1. Einstein (1905): Energy of e.m.r. is quantised E photon = h = hc/ (1) E.m.r.stream of ” particles “ called photons Also:Energy has mass! i.e. E = mc 2 or m = E/c 2 (2) Combining (1) and (2) : “ Mass “ of e.m.r. photon = m = h/ c Overall conclusion: Light (e.m.r.) has both wave and particulate properties Question: Does matter have wave-like properties?

2. Does matter have wave-like properties?

3. Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses

4. Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle)

5. Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle) For an electron (m = 9.1 x 10 -28 g, v = 1.2 x 10 5 m/s): = 6.1 x 10 -9 m

6. Does matter have wave-like properties? Wave Properties of Matter De Broglie (1925): Yes, particularly for small masses Similar equation applies as for light: m = h/.v or = h/mv (v = velocity of particle) For an electron (m = 9.1 x 10 -28 g, v = 1.2 x 10 5 m/s): = 6.1 x 10 -9 m For a cyclist (m = 1 x 10 5 g, v = 3m/s): = 3.5 x 10 -40 m

7. Energy of photon E = h = hc/

8. Energy of photon E = h = hc/ = 500 nm

9. Energy of photon E = h = hc/ = 500 nm E = 6.626 x 10 -34 x 3.00 x 10 9 / 500 x 10 -9 = 3.97 x 10 -18 J s m s -1 m -1 = 3.97 x 10 -18 J

10. Energy of photon E = h = hc/ = 500 nm E = 6.626 x 10 -34 x 3.00 x 10 9 / 500 x 10 -9 = 3.97 x 10 -18 J s m s -1 m -1 = 3.97 x 10 -18 J 100 W bulb emits 100 J s -1  100/ 3.97 x 10 -18 = 2.5 x 10 19 photons s -1.

11. Summary so far: l On the atomic scale energy is transferred in discrete quantities (quanta) l Light (e.m.r.) exhibits both wave and particulate behaviour

12. Summary so far: l On the atomic scale energy is transferred in discrete quantities (quanta) l Light (e.m.r.) exhibits both wave and particulate behaviour l Matter, if small enough in mass, exhibits wave behaviour which is measurable

13. 400550700 nm Line spectrum of sodium

14. 400550700 nm Line spectrum of sodium 400550700 nm Line spectrum of hydrogen

15. 400550700 nm Line spectrum of sodium 400550700 nm Line spectrum of hydrogen Balmer-Rydberg equation: 1/ = R[1/m 2 - 1/n 2 ] m = 1, n > 1

16. Bohr Theory of the Atom (1913) l Developed a quantum model for the H atom

17. Bohr Theory of the Atom (1913) l Developed a quantum model for the H atom l Electron moves around nucleus only in certain allowed circular orbits

18. Bohr Theory of the Atom (1913) l Developed a quantum model for the H atom l Electron moves around nucleus only in certain allowed circular orbits Nucleus Allowed electron energy levels: n = 1 n = 2 n = 3, etc. Electron in ground state atom

20. Energy put in Energy given out

21. Energy given out

22. Energy given out As a photon of given

23. Energy given out As a photon of given Other allowed e - transitions Hydrogen emission spectrum

24. Energy put in As a photon of given Other allowed e - transitions Hydrogen absorption spectrum

25. Bohr theory: l Was based on a particulate view of the electron l Worked reasonably well for the hydrogen atom l But failed for all other elements

26. Bohr theory: l Was based on a particulate view of the electron l Worked reasonably well for the hydrogen atom l But failed for all other elements Wave (Quantum) Mechanical Model of the Atom Schroedinger (1925): l Applied de Broglie’s idea of a wave-like electron to electrons in atoms

27. Bohr theory: l Was based on a particulate view of the electron l Worked reasonably well for the hydrogen atom l But failed for all other elements Wave (Quantum) Mechanical Model of the Atom Schroedinger (1925): l Applied de Broglie’s idea of a wave-like electron to electrons in atoms l Visualised electrons as standing waves, existing around the nucleus

