Quantum Theory and the Electronic Structure of Atoms

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

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 l x n = c 7.1

5. 7.1

6. 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 n Radio wave l x n = c l = c/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 7.1

7. Mystery #1, “Black Body Problem”Solved by Planck in 1900 Energy (light) is emitted or absorbed in discrete units (quantum). E = h x n Planck’s constant (h) h = 6.63 x 10-34 J•s 7.1

8. Mystery #2, “Photoelectric Effect”Solved by Einstein in 1905 hn • Light has both: • wave nature • particle nature KE e- Photon is a “particle” of light hn = KE + BE KE = hn - BE 7.2

9. 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 = h x n E = h x c / 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

10. Line Emission Spectrum of Hydrogen Atoms 7.3

11. 7.3

12. ( ) En = -RH 1 n2 Bohr’s Model of the Atom (1913) THIS CALCULATION HAS BEEN REMOVED • e- can only have specific (quantized) energy values • light is emitted as e- moves from one energy level to a lower energy level n (principal quantum number) = 1,2,3,… RH (Rydberg constant) = 2.18 x 10-18J 7.3

13. E = hn E = hn 7.3

14. ni = 3 ni = 3 f i ni = 2 nf = 2 DE = RH i f ( ) ( ) ( ) Ef = -RH Ei = -RH nf = 1 nf = 1 1 1 1 1 n2 n2 n2 n2 THIS CALCULATION HAS BEEN REMOVED Ephoton = DE = Ef - Ei 7.3

15. 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. Ephoton = DE = RH i f ( ) 1 1 n2 n2 THIS CALCULATION HAS BEEN REMOVED Ephoton = 2.18 x 10-18 J x (1/25 - 1/9) Ephoton = DE = -1.55 x 10-19 J Ephoton = h x c / l l = h x c / Ephoton l= 6.63 x 10-34 (J•s) x 3.00 x 108 (m/s)/1.55 x 10-19J l=1280 nm 7.3

16. Why is e- energy quantized? De Broglie (1924) reasoned that e- is both particle and wave. 2pr = nll = h/mu u = velocity of e- m = mass of e- THIS CALCULATION HAS BEEN REMOVED 7.4

17. What is the de Broglie wavelength (in nm) associated with a 2.5 g Ping-Pong ball traveling at 15.6 m/s? THIS CALCULATION HAS BEEN REMOVED 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

18. 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).

19. Chemistry in Action: Laser – The Splendid Light Laser light is (1) intense, (2) monoenergetic, and (3) coherent

20. Chemistry in Action: Electron Microscopy le = 0.004 nm STM image of iron atoms on copper surface

21. Schrodinger Wave Equation • In 1926 Schrodinger wrote an equation that described both the particle and wave nature of the e- • Wave function (Y) describes: • . energy of e- with a given Y • . 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

22. 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 ml(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)

23. n=1 n=2 n=3 Schrodinger Wave Equation Y = fn(n, l, ml, ms) principal quantum number n n = 1, 2, 3, 4, …. distance of e- from the nucleus 7.6

24. 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

25. 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 l = 1 p orbital l = 2 d orbital l = 3 f orbital 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 7.6

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

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

28. p Orbitals 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 3py orbital

29. p Orbitals • The three p orbitals lie 90o apart in space

30. l = 2 (d orbitals) 7.6

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

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, or1 if l = 2 (d orbital), ml = -2, -1, 0, 1, or2 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 = +½or -½ ms = +½ ms = -½ 7.6

35. 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

36. 7.6

37. How many electrons can an orbital hold? Schrodinger Wave Equation Y = fn(n, l, ml, ms) Shell – electrons with the same value of n Subshell – electrons with the same values of nandl Orbital – electrons with the same values of n, l, andml If n, l, and mlare 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

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

39. 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 Energy depends on n and l 7.7

40. ? ? “Fill up” electrons in lowest energy orbitals (Aufbau principle) Li 3 electrons Be 4 electrons B 5 electrons C 6 electrons B 1s22s22p1 Be 1s22s2 Li 1s22s1 H 1 electron He 2 electrons He 1s2 H 1s1 7.7

41. The most stable arrangement of electrons in subshells is the one with the greatest number of parallel spins (Hund’s rule). C 6 electrons N 7 electrons O 8 electrons F 9 electrons Ne 10 electrons Ne 1s22s22p6 C 1s22s22p2 N 1s22s22p3 O 1s22s22p4 F 1s22s22p5 7.7

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

43. 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!

44. number of electrons in the orbital or subshell principal quantum number n angular momentum quantum number l 1s1 Electron configuration is how the electrons are distributed among the various atomic orbitals in an atom. 1s1 Orbital diagram H 7.8

45. What is the electron configuration of Mg? What are the possible quantum numbers for the last (outermost) electron in Cl? Mg 12 electrons 1s < 2s < 2p < 3s < 3p < 4s 1s22s22p63s2 2 + 2 + 6 + 2 = 12 electrons Abbreviated as [Ne]3s2 [Ne] 1s22s22p6 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

46. Outermost subshell being filled with electrons 7.8

47. 2p 2p Paramagnetic Diamagnetic unpaired electrons all electrons paired 7.8

48. 7.8

49. 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 d4 and d9 For the purposes of this class, we are going to assume that ALL atoms (or ions) that end in d4 or d9 are exceptions to the rule. This may or may not be true, it just depends on the atom.

50. Exceptions to the Aufbau Principle d4 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 d5 instead of d4. For example: Cr would be [Ar] 4s2 3d4, but since this ends exactly with a d4 it is an exception to the rule. Thus, Cr should be [Ar] 4s1 3d5. Procedure: Find the closest s orbital. Steal one electron from it, and add it to the d.