can be emitted (or absorbed) in the form of electromagnetic radiation Energy is in multiples of h h  h  h  h  but not 1.23h or 5.15h

11. Light has both: 1.wave nature 2.particle nature h = KE + BE Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light KE = h - BE h KE e - Convert radiant energy (light) to electrical energy. KE = Kinetic Energy BE = Binding Energy Energy of incident light KE BE Cannot generate an e - unless h >BE

12. E = h x 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 E = h x c /  7.2 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.

13. Einstein’s work helped explain another “mystery”. Emission spectra of atoms Newton showed that sunlight is composed of various color components  Emission spectrum is either a continuous or line spectrum of radiation emitted by substances Ex: Energize Fe sample with heat Fe “red hot” “white hot” Visible glow  Visible light Heat  IR light

14. Emission spectrum from atoms in gas phase emit light at certain wavelengths  Line spectra

15. 7.3 Line Emission Spectrum of Hydrogen Atoms

16. 7.3

17. 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 Bohr’s Model of the Atom (1913) E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… R H (Rydberg constant) = 2.18 x 10 -18 J Model takes ideas from astronomy – planets moving around the sun => electrons in orbit around the nucleus Negative sign signifies that energy is lower than a “free electron” Smallest quantum number is nearest the nucleus

18. Bohr’s Model of the Atom Has Electrons Moving in Circular Orbits Each orbit has a particular energy  Energy is quantized

19. E n = -R H ( ) 1 n2n2 n (principal quantum number) = 1,2,3,… n = 1 has the most negative value  Ground state  Ground level  Lowest energy state n = 2 = 3 = 4   Higher energy state  Excited level  Excited state E  = -R H ( ) 1 22 0 = “free electron”

20. E = h Energy can be emitted or absorbed  Must go between energy levels  E = E f - E i = h = -R H ( - ) 1 nf2nf2 1 ni2ni2 n = 1 n = 2 n = 3 n = 4 n = 5 Energy levels get closer together as n increases.

21. E = h Energy can be emitted or absorbed  Must go between energy levels  E = E f - E i = h = -R H ( - ) 1 nf2nf2 1 ni2ni2 n = 1 n = 2 n = 3 n = 4 n = 5  E = E 4 – E 5 = h = -R H ( - ) 1 4242 1 5252 R H = 2.18 x 10 -18 J = - 0.0225 R H = - 4.91 x 10 -20 J emit light  E = E 1 – E 3 = h = -R H ( - ) 1 1212 1 3232 = - 0.888 R H = - 1.94 x 10 -18 J emit light

22. E photon =  E = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f  E = R H ( ) 1 n2n2 1 n2n2 n f = 1 n i = 2 n f = 1 n i = 3 n f = 2 n i = 3 7.3

23. E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon =  E = -1.55 x 10 -19 J = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J = 1280 nm Calculate the wavelength (in nm) of a photon emitted by a hydrogen atom when its electron drops from the n = 5 state to the n = 3 state. E photon = h x c /  = h x c / E photon i f  E = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3

24. According to de Broglie, electrons bound to a nucleus behave as a standing wave. Think of a violin or guitar string. node = point that does not move # of nodes depends on the length (l) and the wavelength(λ)  See only half or whole waves

25. De Broglie (1924) reasoned that e - is both particle and wave. 2  r = n = h/mu u = velocity of e - m = mass of e - Why is e - energy quantized? Wave property Particle property Mismatch in waves

26. = h/mu = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) = 1.7 x 10 -32 m = 1.7 x 10 -23 nm What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? m in kgh in J su in (m/s) 7.4

27. Bohr’s model did not explain all the properties of emission spectra  more complicated atoms  extra lines for H in a magnetic field  how do you specify the “position” of a wave? Heisenberg Uncertainty Principle: It is impossible to know both the momentum (p = mass x velocity) and position of a particle simultaneously with certainty. (Δx)(Δp) > h / 4π position momentum  If x is known precisely then p is not known precisely Δx  Δp 

28. Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (  ) describes: 1. energy of e - with a given  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. Atoms are mostly space  electron density  where do we find electrons? Atomic orbital can be throught of as the wavefunction of an electron in an atom.

29. Schrodinger Wave Equation  fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1n=2 n=3 7.6 distance of e - from the nucleus

30. e - density (1s orbital) falls off rapidly as distance from nucleus increases Where 90% of the e - density is found for the 1s orbital 7.6

32.  = fn(n, l, m l, m s ) angular momentum quantum number l for a given value of n, l = 0, 1, 2, 3, … n-1 n = 1, l = 0 n = 2, l = 0 or 1 n = 3, l = 0, 1, or 2 Shape of the “volume” of space that the e - occupies l = 0 s orbital l = 1 p orbital l = 2 d orbital l = 3 f orbital Schrodinger Wave Equation A collection of orbitals with the same n value are referred to as a shell, the different l ’s refer to subshells.

