2. New Way Chemistry for Hong Kong A-Level Book 11 The Electronic Structure of Atoms 4.1The Electromagnetic Spectrum 4.2Deduction of Electronic Structure from Ionization Enthalpies 4.3The Wave-mechanical Model of the Atom 4.4 Atomic Orbitals 4

3. New Way Chemistry for Hong Kong A-Level Book 12 The electronic structure of atoms Niels Bohr Bohr’s model of H atom Chapter 4 The electronic structure of atoms (SB p.80)

4. New Way Chemistry for Hong Kong A-Level Book 13 The electronic structure of atoms Niels Bohr Bohr’s model of H atom Chapter 4 The electronic structure of atoms (SB p.80)

5. New Way Chemistry for Hong Kong A-Level Book 14 The electromagnetic spectrum 4.1 The electromagnetic spectrum (SB p.82)

6. New Way Chemistry for Hong Kong A-Level Book 15 Continuous spectrum of white light Fig.4-5(a) 4.1 The electromagnetic spectrum (SB p.82)

7. New Way Chemistry for Hong Kong A-Level Book 16 Line spectrum of hydrogen Fig.4-5(b) 4.1 The electromagnetic spectrum (SB p.83)

8. New Way Chemistry for Hong Kong A-Level Book 17 The emission spectrum of atomic hydrogen UVVisible IR 4.1 The electromagnetic spectrum (SB p.83)

9. New Way Chemistry for Hong Kong A-Level Book 18 Interpretation of the atomic hydrogen spectrum 4.1 The electromagnetic spectrum (SB p.84)

10. New Way Chemistry for Hong Kong A-Level Book 19 4.1 The electromagnetic spectrum (SB p.84) Interpretation of the atomic hydrogen spectrum

11. New Way Chemistry for Hong Kong A-Level Book 110 4.1 The electromagnetic spectrum (SB p.84) Interpretation of the atomic hydrogen spectrum

12. New Way Chemistry for Hong Kong A-Level Book 111 Bohr proposed for a hydrogen atom: 1. An electron in an atom can only exist in certain states characterized by definite energy levels (called quantum). 2. Different orbits have different energy levels. An orbit with higher energy is further away from the nucleus. 3.When an electron jumps from a higher energy level (of energy E 1 ) to a lower energy level (of energy E 2 ), the energy emitted is related to the frequency of light recorded in the emission spectrum by:  E = E 1 - E 2 = h 4.1 The electromagnetic spectrum (SB p.85)

13. New Way Chemistry for Hong Kong A-Level Book 112 How can we know the energy levels are getting closer and closer together? 4.1 The electromagnetic spectrum (SB p.86)

14. New Way Chemistry for Hong Kong A-Level Book 113  E = E 1 - E 2 = h Planck ’s constant Frequency of light emitted 4.1 The electromagnetic spectrum (SB p.87)

15. New Way Chemistry for Hong Kong A-Level Book 114 Emission spectrum of hydrogen Absorption spectrum of hydrogen dark background (photographic plate) bright lines bright background (photographic plate) dark lines 4.1 The electromagnetic spectrum (SB p.87)

16. New Way Chemistry for Hong Kong A-Level Book 115 Production of the absorption spectrum Absorption spectrum of hydrogen bright background (photographic plate) dark lines 4.1 The electromagnetic spectrum (SB p.87)

17. New Way Chemistry for Hong Kong A-Level Book 116 Convergence limits and ionization What line in the H spectrum corresponds to this electron transition (n= ∞  n=1)? Last line in the Lyman Series For n=∞  n=1: H (g) H + (g) + e - 4.1 The electromagnetic spectrum (SB p.87)

18. New Way Chemistry for Hong Kong A-Level Book 117 4.1 The electromagnetic spectrum (SB p.87) Example 4-1A Example 4-1A Example 4-1B Example 4-1B

19. New Way Chemistry for Hong Kong A-Level Book 118 The uniqueness of atomic emission spectra No two elements have identical atomic spectra  atomic spectra can be used to identify unknown elements. 4.1 The electromagnetic spectrum (SB p.89) Check Point 4-1 Check Point 4-1

