Chm151 exam 4 slides sp10

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Chm151 exam 4 slides sp10

  1. 1. CHM 151 General Chemistry<br /> EXAM 4 <br /> SLIDES<br />
  2. 2. Electron Energy Levels<br />Electrons are arranged in<br />specific energy levels that<br /><ul><li>are labeled n = 1, n = 2, n = 3, and so on
  3. 3. increase in energy as n increases
  4. 4. have the electrons with the lowest energy in the first energy level (n = 1) closestto the nucleus</li></li></ul><li>Energy Level Changes<br /><ul><li>An electron absorbsenergy to “jump” to a higher energy level.
  5. 5. When an electron falls to a lower energy level, energy is emitted.
  6. 6. In the visible range, the emitted energy appears as a color.</li></li></ul><li>In each of the following energy level changes, indicate<br />if energy is 1) absorbed, 2) emitted, or 3) not changed.<br />A. An electron moves from the first energy level (n = 1) to the third energy level (n = 3).<br />B. An electron falls from the third energy level to the second energy level.<br />C. An electron moves within the third energy level.<br />Learning Check<br />
  7. 7. Sublevels<br />Sublevels<br /><ul><li>contain electrons with the same energy
  8. 8. are found within each energy level.
  9. 9. are designated by the letters s, p, d, and f</li></ul> The number of sublevels is equal to the value of the principal quantum number (n).<br />
  10. 10. Number of Sublevels<br />
  11. 11. Energy of Sublevels<br />In any energy level<br /><ul><li>the s sublevel has the lowest energy
  12. 12. the s sublevel is followed by the p, d, and f sublevels (in order of increasing energy)</li></li></ul><li>Orbitals<br />An orbital <br /><ul><li>is a three-dimensional space around a nucleus where an electron is most likely to be found
  13. 13. has a shape that represents electron density (not a path the electron follows)
  14. 14. can hold up to 2 electrons
  15. 15. contains two electrons that must spin in opposite</li></li></ul><li>sOrbitals<br />An s orbital <br /><ul><li>has a spherical shape around the nucleus
  16. 16. increases in size around the nucleus as the energy level n value increases
  17. 17. is a single orbital found in each s sublevel</li></li></ul><li>p Orbitals<br />A p orbital <br /><ul><li>has a two-lobed shape
  18. 18. is one of three p orbitals that make up each p sublevel
  19. 19. increases in size as the value of n increases</li></li></ul><li>Sublevels and Orbitals<br />Each sublevel consists of a specific number of<br />orbitals. <br /><ul><li>An s sublevel contains one s orbital.
  20. 20. A p sublevel contains three p orbitals.
  21. 21. A d sublevel contains five d orbitals.
  22. 22. An f sublevel contains seven f orbitals.</li></li></ul><li>Electrons in Each Sublevel<br />
  23. 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.<br />Two electrons in the same orbital must have opposite spins.<br />Orbital Diagrams<br />
  24. 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.<br />Two electrons in the 2ssubshell would be 2s2 (spoken as “two-ess-two”).<br />Four electrons in the 3psubshell would be 3p4 (“three-pea-four”).<br />Electron Configuration<br />
  25. 25. Electron Configurations of Elements<br />Hydrogen contains one electron in the 1s subshell.<br />1s1<br />Helium has two electrons in the 1s subshell.<br /> 1s2<br />
  26. 26. Electron Configurations of Elements<br />Lithium has three electrons.<br />1s2 2s1 <br />Beryllium has four electrons.<br />1s2 2s2<br />Boron has five electrons.<br />1s2 2s2 2p1<br />
