Chm151 exam 4 slides sp10

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1. CHM 151 General Chemistry EXAM 4 SLIDES

2. Electron Energy Levels Electrons are arranged in specific energy levels that are labeled n = 1, n = 2, n = 3, and so on increase in energy as n increases have the electrons with the lowest energy in the first energy level ( n = 1) closest to the nucleus

3. Energy Level Changes An electron absorbs energy to “jump” to a higher energy level. When an electron falls to a lower energy level, energy is emitted. In the visible range, the emitted energy appears as a color.

4. In each of the following energy level changes, indicate if energy is 1) absorbed, 2) emitted, or 3) not changed. A. An electron moves from the first energy level ( n = 1) to the third energy level ( n = 3). B. An electron falls from the third energy level to the second energy level. C. An electron moves within the third energy level. Learning Check

5. Sublevels Sublevels contain electrons with the same energy are found within each energy level. 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 ) .

6. Number of Sublevels

7. Energy of Sublevels In any energy level the s sublevel has the lowest energy the s sublevel is followed by the p, d, and f sublevels (in order of increasing energy)

8. Orbitals An orbital is a three-dimensional space around a nucleus where an electron is most likely to be found has a shape that represents electron density ( not a path the electron follows) can hold up to 2 electrons contains two electrons that must spin in opposite

9. s Orbitals An s orbital has a spherical shape around the nucleus increases in size around the nucleus as the energy level n value increases is a single orbital found in each s sublevel

10. p Orbitals A p orbital has a two-lobed shape is one of three p orbitals that make up each p sublevel increases in size as the value of n increases

11. Sublevels and Orbitals Each sublevel consists of a specific number of orbitals . An s sublevel contains one s orbital. A p sublevel contains three p orbitals. A d sublevel contains five d orbitals. An f sublevel contains seven f orbitals.

12. Electrons in Each Sublevel

13. 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 ↑↓

14. An electron configuration lists the occupied subshells using the usual notation (1 s , 2 p , etc.). Each subshell is followed by a superscripted number giving the number of electrons present in that subshell . Two electrons in the 2 s subshell would be 2 s 2 (spoken as “two- ess -two”). Four electrons in the 3 p subshell would be 3 p 4 (“three-pea-four”). Electron Configuration

15. Hydrogen contains one electron in the 1 s subshell. 1 s 1 Helium has two electrons in the 1 s subshell. 1 s 2 Electron Configurations of Elements ↑ ↑↓

16. Electron Configurations of Elements Lithium has three electrons. 1 s 2 2 s 1 Beryllium has four electrons. 1 s 2 2 s 2 Boron has five electrons. 1 s 2 2 s 2 2 p 1 ↑↓ ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑

17. Orbital Diagram of Carbon Carbon , with six electrons, has the electron configuration of 1 s 2 2 s 2 2 p 2 . 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. ↑↓ ↑↓ ↑ ↑

18. Other Elements in the Second Period N 1 s 2 2 s 2 2 p 3 O 1 s 2 2 s 2 2 p 4 F 1 s 2 2 s 2 2 p 5 Ne 1 s 2 2 s 2 2 p 6 ↑↓ ↑↓ ↑ ↑ ↑ ↑↓ ↑↓ ↑↓ ↑ ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

19. Because their electron configurations can get long, larger atoms can use an abbreviated electron configuration , using a noble gas to represent core electrons. Fe: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 → [ Ar ] 4 s 2 3 d 6 Ar Electron Configurations of Heavier Atoms

20. Chapter 7 Visual Summary

21. Chapter 8 The Periodic Table: Structure and Trends

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

23. For anions, the additional electrons fill orbitals following the same rules that applies to atoms. Cl : [Ne] 3 s 2 3 p 5 Cl - : [Ne] 3 s 2 3 p 6 As: [ Ar ] 4 s 2 3 d 10 4 p 3 As 3- : [ Ar ] 4 s 2 3 d 10 4 p 6 Many stable anions have the same electron configuration as a noble gas atom. Electron Configurations of Anions

