Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th...
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![Page 1: Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.](https://reader033.fdocuments.in/reader033/viewer/2022061614/56649d015503460f949d3f12/html5/thumbnails/1.jpg)
Review Chapter 10: Theories of Bonding &
Structure
Chemistry: The Molecular Nature of Matter, 6th editionBy Jesperson, Brady, & Hyslop
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Chapter 10 Concepts
VESPR theory Predict molecular geometry & overall dipole moment
Valence Bond Theory Identify & draw hybridization of orbitals at central atom Predict molecular geometry & overall dipole moment Identify & draw σ and π bonds
Molecular Orbital Theory Draw MO Energy Diagrams to identify
Bond Order Number of Unpaired Electrons
![Page 3: Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.](https://reader033.fdocuments.in/reader033/viewer/2022061614/56649d015503460f949d3f12/html5/thumbnails/3.jpg)
Memorize
Linear
Trigonal Planaror
Planar Triangular
Tetrahedral
Trigonal Bipyramidal
Octahedral
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Memorize
Trigonal Pyramidal
Bent
Seesaw orDistorted
Tetrahedron
T-shaped
SquarePyramid
SquarePlanar
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Determining 3-D VESPR Stuctures
1. Draw Lewis Structure of Molecule– Don't need to compute formal charge– If several resonance structures exist, pick only one
2. Count electron pair domains– Lone pairs and bond pairs around central atom– Multiple bonds count as one set (or one effective pair)
3. Arrange electron pair domains to minimize repulsions• Lone pairs
– Require more space than bonding pairs– May slightly distort bond angles from those predicted.– In trigonal bipyramid lone pairs are equatorial – In octahedron lone pairs are axial
4. Name molecular structure by position of atoms—only bonding electrons
![Page 6: Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.](https://reader033.fdocuments.in/reader033/viewer/2022061614/56649d015503460f949d3f12/html5/thumbnails/6.jpg)
Polarity
If symmetrical polar bonds around a central atom then they cancel and there is no overall dipole moment.
Molecule is usually polar if – All atoms attached to central atom are NOT same, OR– There are one or more lone pairs on central atom
EXCEPT:
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Valence Bond TheoryValence Bond Theory
Individual atoms, each have their own orbitals and orbitals overlap to form bonds. Extent of overlap of atomic orbitals is related to bond strength
Hybridization = mixing atomic orbitals
sp: Linear
sp2: Planar Triangular
sp3: Tetrahedral
sp3d: Trigonal Bipyramidal
sp3d2: Octahedral
![Page 8: Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.](https://reader033.fdocuments.in/reader033/viewer/2022061614/56649d015503460f949d3f12/html5/thumbnails/8.jpg)
Valence Bond Theory1) Draw lewis dot structure
2) How many lone pairs & bonded atoms will repel one another? That is the number of hybrid orbitals needed
2: sp 3: sp2 4: sp3 5: sp3d6: sp3d2
3) Convert the needed number of atomic orbitals into hybridized orbitals.
4) Fill in valence electrons. Remember that any p orbitals will be close enough in energy that they both hybrid & remaining p orbitals will fill half way with electrons before they pair.
5) Draw hybrid orbitals equally spaced around the central atom:a) Form σ bonds using hybrid orbitals with one electronb) Hybrid orbitals with two electrons are likely lone pairsc) If any lone electrons are in p orbitals π bonds will form
6) Describe molecular geometry around central atom(s).
![Page 9: Review Chapter 10: Theories of Bonding & Structure Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop.](https://reader033.fdocuments.in/reader033/viewer/2022061614/56649d015503460f949d3f12/html5/thumbnails/9.jpg)
Valence Bond Theory: Example N2
:NN:
Es
p p
sp
• Lone pair and atom bonded to Nitrogen will repel each other.
• Therefore, need to hybridize 2 orbitals• One hybrid orbital with one electron
will participate in a σ bond with the other nitrogen
• The other hybrid orbital contains a lone pair
• 2 p orbitals each have 1 electron so form 2 π bonds to the other nitrogen
AO of NHybrid & AO of N
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Molecular Orbital TheoryMolecular Orbital Theory
Views molecule as collection of positively charged nuclei having a set of molecular orbitals that are filled with electrons (similar to filling atomic orbitals with electrons).
Doesn't worry about how atoms come together to form molecule
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Molecular Orbital Theory
Li2 N2 2p Lower in energy than 2p
O2, F2 and Higher 2p Lower in energy than 2p
Can ignore filled 1s bonding & antibonding and focus on valence electrons
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Molecular Orbital Theory: Diagrams
E
s
p
AO of one atom AO of second atom
π
π*
σ
σ*
σ
σ*
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ProblemSet A
1. For the following molecules: a. Draw a lewis dot structure.b. Determine the molecular geometry at each central atom.c. Identify the bond angles.d. Identify all polar bonds: δ+ / δ-e. Assess the polarity of the molecule & indicate the overall
dipole moment if one exists
AsF5 AsF3 SeO2
GaH3
ICl2- SiO4-4
TeF6
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ProblemSet B
2. What is the hybridization of oxygen in OCl2? 3. For the species and XeF4O, determine the following:
a. electron domain geometry (geometry including non-bonding pairs)
b. molecular geometryc. Hybridization around central atomd. Polarity
4. How many and bonds are there in CH2CHCHCH2, and what is the hybridization around the carbon atoms?
5. Draw & list the bonding orbitals for HCN.
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ProblemSet C
6. What is the MO Energy Diagram for B2? How many unpaired electrons does B2 have?
7. What is the bond order & number of unpaired electrons in
8. Draw the MO Energy Diagram for BN.
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ProblemSet A
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ProblemSet A
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ProblemSet A
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ProblemSet B
2. sp3
3. XeF4O: octahedral, square pyramid, sp3d2, polar
4. 9, 2, sp2
5. HCN: C will be create a σ bond to H and N with sp2 hybridized orbitals and use 2 p orbitals to participate in 2 π bonds with N. N will participate in the σ bond with C with an sp2 hybridized orbital, the other will hold the N lone pair, and then N will use 2 p orbitals to π bond with C.
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ProblemSet C
6. MO B2: unpaired e- = 2
7. O2: BO = 2 unpaired e- = 2O2
+: BO = 5/2 unpaired e- = 1O2
- : BO = 3/2 unpaired e- = 1
8. MO BN:
π
π*
σ
σ*
σ
σ*
π
π*
σ
σ*
σ
σ*
BNB2
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More Practice Problems:
Analyze the following compounds: 1. Draw the lewis dot structure2. Determine the Geometry using VESPR theory3. Determine the hybridization at any central atoms using VB Theory4. Draw & Describe bonding in VB terms: ie # of sigma vs pi bonds
H2S
SO2
H2C=C=CH2
HCOCl