28. Electron as a standing wave n = 4 wavelengths Nucleus Standing wave must have a whole number of wavelengths to prevent destructive interference

29. l Quantization of e - energy levels by standing wave assumption

30. l Quantization of e - energy levels by standing wave assumption l Derived a mathematical description for e - s as 3-d standing waves in atoms:

31. l Quantization of e - energy levels by standing wave assumption l Derived a mathematical description for e - s as 3-d standing waves in atoms: H  = E 

32. l Quantization of e - energy levels by standing wave assumption l Derived a mathematical description for e - s as 3-d standing waves in atoms: H  = E  , the wave function, is a function spatial position of e - ; H, is a mathematical function which allows the calculation of the e - energy (Hamiltonian operator); and E, is the total energy of the electron

33. Fe atoms (blue) on a Cu crystal. Quantum energy states of Cu electrons trapped inside the circular Fe corral are imaged by STM as waveforms (in red).

34. 12 3 4 Making a quantum corral: image sequence shows how the scientists built the circular corral of iron atoms in order to “trap” surface electron energy states. (Courtesy of IBM)

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

36. 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

37. Compare the energies of photons emitted by two radio stations, operating at 92 MHz (FM) and 1500 kHz (MW)? E = h 92 MHz = 2 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 3 = 9.94 x 10 -31 J

38. 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?

39. 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

40. 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.

41. 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?

42. 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

43. 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

44. 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

45. 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

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

47. 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

48. 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

49. 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

50. 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

51. 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

53. 400550700 nm Line spectrum of sodium 400550700 nm Line spectrum of hydrogen Balmer-Rydberg equation: 1/ = R[1/m 2 - 1/n 2 ] m = 1, n > 1

54. l Quantization of e - energy levels by standing wave assumption l Derived a mathematical description for e - s as 3-d standing waves in atoms: H  = E  , the wave function, is a function spatial position of e - ; H, is a mathematical function which allows the calculation of the e - energy (Hamiltonian operator); and E, is the total energy of the electron

55. What is  2 ? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x, where x is distance along the wave

56. What is  2 ? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x, where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927):

57. What is   ? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x, where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927): "There is a fundamental limit to how precisely we can simultaneously determine both the position (x) and the momentum (m.v) of a particle"

58. What is   ? " Amplitude probability function " e.g. for a linear standing wave: Amplitude = sin x, where x is distance along the wave Why amplitude probability function ? Heisenberg Uncertainty Principle (1927): "There is a fundamental limit to how precisely we can simultaneously determine both the position (x) and the momentum (m.v) of a particle"  x.m  v > h/4  i.e.  x.m  v > 5.272(10 -35 ) m 2 kg s -1

59. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg).

60. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg). (a)  x > 5.272(10 -35 )/9.11(10 -31 )(0.1) = 5.79(10 -4 ) m

61. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg). (a)  x > 5.272(10 -35 )/9.11(10 -31 )(0.1) = 5.79(10 -4 ) m  x of e - is big c.f. size of an atom (2(10 -10 m))!

62. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg). (a)  x > 5.272(10 -35 )/9.11(10 -31 )(0.1) = 5.79(10 -4 ) m  x of e - is big c.f. size of an atom (2(10 -10 m))! (b)  x > 5.272(10 -35 )/0.4(0.1) = 1.31(10 -33 ) m

63. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg). (a)  x > 5.272(10 -35 )/9.11(10 -31 )(0.1) = 5.79(10 -4 ) m  x of e - is big c.f. size of an atom (2(10 -10 m))! (b)  x > 5.272(10 -35 )/0.4(0.1) = 1.31(10 -33 ) m  x of ball is v. small c.f. its size (0.3m)

