2. 1 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. 2 The electronic structure of atoms Chapter 4 The electronic structure of atoms (SB p.80) Two sources of evidence : - (a) Study of atomic emission spectra of the elements (b) Study of ionization enthalpies of the elements

4. 3 Atomic Emission Spectra 原子放射光譜 Spectra  plural of Spectrum Arises from light emitted from individual atoms

5. 4  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum Speed of light (ms  1 ) = Frequency  Wavelength c =  3  10 8 ms  1

6. 5  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum red violet Visible light

7. 6  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum red violet Ultraviolet light  Increasing energy

8. 7  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum red violet X-rays  Increasing energy

9. 8  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum red violet Gamma rays  Increasing energy

10. 9  Frequency (Hz / s  1 ) Wavelength (m)  Decreasing energy  The electromagnetic spectrum red violet Infra-red light

11. 10  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum red violet Microwave & radio waves Decreasing energy 

12. 11 Types of Emission Spectra 4.1 The electromagnetic spectrum (SB p.82) 1.Continuous spectra E.g.Spectra from tungsten filament and sunlight 2.Line Spectra E.g.Spectra from excited samples in discharge tubes

13. 12 Continuous spectrum of white light Fig.4-5(a) 4.1 The electromagnetic spectrum (SB p.82)

14. 13 Line spectrum of hydrogen Fig.4-5(b) 4.1 The electromagnetic spectrum (SB p.83)

15. 14 How Do Atoms Emit Light ? H 2 (g) Electric discharge Or Heating H(g) Hydrogen atom in ground state means its electron has the lowest energy

16. 15 Energy Atom in ground state H(g) Electric discharge Or Heating H*(g) ExcitedGround Atoms in excited state

17. 16 H*(g) H(g) GroundExcited Energy Atom in ground state Atoms in excited state Not Stable h Atom returns to ground state h

18. 17 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h Planck’s equation

19. 18 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h Planck : Nobel laureate Physics, 1918

20. 19 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h E = energy of the emitted light = E 2 – E 1 E2E2 E1E1

21. 20 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h = Frequency of the emitted light E2E2 E1E1

22. 21 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h h (Planck’s constant) = 6.63  10  34 Js E2E2 E1E1

23. 22 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h h = 6.63  10  34 Js E2E2 E1E1 Energy cannot be absorbed or emitted by an atom in any arbitrary amount.

24. 23 Energy Atom in ground state Atoms in excited state Atom returns to ground state h E = h h = 6.63  10  34 Js E2E2 E1E1 Energy can only be absorbed or emitted by an atom in multiples of 6.63  10  34 J.

25. 24 4.1 The electromagnetic spectrum (SB p.84) Characteristic Features of the Hydrogen Emission Line Spectrum 1. The visible region – The Balmer Series greenred violet ultraviolet Convergence limit

26. 25 Q.1 = 3.03  10  19 J

27. 26 energy of one photon emitted

28. 27 Q.2 (a) The spectral lines come closer at higher frequency and eventually merge into a continuum ( 連續體 ) (b)n =   n = 2

29. 28 (c)The electron has been removed from the atom. I.e. the atom has been ionized. Q.2 e1e1 H(g)  H + (g) + e 

30. 29 corresponds to the last spectral line of the Balmer series

31. 30 The Complete Hydrogen Emission Spectrum UVVisible IR 4.1 The electromagnetic spectrum (SB p.83)

32. 31 Q.3 (a)The spectral lines in each series get closer at higher frequency. (b)Since the energy levels converge at higher level, the spectral lines also converge at higher frequency.

