Ap chem unit 8 presentation


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Ap chem unit 8 presentation

  1. 1. Bonding: General Concepts AP Chemistry Unit 8
  2. 2. Types of Chemical Bonds
  3. 3. Ionic BondsIonic Bonds are formed when an atomthat loses electrons relatively easilyreacts with an atom that has a highattraction for electrons. Ionic Compounds results when a metal bonds with a nonmetal.
  4. 4. Bond EnergyBond energy is the energy required to break abond.The energy of interaction between a pair of ionscan be calculated using Coulomb‟s law -19 æ Q1Q2 ö E = (2.31x10 Jinm) ç ÷ è r ør = the distance between the ions in nm. Q1 and Q2 are the numerical ion charges.E is in joules
  5. 5. Bond EnergyWhen the calculated energy betweenions is negative, that indicates anattractive force.A positive energy is a repulsive energy.The distance where the energy isminimal is called the bond length.
  6. 6. Covalent BondsCovalent bonds form betweenmolecules in which electrons are sharedby nuclei. The bonding electrons are typically positioned between the two positively charged nuclei.
  7. 7. Polar Covalent BondsPolar covalent bonds are an intermediatecase in which the electrons are notcompletely transferred but form unequalsharing. A δ- or δ+ is used to show a fractional or partial charge on a molecule with unequal sharing. This is called a dipole.
  8. 8. Electronegativity
  9. 9. ElectronegativityElectronegativity is the ability of an atom in a molecule to attract shared electrons to itself. (electron love) Relative electronegativities are determined by comparing the measured bond energy with the “expected” bond energy. Measured in Paulings. After Linus Pauling the American scientist who won the Nobel Prizes for both chemistry and peace.
  10. 10. ElectronegativityExpected H-X bond energy= H - H bond energy + X - X bond energy 2
  11. 11. ElectronegativityElectronegativity values generally increasegoing left to right across the periodic tableand decrease going top to bottom.
  12. 12. Electronegativity and Bond type
  13. 13. Bond Polarity and Dipole
  14. 14. Dipoles and Dipole MomentsA molecule that has a center of positive charge and a center of negative charge is said to be dipolar or to have a dipole moment. An arrow is used to show this dipole moment by pointing to the negative charge and the tail at the positive charge.
  15. 15. Dipoles and Dipole MomentsElectrostatic potentialdiagram showsvariation in charge.Red is the mostelectron rich regionand blue is the mostelectron poor region.
  16. 16. Dipoles and Dipole Moments
  17. 17. Dipoles and Dipole Moments
  18. 18. Dipoles and Dipole Moments
  19. 19. Dipoles and Dipole MomentsDipole moments are when opposingbond polarities don‟t cancel out.
  20. 20. Dipoles and Dipole Moments
  21. 21. Example ProblemsFor each of the following molecules,show the direction of the bondpolarities and indicate which ones havea dipole moment: HCl, Cl2, SO3, CH4, H2S
  22. 22. HCl
  23. 23. Cl2
  24. 24. SO3
  25. 25. CH4
  26. 26. H2S
  27. 27. Ions: ElectronConfigurations and Sizes
  28. 28. Electron Configurations of Compounds When two nonmetals react to form a covalent bond, they share electrons in a way that completes the valence electron configurations of both atoms. That is, both nonmetals attain noble gas electron configurations.
  29. 29. Electron Configurations of Compounds When a nonmetal and a representative-group metal react to form a binary ionic compounds, the ions form so that the valence electron configuration of the nonmetal achieves the electron configuration of the next noble gas atom and the valence orbitals of the metal are emptied. In this way both ions achieve noble gas electron configurations.
  30. 30. Predicting Ionic FormulasTo predict the formula of the ioniccompound, we simply recognize that thechemical compounds are always electricallyneutral. They have the same quantities ofpositive and negative charges.
  31. 31. Sizes of IonsSize of an ion generally follows the same trend as atomic radius. The big exception to this trend is where the metals become nonmetals and the ions switch charge.
  32. 32. Sizes of IonsA positive ion is formed by removing one or more electrons from a neutral atom, the resulting cation is smaller than the neutral atom. Less electrons allow for less repulsions and the ion gets smaller.
  33. 33. Sizes of IonsAn addition of electrons to a neutral atom produces an anion that is significantly larger than the neutral atom. An addition of an electron causes additional repulsions around the atom and therefore its size increases.