24. Test Your Skill Write the electron configurations of the following ions: (a) N 3- (b) Co 3+ (c) K +

25. Test Your Skill Write the electron configurations of the following ions: (a) N 3- (b) Co 3+ (c) K + Answers : (a) 1 s 2 2 s 2 2 p 6 (b) [ Ar ] 4s 2 3 d 4 (c) [ Ar ]

26. Size Trends for an Isoelectronic Series

27. Sizes of the Atoms and Their Cations Atoms are always larger than their cations.

28. Sizes of the Atoms and Their Cations If an atom makes more than one cation, the higher-charged ion has a smaller size .

29. Anions are always larger than their atoms. Atomic and Ionic Radii

30. Identify the larger species of each pair: (a) Mg or Mg 2+ (b) Se or Se 2- Test Your Skill

31. Identify the larger species of each pair: (a) Mg or Mg 2+ (b) Se or Se 2- Answer : (a) Mg is larger. (b) Se 2- is larger. Test Your Skill

32. Chapter 8 Visual Summary

33. Chapter 9 Chemical Bonds

34. 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

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

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

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

38. 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 )

39. A covalent bond result from the sharing of two electrons between two atoms, as shown here for H 2 . Covalent Bonding

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

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

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

43. Octet Rule Octet Rule : atoms share electrons until each atom is surrounded by eight. Single Bond - sharing one pair of electrons Double Bond - sharing two pairs of electrons Triple Bond - sharing three pairs of electrons

44. 1. Write the skeleton structure. 2 . Sum the valence electrons. 3 . Subtract two electrons for each bond in the skeleton structure. 4 . Count the number of electrons needed to satisfy octet rule for each atom. If the number of electrons needed equals the number remaining, go to 5. If fewer electron remain, add one bond for every two additional electrons needed. 5 . Place remaining electrons as lone pairs to satisfy the octet rule for each atom (not H). Writing Lewis Structures

45. Writing Lewis Structures Write the Lewis structure of fromaldehyde , H 2 CO. The skeleton structure is O H H C

46. 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 Writing Lewis Structures O H H C

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

48. Finish the structure by placing remaining electrons as lone pairs. Check that the final Lewis structure has the correct number of valence electrons (12) and each atom (not H) has 8 electrons. Writing Lewis Structures O H H C

49. Test Your Skill Write the Lewis structure of N 2 H 2 . The skeleton structure is: H N N H

50. Test Your Skill Write the Lewis structure of N 2 H 2 . Answer: H N N H

51. 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

52. In I 2 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

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

54. Electronegativity Trends

55. Electronegativity Trends

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

57. Properties of Compounds

58. Formal charge is a charge assigned to atoms in Lewis structures by assuming the shared electrons are divided equally between the bonded atoms . Equation for Formal Charge: (number of valence electrons in atom) – (number of lone pair electrons) – ½ (number of shared electrons) Formal Charges

59. Atom # Bonds # Bonds # Bonds N O C -1 [phone] N N N O O O C C _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ : : : : : : : : : : Formal Charges

60. Formal Charges Add formal charges to the Lewis structure of HNO 3 shown below. O N O O H

61. 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. Formal Charges O N O O H

62. Test Your Skill Add formal charges to the Lewis structure of HNO 3 shown below. O N O O H

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

64. 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)

65. Test Your Skill Of the two structures shown for HNO 3 , use the stability rules to predict which will be more favored. O N O O H O N O O H

66. 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 . Test Your Skill O N O O H O N O O H

67. 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. O N O O H O N O O H ↔

68. Draw the third possible resonance structure for HNO 3 , the first two are below. Drawing Resonance Structures O N O O H O N O O H ↔

69. First two: In the third, add the double bond between the nitrogen and the oxygen to the bottom left. Complete the octets with the remaining electrons and add formal charges. Drawing Resonance Structures O N O O H O N O O H ↔ O N O O H