  27. 27. Carbon, with six electrons, has the electron configuration of 1s2 2s2 2p2.<br />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.<br />Orbital Diagram of Carbon<br />
  28. 28. Other Elements in the Second Period<br />N 1s2 2s2 2p3<br />O 1s2 2s2 2p4<br />F 1s2 2s2 2p5<br />Ne 1s2 2s2 2p6<br />
  29. 29. Because their electron configurations can get long, larger atoms can use an abbreviated electron configuration, using a noble gas to represent core electrons.<br />Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6-> [Ar] 4s2 3d6<br />Ar<br />Electron Configurations of Heavier Atoms<br />
  30. 30. Chapter 7 Visual Summary<br />
  31. 31. Chapter 8The Periodic Table: Structure and Trends<br />
  32. 32. The 4s orbital is lower in energy than the 3d orbital and fills first, starting the fourth period at potassium.<br />The 3dorbitals fill after the 4s.<br />Similar inversions occur in the remaining periods.<br />Electron Configurations<br />
  33. 33. Electron Configurations of Anions<br />For anions, the additional electrons fill orbitals following the same rules that applies to atoms.<br />Cl: [Ne] 3s2 3p5Cl-: [Ne] 3s2 3p6<br /> As: [Ar] 4s2 3d10 4p3 As3-: [Ar] 4s2 3d10 4p6<br />Many stable anions have the same electron configuration as a noble gas atom.<br />
  34. 34. Test Your Skill<br />Write the electron configurations of the following ions: (a) N3- (b) Co3+ (c) K+<br />
  35. 35. Test Your Skill<br />Write the electron configurations of the following ions: (a) N3- (b) Co3+ (c) K+<br />Answers:<br />(a) 1s2 2s2 2p6<br />(b) [Ar] 4s23d4<br />(c) [Ar]<br />
  36. 36. Size Trends for an Isoelectronic Series<br />
  37. 37. Sizes of the Atoms and Their Cations<br /><ul><li>Atoms are always larger than their cations.</li></li></ul><li>Sizes of the Atoms and Their Cations<br /><ul><li>If an atom makes more than one cation, the higher-charged ion has a smaller size.</li></li></ul><li>Anions are always larger than their atoms.<br />Atomic and Ionic Radii<br />
  38. 38. Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2-<br />Test Your Skill<br />
  39. 39. Identify the larger species of each pair: (a) Mg or Mg2+ (b) Se or Se2-<br />Answer: (a) Mg is larger.<br /> (b) Se2- is larger.<br />Test Your Skill<br />
  40. 40. Chapter 8 Visual Summary<br />
  41. 41. Chapter 9Chemical Bonds<br />
  42. 42. Chemical Bonds<br />Chemical bonds are the forces that hold the atoms together in substances.<br />This chapter discusses two limiting types of bonding.<br />Ionic bonding<br />Covalent bonding<br />
  43. 43. Lewis Electron-dot Symbols<br />A Lewis electron-dot symbol consists of the symbol for the element surrounded by dots, one for each valence electron.<br />
  44. 44. Cations of most representative elements have no valence shell electrons shown in the Lewis symbol.<br />Na×® Na+ + e-<br />×Ca×® Ca2+ + 2e-<br />Lewis Symbols for Cations<br />
  45. 45. Lewis Symbols for Anions<br />The Lewis symbols of most monatomic anions show eight valence electrons.<br />Cl + e--> Cl -<br />Se + 2e- -> Se 2-<br />
  46. 46. Ionic Bonding<br />Ionic bonding results from the electrostatic attraction between cations and anions. <br />Formation of an ionic bond can be viewed as a transfer of electrons.<br /> Na + F -> Na+ + F - (or NaF)<br />
  47. 47. A covalent bond result from the sharing of two electrons between two atoms, as shown here for H2.<br />Covalent Bonding<br />
  48. 48. Two hydrogen atoms become more stable as their orbitals, each containing one electron, overlap.<br />Orbital Overlap<br />
