2. Students who haven’t passed the safety quiz 708411 871401

4. Understanding “periodicity” Trends within a period what is changing? what is constant? Trends within a group what is changing? what is constant? 11 Na 12 Mg 11 Na 19 K

5. First Ionization Energy (first ionization potential) The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion

6. Periodicity of First Ionization Energy (IE 1 ) Like Figure 8-18

7. Trends Going down a group, first ionization energy decreases. This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

8. Fig. 8.15

9. Generally, ionization energy increases with atomic number. Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.

10. Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. A dramatic increase occurs when the first electron from the noble-gas core is removed.

11. Trends Going down a group, first ionization energy decreases. This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

12. Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA. Examining the valence configurations for these groups helps us to understand these deviations: IIAns 2 IIIA ns 2 np 1 VA ns 2 np 3 VIA ns 2 np 4 It takes less energy to remove the np 1 electron than the ns 2 electron. It takes less energy to remove the np 4 electron than the np 3 electron.

13. Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb. Sb is larger than As. As is larger than Br. Ionization energies: Sb < As < Br 35 Br 33 As 51 Sb

14. Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasing IE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution: a. Rn>Ar>Ne b. Bi<Po<At c.Na<Mg<Be d.K<Cl<Ar

15. Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. A dramatic increase occurs when the first electron from the noble-gas core is removed.

16. Fig. 8.16 What are the units here????

17. Left of the line, valence shell electrons are being removed. Right of the line, noble-gas core electrons are being removed.

18. Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE 1 IE 2 IE 3 IE 4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution:

19. Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE 1 IE 2 IE 3 IE 4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution: The largest jump in IE occurs after IE 3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is : 1s 2 2s 2 2p 6 3s 2 3p 1

20. Electron affinity (E.A.) The energy change for the process of adding an electron to a neutral atom in the gaseous state to form a negative ion A negative energy change (exothermic) indicates a stable anion is formed. The larger the negative number, the more stable the anion. Small negative energies indicate a less stable anion. A positive energy change (endothermic) indicates the anion is unstable.

21. The Electron affinity of a molecule or atom is the energy change when an electron is added to the neutral atom to form a negative ion.moleculeatom

23. Overall periodic trends Note: Electronegativity has similar trend as electron affinity

24. 23 Reactivity of the Alkali Metals Potassium video Sodium video Lithium video 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2 (g) 2K (s) + 2H 2 O (l)  2KOH (aq) + H 2 (g) 2Li (s) + 2H 2 O (l)  2LiOH (aq) + H 2 (g) Trend?

25. 24 More Sodium Reaction Videos Prepping Na 150 g Na in small pieces 2Na (s) + 2H 2 O (l) → 2NaOH (aq) + H 2 (g) http://www.theodoregray.com/PeriodicTable/ 100 g Na in one piece

26. Electronic Configuration Ions Na 1s 2 2s 2 2p 6 3s 1 Na + 1s 2 2s 2 2p 6 Mg 1s 2 2s 2 2p 6 3s 2 Mg +2 1s 2 2s 2 2p 6 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al +3 1s 2 2s 2 2p 6 O 1s 2 2s 2 2p 4 O - 2 1s 2 2s 2 2p 6 F 1s 2 2s 2 2p 5 F - 1 1s 2 2s 2 2p 6 N 1s 2 2s 2 2p 3 N - 3 1s 2 2s 2 2p 6

28. Isoelectronic Atoms and Ions H - 1 { He } Li + Be +2 N - 3 O - 2 F - { Ne } Na + Mg +2 Al +3 P - 3 S - 2 Cl - { Ar } K + Ca +2 Sc +3 Ti +4 As - 3 Se - 2 Br - { Kr } Rb + Sr +2 Y +3 Zr +4 Sb - 3 Te - 2 I - { Xe } Cs + Ba +2 La +3 Hf +4

