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  1. 1. Ch. 8 Bonding and Molecular Structure Ch. 9 Orbital Hybridization The content of Ch. 8 – 9 will be taught through both this PPT and LM 8-9.1 Text references: 8.1, 8.2, 8.4-8.9, 9.1-9.2
  2. 2. The rest of this PPT… • Focuses on substances with covalent bonds – Molecules – Polyatomic ions • For any molecule or PAion, you will learn to… – – – – – draw the Lewis Dot Structure (LDS) identify the 3D molecular geometry determine the polarity Identify hybridization and sigma/pi bonding (These are all components of LM 8-9.1)
  3. 3. Chapter 8 LM 8-9.1 Part I Lewis Dot Structures
  4. 4. How to draw LDS for molecules and ions: 1. 2. 3. 4. 5. 6. 7. Write formula for compound. Write symbol for least electronegative atom in center and write symbols for all other atoms around it. Add valence e- from all atoms to find out total # of e- you must assign to the compound. Draw one line (single bond) from central atom to all outer atoms. (1 line = 2 e-) Subtract 2e- for each line you’ve drawn from the total # valence e-. Give all OUTER atoms their complete “octet.” Subtract this value from total. If there are ANY remaining e-, assign them in pairs around the central atom. If there are no more e- but the central atoms still needs more to complete the octet, remove a lone pair from an outer atom and turn this into a double bond. Repeat if you need to form a triple bond. Ex. #1 Ex. #2
  5. 5. Example #1 • Draw LDS for nitrogen trifluoride: Formula = NF3 • •• • •F • •• N • •• F• •• •F• •• •• Total valence e- = 5 + 3(7) = 26eRemaining e- = 26 – 6 = 20eRemaining e- = 20 – 18 = 2eClick this link and go to Lewis Dot Structures for more info.
  6. 6. Example #2 • Draw LDS for carbon dioxide: Formula = CO2 O=C=O Since C did not have enough e- to complete the octet, each oxygen must donate a lone pair of e- to make a double bond.
  7. 7. Electron Deficient Atoms • Elements from Groups I, II, & III will not hold 8 e- when bonded valence e• Group I  has 21e- in bonding • Group II  has 4 2 valence ee- in bonding • Group III  has 6 3 valence ee- in bonding
  8. 8. Expanded Octet • Some atoms (usually group V and higher) can POTENTIALLY take on more than 8 e- in bonding – Group IV – VIII must always have at least 8 e-, however some will take more • Ex. AsF5 • As has 5 fluorine atoms bonded to it and thus is sharing 10 total e- with the outer atoms F F As F F F Click on this link and go to Expanded Octet for more info.
  9. 9. Resonance Structures • When there are equivalent LDS for a molecule or ion we say that the structure experiences resonance. • Applies to a symmetric molecule that contains at least one double bond • The double bond could be in multiple places and still have the same structure. Click this link and go to Resonance for more info.
  10. 10. Why Resonance Structures? • Electrons in molecules w/double bonds are often delocalized between two or more atoms. • Electrons in a single Lewis structure are assigned to specific atoms-a single Lewis structure is insufficient to show electron delocalization. • Composite of resonance forms more accurately depicts electron distribution. Molecule or ion doesn’t actually switch between resonance structures. The structure behaves as a blend or composite of all possible structures • Proof of resonance? – Bond lengths in resonance structures shorter than single bonds/longer than double bonds – Bonds are stronger than single bonds/weaker than double bonds
  11. 11. Resonance Structures • Draw all resonance structures for ozone (O3)
  12. 12. Chapter 8 LM 8-9.1 Part II Molecular Geometry Getting a 3D perspective of molecules… For a preview of what’s to come, click on this link and click on VSEPR Model.
  13. 13. Applying VSEPR to Balloons • Valence Shell Electron Pair Repulsion = model for predicting molecular geometry (3D arrangement of BP and LP) • Watch what your instructor can do w/balloons…
  14. 14. VSEPR • Valence Shell Electron Pair Repulsion model for predicting molecular geometry • Model describes the 3D arrangement of atoms in molecule based on # of lone pairs (LP) and bond pairs (BP) of e- on a central atom. • A molecule adopts the geometry that minimizes the repulsive force among the e- pairs on central atom • By minimizing repulsion around central atom you get the most stable geometry of a molecule (minimize repulsion to maximize stability)
  15. 15. For the next slide, draw LDS for: • BF3, O3 • CH4, NH3, H2O • PCl5, SF4, ClF3, XeF2 • What, does 5each grouping have in common? What SF6 BrF , XeF4 changes as you go from one grouping to the next? How might this effect the placement of e- or “geometry” according to VSEPR theory?
