1. Principle quantum number n = 1, 2, 3,….. describes orbital size and energy Angular momentum quantum number l = 0 to n-1 describes orbital shape Magnetic quantum number m l = l, l-1…-l describes orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number m s = +1/2 or -1/2 describes the direction of spin of the e - on its axis Pauli Exclusion Principle: "no two electrons in an atom can have the same set of quantum numbers", or, only two electrons (of opposite spin) per orbital.

2. Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, m l = 0, m s = - 1/2 n = 2, l = 0, m l = 0, m s = ± 1/2 2 combinations

3. Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, m l = 0, m s = - 1/2 n = 2, l = 0, m l = 0, m s = ± 1/2 2 combinations (b) 2 p n = 2, l = 1, m l = -1, m s = - 1/2 n = 2, l = 1, m l = -1, 0 or 1, m s = ± 1/2 6 combinations

4. Write a valid set of quantum numbers for each of the following sub-shells: (a) 2 s n = 2, l = 0, m l = 0, m s = - 1/2 n = 2, l = 0, m l = 0, m s = ± 1/2 2 combinations (b) 2 p n = 2, l = 1, m l = -1, m s = - 1/2 n = 2, l = 1, m l = -1, 0 or 1, m s = ± 1/2 6 combinations (c) 3 d n = 3, l = 2, m l = -2, m s = - 1/2 n = 3, l = 2, m l = -2, -1, 0, 1, or 2, m s = ± 1/2 10 combinations

5. How many orbitals in a subshell? l = 0, 1s1 l = 1, p x, p y, p z 3 l = 2, d xy,, d xz,, d yz,, d x 2 -y 2, d z 2 5

6. How many orbitals in a subshell? l = 0, 1s1 l = 1, p x, p y, p z 3 l = 2, d xy,, d xz,, d yz,, d x 2 -y 2, d z 2 5 2 l + 1 orbitals per subshell

7. How many orbitals in a subshell? l = 0, 1s1 l = 1, p x, p y, p z 3 l = 2, d xy,, d xz,, d yz,, d x 2 -y 2, d z 2 5 2 l + 1 orbitals per subshell How many orbitals in a shell? n = 1, 1s1 n = 2, 2s, 2p x, 2p y, 2p z 4 n = 3, 3s, 3p x, 3p y, 3p z, 3d xy,, 3d xz,, 3d yz,, 3d x 2 -y 2, 3d z 2 9

8. How many orbitals in a subshell? l = 0, 1s1 l = 1, p x, p y, p z 3 l = 2, d xy,, d xz,, d yz,, d x 2 -y 2, d z 2 5 2 l + 1 orbitals per subshell How many orbitals in a shell? n = 1, 1s1 n = 2, 2s, 2p x, 2p y, 2p z 4 n = 3, 3s, 3p x, 3p y, 3p z, 3d xy,, 3d xz,, 3d yz,, 3d x 2 -y 2, 3d z 2 9 n 2 orbitals per principal quantum level

9. Hydrogen atom- l all orbitals within a shell have the same energy l electrostatic interaction between e - and proton

10. Hydrogen atom- l all orbitals within a shell have the same energy l electrostatic interaction between e - and proton Multi-electron atoms- l the energy level of an orbital depends not only on the shell but also on the subshell l electrostatic interactions between e - and proton and other e -

11. Quantum Mechanical Model for Multi-electron Atoms l electron repulsions HeHe + + e - E = 2372 kJ mol -1 He has two electron which repel each other

12. Quantum Mechanical Model for Multi-electron Atoms l electron repulsions HeHe + + e - E = 2372 kJ mol -1 He has two electron which repel each other He + He 2+ + e - E = 5248 kJ mol -1 He + has one electron, no electrostatic repulsion

13. Quantum Mechanical Model for Multi-electron Atoms l electron repulsions HeHe + + e - E = 2372 kJ mol -1 He has two electron which repel each other He + He 2+ + e - E = 5248 kJ mol -1 He + has one electron, no electrostatic repulsion Less energy required to remove e - from He than from He + l Shielding of outer orbital electrons from +ve nuclear charge by inner orbital electrons => outer orbital electrons have higher energies

14. Quantum Mechanical Model for Multi-electron Atoms l Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d

15. Quantum Mechanical Model for Multi-electron Atoms l Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d l Effective nuclear charge (Z eff ) experienced by an electron is used to quantify these additional effects.