29. Trends when atoms form chemical bonds Empirical Observation “when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas” Metals tend to lose e - ’s Nonmetals tend to gain e - ’s

30. Are ions bigger or smaller than atoms? Representative cation Na → Na + + e  Representative anion F + e  → F 

32. Cations are always smaller than parent atom - decreased e - repulsion (clouds contract) - if emptying valence shell, “n” decreases Anions are always larger than parent atoms - increased e - repulsion (clouds expand) C ations C ontract A nions A dd Trends in ion size

33. Trends in atom & ion size

34. Trends in ion size

35. Ranking Ions According to Size Problem: Rank each set of Ions in order of increasing size. a) K +, Rb +, Na + b) Na +, O 2-, F - c) Fe +2, Fe +3 Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period but increases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series. iv) cations of the same element decreases in size as the charge increases. Solution: a) since K +, Rb +, and Na + are from the same group (1A), they increase in size down the group: Na + < K + < Rb + b) the ions Na +, O 2-, and F - are isoelectronic. O 2- has lower Z eff than F -, so it is larger. Na + is a cation, and has the highest Z eff, so it is smaller: Na + < F - < O 2- c) Fe +2 has a lower charge than Fe +3, so it is larger: Fe +3 < Fe +2

36. Chapter #9 - Models of Chemical Bonding 9.1) Atomic Properties and Chemical Bonds 9.2) The Ionic Bonding Model 9.3) The Covalent Bonding Model 9.4) Between the Extremes: Electronegativity and Bond Polarity 9.5) An Introduction to Metallic Bonding

38. Sodium Chloride

39. Depicting Ion Formation with Orbital Diagrams and Electron Dot Symbols - I Problem: Use orbital diagrams and Lewis structures to show the formation of magnesium and chloride ions from the atoms, and determine the formula of the compound. Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we need two Cl atoms for every Mg atom. Solution: 2 Cl Mg +2 + 2 Cl - Mg + Mg +.. Cl..... Mg +2 + 2 Cl....

40. Depicting Ion Formation from Orbital Diagrams and Electron Dot Symbols - II Problem: Use Lewis structures and orbital diagrams to show the formation of potassium and sulfide ions from the atoms, and determine the formula of the compound. Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals, sulfur must gain two electrons, and potassium must lose one electron. Solution: 2 K S 2 K + + S - 2 + K K.. +S.... 2 K + + S.. 2 -

41. Three Ways of Showing the Formation of Li + and F - through Electron Transfer

42. Lewis Electron-Dot Symbols for Elements in Periods 2 & 3

43. The Reaction between Na and Br to Form NaBr The Elements The Reaction!

44. Melting and Boiling Points of Some Ionic Compounds Compound mp( o C) bp( o C) CsBr 636 1300 NaI 661 1304 MgCl 2 714 1412 KBr 734 1435 CaCl 2 782 >1600 NaCl 801 1413 LiF 845 1676 KF 858 1505 MgO 2852 3600

46. Figure 9.11: Potential-energy curve for H 2.

47. Covalent Bonding in Hydrogen, H 2

49. Figure 9.10: The electron probability distribution for the H 2 molecule.

50. Covalent bonds animation http://www.chem1.com/acad/webtext/chembond/cb03.html This is a good overview. http://www.chem.ox.ac.uk/vrchemistry/electronsandbonds/intro1.htm

51. For elements larger than Boron, atoms usually react to develop octets by sharing electrons. H, Li and Be strive to “look” like He. B is an exception to the noble gas paradigm. It’s happy surrounded by 6 electrons so the compound BH 3 is stable. Try drawing a Lewis structure for methane.

52. Draw Lewis dot structures for the halogens. Try oxygen and nitrogen. Notice that these all follow the octet rule! These also follow the octet rule!