  34. 34. Energy Effects in Binary Ionic Compounds
  35. 35. Lattice EnergyLattice energy is the change in energy that takes place when separated gaseous ions are packed together to form an ionic solid. The lattice energy is often defined as the energy released when an ionic solid forms from its ions. Lattice energy has a negative sign to show that the energy is released.
  36. 36. Lattice Energy ExampleEstimate the enthalpy of lithium fluoride and the changes of energy and lattice energy during formation: Li+(g) + F-(g)  LiF(s)1. Break down LiF into its standard state elements (use formation reaction): Li(s) + ½F2(g)  LiF(s)
  37. 37. Lattice Energy Example Li(s) + ½F2(g)  LiF(s) Li+(g) + F-(g)  LiF(s)2. Use sublimation and evaporation reactions to get reactants into gas form (since lattice energy depends on gaseous state). Find the enthalpies to these reactions: Li(s)  Li(g) 161 kJ/mol Li(g) + ½F2(g)  LiF(s)
  38. 38. Lattice Energy Example Li(g) + ½F2(g)  LiF(s) Li+(g) + F-(g)  LiF(s)3. Ionize cation to form ions for bonding. Use Ionization energy for the enthalpy of the reaction. Li(g)  Li+(g) + e- Ionization energy: 520 kJ/mol Li+(g) + ½F2(g)  LiF(s)
  39. 39. Lattice Energy Example Li+(g) + ½F2(g)  LiF(s) Li+(g) + F-(g)  LiF(s)4. Dissociate diatomic gas to individual atoms: ½F2(g)  F(g) ½ Bond dissociation energy of F-F = 154 kJ/ 2 = 77 kJ/mol Li+(g) + F(g)  LiF(s)
  40. 40. Lattice Energy Example Li+(g) + F(g)  LiF(s) Li+(g) + F-(g)  LiF(s)5. Electron addition to fluorine is the electron affinity of fluorine: F(g) + e-  F-(g) -328 kJ/mol Li+(g) + F-(g)  LiF(s)
  41. 41. Lattice Energy Example Li+(g) + F-(g)  LiF(s) Li+(g) + F-(g)  LiF(s)6. Formation of solid lithium fluoride from the gaseous ions corresponds to its lattice energy: Li+(g) + F-(g)  LiF(s) -1047 kJ/mol
  42. 42. Lattice Energy ExampleThe sum of these five processes yields the overall reaction and the sum of the individual energy changes gives the overall energy change and the enthalpy of formation: Li(s)  Li(g) 161 kJ Li(g)  Li+(g) + e- 520 kJ ½F2(g)  F(g) 77 kJ F(g) + e-  F-(g) -328 kJ Li+(g) + F-(g)  LiF(s) -1047 kJ Total = -617 kJ/mol
  43. 43. Lattice Energy
  44. 44. Lattice EnergyLattice energy can be calculated with at form of Coulomb‟s law: æ Q1Q2 ö LatticeEnergy = k ç è r ÷ ø Q is the charges on the ions and r is the shortest distance between the centers of the cations and anions. k is a constant that depends on the structure of the solid and the electron configurations of the ions.
  45. 45. Partial Ionic Character of Covalent Bonds
  46. 46. Bond Character Calculations of ionic character: æ dipole moment of x - y öPercent ionic character of a bond = ç + y ÷ x100% è dipole moment of x y ø Even compounds with the maximum possible electronegativity differences are not 100% ionic in the gas phase. Therefore the operational definition of ionic is any compound that conducts an electric current when melted will be classified as ionic.
  47. 47. Bond Character
  48. 48. The Covalent Chemical Bond
  49. 49. Chemical Bond ModelA chemical bond can be viewed as forces that cause a group of atoms to behave as a unit. Bonds result from the tendency of a system to seek its lowest possible energy. Individual bonds act relatively independent.
  50. 50. ExampleIt takes 1652 kJ of energy required to break thebonds in 1 mole of methane.1652 kJ of energy is released when 1 mole ofmethane is formed from gaseous atoms.Therefore, 1 mole of methane in gas phase has1652 kJ lower energy than the total of theindividual atoms.One mole of methane is held together with 1652kJ of energy.Each of the four C-H bonds contains 413 kJ ofenergy.