33. 32 Rydberg Equation Relates wavelength of the emitted light of hydrogen atom with the electron transition

34. 33 = wave number of the emitted light = number of waves in a unit length e.g.  100 waves in 1 meter

35. 34

36. 35 Electron transition : b  a, a, b are integers and b > a a represents the lower energy level to which the electron is dropping back b represents the higher energy levels from which the electron is dropping back

37. 36 Balmer series, a = 2 b = 3, 4, 5,…

38. 37 Lyman series, a = 1 b = 2, 3, 4,… 2 3 4 5 6 

39. 38 Paschen series, a = 3 b = 4, 5, 6,… 4 5 6 7 

40. 39 R H is the Rydberg constant

41. 40 Q.4 0.0000.0120.0160.0200.0280.0400.0630.111 2.742.602.572.522.442.302.061.52 (10 6 m  1 )

42. 41 10 6 m  1 y-intercept R H = 1.096  10 7 m  1

43. 42 Interpretation of the atomic hydrogen spectrum 4.1 The electromagnetic spectrum (SB p.84) Discrete spectral lines  energy possessed by electrons within hydrogen atoms cannot be in any arbitrary quantities but only in specified amounts called quanta.

44. 43 Interpretation of the atomic hydrogen spectrum 4.1 The electromagnetic spectrum (SB p.84) Only certain energy levels are allowed for the electron in a hydrogen atom. The energy of the electron in a hydrogen atom is quantized.

45. 44 Interpretation of the atomic hydrogen spectrum 4.1 The electromagnetic spectrum (SB p.84)

46. 45 Niels Bohr (1885-1962) Interpretation of the atomic hydrogen spectrum Bohr’s Atomic Model of Hydrogen Nobel Prize Laureate in Physics, 1922

47. 46 Interpretation of the atomic hydrogen spectrum Nobel Prize Laureate in Physics, 1922 for his services in the investigation of the structure of atoms and of the radiation emanating from them

48. 47 Interpretation of the atomic hydrogen spectrum Bohr’s model of H atom

49. 48 Interpretation of the atomic hydrogen spectrum 1.The electron can only move around the nucleus of a hydrogen atom in certain circular orbits with fixed radii. Each allowed orbit is assigned an integer, n, known as the principal quantum number.

50. 49 Interpretation of the atomic hydrogen spectrum 2. Different orbits have different energy levels. An orbit with higher energy is further away from the nucleus.

51. 50 Interpretation of the atomic hydrogen spectrum 3.Spectral lines arise from electron transitions from higher orbits to lower orbits. E2E2 E1E1 n = 2 n = 1

52. 51 Interpretation of the atomic hydrogen spectrum For the electron transition E 2  E 1,  E = E 2 – E 1 = h E2E2 E1E1 n = 2 n = 1 the energy emitted is related to the frequency of light emitted by the Plank’s equation :

53. 52 Interpretation of the atomic hydrogen spectrum 4.In a sample containing numerous excited H * atoms, different H * atoms may undergo different kinds of electron transitions to give a complete emission spectrum.

54. 53 4  3 5  2 4  2 3  2 4  1 3  1 1876 434.1 486.1 656.7 97.32 102.6 121.7 1876 434.3 486.5 656.6 97.28 102.6 121.62  1 Wavelength Determined by Experiment (nm) Wavelength Predicted by Bohr (nm) Electronic Transition

55. 54 Interpretation of the atomic hydrogen spectrum 5.The theory failed when applied to elements other than hydrogen (multi- electron systems)

56. 55 Illustrating Bohr’s Theory First line of Lyman series, n = 2  n = 1 E = E 2 – E 1 E2E2 E1E1 n = 2 n = 1 By Planck’s equation,

57. 56 E = E 2 – E 1 E2E2 E1E1 n = 2 n = 1 By Rydberg’s equation,

58. 57 E = E 2 – E 1 E2E2 E1E1 n = 2 n = 1

59. 58 All electric potential energies have negative signs except E  E 1 < E 2 < E 3 < E 4 < E 5 < …… < E  = 0 All are negative

60. 59 Balmer series, Transitions from higher levels to n = 2 n = 3, 4, 5, … Visible region

61. 60 Lyman series, Transitions from higher levels to n = 1 n = 2, 3, 4,… More energy released  Ultraviolet region

62. 61 Paschen series, Transitions from higher levels to n = 3 n = 4, 5, 6,… Less energy released  Infra-red region

63. 62  E = E 2 – E 1 = h 4.1 The electromagnetic spectrum (SB p.87) UV visible IR

64. 63 Q.5 n = 100  n = 99 Energy of the light emitted is extremely small.