  16. 16. Applying VSEPR Theory • Lowest energy arrangement when there are: Linear 180o – 2BP, 0LP = 120o – 3BP, 0LP = Trigonal Planar – 4BP, 0LP = Tetrahedral 109.5o – 5BP, 0LP = Trigonal bipyramidal 120o, 90o – 6BP, 0LP = Octahedral 90o • Bond angle = angle formed when 3 atoms bond together
  17. 17. Linear Shapes Examples; Examples; BeH2 HF CO2 HCl HCN H2 Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
  18. 18. Trigonal Planer Geometry Examples; BF3 CO32COCl2 Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
  19. 19. Tetrahedral Geometry Examples; CH4 CCl4 SO42- Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
  20. 20. Trigonal Bipyramidal Geometry Examples; PF5 PCl5 Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
  21. 21. Octahedral Geometry Examples; SF6 SiF63- Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html
  22. 22. Applying VSEPR Theory (cont’d) • Lowest energy arrangement when there are lone pairs??? Leads to molecular geometry • LP’s push harder on surrounding BP’s – LP can spread out more in nonbonding orbital than BP can (BP more localized/constrained in space) – LP’s make bond angles slightly smaller than expected – The bond angle tends to decrease as # of LP increases • Steric # (total # of e- pairs) determines e- pair arrangement to minimize repulsion; #BP vs. LP determines name of molecular geometry (as opposed to electronic geometry); LP influence bond angles • See how this works…try link above
  23. 23. Steric # = 3 (3 Total Pair of E-) • 3BP, 0LP • 2BP, 1LP Trigonal Planar Bent <120o Examples; NO2SO2 O3
  24. 24. Steric # = 4 (4 Total Pair of E-) • 4BP, 0LP Tetrahedral • 3BP, 1LP Trigonal pyramidal <109.5o • 2BP, 2LP Bent <<109.5o Examples; NH3 NF3 PCl3 Examples; H2O ClO2 OF2
  25. 25. Steric # = 5 (5 Total Pair of E-) • 5BP, 0LP Trigonal bipyramidal • 4BP, 1LP Seesaw <90o Examples; SF4 <120o BrF4+
  26. 26. Steric # = 5 (contd) • 3BP, 2LP T-Shaped Examples; ClF3 BrF3 SeO32- <90o • 2BP, 3LP 180o Linear Examples; XeF2 ICl2I3 -
  27. 27. Steric # = 6 (6 Total Pair of E-) • 6BP, 0LP Octahedral • 5BP, 1LP <90o 4BP, 2LP Examples; BrF5 IF5 Square Pyramidal 90o Examples; XeF4 ICl4- Square Planar
  28. 28. What if there’s >1 central atom? 1 2 3 4 • Number each atom in the chain (this one has 4 central atoms) • Count the number of BP and LP on each central atom • Identify the geometry of each central atom (there’s no geometry for the whole structure) • C1: 4BP, 0LP  tetrahedral • C2, 3, 4: 3BP, 0LP  trigonal planar (each one)
  29. 29. Chapter 8 Lab 8-9.1 Part III Molecular Polarity To be polar or not to be polar, that is the question!
  30. 30. Bond Polarity vs. Molecular Polarity • Bond polarity – refers to distribution of e- cloud around 2 atoms in a bond – Polar if unequal distribution (dipole-one end of bond slightly pos and the other slightly neg) (has a dipole moment) – Nonpolar if equal distribution (has no dipole moment) • Determined by difference in EN values of atoms More info on polarity at this link and select Partial Charges and Bond Dipoles
  31. 31. Chemical Bond Formation • Chemical bond – results when a chem rxn occurs between 2 or more atoms and the valence e- reorganize so that a net attractive force occurs between them – Ionic Bonds • Transfer of valence e• Electrostatic attractn of opp. charged ions • Usu. btwn M + NM, M + PAion or PAion + NM – Covalent Bonds • Sharing of valence e• Electrostatic attractn of e- of one atom to nucleus of another • Usu. btwn NM + NM
  32. 32. Covalent Bond Formation
  33. 33. Electronegativity (EN) • Definition: a value that represents the relative ability of an atom to attract etowards itself in a bond – EN is NOT an energy – Scale ranges from 0 - 4 • What this means: larger EN = atom has greater attraction for e- in a bond
  34. 34. Back
  35. 35. Electronegativity (EN) • Trend: – Down Group = decrease – L to R in Period = increase • Explain Trend: – Down Group = inc shielding e- = decrease pull on own valence e- = decrease pull on any ein bond – L to R in Period = inc Z = increase pull on own valence e- = increase pull on any e- in bond
  36. 36. Electronegativity (EN) • Deviations: Group 8: Most Grp 8 elements tend to have EN = 0 b/c they have full outer en levels. This means they tend not to form bonds like other elements. Some Grp 8 elements do have EN values, though, b/c they are large enough to form bonds.