  49. 49. Lewis Structures<br />Bonding Pair<br />H Cl<br />Lone Pair<br />Lewis structures represent covalent bonding by showing how the valence electrons are present in a molecule.<br />Bonding pairsare shared between two atoms and are represented by lines .<br />Lone pairs are entirely on one atom and are represented by two dots.<br />
  50. 50. The number of Covalent Bonds<br />The number of covalent bonds can be determined from<br />the number of electrons needed to complete an octet.<br />
  51. 51. Octet Rule<br />Octet Rule:atoms share electrons until each atom is surrounded by eight.<br /><ul><li>Single Bond- sharing one pair of electrons
  52. 52. Double Bond- sharing two pairs of electrons
  53. 53. Triple Bond- sharing three pairs of electrons</li></li></ul><li>1. Write the skeleton structure. <br />2. Sum the valence electrons.<br />3. Subtract two electrons for each bond in the skeleton structure.<br />4. Count the number of electrons needed to satisfy octet rule for each atom.<br />If the number of electrons needed equals the number remaining, go to 5.<br />If fewer electron remain, add one bond for every twoadditional electrons needed.<br />5. Place remaining electrons as lone pairs to satisfy the octet rule for each atom (not H).<br />Writing Lewis Structures<br />
  54. 54. Writing Lewis Structures<br />Write the Lewis structure of fromaldehyde, H2CO. The skeleton structure is<br />O<br />C<br />H<br />H<br />
  55. 55. Writing Lewis Structures<br />O<br />C<br />H<br />H<br />The total number of valence electrons is<br />1(C) 1 x 4 = 4<br />1(O) 1 x 6 = 6<br />2(H) 2 x 1 = 2<br /> 12<br />
  56. 56. Writing Lewis Structures<br />needs 6e- to complete octet<br />O<br />C<br />needs 2e- to complete octet<br />H<br />H<br />O<br />C<br />H<br />H<br />Remaining valence electrons = 6<br />8 electrons needed to obey the octet rule<br />Add one bond because 2 more electrons are needed than are available.<br />
  57. 57. Writing Lewis Structures<br />O<br />C<br />H<br />H<br /><ul><li>Finish the structure by placing remaining electrons as lone pairs.</li></ul>Check that the final Lewis structure has the correct number of valence electrons (12) and each atom (not H) has 8 electrons.<br />
  58. 58. Test Your Skill<br />Write the Lewis structure of N2H2. The skeleton structure is:<br />H N N H<br />
  59. 59. Test Your Skill<br />Write the Lewis structure of N2H2. <br />Answer:<br />H N N H<br />
  60. 60. Polar Bond<br />Polar Bond –<br /> A covalent bond in which the two atoms do not share the bonding electrons equally.<br /> Note: The higher electronegative element takes on a partial negative charge and the lower electronegative element takes on a partial positive charge.<br />d+ d-<br /> I—Br (a polar covalent bond)<br />  arrow indicates direction of more electronegative element<br />
  61. 61. In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not.<br />Dipole moment is a measure of the unequal sharing of electrons.<br />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-<br />Cl-F<br />Bond Polarity<br />
  62. 62. Electronegativityis a measure of the ability of an atom to attract the shared electrons in a chemical bond.<br />Electronegativity<br />
  63. 63. Electronegativity Trends<br />
  64. 64. Electronegativity Trends<br />
  65. 65. Example: Electronegativity<br />Select the most polar bond.<br />Cl-F O-F P-F<br />
  66. 66. Properties of Compounds<br />
  67. 67. <ul><li>Formal charge is a charge assigned to atoms in Lewis structures by assuming the shared electrons are divided equally between the bonded atoms.