65. 64  Frequency (Hz / s  1 ) Wavelength (m)  The electromagnetic spectrum Microwave

66. 65 Q.5 n = 100  n = 99  Microwave region Energy of the light emitted is extremely small.

67. 66 Q.6

68. 67 Energy of the first line n = 3 n = 2 n = 1

69. 68 Energy of the first line = 45.7  10 13 Hz

70. 69 Energy of the second line n = 4 n = 2 n = 1

71. 70 = 61.7  10 13 Hz Energy of the second line

72. 71 n = 5 n = 2 n = 1 Energy of the third line

73. 72 = 69.1  10 13 Hz Energy of the third line

74. 73 Convergence Limits and Ionization Enthalpies Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  ) X(g)  X + (g) + e  Units : kJ mol  1

75. 74 Convergence Limits and Ionization Enthalpies Convergence Limits The frequency at which the spectral lines of a series merge.

76. 75 Q.7 Ionization enthalpy Convergence limit of Lyman series H(g)  H + (g) + e  n = 1n = 

77. 76 Na Li He n = 3  n =  n = 2  n =  n = 1  n =  n = 3 n = 2 n = 1 H Electron transition for ionization of atom Energy Level of e  to be removed in ground state of atom Atom Lyman series ? Balmer series ? Paschen series ?

78. 77 Q.9 Ionization enthalpy of helium = (6.626  10  34 Js)(5.29  10 15 s  1 )(6.02  10 23 mol  1 ) = 2110 KJ mol  1 > Ionization enthaply of hydrogen

79. 78 EE E2E2 E1E1 E2’E2’ E1’E1’ HeH Ionization enthalpy : He > H Relative positions of energy levels depend on the nuclear charge of the atom.

80. 79 The uniqueness of atomic emission spectra No two elements have identical atomic spectra 4.1 The electromagnetic spectrum (SB p.89)

81. 80 atomic spectra can be used to identify unknown elements. 4.1 The electromagnetic spectrum (SB p.89) The uniqueness of atomic emission spectra

82. 81 Flame Test NaCl(s) Na(g) + Cl(g) heat atomization Na(g) + Cl(g) Na * (g) + Cl * (g) heat Na * (g) + Cl * (g) Na(g) + Cl(g) h Na + h Cl

83. 82 Flame Test The most intense yellow light is observed.

84. 83 Q.10 = 4.42  10  19 J

85. 84 Bright lines against a dark background Dark lines against a bright background Emission vs Absorption Emission spectrum of hydrogen (nm) Absorption spectrum of hydrogen (nm) visible

86. 85 Absorption spectra are used to determine the distances and chemical compositions of the invisible clouds.

87. 86 The Doppler effect Lower pitch heard Higher pitch heard The frequency of sound waves from a moving object (a)increases when the object moves towards the observer. (b)decreases when the object moves away from the observer.

88. 87 Redshift and the Doppler effect Frequency of light waves emanating from a moving object decreases when the light source moves away from the observer.  wavelength increases  spectral lines shift to the red side with longer wavelength  redshift

89. 88 Atomic Absorption Spectra redshift Moving at higher speed Moving at lower speed

90. 89 Redshift Left : - Absorption spectrum from sunlight. Right : - Absorption spectrum from a supercluster of distant galaxies

91. 90 Atomic Absorption Spectra Only atomic emission spectrum of hydrogen is required in A-level syllabus !!!

92. 91 4.2 Deduction of Electronic Structure from Ionization Enthalpies

93. 92 Evidence of the Existence of Shells For multi-electron systems, Two questions need to be answered

94. 93 Evidence of the Existence of Shells 1.How many electrons are allowed to occupy each electron shells ? 2.How are the electrons arranged in each electron shell ? Existence of Shells Existence of Subshells

95. 94 Ionization enthalpy is the energy needed to remove one mole of electrons from one mole of gaseous atoms in ground state to give one moles of gaseous ions (n =  ) X(g)  X + (g) + e  Ionization enthalpy

96. 95 Evidence of the Existence of Shells Successive ionization enthalpies Q. 11

97. 96 Q.11(a) Be(g)  Be + (g) + e  IE 1 Be + (g)  Be 2+ (g) + e  IE 2 Be 2+ (g)  Be 3+ (g) + e  IE 3 Be 3+ (g)  Be 4+ (g) + e  IE 4

98. 97 Q.11(b) IE 1 < IE 2 < IE 3 < IE 4 Positive ions with higher charges attract electrons more strongly. Thus, more energy is needed to remove an electron from positive ions with higher charges.