20. New Way Chemistry for Hong Kong A-Level Book 119 4.2 Deduction of Electronic Structure from Ionization Enthalpies

21. New Way Chemistry for Hong Kong A-Level Book 120 Ionization enthalpy Ionization enthalpy (ionization energy) of an atom is the energy required to remove one mole of electrons from one mole of its gaseous atoms to form one mole of gaseous positive ions. The first ionization enthalpy M(g)  M + (g) + e -  H = 1st I.E. The second ionization enthalpy M + (g)  M 2+ (g) + e -  H = 2nd I.E. 4.2 Deduction of electronic structure from ionization enthalpies (p.91)

22. New Way Chemistry for Hong Kong A-Level Book 121 Evidence of shells  shells 4.2 Deduction of electronic structure from ionization enthalpies (p.91)

23. New Way Chemistry for Hong Kong A-Level Book 122 Evidence of sub-shells 2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8  subshells 4.2 Deduction of electronic structure from ionization enthalpies (p.91) Check Point 4-2 Check Point 4-2

24. New Way Chemistry for Hong Kong A-Level Book 123 4.3 The Wave- mechanical Model of the Atom

25. New Way Chemistry for Hong Kong A-Level Book 124 Bohr’s atomic model and its limitations Bohr considered the electron in the H atom (a one-electron system) moves around the nucleus in circular orbits. Basing on classical mechanics, Bohr calculated values of frequencies of light emitted for electron transitions between such ‘orbits’. The calculated values for the frequencies of light matched with the data in the emission spectrum of H. 4.3 The Wave-mechanical model of the atom (p.94)

26. New Way Chemistry for Hong Kong A-Level Book 125 Bohr tried to apply similar models to atoms of other elements (many-electron system), e.g. Na atom. Basing on classical mechanics, Bohr calculated values of frequencies of light emitted for electron transitions between such ‘orbits’. The calculated values for the frequencies of light did NOT match with the data in the emission spectra of the elements.  The electron orbits in atoms may NOT be simple circular path. 4.3 The Wave-mechanical model of the atom (p.94)

27. New Way Chemistry for Hong Kong A-Level Book 126 Wave nature of electrons A beam of electrons shows diffraction phenomenon  Electrons possess wave properties (as well as particle properties). 4.3 The Wave-mechanical model of the atom (p.95)

28. New Way Chemistry for Hong Kong A-Level Book 127 Wave nature of electrons Schrödinger used complex differential equations/wave fucntions to describe the wave nature of the electrons inside atoms (wave mechanic model). The solutions to the differential equations describes the orbitals of the electrons inside the concerned atom. An orbital is a region of space having a high probability of finding the electron. 4.3 The Wave-mechanical model of the atom (p.95)

29. New Way Chemistry for Hong Kong A-Level Book 128 Quantum numbers Electrons in orbitals are specified with a set of numbers called Quantum Numbers: 1. Principal quantum number (n) n = 1, 2, 3, 4, …... 2. Subsidiary quantum number (l) l = 0, 1, 2, 3…, n-1 s p d f 3. Magnetic quantum number (m) m = -l, …, 0, …l 4. Spin quantum number (s) s= +½, -½ The solutions of the wave functions are the orbitals -- which are themselves equations describing the electrons. 4.3 The Wave-mechanical model of the atom (p.95)

30. New Way Chemistry for Hong Kong A-Level Book 129 Principal quantum number (n) Subsidiary quantum number (l) Number of orbitals (2l+1) Symbol of orbitals Maximum number of electrons held 1011s1s2 20101 1313 2s2p2s2p 2626 3012012 135135 3s3p3d3s3p3d 5 6 10 401230123 13571357 4s4p4d4f4s4p4d4f 2 6 10 14 8 18 32 4.3 The Wave-mechanical model of the atom (p.96)

31. New Way Chemistry for Hong Kong A-Level Book 130 Each orbital can accommodate 2 electrons with opposite spin. 1s 2s 2p 3s 3p 3d 4s 4.3 The Wave-mechanical model of the atom (p.97) Check Point 4-3 Check Point 4-3

32. New Way Chemistry for Hong Kong A-Level Book 131 4.4 Atomic Orbitals

33. New Way Chemistry for Hong Kong A-Level Book 132 s Orbitals 4.4 Atomic orbitals (p.98) Graph of probability of finding an electron against distance from nucleus

34. New Way Chemistry for Hong Kong A-Level Book 133 s Orbitals 4.4 Atomic orbitals (p.98)

35. New Way Chemistry for Hong Kong A-Level Book 134 p Orbitals The shapes and orientations of 2p x, 2p y and 2p z orbitals 4.4 Atomic orbitals (p.100)