64. Example: For a  v of 0.1m/s, calculate  x for (a) an electron (m=9.11 x 10 -31 kg) and (b) a football (m=0.4kg). (a)  x > 5.272(10 -35 )/9.11(10 -31 )(0.1) = 5.79(10 -4 ) m  x of e - is big c.f. size of an atom (2(10 -10 m))! (b)  x > 5.272(10 -35 )/0.4(0.1) = 1.31(10 -33 ) m  x of ball is v. small c.f. its size (0.3m) Conclusion H.U.P. only important for v. small particles such as e -

65. Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e - in an atom

66. Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e - in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot

67. Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e - in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e - density is given by the term  2

68. Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e - in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e - density is given by the term  2 Probability maps for a particular wavefunction are called electron orbitals

69. Major implication of H.U.P.: It is not possible to find out the exact position & velocity of an e - in an atom Can only define the shapes of electron "clouds" in terms of probability of finding electron at a given spot Probability or e - density is given by the term  2 Probability maps for a particular wavefunction are called electron orbitals Orbitals usually drawn as shape which encloses 90% of the total e - density for that wavefunction

70. Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom.

71. Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital

72. Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital:

73. Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital: Principle quantum number (n) has integral values (1,2,3, etc.) & describes orbital size and energy

74. Characterisation of Electron Orbitals Many different solutions to the Schroedinger Equation for an electron in an atom. Each solution represented by an orbital A series of quantum numbers are used to describe the various properties of an orbital: Principle quantum number (n) has integral values (1,2,3, etc.) & describes orbital size and energy Angular momentum quantum number (l) has integral values from 0 to n-1 for each value of n & describes the orbital shape

75. Magnetic quantum number (m l ) has integral values between l and - l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom

76. Magnetic quantum number (m l ) has integral values between l and - l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number (m s ) is either +1/2 or -1/2 for a given e - & describes the direction of spin of the e - on its axis

77. Magnetic quantum number (m l ) has integral values between l and - l, including zero & describes the orientation in space of the orbital relative to the other orbitals in the atom Spin quantum number (m s ) is either +1/2 or -1/2 for a given e - & 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.

78. Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designationm l Orbitals No. of e -

79. Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designationm l Orbitals No. of e - 1 0 1s0 1 2

80. Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designationm l Orbitals No. of e - 1 0 1s0 1 2 2 0 2s0 1 2 1 2p -1, 0, +1 36

81. Some Quantum Nos. & Orbitals in the H Atom Sub-shell No. of n l Designationm l Orbitals No. of e - 1 0 1s0 1 2 2 0 2s0 1 2 1 2p -1, 0, +1 36 3 0 3s0 12 1 3p -1, 0, +1 36 2 3d-2, -1, 0, +1, +2 510 4...etc...

83. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2

84. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid

85. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid (c) n =2, l =0, m l = -1, m s = 1/2 invalid m l = -l to l

86. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid (c) n =2, l =0, m l = -1, m s = 1/2 invalid m l = -l to l (d) n =3, l =1, m l = 0, m s = 0 invalid m s = -1/2 or +1/2

87. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid (c) n =2, l =0, m l = -1, m s = 1/2 invalid m l = -l to l (d) n =3, l =1, m l = 0, m s = 0 invalid m s = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, m l = 0, m s = 1/2

88. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid (c) n =2, l =0, m l = -1, m s = 1/2 invalid m l = -l to l (d) n =3, l =1, m l = 0, m s = 0 invalid m s = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, m l = 0, m s = 1/2 (b) 2p n = 2, l = 1, m l = -1, m s = 1/2

89. Which of the following are valid sets of quantum numbers? (a) n =1, l =1, m l = 0, m s = 1/2 invalid l = 0 to n-1 (b) n =2, l =1, m l = 0, m s = 1/2 valid (c) n =2, l =0, m l = -1, m s = 1/2 invalid m l = -l to l (d) n =3, l =1, m l = 0, m s = 0 invalid m s = -1/2 or +1/2 Write a valid set of quantum numbers for each of the following sub-shells: (a) 2s n = 2, l = 0, m l = 0, m s = 1/2 (b) 2p n = 2, l = 1, m l = -1, m s = 1/2 (c) 3d n = 3, l = 2, m l = -2, m s = 1/2

