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1. CHM 151 General Chemistry EXAM 4 SLIDES

3. increase in energy as n increases

5. When an electron falls to a lower energy level, energy is emitted.

8. are found within each energy level.

9. are designated by the letters s, p, d, and f The number of sublevels is equal to the value of the principal quantum number (n).

10. Number of Sublevels

13. has a shape that represents electron density (not a path the electron follows)

14. can hold up to 2 electrons

16. increases in size around the nucleus as the energy level n value increases

18. is one of three p orbitals that make up each p sublevel

20. A p sublevel contains three p orbitals.

21. A d sublevel contains five d orbitals.

23. An orbital diagram represents each orbital with a box, with orbitals in the same subshell in connected boxes; electrons are shown as arrows in the boxes, pointing up or down to indicate their spins. Two electrons in the same orbital must have opposite spins. Orbital Diagrams

24. An electron configuration lists the occupied subshells using the usual notation (1s, 2p, etc.). Each subshell is followed by a superscripted number giving the number of electrons present in that subshell. Two electrons in the 2ssubshell would be 2s2 (spoken as “two-ess-two”). Four electrons in the 3psubshell would be 3p4 (“three-pea-four”). Electron Configuration

25. Electron Configurations of Elements Hydrogen contains one electron in the 1s subshell. 1s1 Helium has two electrons in the 1s subshell. 1s2

26. Electron Configurations of Elements Lithium has three electrons. 1s2 2s1 Beryllium has four electrons. 1s2 2s2 Boron has five electrons. 1s2 2s2 2p1

27. Carbon, with six electrons, has the electron configuration of 1s2 2s2 2p2. The lowest energy arrangement of electrons in degenerate (same-energy) orbitals is given by Hund’s rule: one electron occupies each degenerate orbital with the same spin before a second electron is placed in an orbital. Orbital Diagram of Carbon

28. Other Elements in the Second Period N 1s2 2s2 2p3 O 1s2 2s2 2p4 F 1s2 2s2 2p5 Ne 1s2 2s2 2p6

29. Because their electron configurations can get long, larger atoms can use an abbreviated electron configuration, using a noble gas to represent core electrons. Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6-> [Ar] 4s2 3d6 Ar Electron Configurations of Heavier Atoms

30. Chapter 7 Visual Summary

31. Chapter 8The Periodic Table: Structure and Trends

32. The 4s orbital is lower in energy than the 3d orbital and fills first, starting the fourth period at potassium. The 3dorbitals fill after the 4s. Similar inversions occur in the remaining periods. Electron Configurations

33. Electron Configurations of Anions For anions, the additional electrons fill orbitals following the same rules that applies to atoms. Cl: [Ne] 3s2 3p5Cl-: [Ne] 3s2 3p6 As: [Ar] 4s2 3d10 4p3 As3-: [Ar] 4s2 3d10 4p6 Many stable anions have the same electron configuration as a noble gas atom.

34. Test Your Skill Write the electron configurations of the following ions: (a) N3- (b) Co3+ (c) K+

35. Test Your Skill Write the electron configurations of the following ions: (a) N3- (b) Co3+ (c) K+ Answers: (a) 1s2 2s2 2p6 (b) [Ar] 4s23d4 (c) [Ar]

36. Size Trends for an Isoelectronic Series

38. Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2- Test Your Skill

39. Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2- Answer: (a) Mg is larger. (b) Se2- is larger. Test Your Skill

40. Chapter 8 Visual Summary

41. Chapter 9Chemical Bonds

42. Chemical Bonds Chemical bonds are the forces that hold the atoms together in substances. This chapter discusses two limiting types of bonding. Ionic bonding Covalent bonding

43. Lewis Electron-dot Symbols A Lewis electron-dot symbol consists of the symbol for the element surrounded by dots, one for each valence electron.

44. Cations of most representative elements have no valence shell electrons shown in the Lewis symbol. Na×® Na+ + e- ×Ca×® Ca2+ + 2e- Lewis Symbols for Cations

45. Lewis Symbols for Anions The Lewis symbols of most monatomic anions show eight valence electrons. Cl + e--> Cl - Se + 2e- -> Se 2-

46. Ionic Bonding Ionic bonding results from the electrostatic attraction between cations and anions. Formation of an ionic bond can be viewed as a transfer of electrons. Na + F -> Na+ + F - (or NaF)

47. A covalent bond result from the sharing of two electrons between two atoms, as shown here for H2. Covalent Bonding

48. Two hydrogen atoms become more stable as their orbitals, each containing one electron, overlap. Orbital Overlap

49. Lewis Structures Bonding Pair H Cl Lone Pair Lewis structures represent covalent bonding by showing how the valence electrons are present in a molecule. Bonding pairsare shared between two atoms and are represented by lines . Lone pairs are entirely on one atom and are represented by two dots.