  37. 37. Is bond ionic or polar? • Difference in EN values between two atoms will determine if bond is Ionic or Covalent (found in Ch. 8.7) • On a continuum of ∆EN: 100% covalent 0 0.4 Nonpolar 0% ionic covalent 0% covalent 1.7 Polar covalent 4 Ionic 100% ionic • Can also use distance on PT as indicator of bond type: the closer two atoms are, the more covalent the bond; the farther apart they are, the more ionic the bond
  38. 38. Is bond ionic or polar?
  39. 39. Bond Polarity vs. Molecular Polarity • Molecular polarity – related to distribution of e- cloud around entire molecule – Polar molecule – experiences unequal distribution of e- cloud (aka: dipole – one end of molecule slightly + and other end slightly -) (has a dipole moment) – Nonpolar molecule – experiences equal distribution of e- cloud (no dipole moment) • Determined by symmetry of molecule (geometry and arrangement of LP) • LP tend to cause asymmetry in structure which results in a shift in the e- cloud
  40. 40. What is a dipole moment? • A substance possesses a dipole moment if its centers of positive and negative charge do not coincide and cancel each other out. µ=exd — + (expressed in Debye units) (you won’t need to use this) polar • The larger the dipole moment, the more polar is the bond or molecule (due to a greater asymmetry in the distribution of electrons around the bond or molecule) • The greater the ∆EN, the larger the dipole moment for a bond.
  41. 41. Why is water polar? • H—O bonds (polar bonds) are not “balanced” around the structure (the pull of each bond does not cancel the other) • Lone pairs cause asymmetry (imbalance) • E- cloud of molecule is shifted such that there is greater neg charge near O and greater pos charge by H
  42. 42. General Rule for Molecular Polarity • If all outer atoms are same and structure is symmetrical (dipole moments of bonds “cancel” each other out b/c same magnitude but opposite direction)  molecule is NP (no overall dipole moment) – If 0 LP then NP (if all outer atoms same) • If structure is asymmetrically arranged (has LP or different outer atoms) molecule is Polar (has dipole moment) – If has LP* then polar – *Exception: Square planar (4 BP, 2LP) and Linear (2 BP, 3 LP) are NP even though they have LP – LP’s cancel each other out
  43. 43. Attractive forces in & btwn molecules • Intramolecular forces = attractive forces within the molecule (bonds) • Intermolecular forces (IMF) = attractive forces between molecules (sticky factors) • Properties of covalent compounds are attributed to their IMF – Ex. Polar cmpds tend to have higher MP and BP due to stronger IMF
  44. 44. Chapter 9 Lab 8-9.1 Part IV Hybridization The “real story” behind bonding… For a preview of what’s to come, click on this link and select Hybridization.
  45. 45. Draw the LDS of CH4: What is the e- configuration for C? for H? Carbon = 2s2 2p2 s s s s px py pz H = 1s1 s Is there really room for each H to share e- with Carbon?
  46. 46. Is there really room for Hydrogen to share 4 ewith Carbon? Carbon = 2s2 2p2 s s s s px py pz H = 1s1 s No! Carbon must make room for 2 more e- so it promotes an e- (moves it to a p orbital) s px py pz
  47. 47. Here’s what it would look like so far: s p p p H = 1s1 Carbon = 2s2 2p2
  48. 48. And then if Hydrogen bonds with Carbon: Remember that each p orbital is on an X, Y or Z axis at 90 o angles from one another. Also remember that the s orbital is smaller in radius than the p orbitals.
