Ch9 z5e orbitals unhidden

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  • Section 9.1 Hybridization and the Localized e- model Z5e 415
  • The 4 Hydrogen “s” orbital e- share with carbon’s 2s and its three 2p orbitals. The p orbitals are 90 degrees perpendicular in space along the x, y and z axis. So, would expect 90 degree bond angels. But, the bond angles are actually 109.5. So, the s and p orbitals must hybridize to something giving equal bond angles of 109.5 degrees.
  • Fig 9.3 Z5e 417; 1st half
  • Fig 9.3 Z5e 417; 2nd half
  • Fig 9.3 Z5e 417
  • Fig 9.5 Z5e 417 Hybridization is intermediate in energy
  • Fig 9.8 1st half Z5e 419
  • Fig 9.8 2nd half Z5e 419
  • Fig 9.8 Z5e 419
  • Fig 9.9 Z5e 419
  • Figs 9.10 & 9.11 Z5e 420
  • Fig. 9.13 Z5e 420 Note the sp 2 carbon orbitals with the 1s hydrogen orbitals
  • Fig. 9.16 Z5e 421
  • Z5e 420
  • Z5e 422
  • Z5e 423
  • Fig 9.21 Z5e 423
  • Although assume Cl is sp 3 , P is dsp 3
  • Z5e 424
  • Z5e 425
  • Fig 9.24 Z5e 426
  • Section 9.2 The Molecular Orbital Model; Z5e 428 Bond order is the important thing here for AP exam (a few MC questions)
  • Z5e 429 Fig. 9.25 The combination of hydrogen 1 s atomic orbitals to form molecular orbitals.
  • Z5e 429
  • Z5e 430. The 1s on either side are the atomic orbitals in a free H atom.
  • Z5e 431 Fig. 9.29. The molecular orbital energy-level diagram for the H 2 1- ion.
  • Go to Web: http://learn.chem.vt.edu/tutorials/bonding/mo2.html (defunct?) Try http://www.wwnorton.com/chemistry/tutorials/ch6.htm
  • Z5e 431
  • Z5e 431
  • Z5e 431
  • Z5e section 9.3 Bonding in Homonuclear Diatomic Molecules
  • Z5e 438 Section 9.4 Bonding in Heternoculear Diatomic Molecules
  • Z5e 440. Section 9.5 Combining the Localized Electron and Molecular Orbital Models.
  • Ch9 z5e orbitals unhidden

    1. 1. Chapter 9 Orbitals and Covalent Bonds pp
    2. 2. 9.1 Atomic Orbitals Don’t Work <ul><li>They don’t explain molecular geometry. </li></ul><ul><li>In methane, CH 4 , the shape is tetrahedral. </li></ul><ul><li>The valence electrons of carbon should be two in s , and two in p ( if atomic orbitals “worked.” </li></ul><ul><li>If so, the p orbitals would have to be at right angles . But bond angles are 109.5 o </li></ul><ul><li>So, the atomic orbitals change when making a molecule </li></ul>
    3. 3. Hybridization <ul><li>We must blend the s and p orbitals of the valence electrons to get 109.5º angles </li></ul><ul><li>We end up with the tetrahedral geometry. </li></ul><ul><li>We combine one s orbital and 3 p orbitals to get sp 3 molecular orbitals. </li></ul><ul><li>sp 3 hybridization has tetrahedral geometry. </li></ul>
    4. 4. Figure 9.3 p. 417
    5. 5. Fig. 9.3 p. 417 - Tetrahedral
    6. 6. Figure 9.3 p. 417 The Formation of sp 3 Hybrid Orbitals
    7. 7. In terms of energy (LD 2:6.65) Energy 2p 2s Hybridization sp 3
    8. 8. How we get to hybridization <ul><li>We know the geometry from experiment. </li></ul><ul><li>We know the orbitals of the atom </li></ul><ul><li>hybridizing atomic orbitals can explain the geometry . </li></ul><ul><li>So, if the geometry requires a tetra hedral shape, it is sp 3 hybridized. </li></ul><ul><li>This includes bent and trigonal pyramidal molecules because one or more of the sp 3 lobes holds the lone pair(s). </li></ul>