53. Bond Lengths and Covalent Radius

54. Figure 9.14: The HCl molecule.

55. Figure 9.12: Molecular model of nitro- glycerin. What is the formula for this compound?

56. Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He)

57. Make some Lewis Dot Structures with other elements: SiH 4 NH 3 H2OH2O C2H6C2H6 C2H6OC2H6O CH 2 O

58. Figure 9.9: Model of CHI 3 Courtesy of Frank Cox.

59. Make some Lewis Dot Structures with other elements: CH 4 NH 3 H2OH2O C2H6C2H6 C2H6OC2H6O CH 2 O

60. Look at all these structures and make some bonding rules:

62. Rules for drawing Lewis structures 1. Count up all the valence electrons 2. Arrange the atoms in a skeleton 3. Have all atoms develop octets (except those around He) 4. Satisfy bonding preferences!

63. A model of ethylene.

64. A model of acetylene.

65. A model of COCl 2.

68. The Relation of Bond Order, Bond Length and Bond Energy Bond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol) C O 1 143 358 C O 2 123 745 C O 3 113 1070 C C 1 154 347 C C 2 134 614 C C 3 121 839 N N 1 146 160 N N 2 122 418 N N 3 110 945 Table 9.4

69. Conceptual Problem 9.103

70. Fig. 9.14

71. Figure 9.15: Electronegatives of the elements.

72. The Periodic Table of the Elements 2.1 0.91.5 0.91.2 0.81.01.3 0.8 0.7 1.0 0.9 1.51.6 1.51.8 1.2 1.1 1.8 1.91.6 1.41.6 1.5 1.8 1.7 1.9 2.2 1.9 2.4 1.7 1.9 2.02.53.03.54.0 He Ne Ar1.51.82.12.53.0 1.61.82.02.42.8Kr Xe Rn 2.52.1 2.2 1.9 2.01.9 1.81.7 1.8 1.1 1.3 1.2 1.3 1.51.71.3 1.5 0.9 1.32.2 Electronegativity 1.1 Th Pa U Np No Lr 1.3 Ce Pr Nd PmYb Lu

73. Fig. 9.16

74. Fig. 9.17

75. Determining Bond Polarity from Electronegativity Values Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity. Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values. Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5

76. Fig. 9.18

77. Percent Ionic Character as a Function of Electronegativity Difference (  En) Fig. 9.19

78. Lewis Structures of Simple Molecules C H HH H Cl O OO K+K+ KClO 3 CF 4.. HCOH H H H H C Ethyl Alcohol (Ethanol) Potassium ChlorateCarbon Tetrafluoride.. C F FF F CH 4 Methane

79. Resonance: Delocalized Electron-Pair Bonding - I Ozone : O 3... OO O OO O I II O O O.. Resonance Hybrid Structure One pair of electron’s resonances between the two locations!!

80. Resonance: Delocalized Electron-Pair Bonding - II C C C CC C C C C C C C C C C CC C H H H H HH H H H H H H H HH H H H Resonance Structure Benzene

81. Lewis Structures of Simple Molecules Resonance Structures -IIINitrate N O OO N O O O.. N O OO

82. Lewis Structures for Octet Rule Exceptions Cl F F F.. B Cl.. Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons! Each chlorine atom has 8 electrons associated. Boron has only 6! Cl Be.. Each chlorine atom has 8 electrons associated. The beryllium has only 4 electrons. N O O... NO 2 is an odd electron atom. The nitrogen has 7 electrons.

83. Resonance Structures - Expanded Valence Shells.. S F F F FF F Sulfur hexafluoride.. P F F F FF Phosphorous pentafluoride O S O OOHH.. O S O OOHH Sulfuric acid S = 12e - p = 10e - S = 12e - Resonance Structures

84. Lewis Structures of Simple Molecules Resonance Structures-V SOO O O SOO O O. -2. -2 Sulfate S O O OO x x x = Sulfur electronso = Oxygen electrons oo o x o x x o o o * -2 o Plus 4 others for a total of 6.

85. VSEPR: Valence Shell Electron Pair Repulsion: A way to predict the shapes of molecules Pairs of valence electrons want to get as far away from each other as possible in 3-dimensional space.