16. Quantum Mechanical Model for Multi-electron Atoms l Penetration effect of outer orbitals within inner orbitals: ns > np > nd For a given n, energy of s < energy of p < energy of d l Effective nuclear charge (Z eff ) experienced by an electron is used to quantify these additional effects. Example: Sodium, Na, Z = 11 Na 1s e - : Z eff = 10.3 shielding effect is small Na 3s e - : Z eff = 1.84 large shielding effect by inner e - ’s penetration effect counteracts this to a small extent

17. Z

18. Z

20. Orbital Energies Energy 1s 2s 2p x 2p y 2p z 3s 3p x 3p y 3p z 3d xy 3d xz 3d yz 3d x2-y2 3d z2

21. Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle

22. Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle)

23. Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule)

24. Energy 1s 2s 2p Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) H (Z = 1)1s 1

25. Energy 1s 2s 2p Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) N (Z = 7)1s 2, 2s 2, 2p 3,

26. Energy 1s 2s 2p Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) B (Z = 5)1s 2, 2s 2, 2p 1

27. Energy 1s 2s 2p Electronic Configuration: Filling-in of Atomic Orbitals Rules: 1. Pauli Principle 2. Fill in e-'s from lowest energy orbital upwards (Aufbau Principle) 3. Try to attain maximum number of unpaired e- spins in a given sub-shell (Hund's Rule) F (Z = 9)1s 2, 2s 2, 2p 5

28. Hydrogen 2s3s4s 1s 2p3p4p 3d4d 4f Multi-electron atoms 1s 2s3s4s5 s 2p3p4p 3d4d

29. H1s 1 He1s 2 Li1s 2, 2s 1 Be1s 2, 2s 2 B1s 2, 2s 2, 2p x 1 C1s 2, 2s 2, 2p x 1, 2p y 1 N 1s 2, 2s 2, 2p x 1, 2p y 1, 2p z 1 O 1s 2, 2s 2, 2p x 2, 2p y 1, 2p z 1 F 1s 2, 2s 2, 2p x 2, 2p y 2, 2p z 1 Ne 1s 2, 2s 2, 2p x 2, 2p y 2, 2p z 2 1s 2s 2p x 2p y 2p z

30. H1s 1 He1s 2 Li[He], 2s 1 Be[He], 2s 2

31. H1s 1 He1s 2 Li[He], 2s 1 Be[He], 2s 2 B[He], 2s 2, 2p 1 Ne [He], 2s 2, 2p 6 Na [He], 2s 2, 2p 6, 3s 1  [Ne], 3s 1

32. H1s 1 He1s 2 Li[He], 2s 1 Be[He], 2s 2 B[He], 2s 2, 2p 1 Ne [He], 2s 2, 2p 6 Na [He], 2s 2, 2p 6, 3s 1  [Ne], 3s 1 Mg [He], 2s 2, 2p 6, 3s 2  [Ne], 3s 2 Al [Ne], 3s 2, 3p 1 Si [Ne], 3s 2, 3p 2

33. H1s 1 He1s 2 Li[He], 2s 1 Be[He], 2s 2 B[He], 2s 2, 2p 1 Ne [He], 2s 2, 2p 6 Na [He], 2s 2, 2p 6, 3s 1  [Ne], 3s 1 Mg [He], 2s 2, 2p 6, 3s 2  [Ne], 3s 2 Al [Ne], 3s 2, 3p 1 Si [Ne], 3s 2, 3p 2 P [Ne], 3s 2, 3p 3 S [Ne], 3s 2, 3p 4 Cl [Ne], 3s 2, 3p 5 Ar [Ne], 3s 2, 3p 6