99. 98 Q.11(c) 21060149051758900 IE 4 (kJ mol  1 ) IE 3 (kJ mol  1 ) IE 2 (kJ mol  1 ) IE 1 (kJ mol  1 )

100. 99 Q.11(c) (i)The first two electrons are relatively easy to be removed.  they experience less attraction from the nucleus,  they are further away from the nucleus and occupy the n = 2 electron shell.

101. 100 Q.11(c) (ii)The last two electrons are very difficult to be removed.  they experience stronger attraction from the nucleus,  they are close to the nucleus and occupy the n = 1 electron shell.

102. 101 Electron Diagram of Beryllium Second Shell First Shell

103. 102 Energy Level Diagram of Beryllium n =2 n= 1

104. 103 EE IE 1 = E  - E 2 E2E2 E1E1 Be

105. 104 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + IE 2 IE 1 < IE 2

106. 105 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + E 2 ’’ E 1 ’’ Be 2+ IE 2 IE 1 < IE 2 << IE 3 IE 3

107. 106 EE IE 1 E2E2 E1E1 Be E2’E2’ E1’E1’ Be + E 2 ’’ E 1 ’’ Be 2+ E 2 ’’’ E 1 ’’’ Be 3+ IE 1 < IE 2 << IE 3 < IE 4 IE 3 IE 2 IE 4

108. 107 The Concept of Spin( 自旋 ) Spin is the angular momentum intrinsic to a body. E.g.Earth’s spin is the angular momentum associated with Earth’s rotation about its own axis. 自轉

109. 108 On the other hand, orbital angular momentum of the Earth is associated with its annual motion around the Sun ( 公轉 )

110. 109 Subatomic particles like protons and electrons possess spin properties. i.e. they have spin angular momentum. But their spins cannot be associated with rotation since they display both particle-like and wave-like behaviours.

111. 110 Paired electrons in an energy level should have opposite spins. Electrons with opposite spins are represented by arrows in opposite directions. Q.12

112. 111 4 groups of electrons Q.12

113. 112 n = 4 n = 3 n = 2 n = 1 Which group of electrons is in the first shell ? Q.12

114. 113 2, 8, 8, 2 2 8 8 2 Q.12

115. 114 n = 4 n = 2 n = 3 n = 1 Q.12

116. 115 Evidence of the Existence of Shells 4.2 Deduction of electronic structure from ionization enthalpies (p.91) 2, 8, 1

117. 116 n = 2 n = 3 n = 1 Na

118. 117 Variation of IE 1 with Atomic Number Evidence for Subshell

119. 118 Only patterns across periods are discussed Refer to pp.27-29 for further discussion

120. 119 1.A general  in IE 1 with atomic number across Periods 2 and 3. 13.(a)

121. 120 2 – 3 - 3 13.(a)

122. 121 2.IE 1 value : Group 2 > Group 3 3. IE 1 value : Group 5 > Group 6 13.(a)

123. 122 2.Peaks appear at Groups 2 & 5 3. Troughs appear at Groups 3 & 6 13.(a)

124. 123 2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8 2,8,1 2,8,2 2,8,3 2,8,4 2,8,5 2,8,6 2,8,7 2,8,8 On moving across a period from left to right, 1. the nuclear charge of the atoms  (from +3 to +10 or +11 to +18) 2. electrons are being removed from the same shell  e  s removed experience stronger attraction from the nucleus. 13.(b)

125. 124 IE 1 : B(2,3) < Be(2,2) The electron removed from B occupies a subshell of higher energy within the n = 2 quantum shell 13.(b)

126. 125 2s 2p 1s n =  IE 2s IE 2p IE 2s (Be) > IE 2p (B) IE 1 : Be > B 13.(b)

