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See Handout on Website

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Trigonal Planar. sp2. 4. Tetrahedral. sp3. 5. Trigonal bypramidal. dsp3. 6. Octahedral. d2sp3 ... Number of Electron Pairs, not necessarily Molecular Geometry ... – PowerPoint PPT presentation

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Title: See Handout on Website


1
See Handout on Website
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See Handout on Website
3
To Determine Hybridization
  • Steps (page 670)
  • Draw Lewis Structures
  • Determine Arrange of Electron Pairs
  • Specify Hybrid Orbitals

Amount of hybridization matches number of
effective pairs i.e. s p2 3 pairs
4
Molecular Orbital Model
  • A different model for describing bonding
  • Hybridization an extension of Lewis Structures
    and VSEPR theory
  • Molecular Orbitals an extension of atomic
    orbitals (chapter 12)

5
Atomic Orbitals A Refresher
  • Shapes that describe the probability of finding
    an electron
  • Orbitals have different energies
  • Electrons fill lower energy orbitals first (in
    pairs)

6
Atomic Orbital Energies
7
Molecular Orbitals
  • Different combinations of atomic orbitals
  • Describe the probability of finding an electron
    in a molecular bond
  • Each can hold two electrons

8
Molecular Orbitals of H2
  • Two atomic orbitals form two molecular orbitals,
    by addition and subtraction

9
  • Symmetric around line through nuclei (s)
  • 1s orbitals no longer exist replaced by MOs
  • MO1 is lower energy than AO, MO2 is higher
  • MO1 is the Bonding MO (s1s)
  • MO2 is the Antibonding MO (s1s)

10
Bond Order
  • Bond Order The number of bonds between a pair of
    atoms
  • From Lewis Structures O O bond order 2
  • O S O ?? O S O bond order 1.5
  • From MO Theory
  • bonding electrons antibonding electrons
  • 2

Bond Order
11
  • Larger bond order greater bond strength
  • Bond Order for H2?

12
  • Bond Order for He2?

13
Li2
  • Do we need both 1s and 2s orbitals?
  • No. Make MOs from 2s AOs

14
Molecular Orbitals from p AOs
15
s MOs
p MOs
16
Expected Energy Diagram
  • The Idea electrons prefer to be directly between
    nuclei, so s MO is lowest energy
  • Note that this is not actually true for B2
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