34. H1s 1 He1s 2 Li[He], 2s 1 Be[He], 2s 2 B[He], 2s 2, 2p 1 Ne [He], 2s 2, 2p 6 Na [He], 2s 2, 2p 6, 3s 1  [Ne], 3s 1 Mg [He], 2s 2, 2p 6, 3s 2  [Ne], 3s 2 Al [Ne], 3s 2, 3p 1 Si [Ne], 3s 2, 3p 2 P [Ne], 3s 2, 3p 3 S [Ne], 3s 2, 3p 4 Cl [Ne], 3s 2, 3p 5 Ar [Ne], 3s 2, 3p 6 l outermost shell - valence shell l most loosely held electron and are the most important in determining an element’s properties

35. K [Ar], 4s 1 Ca [Ar], 4s 2 Sc [Ar], 4s 2, 3d 1 Ti [Ar], 4s 2, 3d 2

36. K [Ar], 4s 1 Ca [Ar], 4s 2 Sc [Ar], 4s 2, 3d 1 Ca [Ar], 4s 2, 3d 2 Zn [Ar], 4s 2, 3d 10 Ga [Ar], 4s 2, 3d 10, 3p 1 Kr [Ar], 4s 2, 3d 10, 3p 6

37. K [Ar], 4s 1 Ca [Ar], 4s 2 Sc [Ar], 4s 2, 3d 1 Ca [Ar], 4s 2, 3d 2 Zn [Ar], 4s 2, 3d 10 Ga [Ar], 4s 2, 3d 10, 3p 1 Kr [Ar], 4s 2, 3d 10, 3p 6 Anomalous electron configurations d 5 and d 10 are lower in energy than expected Cr [Ar], 4s 1, 3d 5 not [Ar], 4s 2, 3d 4 Cu[Ar], 4s 1, 3d 10 not [Ar], 4s 2, 3d 9

38. Electron Configuration of Ions Electrons lost from the highest energy occupied orbital of the donor and placed into the lowest unoccupied orbital of the acceptor (placed according to the Aufbau principle)

39. Electron Configuration of Ions Electrons lost from the highest energy occupied orbital of the donor and placed into the lowest unoccupied orbital of the acceptor (placed according to the Aufbau principle) Examples: Na [Ne], 3s 1 Na + [Ne] + e - Cl [Ne], 3s 2, 3p 5 + e - Cl - [Ne], 3s 2, 3p 6 Mg [Ne], 3s 2 Mg 2+ [Ne] O [He], 2s 2, 2p 4 O 2- [He], 2s 2, 2p 6

40. Modern Theories of the Atom - Summary Wave-particle duality of light and matter Bohr theory Quantum (wave) mechanical model Orbital shapes and energies Quantum numbers Electronic configuration in atoms

41. Compare the energies of photons emitted by two radio stations, operating at 92 MHz (FM) and 1500 kHz (MW)?

42. Compare the energies of photons emitted by two radio stations, operating at 92 MHz (FM) and 1500 kHz (MW)? E = h 92 MHz = 92 x 10 6 Hz => E = 6.626 x 10 -34 x 92 x 10 6 = 6.1 x 10 -26 J

43. Compare the energies of photons emitted by two radio stations, operating at 92 MHz (FM) and 1500 kHz (MW)? E = h 92 MHz = 92 x 10 6 Hz => E = 6.626 x 10 -34 x 2 x 10 6 = 1.33 x 10 -27 J 1500 kHz E = 6.626 x 10 -34 x 1.5 x 10 6 = 9.94 x 10 -28 J

44. The energy from radiation can be used to break chemical bonds. Energy of at least 495 kJ mol -1 is required to break the oxygen-oxygen bond. What is the wavelength of this radiation?