127. 126 IE 1 : Al(2,8,3) < Mg(2,8,2) The electron removed from Al occupies a subshell of higher energy within the n = 3 quantum shell 13.(b)

128. 127 3s 3p 2p n =  IE 3s IE 3p IE 3s (Mg) > IE 3p (Al) IE 1 : Mg > Al 13.(b)

129. 128 3s 3p 2p n =  Mg(12)Al(13) IE 3s IE 3p

130. 129 It is more difficult to remove an electron from a half-filled p subshell 13.(b) IE 1 : N(2,5) > O(2,6) ; P(2,8,5) > S(2,8,6)

131. 130 2s 2p n =  IE N O(8) N(7) IE O > First IE

132. 131 2s 2p n =  IE N O(8) N(7) IE O > First IE The three electrons in the half-filled 2p subshell occupy three different orbitals (2p x, 2p y, 2p z ).  repulsion between electrons is minimized.

133. 132 3s 3p n =  IE P S(16) P(15) IE S > First IE The three electrons in the half-filled 3p subshell occupy three different orbitals (3p x, 3p y, 3p z ).  repulsion between electrons is minimized.

134. 133 The removal of an electron from O or S results in a half-filled p subshell with extra stability.(p.28, part 3) Misleading !!!

135. 134 O(2,6)  O + (2,5) 2p Is the 2p energy level of O + lower or higher than that of O ? Electrons in O + experience a stronger attractive force from the nucleus. Not because of the half-filled 2p subshell

136. 135 2p As a whole, O + is always less stable than O. It is because O(g) has one more electron than O + (g) and this extra electron has a negative potential energy. O(2,6)  O + (2,5)

137. 136 Conclusions : - (a)Each electron in an atom is described by a set of four quantum numbers. (b)No two electrons in the same atom can have the same set of quantum numbers. The quantum numbers can be obtained by solving the Schrodinger equation (p.16).

138. 137 Quantum Numbers 1.Principal quantum number (n) n = 1, 2, 3,… related to the size and energy of the principal quantum shell. E.g. n = 1 shell has the smallest size and electrons in it possess the lowest energy

139. 138 Quantum Numbers = 0, 1, 2,…,n-12.Subsidiary quantum number ( ) related to the shape of the subshells.

140. 139 Quantum Numbers 2.Subsidiary quantum number ( ) = 0, 1, 2,…,n-1 = 0  spherical s subshell= 1  dumb-bell p subshell = 2  d subshell = 3  f subshell complicated shapes

141. 140 Each principal quantum shell can have one or more subshells depending on the value of n. If n = 1, possible range of : name of subshell : 1s 0 to (1-1)  0 No. of subshell : 1

142. 141 Each principal quantum shell can have one or more subshells depending on the value of n. 0 to (2-1)  0, 1 If n = 2, possible range of : names of subshells : 2s, 2p No. of subshells : 2

143. 142 Each principal quantum shell can have one or more subshells depending on the value of n. names of subshells : 3s, 3p, 3d 0 to (3-1)  0, 1, 2 If n = 3, possible range of : No. of subshells : 3

144. 143 Each principal quantum shell can have one or more subshells depending on the value of n. names of subshells : 4s, 4p, 4d, 4f 0 to (4-1)  0, 1, 2, 3 If n = 4, possible range of : No. of subshells : 4

145. 144 Quantum Numbers 3.Magnetic quantum number (m) related to the spatial orientation of the orbitals in a magnetic field. m = ,…0,…+

146. 145 Each subshell consists of one or more orbitals depending on the value of Possible range of m :  0 to +0  0 No. of orbital : 1 Name of orbital : s If = 0 (s subshell),

147. 146 Possible range of m :  1 to +1   1,0,+1 No. of orbitals : 3 Names of orbitals : p x, p y, p z Each subshell consists of one or more orbitals depending on the value of If = 1 (p subshell),

148. 147 Possible range of m :  2 to +2   2,  1, 0, +1, +2 No. of orbitals : 5 Names of orbitals : Each subshell consists of one or more orbitals depending on the value of If = 2 (d subshell),