  68. 68. Equation for Formal Charge:</li></ul>(number of valence electrons in atom) – <br />(number of lone pair electrons) – ½ (number of shared electrons)<br />Formal Charges<br />
  69. 69. Formal Charges<br /># Bonds<br /># Bonds<br /># Bonds<br />-1<br />0<br />+1<br />Atom<br />_<br />_<br />:<br />_<br />_<br />_<br />_<br />_<br />_<br />N<br />2<br />3<br />4<br />N<br />N<br />N<br />:<br />:<br />_<br />_<br />:<br />:<br />_<br />_<br />_<br />_<br />_<br />3<br />O<br />2<br />1<br />O<br />O<br />O<br />:<br />:<br />:<br />:<br />_<br />_<br />_<br />_<br />_<br />_<br />3<br />4<br />C<br />C<br />C<br />:<br />_<br />
  70. 70. Formal Charges<br />Add formal charges to the Lewis structure of HNO3 shown below.<br />H<br />O<br />N O<br /> O<br />
  71. 71. Formal Charges<br />H<br />O<br />N O<br />O<br />The nitrogen atom has 4 bonds giving it a +1 formal charge.<br />The oxygen atom on the bottom left only has 1 bond giving it a -1 formal charge.<br />The sum of formal charges equals the charge of the species.<br />
  72. 72. Test Your Skill<br />Add formal charges to the Lewis structure of HNO3 shown below.<br />H<br />O<br />N O<br />O<br />
  73. 73. Test Your Skill<br />H<br />O<br />N O<br />O<br />Answer:<br />
  74. 74. Lewis structures that show the smallest formal charges are favored.<br />Lewis structures that have adjacent atoms with formal charges of the same sign are much less favorable.<br />Lewis structures that place negative formal charges on the more electronegative atoms are favored.<br />Formal charges of opposite sign are usually on adjacent atoms.<br />Fewer formal charges overall.<br />Structure Stability (which structure is more favored)<br />
  75. 75. Test Your Skill<br />Of the two structures shown for HNO3, use the stability rules to predict which will be more favored.<br />H<br />H<br />O<br />O<br />N O<br />N O<br />O<br />O<br />
  76. 76. Test Your Skill<br />H<br />O<br />H<br />O<br />N O<br />N O<br />O<br />O<br />Answer:<br />The structure on the left is favored because <br />it has fewer formal charges.<br />it does not have adjacent atoms with the same formal charge.<br />
  77. 77. Resonance in Lewis Structures<br />Resonance structures differ only in the distribution of the valence electrons. <br />All resonance structures follow the rules for writing Lewis structures.<br />Resonance structures are indicated by a double headed arrow.<br />H<br />O<br />H<br />O<br />N O<br />N O<br />O<br />↔<br />O<br />
  78. 78. Drawing Resonance Structures <br />H<br />O<br />H<br />O<br />N O<br />N O<br />O<br />↔<br />O<br />Draw the third possible resonance structure for HNO3, the first two are below.<br />
  79. 79. H<br />O<br />H<br />O<br />N O<br />N O<br />O<br />↔<br />O<br />H<br />O<br />N O<br />O<br />Drawing Resonance Structures <br /><ul><li>First two:
  80. 80. 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.</li></li></ul><li>HNO3 Resonance Forms<br />A total of three resonance forms can be written for HNO3.<br />The first and last structures are equally favored because of fewer formal charges; the middle structure less favored.<br />↔<br />↔<br />H<br />O<br />H<br />O<br />N O<br />N O<br />O<br />O<br />H<br />O<br />N O<br />O<br />
  81. 81. Test Your Skill<br />Write all resonance structures, including formal charges, for O3, O-O-O skeleton structure.<br />
  82. 82. Test Your Skill<br />Write all resonance structures, including formal charges, for O3, O-O-O skeleton structure.<br />Answer:<br /> O<br /> O<br />O ↔O<br />O<br />O<br />
  83. 83. No resonance structure is correct by itself; the correct structure is an average of all resonance structures.<br />Average Structure<br />
  84. 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.<br />O<br />O<br />O ↔ O<br />O<br />O<br />Contribution of Resonance Structures<br />
  85. 85. Test Your Skill<br />Draw the Lewis structure of IF3.<br />Answer:<br />F<br /> I<br />F<br />F<br />
