2. Quantum Theory and the Electronic Structure of Atoms Chapter 7

3. Properties of Waves Wavelength ( λ ) 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

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

5. 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  c 7.1

6. What will an electron do? An electron does go though both, and makes an interference pattern. It behaves like a wave. Other matter has wavelengths too short to notice. Image

8. 7.1

9. λν = c λ = c/ν λ = 3.00 x 10 8 m/s / 6.0 x 10 4 Hz λ = 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? λ = 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 ν Planck’s constant (h) h = 6.63 x 10 -34 J s 7.1

11. Light has both: 1.wave nature 2.particle nature h ν= KE + W Mystery #2, “Photoelectric Effect” Solved by Einstein in 1905 Photon is a “particle” of light KE = hν - W hν KE e - 7.2 KE: kinetic energy of the ejected electron W:is the work function

12. E = h ν 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 c / λ 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. 7.3 Line Emission Spectrum of Hydrogen Atoms

14. 7.3

15. 1.e - can only have specific (quantized) energy values No orange or yellow light in H2 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 7.3 THIS CALCULATION HAS BEEN REMOVED

16. E = hν 7.3

17. E photon = ΔE = E f - E i E f = -R H ( ) 1 n2n2 f E i = -R H ( ) 1 n2n2 i i f ΔE = 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 THIS CALCULATION HAS BEEN REMOVED Balmer Lymann

18. E photon = 2.18 x 10 -18 J x (1/25 - 1/9) E photon = ΔE = -1.55 x 10 -19 J λ = 6.63 x 10 -34 (Js) x 3.00 x 10 8 (m/s)/1.55 x 10 -19 J λ = 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 /  λ = h x c / E photon i f ΔE = R H ( ) 1 n2n2 1 n2n2 E photon = 7.3 + E only indicates wether or not energy has been added or lost from the system

19. De Broglie (1924) reasoned that e - is both particle and wave. 2  r = νλ λ = h/mu u = velocity of e - m = mass of e - Why is e - energy quantized? 7.4

20. Wave interference Animation

21. λ = h/mu λ = 6.63 x 10 -34 / (2.5 x 10 -3 x 15.6) λ = 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

22. 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 In 1895, William Ramsey discovered helium in a mineral of uranium (from alpha decay). Mystery element was named Helium

23. Chemistry in Action: Laser – The Splendid Light Light Amplification by Stimulated Emission of Radiation Laser light is (1) intense, (2) monoenergetic, and (3) coherent

24. Chemistry in Action: Electron Microscopy STM image of iron atoms on copper surface e = 0.004 nm Scanning tunneling microscope

25. Heisenberg Uncertainty Principal: It is impossible to know simultaneously both the momentum p (defined as mass time velocity) and the position of a particle with certainty. Δx Δp ≥ h/4π Δx position, Δp momentum, ≥ crude exp > rules, the probability is never less than h/4π This proves that Bohr was not completely right so that brings us to Erwin Schrodinger.

26. Schrodinger Wave Equation In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e - Wave function (Ψ)psi describes: 1. energy of e - with a given Ψ 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. It requires advanced calculus to solve 7.5

27. Schrodingers Equation It is important to know that the equation incorporates both particle behavior (mass) and wave behavior Ψ

28. QUANTUM NUMBERS The shape, size, and energy of each orbital is a function of 3 quantum numbers which describe the location of an electron within an atom or ion n (principal) ---> energy level l (orbital) ---> shape of orbital m l (magnetic) ---> designates a particular suborbital The fourth quantum number is not derived from the wave function s(spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½) s (spin) ---> spin of the electron (clockwise or counterclockwise: ½ or – ½)

29. Schrodinger Wave Equation  fn(n, l, m l, m s ) principal quantum number n n = 1, 2, 3, 4, …. n=1 n=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

31.  = 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 7.6

32. Types of Orbitals ( l ) s orbital p orbital d orbital

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

34. p Orbitals this is a p sublevel with 3 orbitals These are called x, y, and z this is a p sublevel with 3 orbitals These are called x, y, and z There is a PLANAR NODE thru the nucleus, which is an area of zero probability of finding an electron 3p y orbital

35. p Orbitals The three p orbitals lie 90 o apart in spaceThe three p orbitals lie 90 o apart in space

36. l = 2 (d orbitals) 7.6

37. f Orbitals For l = 3, f sublevel with 7 orbitals

38.  = 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

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

40.  = 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

41. Existence (and energy) of electron in atom is described by its unique wave function Ψ. 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

43. 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

44. 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

45. Orbital Diagrams Energy of orbitals in a multi-electron atom 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 Energy depends on n and l

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

47. 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 electrons Ne 1s 2 2s 2 2p 6 7.7

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

49. Outermost subshell being filled with electrons 7.8

50. Why are d and f orbitals always in lower energy levels? d and f orbitals require LARGE amounts of energy It’s better (lower in energy) to skip a sublevel that requires a large amount of energy (d and f orbtials) for one in a higher level but lower energy This is the reason for the diagonal rule! BE SURE TO FOLLOW THE ARROWS IN ORDER!

51. 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

52. 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

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

55. Exceptions to the Aufbau Principle Remember d and f orbitals require LARGE amounts of energy If we can’t fill these sublevels, then the next best thing is to be HALF full (one electron in each orbital in the sublevel) There are many exceptions, but the most common ones are d 4 and d 9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d 4 or d 9 are exceptions to the rule. This may or may not be true, it just depends on the atom.

56. Exceptions to the Aufbau Principle d 4 is one electron short of being HALF full In order to become more stable (require less energy), one of the closest s electrons will actually go into the d, making it d 5 instead of d 4. For example: Cr would be [Ar] 4s 2 3d 4, but since this ends exactly with a d 4 it is an exception to the rule. Thus, Cr should be [Ar] 4s 1 3d 5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d.

57. Try These! Write the shorthand notation for: Cu W Au [Ar] 4s 1 3d 10 [Xe] 6s 1 4f 14 5d 5 [Xe] 6s 1 4f 14 5d 10

58. Exceptions to the Aufbau Principle The next most common are f 1 and f 8 The electron goes into the next d orbital Example: –La [Xe]6s 2 5d 1 –Gd [Xe]6s 2 4f 7 5d 1

59. Keep an Eye On Those Ions! Electrons are lost or gained like they always are with ions… negative ions have gained electrons, positive ions have lost electrons The electrons that are lost or gained should be added/removed from the highest energy level (not the highest orbital in energy!)

60. Keep an Eye On Those Ions! Tin Atom: [Kr] 5s 2 4d 10 5p 2 Sn +4 ion: [Kr] 4d 10 Sn +2 ion: [Kr] 5s 2 4d 10 Note that the electrons came out of the highest energy level, not the highest energy orbital!