  51. 51. ExampleEach of the four C-H bonds contains 413 kJof energy.CH3Cl contains 1578 kJ of energy: 1 mol of C-Cl bonds + 3 mol (C-H bonds)=1578 kJ C-Cl bond energy + 3 (413 kJ/mol) = 1578 kJ C-Cl bond energy = 339 kJ/mol
  52. 52. Properties of ModelsA model doesn‟t equal reality; they are usedto explain incomplete understanding of hownature works.Models are often oversimplified and aresometimes wrong.Models over time tend to get overcomplicated due to “repairs”.
  53. 53. Properties of ModelsRemember that simple models often requirerestrictive assumptions. Best way to usemodels is to understand their strengths andweaknesses.We often learn more when models areincorrect than when they are right. Cu and Cr.
  54. 54. Covalent Bond Energiesand Chemical Reactions
  55. 55. Bond EnergiesBond energy averages are used for individual bond dissociation energies to give approximate energies in a particular bond. Bond energies vary due to several reasons: multiple bonds, 4 C-H bonds in methane different elements in the molecule, C-H bond in C2H6 or C-H bond in HCCl3
  56. 56. Bond Energy ExampleCH4(g)CH3(g) + H(g) 435 kJCH3(g)CH2(g) + H(g) 453 kJCH2(g)CH(g) + H(g) 425 kJCH(g)C(g) + H(g) 339 kJ Total 1652 kJ Average 413 kJ
  57. 57. Bond Energy ExampleHCBr3 380 kJHCCl3 380 kJHCF3 430 kJC2H6 410 kJ
  58. 58. Average Bond Energies
  59. 59. Bond EnergyA relationship also exists between the number of shared electron pairs. single bond – 2 electrons double bond – 4 electrons triple bond – 6 electrons
  60. 60. Bond EnergyBond energy values can be used to calculate approximate energies for reactions. Energy associated with bond breaking have positive signs Endothermic process Energy associated with forming bonds releases energy and is negative. Exothermic process
  61. 61. Bond EnergyA relationship exists between the number of shared electron pairs and the bond length.• As the number of electrons shared goes up the bond length shortens.
  62. 62. Bond Energy
  63. 63. Bond EnergyΔH = sum of the energies required to break old bonds (positive signs) plus the sum of the energies released in the formation of new bonds (negative signs).DH = Sn x D(bonds broken) - ån x D(bonds formed)• D represents bond energies per mole and always has positive sign• n is number of moles
  64. 64. Bond Energy ExampleH2(g) + F2(g) 2HF(g) 1 H-H bond, F-F bond and 2 H-F bonds ΔH = DH-H + DF-F – 2DH-F ΔH= (1mol x 432 kJ/mol) + (1mol x 154 kJ/mol) – (2mol x 565 kJ/mol) ΔH = -544 kJ
  65. 65. The Localized Electron Bonding Model
  66. 66. Localized Electron ModelThe localized electron model assumes that amolecule is composed of atoms that arebound together by sharing pairs of electronsusing the atomic orbitals of the bound atoms.Electrons are assumed to be localized on aparticular atom individually or in the spacebetween atoms.
  67. 67. Localized Electron ModelPairs of electrons that are localized on anatom are called lone pairs.Pairs of electrons that are found in the spacebetween the atoms are called bonding pairs
  68. 68. Localized Electron ModelThree parts of the LE Model:1. Description of the valence electron arrangement in the molecule using Lewis structures.2. Prediction of the geometry of the molecule using VSEPR model3. Description of the type of atomic orbitals used by the atoms to share electrons or hold lone pairs.
  69. 69. Lewis Structures
  70. 70. Lewis StructuresThe Lewis structure of a molecule show how the valence electrons are arranged among the atoms in the molecule. Named after G. N. Lewis Rules are based on observations of thousands of molecules. Most important requirement for the formation of a stable compound is that the atoms achieve noble gas electron configurations.
  71. 71. Lewis StructuresOnly the valence electrons are included.The duet rule: diatomic molecules can findstability in the sharing of two electrons.The octet rule: since eight electrons arerequired to fill these orbitals, these elementstypically are surrounded by eight electrons.
  72. 72. Lewis Structure Steps1. Sum the valence electrons from all the atoms. Total valence electrons.2. Use a pair of electrons to form a bond between each pair of bound atoms.3. Arrange the remaining electrons to satisfy the duet rule for hydrogen and the octet rule for the others. a) Terminal atoms first. b) Check for happiness
  73. 73. ExamplesHFN2NH3CH4CF4NO+
  74. 74. Exceptions to the Octet Rule
  75. 75. Exceptions to the Octet RuleIncomplete: An odd number of electrons areavailable for bonding. One lone electron is leftunpaired.Suboctet: Less than 4 pairs of electrons areassigned to the central atom Suboctets tend to form coordinate covalentbonds BH3 + NH3
  76. 76. Exceptions to the Octet RuleExtended: The central atom has more than 4 pairsof electrons. At the third energy level and higher, atoms may have empty d orbitals that can be used for bonding.