50. The number of Covalent Bonds The number of covalent bonds can be determined from the number of electrons needed to complete an octet.

52. Double Bond- sharing two pairs of electrons

54. Writing Lewis Structures Write the Lewis structure of fromaldehyde, H2CO. The skeleton structure is O C H H

55. Writing Lewis Structures O C H H The total number of valence electrons is 1(C) 1 x 4 = 4 1(O) 1 x 6 = 6 2(H) 2 x 1 = 2 12

56. Writing Lewis Structures needs 6e- to complete octet O C needs 2e- to complete octet H H O C H H Remaining valence electrons = 6 8 electrons needed to obey the octet rule Add one bond because 2 more electrons are needed than are available.

58. Test Your Skill Write the Lewis structure of N2H2. The skeleton structure is: H N N H

59. Test Your Skill Write the Lewis structure of N2H2. Answer: H N N H

60. Polar Bond Polar Bond – A covalent bond in which the two atoms do not share the bonding electrons equally. Note: The higher electronegative element takes on a partial negative charge and the lower electronegative element takes on a partial positive charge. d+ d- I—Br (a polar covalent bond)  arrow indicates direction of more electronegative element

61. In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not. Dipole moment is a measure of the unequal sharing of electrons. The unequal sharing leads to a polar covalent bond that is indicated with the symbol d followed by a sign to show partial charges. d+ d- Cl-F Bond Polarity

62. Electronegativityis a measure of the ability of an atom to attract the shared electrons in a chemical bond. Electronegativity

63. Electronegativity Trends

64. Electronegativity Trends

65. Example: Electronegativity Select the most polar bond. Cl-F O-F P-F

66. Properties of Compounds

68. Equation for Formal Charge:(number of valence electrons in atom) – (number of lone pair electrons) – ½ (number of shared electrons) Formal Charges

69. Formal Charges # Bonds # Bonds # Bonds -1 0 +1 Atom _ _ : _ _ _ _ _ _ N 2 3 4 N N N : : _ _ : : _ _ _ _ _ 3 O 2 1 O O O : : : : _ _ _ _ _ _ 3 4 C C C : _

70. Formal Charges Add formal charges to the Lewis structure of HNO3 shown below. H O N O O

71. Formal Charges H O N O O The nitrogen atom has 4 bonds giving it a +1 formal charge. The oxygen atom on the bottom left only has 1 bond giving it a -1 formal charge. The sum of formal charges equals the charge of the species.

72. Test Your Skill Add formal charges to the Lewis structure of HNO3 shown below. H O N O O

73. Test Your Skill H O N O O Answer:

74. Lewis structures that show the smallest formal charges are favored. Lewis structures that have adjacent atoms with formal charges of the same sign are much less favorable. Lewis structures that place negative formal charges on the more electronegative atoms are favored. Formal charges of opposite sign are usually on adjacent atoms. Fewer formal charges overall. Structure Stability (which structure is more favored)

75. Test Your Skill Of the two structures shown for HNO3, use the stability rules to predict which will be more favored. H H O O N O N O O O

76. Test Your Skill H O H O N O N O O O Answer: The structure on the left is favored because it has fewer formal charges. it does not have adjacent atoms with the same formal charge.

77. Resonance in Lewis Structures Resonance structures differ only in the distribution of the valence electrons. All resonance structures follow the rules for writing Lewis structures. Resonance structures are indicated by a double headed arrow. H O H O N O N O O ↔ O

78. Drawing Resonance Structures H O H O N O N O O ↔ O Draw the third possible resonance structure for HNO3, the first two are below.