  86. 86. Chapter 9 Visual Summary<br />
  87. 87. Chapter 10 Molecular Structure and Bonding Theories<br />
  88. 88. Valence-Shell Electron-Pair Repulsion Model (VSEPR) predicts shape from Lewis Structures.<br />VSEPR Rule 1: <br />A molecule has a shape that minimizes electrostatic repulsions between valence-shell electron pairs. <br />Minimum repulsion results when the electron pairs are as far apart as possible.<br />VSEPR<br />
  89. 89. Steric number = <br />(number of lone pairs on central atom) + (number of atoms bonded to central atom)<br />The steric number is determined from the Lewis structure.<br />Stericnumber determines the bonded-atom lone-pair arrangement, the shape that maximizes the distances between the valence-shell electron pairs.<br />Steric Number<br />
  90. 90. Geometric Arrangements<br />
  91. 91. Geometric Arrangements<br />
  92. 92. In the Lewis structure of BeCl2, <br />beryllium has two bonded atoms and no lone pairs, <br />stericnumber = 2.<br />Alinear geometry places the two pairs of electrons on the central beryllium atom as far apart as possible.<br />Steric Number = 2<br />
  93. 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.<br />The bonded-atom lone-pair arrangement is linear. <br />The number of bonded atoms, not the number of bonds, determines the steric number.<br />Molecules with Multiple Bonds<br />
  94. 94. The Lewis structure of BF3<br />shows the boron atom has a steric number = 3; the bonded-atom lone-pair arrangement is trigonal planar.<br />Steric Number = 3<br />
  95. 95. The Lewis structure of CH4<br /> shows the carbon atom has a steric number = 4; the bonded-atom lone-pair arrangement is tetrahedral.<br />Steric Number = 4<br />
  96. 96. The phosphorus atom in PF5 has a steric number = 5; the bonded-atom lone-pair arrangement is trigonal bipyramidal.<br />Steric Number = 5<br />
  97. 97. The sulfur atom in SF6 has a steric number = 6; the bonded-atom lone-pair arrangement is octahedral.<br />Steric Number = 6<br />
  98. 98. Central Atoms with Lone Pairs<br />O<br />H<br />H<br />The Lewis structure of H2O is<br /><ul><li>Stericnumber = 4, 2 bonded atoms and 2 lone pairs.</li></ul>The bonded-atom lone-pair arrangement is tetrahedral.<br />
  99. 99. Molecular shapeis the arrangement of the atoms in a species.<br />The bonded-atom lone-pair arrangement of H2O is tetrahedral (top); the molecular shape is bent or V-shaped (bottom).<br />Molecular Shape of H2O<br />
  100. 100. What is the electron pair geometry and molecular shape of NH3?<br />Molecular Shape of NH3<br />
  101. 101. Molecular Shape of NH3<br />N<br />H<br />H<br />H<br />First, draw the Lewis structure.<br />The nitrogen has 3 bonded atoms and 1 lone pair; the steric number = 4 and the bonded-atom lone-pair arrangement is tetrahedral.<br />
  102. 102. The bonded-atom lone-pair arrangement of NH3 is tetrahedral (top), molecular shape is a trigonal pyramidal (bottom).<br />Molecular Shape of NH3<br />
  103. 103. The measured bond angle in H2O (104.5o) is smaller than the predicted angle (109.5o)<br />Explanation (VESPR Rule #2) Forces between electron pairs vary as: <br />lone pair-lone pair replusion> lone pair-bonding pair replusion> bonding pair-bonding pair repulsion<br />Electron Pair Repulsions<br />
  104. 104. What is the steric number, the bonded-atom lone-pair arrangement, and the molecular shape of ClF3?<br />Test Your Skill<br />
  105. 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.<br />Test Your Skill<br />
  106. 106. The geometry of each central atom is determined separately.<br />The CH3 carbon in CH3CN has tetrahedral geometry and the other carbon has linear geometry.<br />Multiple Central Atoms<br />
  107. 107. H<br />H<br />N S<br />H<br />Shapes of Molecules<br /><ul><li>What are the bonded-atom lone-pairarrangements and the shapes about each central atom in NH2SH?
  108. 108. Draw the Lewis structure.