36. New Way Chemistry for Hong Kong A-Level Book 135 4.4 Atomic orbitals (p.101) d Orbitals The shapes and orientations of 3d xy, 3d yz, 3d x2-y2 and 3d z2 orbitals Check Point 4-4 Check Point 4-4

37. New Way Chemistry for Hong Kong A-Level Book 136 The END

38. New Way Chemistry for Hong Kong A-Level Book 137 Some insects, such as bees, can see light of shorter wavelengths than humans can. What kind of radiation do you think a bee sees? Back 4.1 The electromagnetic spectrum (SB p.82) Ultraviolet radiation Answer

39. New Way Chemistry for Hong Kong A-Level Book 138 What does the convergence limit in the Balmer series correspond to? Back 4.1 The electromagnetic spectrum (SB p.87) The convergence limit in the Balmer series corresponds to the energy required for the transition of an electron from n =2 to n = . Answer

40. New Way Chemistry for Hong Kong A-Level Book 139 4.1 The electromagnetic spectrum (SB p.88) Given the frequency of the convergence limit of the Lyman series of hydrogen, find the ionization enthalpy of hydrogen. Frequency of the convergence limit = 3.29  10 15 Hz Planck constant = 6.626  10 -34 J s Avogadro constant = 6.02  10 23 mol -1 Answer

41. New Way Chemistry for Hong Kong A-Level Book 140 4.1 The electromagnetic spectrum (SB p.88) Back For one hydrogen atom, E = h = 6.626  10 -34 J s  3.29  10 15 s -1 = 2.18  10 -18 J For one mole of hydrogen atoms, E = 2.18  10 -18 J  6.02  10 23 mol -1 = 1312360 J mol -1 = 1312 kJ mol -1 The ionization enthalpy of hydrogen is 1312 kJ mol -1.

42. New Way Chemistry for Hong Kong A-Level Book 141 4.1 The electromagnetic spectrum (SB p.88) The emission spectrum of atomic sodium is studied. The wavelength of the convergence limit corresponding to the ionization of a sodium atom is found. Based on this wavelength, find the ionization enthalpy of sodium. Wavelength of the convergence limit = 242 nm Planck constant = 6.626  10 -34 J s Avogadro constant = 6.02  10 23 mol -1 Speed of light = 3  108 m s -1 1 nm = 10 -9 m Answer

43. New Way Chemistry for Hong Kong A-Level Book 142 4.1 The electromagnetic spectrum (SB p.88) Back For one mole of sodium atoms, E = h L = = 494486 J mol -1 = 494 kJ mol -1 The ionization enthalpy of sodium of 494 kJ mol -1.

44. New Way Chemistry for Hong Kong A-Level Book 143 4.1 The electromagnetic spectrum (SB p.90) (a)The first line of the Balmer series of the emission spectrum of atomic hydrogen corresponds to the energy emitted in the transition of an electron from the third energy level to the second energy level. It has a wavelength of 656.3 nm. What is the energy difference between the second and the third energy levels? (Planck constant = 6.626  10 -34 Js, Avogadro constant = 6.02  10 23 mol -1 ) Answer

45. New Way Chemistry for Hong Kong A-Level Book 144 4.1 The electromagnetic spectrum (SB p.90) (a)  E = hv =  E = 6.626  10 -34 J s  = 3.03  10 -19 J (for one electron) For 1 mole of electrons,  E = 3.03  10 -19 J  6.02  10 23 mol -1 = 182406 J mol -1 = 182 kJ mol -1

46. New Way Chemistry for Hong Kong A-Level Book 145 4.1 The electromagnetic spectrum (SB p.90) (b)Given that the frequency of the convergence limit corresponding to the ionization of helium is 5.29  10 15 Hz, calculate the ionization enthalpy of helium. (Planck constant = 6.626  10 -34 Js, Avogadro constant = 6.02  10 23 mol -1 ) Answer (b)For 1 mole of helium atoms, I.E. = hvL = 6.626  10 -34 J s  5.29  10 15 s -1  6.02  10 23 mol -1 = 2.11  106 J mol -1 = 2110 kJ mol -1