    9. 9. sp 3 geometry <ul><li>This leads to tetrahedral shape. </li></ul><ul><li>Every molecule with a total of 4 things coming off the central atom is sp 3 hybridized. </li></ul><ul><li>Gives us trigonal pyramidal and bent linear shapes also . </li></ul>109.5º
    10. 10. sp 2 hybridization <ul><li>When three effective things come off atom (either a bonding pair or non-bonding pair) </li></ul><ul><li>Electron arrangement is (gumdrops) . . . </li></ul><ul><li>Trigonal planar. Angles are . . . </li></ul><ul><li>120º </li></ul><ul><li>One  and one  bond (from the extra p orbital) </li></ul>
    11. 11. sp 2 hybridization <ul><li>C 2 H 4 (you draw Lewis structure . . .) </li></ul><ul><li>double bond acts as one “effective” pair </li></ul><ul><li>Molecular geometry is . . . </li></ul><ul><li>So, trigonal planar. </li></ul><ul><li>Have to end up with three blended (hybridized) orbitals. </li></ul><ul><li>use one s and two p orbitals to make sp 2 orbitals. </li></ul><ul><li>leaves one p orbital perpendicular </li></ul>
    12. 12. Figure 9.8, p. 419
    13. 13. Fig. 9.8 p. 419
    14. 14. Figure 9.8 The Hybridization of the s , p x , and p y Atomic Orbitals
    15. 15. In terms of energy (fig. 9.9 p. 419) Energy 2p 2s sp 2 Hybridization 2p
    16. 16. Where is the P orbital? <ul><li>Perpendicular. </li></ul><ul><li>The overlap of sp 2 orbitals makes a sigma bond (  bond). </li></ul>
    17. 17. Two types of Bonds <ul><li>Sigma (  ) bonds from overlap of orbitals </li></ul><ul><li>They are between the atoms </li></ul><ul><li>Pi bond (  bond) above and below atoms and </li></ul><ul><li>Between adjacent p orbitals . </li></ul><ul><li>A  bond makes the first and a  bond makes the second bond of a double bond </li></ul>
    18. 18. Fig. 9.13, p. 420 C C H H H H
    19. 19. What about two <ul><li>When two things come off </li></ul><ul><li>one s and one p hybridize (leaving two p orbitals). Get sp hybridization </li></ul><ul><li>linear </li></ul>
    20. 20. sp hybridization <ul><li>End up with two lobes 180º apart. </li></ul><ul><li>The two remaining p orbitals are at right angles </li></ul><ul><li>Makes room for two  bonds. </li></ul><ul><li>So, get a triple bond or two double bonds. </li></ul>
    21. 21. Summary of sigma & pi bonds pp <ul><li>Single bond - composed of one  comprised of overlapping orbitals  </li></ul><ul><li>Double bond - composed of one  and one  is between p orbitals)  </li></ul><ul><li>Triple bond - composed of one  and two  bonds. </li></ul><ul><li>Know this for the online homework! </li></ul>
    22. 22. In terms of energy Energy 2p 2s Hybridization sp 2p
    23. 23. CO 2 <ul><li>C can make two  and two  </li></ul><ul><li>O can make one  and one  </li></ul>C O O
    24. 24. N 2
    25. 25. N 2
    26. 26. Breaking the octet <ul><li>PCl 5 </li></ul><ul><li>The model predicts that we must use the d orbitals (ran out of s and p ). </li></ul><ul><li>Get dsp 3 hybridization </li></ul><ul><li>There is some controversy about how involved the d orbitals are. </li></ul>
    27. 27. dsp 3 <ul><li>Trigonal bipyrimidal </li></ul><ul><li>can only  bond. </li></ul><ul><li>can’t  bond (no unused p orbitals). </li></ul><ul><li>The basic shape for five “things.” </li></ul>
    28. 28. PCl 5 Can’t tell the hybridization of Cl Assume it’s sp 3 to minimize repulsion of electron pairs .
