II. Molecular Geometry (p. 183

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II. Molecular Geometry(p. 183 187)

• Ch. 6 Molecular Structure

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A. VSEPR Theory

• Valence Shell Electron Pair Repulsion Theory
• Electron pairs orient themselves in order to
  minimize repulsive forces.
• Used to predict the geometry of complicated
  molecules.

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A. VSEPR Theory

• Types of e- Pairs
• Bonding pairs - form bonds to central atom
• Lone pairs nonbonding e- around central atom

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A. VSEPR Theory

• Repulsive forces between electrons cause them to
  stay as far away from each other as possible.
  H2O

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How pairs of electrons behave

• The number of paired and unpaired electrons
  determine the shape of the molecule.
• Lone pairs occupy the space in the same way
  paired e- do.

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B. Determining Molecular Shape

• Draw the Lewis Diagram.
• Tally up e- pairs on central atom.
• double/triple bonds ONE pair
• Shape is determined by the of bonding pairs and
  lone pairs.
• Chart on page 186

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Chart

• A Central atom
• B Atoms bonded to central atom
• E Unshared pairs of electrons
• Subscripts refer to the number of each.

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C. Common Molecular Shapes

• 2 total
• 2 bond e-
• 0 lone e-

LINEAR 180
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C. Common Molecular Shapes

• 3 total
• 3 bond e-
• 0 lone e-

TRIGONAL PLANAR 120
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C. Common Molecular Shapes

• 3 total
• 2 bond e-
• 1 lone e-

BENT lt120
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C. Common Molecular Shapes

• 4 total
• 4 bond e-
• 0 lone e-

TETRAHEDRAL 109.5
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C. Common Molecular Shapes

• 4 total
• 3 bond e-
• 1 lone e-

TRIGONAL PYRAMIDAL 107
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C. Common Molecular Shapes

• 4 total
• 2 bond e-
• 2 lone e-

BENT 104.5
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C. Common Molecular Shapes

• 5 total
• 5 bond e-
• 0 lone e-

TRIGONAL BIPYRAMIDAL 120/90
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C. Common Molecular Shapes

• 6 total e-
• 6 bond e-
• 0 lone

OCTAHEDRAL 90
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D. Examples

• PF3

4 total 3 bond 1 lone
TRIGONAL PYRAMIDAL 107
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D. Examples

• CO2

2 total 2 bond 0 lone
LINEAR 180
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Predict the shape of the following

• SO2
• CI4 (carbon and iodine)
• BCl3

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Answers

• Bent or angular
• Tetrahedral
• Trigonal-planar

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Two ways that atoms will not be able to obey the
octet rule.

• 1.Having too few valence electrons to ever obtain
  an octet.
• Hydrogen, Beryllium and Boron have two few
  valence electrons to ever obtain a full octet.

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See This

• Hydrogen can have at most 2 valence electrons
  after it shares its electron with another atom.
• Beryllium will have 4 valence electrons after it
  has finished bonding.
• Boron will have 6 valence electrons after it
  shares its valence electrons with other atoms.

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• 2.Expanding the octet to have 10, 12 or 14
  valence electrons instead of 8.
• Elements in periods 3, 4, 5, 6 and 7 can expand
  their octet to have 10, 12, or 14 valence
  electrons.
• See how this can happen.