  77. 77. General RulesThe second row elements C, N, O, and F alwaysobey the octet ruleThe second row elements B and Be often havefewer than eight electrons around them in theircompounds. They are electron deficient and veryreactive.The second row elements never exceed the octetrule, since their valence orbitals can only hold 8.
  78. 78. General RulesThird-row and heavier elements often satisfy theoctet rule but can exceed the octet rule by usingtheir empty valence d orbitals.When writing the Lewis structure for a molecule,satisfy the octet rule for the atoms first. Ifelectrons remain after the octet rule has beensatisfied, then place them on the elements havingavailable d orbitals
  79. 79. Resonance
  80. 80. ResonanceResonance is when more than on valid Lewisstructure can be written for a particularmolecule. The resulting electron structure ofthe molecule is given by the average of theseresonance structures.
  81. 81. ResonanceThe concept of resonance is necessary becausethe localized electron model postulates thatelectrons are localized between a given pair ofatoms. However, nature does not really operatethis way. Electrons are really delocalized- theymove around the entire molecule. The valenceelectrons in a resonance structure distributethemselves equally and produce equal bonds.
  82. 82. Formal ChargeSome molecules or polyatomic ions can have several non-equivalent Lewis structures.• Example: SO42-Because of this we assign atomic charges to the molecules in order to find the right structure.
  83. 83. Formal ChargeThe formal charge of an atom in a molecule is the difference between the number of valence electrons on the free atom and the number of valence electrons assigned to the atom in the moleculeFormal charge = (# of valence electrons on neutral „free atom‟) – (# of valence electrons assigned to the atom in the molecule)
  84. 84. Formal ChargeAssumptions on electron assignment: Lone pair electrons belong entirely to the atom in question. Shared electrons are divided equally between the two sharing atoms.
  85. 85. Formal Charge ExampleSO42-: All single bondsFormal charge on each O is -1Formal charge on S is 2
  86. 86. Formal Charge ExampleSO42-: two double bonds, two singleFormal charge on single bonded O is -1Formal charge on double bonded O is 0Formal charge on S is 0
  87. 87. Formal Charges1. Atoms in molecules try to achieve formal charges as close to zero as possible.2. Any negative formal charges are expected to reside on the most electronegative atoms.If nonequivalent Lewis structures exist for a species, those with formal charges closest to zero and with any negative formal charges on the most electronegative atoms are considered to best describe the bonding in the molecule or ion.
  88. 88. Molecular Structure: The VSEPR Model
  89. 89. VSEPRValence shell electron repulsion model is useful in predicting the geometries of molecules formed from nonmetals. The structure around a given atom is determined principally by minimizing electron – pair repulsion.
  90. 90. VSEPRFrom the Lewis structure, count the electronpairs around the central atom.Lone pairs require more room than bonding pairsand tend to compress the angles between thebonding pairs.Multiple bonds should be counted as oneeffective pair.With a molecule with resonance, all structuresshould yield the same shape.
  91. 91. Linear180°3-D linear
  92. 92. Trigonal Planer120°3-D trigonal planar3-D trigonal w/lone pair
  93. 93. Tetrahedral109.5°3-D tetrahedral
  94. 94. Trigonal Pyramidal107°3-D tetrahedral 1 lone pair / trigonal pyramidal
  95. 95. Bent/V104.5°3-D tetrahedral 2 lone pair / bent
  96. 96. Tetrahedral Arrangements
  97. 97. Bipyramidal Arrangementstrigonal bipyramidalbipyramidal 1 lone pair / see sawbipyramidal 2 lone pair / T shapebipyramidal 3 lone pair / linear
  98. 98. Octahedral Arrangementsoctahedraloctahedral 1 lone pair / square pyramidaloctahedral 2 lone pair / square planar
  99. 99. Molecules without a central atomThe molecular structure of more complicatedatoms can be predicted from thearrangement of pairs around the centeratoms. A combination of shapes will resultthat allows for minimum repulsionthroughout.
  100. 100. Molecules without a central atom
  101. 101. The End