    29. 29. d 2 sp 3 <ul><li>Gets us to six things around the central atom. </li></ul><ul><li>Octahedral. </li></ul>
    30. 30. Localized e- model: Summary pp <ul><li>Draw the Lewis structure(s) (resonance) </li></ul><ul><li>Determine arrangement of e - pairs using the VSEPR model. </li></ul><ul><li>Specify hybrid molecular orbitals need to accommodate the e - pairs. </li></ul><ul><li>Do the steps in this order! </li></ul><ul><li>http://www.bluffton.edu/~bergerd/classes/CEM222/Handouts/spanimation.html </li></ul>
    31. 31. p. 426 Figure 9.24 The Relationship of the Number of Effective Pairs, Their Spatial Arrangement, and the Hybrid Orbital Set Required (end 9.1) Effective pairs are also known as RHED (Regions of High Electron Density) - use for online HW
    32. 32. (Partial) 9.2 Bond Order <ul><li>The number of bonding e - pairs shared by two atoms in a molecule. </li></ul><ul><li>Can be 1, 2, 3 or fractional. </li></ul><ul><li>E.g., CH 4 = 1, CO 2 = 2, N 2 = 3. </li></ul><ul><li>What about ozone (O 3 ) (wtbd structure) </li></ul><ul><li>B.O. = # shared pairs linking X & Y # X--Y links in the compound </li></ul><ul><li>For ozone, this = 3/2 = 1.5 </li></ul><ul><li>Higher bond order means shorter , stronger bond. </li></ul>
    33. 33. 9.2 Molecular Orbital Model <ul><li>The Localized Model we have learned explains much about bonding. </li></ul><ul><li>Doesn’t do well with the ideas of resonance , un paired electrons, and bond energy . </li></ul><ul><li>The MO model is a parallel of the atomic orbital, using quantum mechanics. </li></ul><ul><li>Each MO can hold two electrons with opposite spins. </li></ul><ul><li>Square of wave function tells probability. </li></ul><ul><li>Bond order is what’s important for AP test (a few MC questions) </li></ul>
    34. 34. What do you get? <ul><li>Solve the two equations for H 2 </li></ul><ul><li>Combine Hydrogen 1s orbitals to get molecular orbitals </li></ul><ul><li>H A H B </li></ul><ul><li>Get the possibility of two orbitals. </li></ul><ul><li>MO 2 = 1s A - 1s B </li></ul><ul><li>MO 1 = 1s A + 1s B </li></ul><ul><li>Which is it? (next slide) </li></ul>
    35. 35. The Molecular Orbital Model <ul><li>The molecular orbitals are centered on a line through the nuclei. For: </li></ul><ul><ul><li>MO 2 the greatest probability is on either side of the nuclei </li></ul></ul><ul><ul><li>MO 1 it is between the nuclei </li></ul></ul><ul><ul><li>Either type of e - distribution is called a sigma molecular orbital. </li></ul></ul>
    36. 36. The Molecular Orbital Model <ul><li>In the molecule only the molecular orbitals exist, the atomic orbitals are gone . </li></ul><ul><li>MO 1 is lower in energy than the 1s orbitals it came from. </li></ul><ul><ul><li>So, this favors molecule formation (top figure) </li></ul></ul><ul><ul><li>Called a bonding orbital </li></ul></ul><ul><li>MO 2 is higher in energy </li></ul><ul><ul><li>This goes against bonding </li></ul></ul><ul><ul><li>anti bonding orbital (bottom figure) </li></ul></ul>
    37. 37. The H 2 Molecular Orbital Model Energy MO 2 MO 1  s  s