33. l = 0 (s orbitals) l = 1 (p orbitals) 7.6

34. l = 2 (d orbitals) 7.6

35.  = fn(n, l, m l, m s ) magnetic quantum number m l for a given value of l m l = -l, …., 0, …. +l orientation of the orbital in space if l = 1 (p orbital), m l = -1, 0, or 1 if l = 2 (d orbital), m l = -2, -1, 0, 1, or 2 Schrodinger Wave Equation 7.6

36. m l = -1m l = 0m l = 1 m l = -2m l = -1m l = 0m l = 1m l = 2 7.6

37.  = fn(n, l, m l, m s ) spin quantum number m s m s = +½ or -½ Schrodinger Wave Equation m s = -½m s = +½ 7.6

38. Existence (and energy) of electron in atom is described by its unique wave function . Pauli exclusion principle - no two electrons in an atom can have the same four quantum numbers. Schrodinger Wave Equation  = fn(n, l, m l, m s ) Each seat is uniquely identified (E, R12, S8) Each seat can hold only one individual at a time 7.6

40. Schrodinger Wave Equation  = fn(n, l, m l, m s ) 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 m l How many electrons can an orbital hold? If n, l, and m l are fixed, then m s = ½ or - ½  = (n, l, m l, ½ ) or  = (n, l, m l, - ½ ) An orbital can hold 2 electrons 7.6

41. How many 2p orbitals are there in an atom? 2p n=2 l = 1 If l = 1, then m l = -1, 0, or +1 3 orbitals How many electrons can be placed in the 3d subshell? 3d n=3 l = 2 If l = 2, then m l = -2, -1, 0, +1, or +2 5 orbitals which can hold a total of 10 e - 7.6

42. Energy of orbitals in a single electron atom Energy only depends on principal quantum number n E n = -R H ( ) 1 n2n2 n=1 n=2 n=3 7.7

43. Energy of orbitals in a multi-electron atom Energy depends on n and l n=1 l = 0 n=2 l = 0 n=2 l = 1 n=3 l = 0 n=3 l = 1 n=3 l = 2 7.7

44. “Fill up” electrons in lowest energy orbitals (Aufbau principle) H 1 electronH 1s 1 He 2 electrons He 1s 2 Li 3 electronsLi 1s 2 2s 1 Be 4 electronsBe 1s 2 2s 2 B 5 electronsB 1s 2 2s 2 2p 1 C 6 electrons ?? 7.7

45. C 6 electrons The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 1s 2 2s 2 2p 2 N 7 electrons N 1s 2 2s 2 2p 3 O 8 electrons O 1s 2 2s 2 2p 4 F 9 electrons F 1s 2 2s 2 2p 5 Ne 10 electronsNe 1s 2 2s 2 2p 6 7.7

46. Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s 7.7

47. Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s 1 principal quantum number n angular momentum quantum number l number of electrons in the orbital or subshell Orbital diagram H 1s 1 7.8

48. What is the electron configuration of Mg? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 2 + 2 + 6 + 2 = 12 electrons 7.8 Abbreviated as [Ne]3s 2 [Ne] 1s 2 2s 2 2p 6 What are the possible quantum numbers for the last (outermost) electron in Cl? Cl 17 electrons1s < 2s < 2p < 3s < 3p < 4s 1s 2 2s 2 2p 6 3s 2 3p 5 2 + 2 + 6 + 2 + 5 = 17 electrons Last electron added to 3p orbital n = 3l = 1m l = -1, 0, or +1m s = ½ or -½

49. Outermost subshell being filled with electrons 7.8

50. 8.2 ns 1 ns 2 ns 2 np 1 ns 2 np 2 ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 d1d1 d5d5 d 10 4f 5f Ground State Electron Configurations of the Elements

51. 7.8

52. Are there any exceptions to the filling rules for the orbitals? Cr, Cu, Mo, Ag and Au do not follow the rules Since filled orbitals and half-filled orbitals are the most stable and the energy levels are so close in energy, electrons go into d orbitals instead of s orbitals Cu 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 Mo 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 Cr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 Ag 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 10 Au 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 1 4f 14 5d 10

53. 8.2 Ground State Electron Configurations of the Elements Exceptions

54. Paramagnetic unpaired electrons 2p Diamagnetic all electrons paired 2p 7.8