149. 148 Possible range of m :  3 to +3   3,  2,  1, 0, +1, +2, +3 No. of orbitals : 7 Names of orbitals : Not required in AL Each subshell consists of one or more orbitals depending on the value of If = 3 (f subshell),

150. 149 Total no. of orbitals in a subshell = 2 + 1 Energy of subshells : - s < p < d < f

151. 150 4.Spin quantum number (m s ) or m s = They describe the spin property of the electron, either clockwise, or anti-clockwise

152. 151 4.Spin quantum number (m s ) Each orbital can accommodate a maximum of two electrons with opposite spins

153. 152 Q.14(a) 32n = 4 18n = 3 8 1+3+5+7=16 1+3+5=9 1+3=4 4s, 4p, 4d, 4f 3s, 3p, 3d 2s, 2pn = 2 21(1s)1sn = 1 Total no. of electrons Total no. of orbitals Subshells Principal quantum shell

154. 153 14(b) The total number of electrons in a principal quantum shell = 2n 2

155. 154 Q.15(a) The two electrons of helium are in the 1s orbital of the 1s subshell of the first principal quantum shell. 1 st electron : n = 1, l = 0, m = 0, m s = 2 nd electron : n = 1, l = 0, m = 0, m s =

156. 155 Q.15(b) There are only 2 electrons in the 3 rd quantum shell. In the ground state, these two electrons should occupy the 3s subshell since electrons in it have the lowest energy.

157. 156 Q.15(b) The two outermost electrons of magnesium are in the 3s orbital of the 3s subshell of the third principal quantum shell. 1 st electron : n = 3, l = 0, m = 0, m s = 2 nd electron : n = 3, l = 0, m = 0, m s =

158. 157 4.3 The Wave- mechanical Model of the Atom

159. 158 Models of the Atoms 1.Plum-pudding model by J.J. Thomson (1899) 2.Planetary(orbit)model by Niels Bohr (1913) 3.Orbital model by E. Schrodinger (1920s)

160. 159 Electrons display both particle nature and wave nature. Particle nature : mass, momemtum,… Wave nature : frequency, wavelength, diffraction,…

161. 160 Wave as particles : photons By Planck (1) By Einstein (2) (3) Wave property Particle property : momentum of a photon

162. 161 Evidence : photoelectric effect by Albert Einstein (1905) Nobel Prize Laureate in Physics, 1921

163. 162 Particles as waves L. De Broglie (1924) Nobel Prize Laureate in Physics, 1929 (3)(4)

164. 163 Any particle (not only photon) in motion (with a momentum, mv) is associated with a wavelength (3) (4)

165. 164 Q.16 Electron

166. 165 Q.16 Helium atom

167. 166 Q.16 100m world record holder

168. 167 Wave nature of electrons (1927) 4.3 The Wave-mechanical model of the atom (p.95) Evidence of electron  inter-atomic spacing in metallic crystals (10  10 m)

169. 168 For very massive ‘particles’, The wavelength associated with them (10  37 m) are much smaller than the dimensions of any physical system. Wave properties cannot be observed.

170. 169 L Standing waves only have certain allowable modes of vibration Similarly, electrons as waves only have certain allowable energies. = 2L = L Standing waves in a cavity

171. 170 The uncertainty principle Heinsenberg Nobel Prize Laureate in Physics, 1932

172. 171 The uncertainty principle It is impossible to simultaneously determine the exact position and the exact momentum of an electron. uncertainty in position measurement uncertainty in momentum measurement

173. 172 high energy photon Small and light electron The momentum of electron would change greatly after collision The uncertainty principle

174. 173 low energy photon Small and light electron The position of electron cannot be located accurately The uncertainty principle

175. 174 high energy photon No problem in macroscopic world ! The uncertainty principle

176. 175 Implications : - The concept of well-defined orbits in Bohr’s model has to be abandoned. We can only consider the probability of finding an electron of a ‘certain’ energy and momentum within a given space.