  109. 109. The bonded-atom lone-pairarrangements of both are tetrahedral, the nitrogen shape is trigonal pyramidal and sulfur is “V” shaped.</li></li></ul><li>Polar Bond<br />Polar Bond –<br /> A covalent bond in which the two atoms do not share the bonding electrons equally.<br /> Note: The higher electronegative element takes on a partial negative charge and the lower electronegative element takes on a partial positive charge.<br />d+ d-<br /> I—Br (a polar covalent bond)<br />  arrow indicates direction of more electronegative element<br />
  110. 110. In I2 the sharing of the electrons in the covalent bond is equal; in ClF it is not.<br />Dipole moment is a measure of the unequal sharing of electrons.<br />Equals the magnitude of the separated charges X the distance between them<br />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-<br />Cl-F<br />Bond Polarity<br />
  111. 111. The bond dipoles in CO2 cancel because the linear shape orients the equal magnitude bond dipoles in exactly opposite directions. <br />Bond dipole determined by difference in electronegativities of bonded atoms. <br />Polarity of Molecules<br />
  112. 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.<br />Polarity of Molecules<br />
  113. 113. The bond dipoles in BCl3 and CCl4 cancel because of the regular shape and equal magnitude.<br />Polarity of Molecules<br />
  114. 114. The bond dipoles in BCl2F and CHCl3 do not cancel because they are not of the same magnitude.<br />Polarity of Molecules<br />
  115. 115. Test Your Skill<br /><ul><li>Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?</li></li></ul><li>Are the following molecules polar or nonpolar: H2S, SiF4, CH2Cl2?<br />Answer: H2S and CH2Cl2 are polar, SiF4 is nonpolar.<br />Test Your Skill<br />
  116. 116. Hybrid orbitalsare:<br />Orbitalsobtained by mixing two or more atomic orbitals on the same central atom.<br />Hybrid Orbitals<br />
  117. 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.<br />Bonding in BeCl2<br />
  118. 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.<br />Bonding in BF3<br />
  119. 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.<br />Bonding in CH4<br />
  120. 120. Hybrid orbitals can hold lone pairs as well as make bonds.<br />Lone Pairs and Hybrid Orbitals<br />
  121. 121. Hybrid Orbitals<br />
  122. 122. Test Your Skill<br /><ul><li>Identify the hybrid orbitals on the central atoms in SiH4 and HCN.</li></li></ul><li>Identify the hybrid orbitals on the central atoms in SiH4 and HCN.<br />Answer: sp3hybrid orbitals on silicon; sp hybrid orbitals on carbon.<br />Test Your Skill<br />
  123. 123. Sigma bonds (s): the shared pair of electrons is symmetric about the line joining the two nuclei of the bonded atoms.<br />Types of Bonds: Sigma Bonds<br />
  124. 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.<br />The second C-C bond forms from sideways overlap of p orbitals.<br />Bonding in C2H4<br />
  125. 125. Pi bonds (p) places electron density above and below the line joining the bonded atoms – they form by sideways overlap of p orbitals.<br />Types of Bonds: Pi Bonds<br />
  126. 126. The double bond in C2H4 is one sigma bond and one pi bond – each bond is of similar strength.<br />Bonding in C2H4<br />
  127. 127. Test Your Skill<br /><ul><li>Describe the bonds made by the carbon atom in HCN.</li></li></ul><li>Describe the bonds made by the carbon atom in HCN.<br />Answer: The carbon is sp hybridized. <br />C sp hybrid forms s bond with H 1s.<br />C sp hybrid forms s bond. N has s, p (along bonding direction) or sp hybrid of the two available for this s bond. <br /> Two pi bonds form by sideways overlap of two p orbitals on both C and N.<br />Test Your Skill<br />
  128. 128. Chapter 10 Visual Summary<br />
  129. 129. Chapter 11Liquids and Solids<br />
  130. 130. Characteristic Properties of Gases, Liquids, and Solids<br />Intermolecular forces are the attractions that hold molecules together in the liquid and solid states.<br />
  131. 131. Physical State and Energy of Attraction<br />
  132. 132. Boiling Point<br />The boiling point of a liquid is the temperature at which the vapor pressure is equal to the external pressure.<br />The normal boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to 1 atmosphere.<br />At the boiling point, bubbles filled with vapor form below the surface of the liquid.<br />