81. Test Your Skill Write all resonance structures, including formal charges, for O3, O-O-O skeleton structure.

82. Test Your Skill Write all resonance structures, including formal charges, for O3, O-O-O skeleton structure. Answer: O O O ↔O O O

83. No resonance structure is correct by itself; the correct structure is an average of all resonance structures. Average Structure

84. Equivalent resonance structures, such as the two for O3, contributeequally to the average structure. Bond order in O3 is the average of a double bond and a single bond = 1.5. O O O ↔ O O O Contribution of Resonance Structures

85. Test Your Skill Draw the Lewis structure of IF3. Answer: F I F F

86. Chapter 9 Visual Summary

87. Chapter 10 Molecular Structure and Bonding Theories

88. Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures. VSEPR Rule 1: A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs. Minimum repulsion results when the electron pairs are as far apart as possible. VSEPR

89. Steric number = (number of lone pairs on central atom) + (number of atoms bonded to central atom) The steric number is determined from the Lewis structure. Stericnumber determines the bonded-atom lone-pair arrangement, the shape that maximizes the distances between the valence-shell electron pairs. Steric Number

90. Geometric Arrangements

91. Geometric Arrangements

92. In the Lewis structure of BeCl2, beryllium has two bonded atoms and no lone pairs, stericnumber = 2. Alinear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible. Steric Number = 2

93. The Lewis structure of HCN (H-Cº N:) shows that the carbon atom is bonded to two atoms and has no lone pairs, steric number = 2. The bonded-atom lone-pair arrangement is linear. The number of bonded atoms, not the number of bonds, determines the steric number. Molecules with Multiple Bonds

94. The Lewis structure of BF3 shows the boron atom has a steric number = 3; the bonded-atom lone-pair arrangement is trigonal planar. Steric Number = 3

95. The Lewis structure of CH4 shows the carbon atom has a steric number = 4; the bonded-atom lone-pair arrangement is tetrahedral. Steric Number = 4

96. The phosphorus atom in PF5 has a steric number = 5; the bonded-atom lone-pair arrangement is trigonal bipyramidal. Steric Number = 5

97. The sulfur atom in SF6 has a steric number = 6; the bonded-atom lone-pair arrangement is octahedral. Steric Number = 6

99. Molecular shapeis the arrangement of the atoms in a species. The bonded-atom lone-pair arrangement of H2O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom). Molecular Shape of H2O

100. What is the electron pair geometry and molecular shape of NH3? Molecular Shape of NH3

101. Molecular Shape of NH3 N H H H First, draw the Lewis structure. The nitrogen has 3 bonded atoms and 1 lone pair; the steric number = 4 and the bonded-atom lone-pair arrangement is tetrahedral.

102. The bonded-atom lone-pair arrangement of NH3 is tetrahedral (top), molecular shape is a trigonal pyramidal (bottom). Molecular Shape of NH3

103. The measured bond angle in H2O (104.5o) is smaller than the predicted angle (109.5o) Explanation (VESPR Rule #2) Forces between electron pairs vary as: lone pair-lone pair replusion> lone pair-bonding pair replusion> bonding pair-bonding pair repulsion Electron Pair Repulsions

104. What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF3? Test Your Skill

105. Answer: The steric number is 5, the bonded-atom lone-pair arrangement is trigonal bipyramidal and the molecule is “T” shaped with the two lone pairs in equatorial positions. Test Your Skill

106. The geometry of each central atom is determined separately. The CH3 carbon in CH3CN has tetrahedral geometry and the other carbon has linear geometry. Multiple Central Atoms

108. Draw the Lewis structure.

110. In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not. Dipole moment is a measure of the unequal sharing of electrons. Equals the magnitude of the separated charges X the distance between them The unequal sharing leads to a polar covalent bond that is indicated with the symbol d followed by a sign to show partial charges. d+ d- Cl-F Bond Polarity

111. The bond dipoles in CO2 cancel because the linear shape orients the equal magnitude bond dipoles in exactly opposite directions. Bond dipole determined by difference in electronegativities of bonded atoms. Polarity of Molecules

112. The bond dipoles do not cancel in COSe; they are oriented in the same direction and are of unequal length. They do not cancel in OF2 because the V-shape of the molecule does not orient them in opposite directions. Polarity of Molecules

113. The bond dipoles in BCl3 and CCl4 cancel because of the regular shape and equal magnitude. Polarity of Molecules

114. The bond dipoles in BCl2F and CHCl3 do not cancel because they are not of the same magnitude. Polarity of Molecules

116. Hybrid orbitalsare: Orbitalsobtained by mixing two or more atomic orbitals on the same central atom. Hybrid Orbitals

117. The bonds in BeCl2 arise from the overlap of two sp hybrid orbitals on the beryllium atom with the 3p orbitals on the two chlorine atoms. Bonding in BeCl2

118. The bonds in BF3 arise from the overlap of three sp2 hybrid orbitals on the boron atom with 2p orbitals on the three fluorine atoms. Bonding in BF3

119. The bonds in CH4 arise from the overlap of four sp3 hybrid orbitals on the carbon atom with 1s orbitals on the four hydrogen atoms. Bonding in CH4