    38. 38. The Molecular Orbital Model <ul><li>We use labels to indicate shapes and if the MO’s are bonding or antibonding. </li></ul><ul><ul><li>MO 1 =  1s </li></ul></ul><ul><ul><li>MO 2 =  1s * (* indicates anti bonding) </li></ul></ul><ul><li>Can write them the same way as atomic orbitals </li></ul><ul><ul><li>H 2 =  1s 2 </li></ul></ul>
    39. 39. The Molecular Orbital Model <ul><li>Each MO can hold two electrons, but they must have opposite spins. </li></ul><ul><li>Orbitals are conserved. The number of molecular orbitals must equal the number of atomic orbitals that are used to make them. </li></ul>
    40. 40. H 2 1- ion Energy  1s  1s *  s  s
    41. 41. Molecular Orbital Summary pp <ul><li>1. Molecular orbitals are either bonding or antibonding; </li></ul><ul><li>2. Bonding orbitals are lower in energy than corresponding atomic orbitals while antibonding orbitals are higher; </li></ul><ul><li>3. MO theory helps explain paramagnetism & diamagnetism; </li></ul><ul><li>4. MO theory obviates &quot;resonance&quot;, replacing it with delocalized pi bonding. </li></ul><ul><li>http://www.wwnorton.com/chemistry/tutorials/ch6.htm </li></ul>
    42. 42. Magnetism pp <ul><li>Magnetism has to do with electrons. </li></ul><ul><li>Paramagnetism attracted by a magnet. </li></ul><ul><li>Associated with un paired electrons (i.e., any lone electrons means “para”) . </li></ul><ul><li>Diamagnetism repelled by a magnet. </li></ul><ul><li>Associated with paired electrons (i.e., all the electrons are in pairs) . . </li></ul><ul><li>B 2 is diamagnetic. </li></ul>
    43. 43. Bond Order pp <ul><li>The difference between the number of bonding electrons and the number of anti bonding electrons divided by two. </li></ul>
    44. 44. Bond Order pp <ul><li>The difference between the number of bonding electrons and the number of anti bonding electrons divided by two. </li></ul><ul><li>E.g., H 2 has 2 bonding electrons (the covalent bond) and 0 anti bonding (no other electrons). </li></ul><ul><li>B.O. = (2 - 0) ÷ 2 = 1 </li></ul><ul><li>H 2 1- ion has 2 bonding + 1 antibonding </li></ul><ul><li>B.O. = (2 - 1) ÷ 2 = 1/2 </li></ul><ul><li>The non-ion is more stable because it has the higher bond order. </li></ul>
    45. 45. Bond Order pp <ul><li>Use bond order to explain why He is monoatamic vs. diatomic. </li></ul><ul><li>He bond order = (2 - 0) ÷ 2 = 1 </li></ul><ul><li>He 2 has 2 bonding and 2 anti-bonding </li></ul><ul><li>(2 - 2) ÷ 2 = 0 </li></ul><ul><li>So He is monoatomic is more stable because it has the higher bond order. </li></ul><ul><li>(End of our part of 9.2, but go on) </li></ul>
    46. 46. Molecular Orbitals Part 2
    47. 47. More on M.O. Theory <ul><li>The 1s orbital is much smaller than the 2s orbital </li></ul><ul><li>When only the 2 s orbitals are involved in bonding </li></ul><ul><li>Don’t use the  1s or  1s s* for He -2 (added 2 e - s) </li></ul><ul><li>He -2 (  2 s ) 2 (  2 s *) 2 </li></ul><ul><li>In order to participate in bonds the orbitals must overlap in space. </li></ul>
    48. 48. 9.3 Bonding in Homonuclear Diatomic Molecules <ul><li>Need to use Homonuclear so that we now can compare the relative energies. </li></ul><ul><li>E.g., Li 2 </li></ul><ul><li>(  2s ) 2 (  2s *) 2 </li></ul><ul><li>What about the p orbitals? </li></ul><ul><li>How do they form orbitals? </li></ul><ul><li>Remember that orbitals must be conserved. </li></ul>