47. New Way Chemistry for Hong Kong A-Level Book 146 4.1 The electromagnetic spectrum (SB p.90) (c)The blue colour in fireworks is often achieved by heating copper(I) chloride (CuCl) to about 1200 o C. The compound then emits blue light with a wavelength of 450 nm. What is the energy released per copper(I) ion at the specified condition? Answer (c)E = = = 4.42  10 -19 J

48. New Way Chemistry for Hong Kong A-Level Book 147 (d)Name the element present in the sample when the following flame colours are observed in flame tests. (i)Golden yellow (ii)Lilac (iii) Brick-red (iv) Bluish green Answer (d)(i) Sodium (ii) Potassium (iii) Calcium (iv) Copper Back 4.1 The electromagnetic spectrum (SB p.90)

49. New Way Chemistry for Hong Kong A-Level Book 148 (a)Given the successive ionization enthalpies of boron, plot a graph of the logarithm of successive ionization enthalpies of boron against the number of electrons removed. Comment on the graph obtained. Successive I.E. (in kJ mol-1): 800, 2400, 3700, 25000, 32800 Answer 4.2 Deduction of electronic structure from ionization enthalpies (p.94)

50. New Way Chemistry for Hong Kong A-Level Book 149 4.2 Deduction of electronic structure from ionization enthalpies (p.94) (a) The first three electrons of boron are easier to be removed because they are in the outermost shell of the atom. As the fourth and fifth electrons are in the inner shell, a larger amount of energy is required to remove them.

51. New Way Chemistry for Hong Kong A-Level Book 150 (b)Give a rough a sketch of the logarithm of successive ionization enthalpies of potassium. Explain your sketch. Answer 4.2 Deduction of electronic structure from ionization enthalpies (p.94)

52. New Way Chemistry for Hong Kong A-Level Book 151 4.2 Deduction of electronic structure from ionization enthalpies (p.94) (b) There are altogether 19 electrons in a potassium atom. They are in four different energy levels. The first electron is removed from the shell of the highest energy level which is the farthest from the nucleus, I.e. the fourth (outermost) shell. It is the most easiest to be removed. The second to ninth electrons are removed from the third shell, and the next eight electrons are removed from the second shell. The last two electrons with highest ionization enthalpy are removed from the first (innermost) shell of the atom. They are the most difficult to be removed.

53. New Way Chemistry for Hong Kong A-Level Book 152 (c)There is always a drastic increase in ionization enthalpy whenever electrons are removed from a completely filled electron shell. Explain briefly. Answer 4.2 Deduction of electronic structure from ionization enthalpies (p.94) (c) A completely filled electron shell has extra stability. Once an electron is removed, the stable electronic configuration will be destroyed. Therefore, a larger amount of energy is required to remove an electron from such a stable electronic configuration. Back

54. New Way Chemistry for Hong Kong A-Level Book 153 (a)What are the limitations of Bohr’s atomic model? (b)Explain the term “dual nature of electrons”. (c) For principal quantum number 4, how many sub-shells are present? What are their symbols? Back 4.3 The Wave-mechanical model of the atom (p.97) (a)It cannot explain the more complicated spectral lines observed in emission spectra other than that of atomic hydrogen. There is no experimental evidence to prove that electrons are moving around the nucleus in fixed orbits. (b)Electrons can behave either as particles or a wave. (c)When n = 4, l = 0, 1, 2 and 3, there are 4 sub-shells. The symbols are 4s, 4p, 4d and 4f respectively. Answer

55. New Way Chemistry for Hong Kong A-Level Book 154 (a)Distinguish between the terms orbit and orbital. (b)Sketch the pictorial representations of an s orbital and a p orbital. What shapes are they? 4.4 Atomic orbitals (p.101) (a)“Orbit” is the track or path where an electron is revolving around the nucleus. “Orbital” is a region of space in which the probability of finding an electron is very high (about 90 %). (b)s orbital is spherical in shape whereas p orbital is dumb-bell in shape. Answer

56. New Way Chemistry for Hong Kong A-Level Book 155 (c)How do the 1 s and 2 s orbitals differ from each other? (d)How do the 2 p orbitals differ from each other? 4.4 Atomic orbitals (p.101) Answer (c)Both 1s and 2s orbitals are spherical in shape, but the 2s orbital consists of areas of high probability known as nodal surfaces. (d)There are three types of p orbitals. All are dumb-bell in shape. They are aligned in three different spatial orientations designated as x, y and z. Hence, the 2p orbitals are designated as 2p x, 2p y and 2p z. Back