  49. 49. Evidence that something “different” is happening… • Each C-H bond in methane has the same length (109 pm) • Each H-C-H bond angle is the same (109.5 o) • How is this possible if : – the s and p orbitals are different sizes from one another (this would lead to bonds having different lengths)? – The s and p orbitals are at 90o not at 109.5o angles
  50. 50. Then answer is….Hybridization • All of the s, p atomic orbitals mix to create new hybrid orbitals • # of new hyrbrid orbitals = # of atomic orbitals that were mixed EX. If s and p atomic orbitals mix get 2 new hybrid orbitals, each called sp hybrid • Each hybrid orbital is exactly like the other and the geometry is based on these (not atomic orbitals)
  51. 51. More on Hybridization… • Identify the hybridization if the following atomic orbitals are mixed: – s + p + p sp2 hybridization – s + p + p + p + d sp3d hybridization • What geometry could you expect from the above hybridization? 3 hybrids (3BP or 2BP, 1LP) 5 hybrids (5BP or 4BP, 1LP, or 3BP, 2 LP or 2BP, 3LP) Hybrids can contain LP or BP. # BP & LP determines geometry. Even if you know hybridization, must know what each orbital contains to determine geometry.
  52. 52. BeF2 http://www.mikeblaber.org/oldwine/chm1045/notes/Geometry/Hybrid/Geom05.htm
  53. 53. BH3
  54. 54. NH3 H2O
  55. 55. Sulfur difluoride 1. 2. 3. 4. Draw the LDS. Identify the geometry. Bent Is the molecule polar? Yes, it is polar What is the hybridization of the central atom? 2BP+2LP = 4 hybrids = sp3 hybridization ••S•• F F
  56. 56. Xenon tetrafluoride 1. 2. 3. 4. Draw the LDS. Identify the geometry. Squar Planar Is the molecule polar? No, it is not polar What is the hybridization of the central atom? 4BP+2LP = 6 hybrids = sp3d2 hybridization F F •• Xe •• F F
  57. 57. Sigma and Pi Bonding • Draw LDS and Label all sigma and pi bonds in each structure: • Ethane (C2H6) • Ethene (Ethylene) (C2H4) • Ethyne (Acetylene) (C2H2) Go to this link for more info about this section.
  58. 58. Sigma and Pi Bonding • Ethane (C2H6)
  59. 59. Sigma and Pi Bonding • Ethene (Ethylene) (C2H4)
  60. 60. Sigma and Pi Bonding • Ethyne (Acetylene) (C2H2)
  61. 61. Sigma and Pi Bonding • Sigma Bonding (σ) – head to head overlap of hybrid or unhybridized orbitals *All single bonds consist of σ bonds. *One of the double bonds is a σ bond. • Pi Bonding (π) – side to side overlap of unhybridized p bonds (contains 2 e- total) *The second and third bond of a double/triple bond are π bonds. *Can only occur when central atom is sp or sp2 hybridized
  62. 62. Benzene (C6H6) b/c of resonance we write…
  63. 63. Covalent Bond Characteristics • Bond Order: The number of bonds between two atoms What’s the BO in a resonance structure? • C—C • C==C •C C BO = 1 BO = 2 BO = 3 • Bond Energy: Energy a bond must absorb to break (sign?) • Bond Length: Avg distance between the centers of two nuclei (Next slide…) • How are they all related?
  64. 64. Determining Bond Length
  65. 65. Factors Affecting Bond Length Larger atoms = longer bonds
  66. 66. How are they all related? ___BL = ___BE indirect ___BO = ___BE ___BO = ___BL direct indirect
  67. 67. Using Bond Energies to find ∆Hrxn • Draw the dot structures of the reactants and the products • Determine the energy needed to BREAK all the bonds in the reactants (Endothermic, positive value) • Determine the energy change to MAKE all the bonds in the products (Exothermic, negative value) • Add them together to get an approximate value for the ∆H of the reaction
  68. 68. Using Bond Energies to find ∆Hrxn • Determine the ∆H for the combustion of methane gas. (use BE values on next slide) • What is the ∆Hrxn when you use ∆Hf values?
  69. 69. CH4 + 2O2  CO2 + 2H2O Notice these are pos energies b/c bonds are breaking!!! BEreactants = 4(C-H) + 2(O=O) = 4(413kJ) + 2(498 kJ) BEreactants = 2648 kJ BEproducts = 2(C=O) + 4(H-O) = 2(-745 kJ) + 4(-463 kJ) Notice these are neg energies b/c BEproducts = -3342 kJ bonds are being made!!! ∆H Rxn = BEproducts + BEreactants = -694 kJ
  70. 70. More links • Click this link for a site that provides an overview of this information • The first 2min 15sec of this video gives a great visual representation of the big picture (where this unit fits into all of chemistry) • A song about ionic and covalent bonds

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