    49. 49. B 2
    50. 50. B 2  2p *  2p  2p *  2p
    51. 51. Expected Energy Diagram Energy 2s 2s 2p 2p  2s  2p *  2p  2s *  2p *  2p *  2p  2p
    52. 52. B 2 Energy 2s 2s 2p 2p
    53. 53. B 2 <ul><li>(  2s ) 2 (  2s *) 2 (  2p ) 2 </li></ul><ul><li>Bond order = (4-2) / 2 </li></ul><ul><li>Should be stable. </li></ul><ul><li>This assumes there is no interaction between the s and p orbitals. </li></ul><ul><li>Hard to believe since they overlap. </li></ul><ul><li>proof comes from magnetism. </li></ul>
    54. 54. Magnetism <ul><li>Magnetism has to do with electrons. </li></ul><ul><li>Paramagnetism attracted by a magnet. </li></ul><ul><li>Associated with un paired electrons (i.e., any lone electrons means “para”) . </li></ul><ul><li>Diamagnetism repelled by a magnet. </li></ul><ul><li>Associated with paired electrons (i.e., all the electrons are in pairs) . . </li></ul><ul><li>B 2 is diamagnetic. </li></ul>
    55. 55. Magnetism <ul><li>The energies of of the  2p and the  2p are reversed by s-p mixing. </li></ul><ul><li>The  2s and the  2s * are no longer equally spaced. </li></ul><ul><li>Here’s what it looks like. </li></ul>
    56. 56. Correct energy diagram 2s 2s 2p 2p  2s  2p *  2p  2s *  2p *  2p *  2p  2p
    57. 57. B 2 2s 2s 2p 2p
    58. 58. Patterns <ul><li>As bond order increases, bond energy increases. </li></ul><ul><li>As bond order increases, bond length decreases. </li></ul><ul><li>Supports basis of MO model. </li></ul><ul><li>There is not a direct correlation of bond order to bond energy. </li></ul><ul><li>O 2 is known to be paramagnetic. </li></ul>
    59. 59. Examples <ul><li>C 2 </li></ul><ul><li>N 2 </li></ul><ul><li>O 2 </li></ul><ul><li>F 2 </li></ul><ul><li>P 2 </li></ul>
    60. 60. Finishing Chapter 9 Details, details, details
    61. 61. 9.4 Heteronuclear Diatomic Species <ul><li>Simple type has them in the same energy level, so can use the orbitals we already know. </li></ul><ul><li>Slight energy differences. </li></ul><ul><li>NO </li></ul>
    62. 62. NO 2s 2s 2p 2p
    63. 63. You try <ul><li>NO + </li></ul><ul><li>CN - </li></ul><ul><li>What if they come from completely different orbitals and energy? </li></ul><ul><li>HF </li></ul><ul><li>Simplify first by assuming that F only uses one if its 2p orbitals. </li></ul><ul><li>2p much higher than 1 s orbital. </li></ul>
    64. 64. 1s 2p  
    65. 65. Consequences <ul><li>Paramagnetic </li></ul><ul><li>Since 2p is lower in energy, favored by electrons. </li></ul><ul><li>Electrons spend time closer to fluorine. </li></ul><ul><li>Compatible with polarity and electroegativity. </li></ul>
    66. 66. 9.5 Names <ul><li>sp orbitals are called the Localized electron model </li></ul><ul><li> and   olecular orbital model </li></ul><ul><li>Localized is good for geometry, doesn’t deal well with resonance. </li></ul><ul><li>seeing  bonds as localized works well </li></ul><ul><li>It is the  bonds in the resonance structures that can move. </li></ul>
    67. 67.  delocalized bonding <ul><li>C 6 H 6 </li></ul>H H H H H H H H H H H H
    68. 68. C 2 H 6
    69. 69. NO 3 -

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