  133. 133. Electrostatic forces account for all types of intermolecular attractions. There are three types of attractions:<br />Dipole-dipole attractions<br />London dispersion forces<br />Hydrogen bonding<br />Intermolecular Attractions<br />
  134. 134. Dipole-dipole attractions result from electronic forces between molecular dipoles:<br />Dipole-Dipole Attractions<br />
  135. 135. London dispersion forces arise from the attractions between instantaneous dipoles and induced dipoles.<br />London Dispersion Forces<br />
  136. 136. Dispersion Forces and Periodic Trends<br />Polarizability is the ease with which a charge distorts the electron cloud in a molecule.<br />Polarizability generally increases with the number of electrons in the molecule.<br />For related series of molecules, London dispersion forces increase going down any group in the periodic table.<br />
  137. 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.<br />Hydrogen bonds are sometimes shown as dotted lines.<br />Hydrogen Bonding<br />
  138. 138. Hydrogen bonding causes ice to have a lower density than liquid water.<br />Structure of Solid Water<br />
  139. 139. Identify the kind of intermolecular forces:<br />(a) BF3, BBr3<br /> (b) C2H5OH, C2H5Cl<br />Example: Intermolecular Forces<br />
  140. 140. Answers <br />London dispersion forces for both.<br />C2H5OH is Hydrogen Bonding<br /> C2H5Cl is Dipole-Dipole<br />
  141. 141. Capillary action causes water to rise in a small diameter glass tube.<br />Capillary action is the result of a competition between:<br />cohesion: the attraction of molecules for other molecules of the same substance.<br />adhesion: the attraction of molecules for other molecules of a different substance.<br />Liquids: Capillary Action<br />
  142. 142. <ul><li>Water rises because adhesion is stronger than cohesion.
  143. 143. Mercury is lowered because cohesion is stronger than adhesion.</li></ul>Capillary Action<br />
  144. 144. Viscosity is the resistance of a fluid to flow.<br />The stronger the intermolecular forces of attraction, the greater the viscosity.<br />Other factors contribute to viscosity as well, like structure, size, and shape of molecules.<br />Liquids: Viscosity<br />
  145. 145. Chapter 11 Visual Summary<br />
  146. 146. Chapter 12Solutions<br />
  147. 147. There are a number of ways to express concentration:<br />Molarity<br />mass percentage<br />ppm and ppb<br />molality<br />Solution Concentration<br />
  148. 148. All concentration units are fractions.<br />The numerator contains the quantity of solute.<br />The denominator is the quantity of either solution or solvent.<br />They differ in the units used to express these two quantities.<br />Solution Concentration<br />
  149. 149. Units of Concentration Used Earlier<br />
  150. 150. Mass Percent Composition<br />
  151. 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.<br />Answer: 1.96 %<br />Example: Percent Composition<br />
  152. 152. Molality (m or molal) is defined as<br />Molality<br />
  153. 153. Example: Calculate Molality<br /><ul><li>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?
  154. 154. Answer: 0.342 m</li></li></ul><li>ppm & ppb<br />ppm (parts per million)<br /> = (g solute / g solution) x 1,000,000<br />ppb (parts per billion)<br /> = (g solute / g solution) x 1,000,000,000<br />
  155. 155. Units for Concentration Conversion<br />
  156. 156. Express the concentration of a 3.00% H2O2 solution as molality.<br /> Answers: <br /> 0.910 molal<br />Example: Concentration Conversion<br />
  157. 157. Test Your Skill<br />Calculate (a) the molality, (C2H5OH; molar mass = 46.07 g/mol) in a wine that has an alcohol concentration of 7.50 mass percent.<br />Answers: 1.76 molal<br />
  158. 158. Example: Conversion to Molarity<br />Conversion of most concentration units to molarity usually involve using the density of the solution to convert units mass to units of volume.<br />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?<br /> Answer = 1.32 M<br />
  159. 159. Chapter 12 Visual Summary<br />

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