177. 176 Schrodinger Equation Nobel Prize Laureate in Physics, 1933 de Broglie : electrons as waves  Use wave functions (  ) to describe electrons Pronounced as psi

178. 177 Schrodinger Equation Nobel Prize Laureate in Physics, 1933 Heisenberg : Uncertainty principle  Probability (  2 ) of finding electron at a certain position < 1.

179. 178  : wave function m : mass of electron h : Planck’s constant E : Total energy of electron V : Potential energy of electron Schrodinger Equation

180. 179 Schrodinger Equation The equation can only be solved for certain  i and E i

181. 180 Schrodinger Equation The wave function of an 1s electron is Z : nuclear charge (Z = 1 for Hydrogen) a 0 : Bohr radius = 0.529Å (1Å = 10  10 m) r : distance of electron from the nucleus Radius of 1s orbit of Bohr’s model

182. 181 Schrodinger Equation The allowed energies of H atom are given by n = 1, 2, 3,… n is the principal quantum number, All other terms in the expression are constants

183. 182 n = 1, 2, 3,… refer to p.6 of notes

184. 183 E = E 2 – E 1 E2E2 E1E1 n = 2 n = 1

185. 184  2 is the of finding an electron at a particular point in space. (electron density)  2 is the probability of finding an electron at a particular point in space. (electron density) 4.4 Atomic orbitals (p.98)  2 (1s) Probability never becomes zero  There is no limit to the size of an atom

186. 185 contour diagram relative probability of finding the electron at the nucleus  as r 

187. 186 In practice, a boundary surface is chosen such that within which there is a high probability (e.g. 90%) of finding the electron.

188. 187 The electron spends 90% of time within the boundary surface

189. 188 A 3-dimensional time exposure diagram. The density of the dots represents the probability of finding the electron at that position.

190. 189 The 3-dimensional region within which there is a high probability of finding an electron in an atom is called an atomic orbital. Each atomic orbital is represented by a specific wave function(  ). The wave function of a specific atomic orbital describes the behaviour of the electron in the orbital.

191. 190 Total probability of finding the electron within the ‘shell’ of thickness dr =  2 4  r 2 dr electron density within the ‘shell’ total volume of the ‘shell’  2 is the probability of finding the electron per unit volume dr is infinitesimally small

192. 191  2  as r , 4  r 2  as r   a maximum at 0.529 Å Orbital Model

193. 192 0.529 Å r (Å) 1 probability Bohr’s Orbit Model

194. 193 Total probability of finding the electron within the ‘shell’ of thickness dr =  2 4  r 2 dr The sum of the probabilities of finding the electron within all ‘shells’ = 1

195. 194 The total area bounded by the curve and the x-axis = 1 Check Point 4-3 Check Point 4-3

196. 195 s Orbitals 4.4 Atomic orbitals (p.98)

197. 196 nodal surface There is no chance of finding the electron on the nodal surface.

198. 197

199. 198 Probabilities of finding the electron at A or B > 0 A B Probability of finding the electron at C = 0 C How can the electron move between A & B ?

200. 199  can be considered as the amplitude of the wave.  2 is always  0

201. 200  2 = 0

202. 201 p Orbitals Two lobes along an axis 4.4 Atomic orbitals (p.100)

203. 202 For each 2p orbital, there is a nodal plane on which the probability of finding the electron is zero. 4.4 Atomic orbitals (p.100) yz planexz plane xy plane

204. 203 4.4 Atomic orbitals (p.101) d Orbitals Four lobes along two axes Two lobes & one belt Four lobes between two axes

205. 204 Grand Orbital Table http://www.orbitals.com/orb/orbtable.htm#table1

206. 205 The END

207. 206 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

208. 207 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 electronic transition from n =  to n = 2. Answer

209. 208 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

210. 209 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.

211. 210 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

212. 211 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.

213. 212 (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)

214. 213 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.

215. 214 (b)Give a sketch of the logarithm of successive ionization enthalpies of potassium against no. of electrons removed. Explain your sketch. Answer 4.2 Deduction of electronic structure from ionization enthalpies (p.94)

216. 215 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.

217. 216 (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

218. 217 (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

219. 218 (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

220. 219 (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 a region of zero probability of finding the electron known as a nodal surface. (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