120. Hybrid orbitals can hold lone pairs as well as make bonds. Lone Pairs and Hybrid Orbitals

121. Hybrid Orbitals

123. Sigma bonds (s): the shared pair of electrons is symmetric about the line joining the two nuclei of the bonded atoms. Types of Bonds: Sigma Bonds

124. The C-C sigma bond in C2H4 arises from overlap of sp2 hybrid orbitals and the four C-H sigma bonds from overlap sp2 hybrid orbitals on C with 1s orbitals on H. The second C-C bond forms from sideways overlap of p orbitals. Bonding in C2H4

125. Pi bonds (p) places electron density above and below the line joining the bonded atoms – they form by sideways overlap of p orbitals. Types of Bonds: Pi Bonds

126. The double bond in C2H4 is one sigma bond and one pi bond – each bond is of similar strength. Bonding in C2H4

128. Chapter 10 Visual Summary

129. Chapter 11Liquids and Solids

130. Characteristic Properties of Gases, Liquids, and Solids Intermolecular forces are the attractions that hold molecules together in the liquid and solid states.

131. Physical State and Energy of Attraction

132. Boiling Point The boiling point of a liquid is the temperature at which the vapor pressure is equal to the external pressure. The normal boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to 1 atmosphere. At the boiling point, bubbles filled with vapor form below the surface of the liquid.

133. Electrostatic forces account for all types of intermolecular attractions. There are three types of attractions: Dipole-dipole attractions London dispersion forces Hydrogen bonding Intermolecular Attractions

134. Dipole-dipole attractions result from electronic forces between molecular dipoles: Dipole-Dipole Attractions

135. London dispersion forces arise from the attractions between instantaneous dipoles and induced dipoles. London Dispersion Forces

136. Dispersion Forces and Periodic Trends Polarizability is the ease with which a charge distorts the electron cloud in a molecule. Polarizability generally increases with the number of electrons in the molecule. For related series of molecules, London dispersion forces increase going down any group in the periodic table.

137. Hydrogen bonding occurs between a hydrogen atom bonded to N, O, or F, and a lone pair of electrons on a second N, O, or F. Hydrogen bonds are sometimes shown as dotted lines. Hydrogen Bonding

138. Hydrogen bonding causes ice to have a lower density than liquid water. Structure of Solid Water

139. Identify the kind of intermolecular forces: (a) BF3, BBr3 (b) C2H5OH, C2H5Cl Example: Intermolecular Forces

140. Answers London dispersion forces for both. C2H5OH is Hydrogen Bonding C2H5Cl is Dipole-Dipole

141. Capillary action causes water to rise in a small diameter glass tube. Capillary action is the result of a competition between: cohesion: the attraction of molecules for other molecules of the same substance. adhesion: the attraction of molecules for other molecules of a different substance. Liquids: Capillary Action

143. Mercury is lowered because cohesion is stronger than adhesion.Capillary Action

144. Viscosity is the resistance of a fluid to flow. The stronger the intermolecular forces of attraction, the greater the viscosity. Other factors contribute to viscosity as well, like structure, size, and shape of molecules. Liquids: Viscosity

145. Chapter 11 Visual Summary

146. Chapter 12Solutions

147. There are a number of ways to express concentration: Molarity mass percentage ppm and ppb molality Solution Concentration

148. All concentration units are fractions. The numerator contains the quantity of solute. The denominator is the quantity of either solution or solvent. They differ in the units used to express these two quantities. Solution Concentration

149. Units of Concentration Used Earlier

150. Mass Percent Composition

151. A solution is prepared by dissolving 3.00 g of NaCl (molar mass = 58.44 g/mol) in 150 g of water. Express its concentration as mass percent. Answer: 1.96 % Example: Percent Composition

152. Molality (m or molal) is defined as Molality

155. Units for Concentration Conversion

156. Express the concentration of a 3.00% H2O2 solution as molality. Answers: 0.910 molal Example: Concentration Conversion

157. Test Your Skill Calculate (a) the molality, (C2H5OH; molar mass = 46.07 g/mol) in a wine that has an alcohol concentration of 7.50 mass percent. Answers: 1.76 molal

158. Example: Conversion to Molarity Conversion of most concentration units to molarity usually involve using the density of the solution to convert units mass to units of volume. The density of a 12.0% sulfuric acid (H2SO4; molar mass = 98.08 g/mol) is 1.080 g/mL. What is the molarity of this solution? Answer = 1.32 M

159. Chapter 12 Visual Summary