45. E = hc/ 495 x 10 3 J mol -1  495 x 10 3 J mol -1 /N A = 8.22 x 10 -19 J per molecule

46. The energy from radiation can be used to break chemical bonds. Energy of at least 495 kJ mol -1 is required to break the oxygen-oxygen bond. What is the wavelength of this radiation? E = hc/ 495 x 10 3 J mol -1  495 x 10 3 J mol -1 /N A = 8.22 x 10 -19 J per molecule = 6.626 x 10 -34 x 3 x 10 8 / 8.22 x 10 -19 = 242 x 10 -9 m = 242 nm.

47. Autumn 1999 2. The best available balances can weigh amounts as small as 10 -5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10 -5 g?

48. Autumn 1999 2. The best available balances can weigh amounts as small as 10 -5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10 -5 g? 1 molecule H 2 O has mass of 16 + 2 = 18 amu 1 mole H 2 O has mass of 18 g  6.022 x 10 23 molecules

49. Autumn 1999 2. The best available balances can weigh amounts as small as 10 -5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10 -5 g? 1 molecule H 2 O has mass of 16 + 2 = 18 amu 1 mole H 2 O has mass of 18 g  6.022 x 10 23 molecules 10 -5 g  10 -5 /18 moles = 5.6 x 10 -7 moles

50. Autumn 1999 2. The best available balances can weigh amounts as small as 10 -5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10 -5 g? 1 molecule H 2 O has mass of 16 + 2 = 18 amu 1 mole H 2 O has mass of 18 g  6.022 x 10 23 molecules 10 -5 g  10 -5 /18 moles = 5.6 x 10 -7 moles  5.6 x 10 -7 x 6.022 x 10 23 molecules = 3.35 x 10 17 molecules

51. Autumn 1999 2. The best available balances can weigh amounts as small as 10 -5 g. If you were to count out water molecules at the rate of one per second, how long would it take to count a pile of molecules large enough to weigh 10 -5 g? 1 molecule H 2 O has mass of 16 + 2 = 18 amu 1 mole H 2 O has mass of 18 g  6.022 x 10 23 molecules 10 -5 g  10 -5 /18 moles = 5.6 x 10 -7 moles  5.6 x 10 -7 x 6.022 x 10 23 molecules = 3.35 x 10 17 molecules 3.35 x 10 17 s

52. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light.

53. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = h = hc/ = 6.626 x 10 -34 x 3 x 10 8 /407 x 10 -9 = J s m s -1 m -1

54. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = h = hc/ = 6.626 x 10 -34 x 3 x 10 8 /407 x 10 -9 = J s m s -1 m -1 = 4.88 x 10 -19 J

55. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = h = hc/ = 6.626 x 10 -34 x 3 x 10 8 /407 x 10 -9 = J s m s -1 m -1 = 4.88 x 10 -19 J 1 millimole = 10 -3 mole = 6.022 x 10 20 photons

56. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = h = hc/ = 6.626 x 10 -34 x 3 x 10 8 /407 x 10 -9 = J s m s -1 m -1 = 4.88 x 10 -19 J 1 millimole = 10 -3 mole = 6.022 x 10 20 photons energy of 1 millimole of photons  6.022 x 10 20 x 4.88 x 10 -19 J

57. Autumn 2000 13. Hemoglobin absorbs light of wavelength 407 nm. Calculate the energy (in J) of one millimole of photons of this light. E = h = hc/ = 6.626 x 10 -34 x 3 x 10 8 /407 x 10 -9 = J s m s -1 m -1 = 4.88 x 10 -19 J 1 millimole = 10 -3 mole = 6.022 x 10 20 photons energy of 1 millimole of photons  6.022 x 10 20 x 4.88 x 10 -19 J = 294 J