70. HNO 3 Resonance Forms A total of three resonance forms can be written for HNO 3 . The first and last structures are equally favored because of fewer formal charges; the middle structure less favored. O N O O H O N O O H ↔ ↔ O N O O H

71. Test Your Skill Write all resonance structures, including formal charges, for O 3 , O-O-O skeleton structure.

72. Test Your Skill Write all resonance structures, including formal charges, for O 3 , O-O-O skeleton structure. Answer: O ↔ O O O O O

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

74. Contribution of Resonance Structures Equivalent resonance structures, such as the two for O 3 , contribute equally to the average structure. Bond order in O 3 is the average of a double bond and a single bond = 1.5. O ↔ O O O O O

75. Test Your Skill Draw the Lewis structure of IF 3 . Answer: I F F F

76. Chapter 9 Visual Summary

77. Chapter 10 Molecular Structure and Bonding Theories

78. 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

79. 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. Steric number determines the bonded-atom lone-pair arrangement , the shape that maximizes the distances between the valence-shell electron pairs. Steric Number

80. Geometric Arrangements

81. Geometric Arrangements

82. In the Lewis structure of BeCl 2 , beryllium has two bonded atoms and no lone pairs, steric number = 2 . A linear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible. Steric Number = 2

83. The Lewis structure of HCN (H-C º 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

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

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

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

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

88. The Lewis structure of H 2 O is Steric number = 4 , 2 bonded atoms and 2 lone pairs. The bonded-atom lone-pair arrangement is tetrahedral . Central Atoms with Lone Pairs O H H

89. Molecular shape is the arrangement of the atoms in a species. The bonded-atom lone-pair arrangement of H 2 O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom). Molecular Shape of H 2 O

90. What is the electron pair geometry and molecular shape of NH 3 ? Molecular Shape of NH 3

91. 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 . Molecular Shape of NH 3 N H H H

92. The bonded-atom lone-pair arrangement of NH 3 is tetrahedral (top), molecular shape is a trigonal pyramidal (bottom). Molecular Shape of NH 3

93. The measured bond angle in H 2 O (104.5 o ) is smaller than the predicted angle (109.5 o ) 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

94. What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF 3 ? Test Your Skill

95. 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

96. The geometry of each central atom is determined separately. The C H 3 carbon in C H 3 CN has tetrahedral geometry and the other carbon has linear geometry. Multiple Central Atoms

97. What are the bonded-atom lone-pair arrangements and the shapes about each central atom in NH 2 SH? Draw the Lewis structure. The bonded-atom lone-pair arrangements of both are tetrahedral, the nitrogen shape is trigonal pyramidal and sulfur is “V” shaped. Shapes of Molecules N S H H H

98. 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

99. In I 2 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

100. The bond dipoles in CO 2 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

101. 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 OF 2 because the V-shape of the molecule does not orient them in opposite directions. Polarity of Molecules

102. The bond dipoles in BCl 3 and CCl 4 cancel because of the regular shape and equal magnitude. Polarity of Molecules

103. The bond dipoles in BCl 2 F and CHCl 3 do not cancel because they are not of the same magnitude. Polarity of Molecules

104. Are the following molecules polar or nonpolar : H 2 S, SiF 4 , CH 2 Cl 2 ? Test Your Skill

105. Are the following molecules polar or nonpolar: H 2 S, SiF 4 , CH 2 Cl 2 ? Answer: H 2 S and CH 2 Cl 2 are polar, SiF 4 is nonpolar. Test Your Skill

106. Hybrid orbitals are: Orbitals obtained by mixing two or more atomic orbitals on the same central atom . Hybrid Orbitals

107. The bonds in BeCl 2 arise from the overlap of two sp hybrid orbitals on the beryllium atom with the 3 p orbitals on the two chlorine atoms. Bonding in BeCl 2

108. The bonds in BF 3 arise from the overlap of three sp 2 hybrid orbitals on the boron atom with 2 p orbitals on the three fluorine atoms. Bonding in BF 3