90. Principal quantum number specifies energy of e -

91. For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz

92. Principal quantum number specifies energy of e - For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz l Can calculate from theory the hydrogen spectrum

93. Principal quantum number specifies energy of e - For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz l Can calculate from theory the hydrogen spectrum l For transition from n = 3 to n = 2 state

94. Principal quantum number specifies energy of e - For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz l Can calculate from theory the hydrogen spectrum l For transition from n = 3 to n = 2 state  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4]

95. Principal quantum number specifies energy of e - For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz l Can calculate from theory the hydrogen spectrum l For transition from n = 3 to n = 2 state  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4] = h x 4.57 x 10 14 Hz = h  and  4.57 x 10 14 Hz

96. Principal quantum number specifies energy of e - For hydrogen, Schroedinger showed that E = - hR/n 2 where R is the Rydberg constant = 3.29 X 10 15 Hz l Can calculate from theory the hydrogen spectrum l for transition from n = 3 to n = 2 state  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4] = h x 4.57 x 10 14 Hz = h  and  4.57 x 10 14 Hz = c/ = 3 x 10 8 ms -1 / 4.57 x 10 14 s -1 = 656 nm

97. 400550700 nm Line spectrum of hydrogen

98. For hydrogen, E = - hR/n 2 l If H atom acquires enough energy e - can be promoted from one level to the next

99. For hydrogen, E = - hR/n 2 l If H atom acquires enough energy e - can be promoted from one level to the next l To promote an e - from n =1 to n = 2 level  E = -hR[1/2 2 - 1/1 2 ] = - h x 3.29 x 10 15 [1/4 - 1] = 1.63 x 10 -18 J

100. For hydrogen, E = - hR/n 2 l If H atom acquires enough energy e - can be promoted from one level to the next l To promote an e - from n =1 to n = 2 level  E = -hR[1/2 2 - 1/1 2 ] = - h x 3.29 x 10 15 [1/4 - 1] = 1.63 x 10 -18 J l To promote an e - from n = 2 to n = 3 level  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4] = 3.03 x 10 -19 J

101. For hydrogen, E = - hR/n 2 l If H atom acquires enough energy e - can be promoted from one level to the next l To promote an e - from n =1 to n = 2 level  E = -hR[1/2 2 - 1/1 2 ] = - h x 3.29 x 10 15 [1/4 - 1] = 1.63 x 10 -18 J l To promote an e - from n = 2 to n = 3 level  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4] = 3.03 x 10 -19 J l To remove e - (ionise) from the atom  E = -hR[1/  2 - 1/1 2 ] = hR = 2.18 x 10 -18 J

102. For hydrogen, E = - hR/n 2 l If H atom acquires enough energy e - can be promoted from one level to the next l To promote an e - from n =1 to n = 2 level  E = -hR[1/2 2 - 1/1 2 ] = - h x 3.29 x 10 15 [1/4 - 1] = 1.63 x 10 -18 J l To promote an e - from n = 2 to n = 3 level  E = -hR[1/3 2 - 1/2 2 ] = - h x 3.29 x 10 15 [1/9 - 1/4] = 3.03 x 10 -19 J l To remove e - (ionise) from the atom  E = -hR[1/  2 - 1/1 2 ] = hR = 2.18 x 10 -18 J l To ionise one mole of hydrogen atoms  E = N A hR = 1.31 x 10 3 kJ H H + + e -

103. 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.

104. 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

105. 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

106. 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

107. 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

108. 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

109. 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

110. 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

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

112. 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 -

113. 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

114. 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

115. 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

116. 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

117. 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.

118. 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

119. 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

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

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

122. 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)

123. 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

124. 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,

125. 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

126. 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

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

128. 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

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

130. 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

131. 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

132. 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

133. 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

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

135. 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

136. 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

137. 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)

138. 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

139. 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