86. Balloon Analogy for the Mutual Repulsion of Electron Groups TwoThree Four Five Six Number of Electron Groups

89. AX 2 Geometry - Linear Cl Be.. 180 0 BeCl 2 Gaseous beryllium chloride is an example of a molecule in which the central atom - Be does not have an octet of electrons, and is electron deficient. Other alkaline earth elements also have the same valence electron configuration, and the same geometry for molecules of this type. Therefore this geometry is common to group II elements. Molecular Geometry = Linear Arrangement COO.. 180 0 Carbon dioxide also has the same geometry, and is a linear molecule, but in this case, the bonds between the carbon and oxygens are double bonds. CO 2

90. The Two Molecular Shapes of the Trigonal Planar Electron-Group Arrangement

91. AX 3 Geometry - Trigonal Planar BF 3 B F FF.. N O OO 120 0 NO 3 - Boron Trifluoride Nitrate Anion All of the boron Family(IIIA) elements have the same geometry. Trigonal Planar ! AX 2 E SO 2.. S O O The AX 2 E molecules have a pair of Electrons where the third atom would appear in the space around the central atom, in the trigonal planar geometry. -

92. The Three Molecular Shapes of the Tetrahedral Electron-Group Arrangement

93. AX 4 Geometry - Tetrahedral C H HH H CH 4 Methane C H H H H 109.5 0 All molecules or ions with four electron groups around a central atom adopt the tetrahedral arrangement H H N H H H HH+H+ + 109.5 0 N H.. 107.3 0 all angles are the same! Ammonia is in a tetrahedral shape, but it has only an electron pair in one location, so the smaller angle! Ammonium Ion

94. The Four Molecular Shapes of the Trigonal Bipyramidal Electron- Group Arrangement

95. AX 5 Geometry - Trigonal Bipyramidal Br F F F.. 86.2 0 AX 3 E 2 - BrF3 I I I.. 180 0 AX 2 E 3 - I 3 - P Cl.......... AX 5 - PCl 5

96. The Three Molecular Shapes of the Octahedral Electron-Group Arrangement

97. AX 6 Geometry - Octahedral S F F F F F F.. AX 6 Sulfur Hexafluoride I F FF F F.. AX 5 E Iodine Pentafluoride Xe F FF F.. Xenon TetrafluorideSquare planar shape

98. Using VSEPR Theory to Determine Molecular Shape 1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups. 2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding. 3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds. 4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.

100. Hybrid Orbital Model

102. The sp Hybrid Orbitals in Gaseous BeCl 2

106. The sp 3 Hybrid Orbitals in NH 3 and H 2 O

107. The sp 3 d Hybrid Orbitals in PCl 5

108. The sp 3 d 2 Hybrid Orbitals in SF 6 Sulfur Hexafluoride -- SF 6

110. Figure 10.26: Sigma and pi bonds.

111. Figure 10.27: Bonding in ethylene.

113. Figure 10.28: Bonding in acetylene.

115. Restricted Rotation of  -Bonded Molecules A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

116. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.

117. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized. Solution: a) For CH 3 NH 2 : The shape is tetrahedral around the C and N atoms. Therefore, each central atom is sp 3 hybridized. The carbon atom has four half-filled sp 3 orbitals: Isolated Carbon Atom 2s2psp 3 Hybridized Carbon Atom

118. The N atom has three half-filled sp 3 orbitals and one filled with a lone pair. 2s 2p2p sp 3 C H H H H H N..

119. b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. 5 s5 p 5 d Hybridized Xe atom: 5 d Isolated Xe atom sp 3 d 2

120. b) continued:For XeF 4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 90 0 to each other, whereas if the two polar positions are chosen, the two electron groups will be 180 0 from each other. Thereby minimizing the repulsion between the two electron groups. Xe F F F F.. Xe F FF F Square planar 180 0