109. The bonds in CH 4 arise from the overlap of four sp 3 hybrid orbitals on the carbon atom with 1 s orbitals on the four hydrogen atoms. Bonding in CH 4

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

111. Hybrid Orbitals

112. Test Your Skill Identify the hybrid orbitals on the central atoms in SiH 4 and HCN.

113. Identify the hybrid orbitals on the central atoms in SiH 4 and HCN. Answer: sp 3 hybrid orbitals on silicon; sp hybrid orbitals on carbon. Test Your Skill

114. 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

115. The C-C sigma bond in C 2 H 4 arises from overlap of sp 2 hybrid orbitals and the four C-H sigma bonds from overlap sp 2 hybrid orbitals on C with 1 s orbitals on H. The second C-C bond forms from sideways overlap of p orbitals. Bonding in C 2 H 4

116. 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

117. The double bond in C 2 H 4 is one sigma bond and one pi bond – each bond is of similar strength. Bonding in C 2 H 4

118. Test Your Skill Describe the bonds made by the carbon atom in HCN.

119. Describe the bonds made by the carbon atom in HCN. Answer: The carbon is sp hybridized. C sp hybrid forms s bond with H 1 s . C sp hybrid forms s bond. N has s , p (along bonding direction) or sp hybrid of the two available for this s bond. Two pi bonds form by sideways overlap of two p orbitals on both C and N. Test Your Skill

120. Chapter 10 Visual Summary

121. Chapter 11 Liquids and Solids

122. Intermolecular forces are the attractions that hold molecules together in the liquid and solid states. State Volume Shape of State Density Compressibility gas assumes shape and volume of container low easily compressed liquid definite volume, assumes shape of container high nearly incompressible solid both definite shape and volume high nearly incompressible Characteristic Properties of Gases, Liquids, and Solids

123. Physical State Relation Between Energy of Attraction and Kinetic Energy of Molecules solid kinetic energy << energy of attraction liquid kinetic energy ≈ energy of attraction gas kinetic energy >> energy of attraction Physical State and Energy of Attraction

124. 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.

125. 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

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

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

128. 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.

129. 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

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

131. Identify the kind of intermolecular forces : (a) BF 3 , BBr 3 ( b) C 2 H 5 OH, C 2 H 5 Cl Example: Intermolecular Forces

132. Answers London dispersion forces for both. C 2 H 5 OH is Hydrogen Bonding C 2 H 5 Cl is Dipole-Dipole

133. Liquids: Capillary Action 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.

134. Capillary Action Water rises because adhesion is stronger than cohesion. Mercury is lowered because cohesion is stronger than adhesion.

135. 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

136. Chapter 11 Visual Summary

137. Chapter 12 Solutions

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

139. 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

140. Units of Concentration Used Earlier

141. Mass Percent Composition

142. 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

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

144. Example: Calculate Molality What is the molality of a solution prepared by dissolving 3.00 g NaCl (molar mass = 58.44 g/mol) in 150 g of water ? Answer: 0.342 m

145. ppm & ppb ppm (parts per million) = (g solute / g solution) x 1,000,000 ppb (parts per billion) = (g solute / g solution) x 1,000,000,000

146. Units for Concentration Conversion Concentration Unit Numerator Units Denominator units Mass % Grams 100 g solution Molarity Moles 1 L solution Molality Moles 1 kg solvent Mole Fraction Moles 1 mol total of solution

147. Express the concentration of a 3.00% H 2 O 2 solution as molality . Answers: 0.910 molal Example: Concentration Conversion

148. Test Your Skill Calculate (a) the molality , ( C 2 H 5 OH; molar mass = 46.07 g/mol) in a wine that has an alcohol concentration of 7.50 mass percent. Answers : 1.76 molal

149. 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 (H 2 SO 4 ; molar mass = 98.08 g/mol) is 1.080 g/ mL. What is the molarity of this solution ? Answer = 1.32 M Example: Conversion to Molarity

150. Chapter 12 Visual Summary