modern chemistry chapter 6 chemical bonding part v molecular geometry and intermolecular forces
TRANSCRIPT
Modern ChemistryChapter 6
Chemical Bonding
Part VMolecular Geometry and Intermolecular
Forces
Chapter 6 Section 5 Molecular Geometry pages 197-207
2
Chapte
r V
oca
bula
ryVSEPR theoryHybridizationHybrid orbitalsDipoleHydrogen bonding London dispersion
forces
VSEPR Theory
• Valence-Shell Electron-Pair Repulsion
• Predicts the shapes of molecules and ions in which valence shell electron pairs are arranged about each atom so that the electron pairs are kept as far apart as possible, thus minimizing repulsions.
VSEPR Theory
• Molecular geometries are predicted by considering the number of regions of electron density.
• Single, double, and triple bonds, as well as lone pairs are all considered one region of electron density.
• Lone pairs require more space than bonding pairs
VSEPR Theory
• To predict the relative positions of atoms around a given central element using VSEPR theory, first note the arrangement of valence electron pairs around the central atom.
• See VSEPR summary sheet on p. 3 of Ch. 6 note packet pt. 2
6
VSE
PR
& M
ole
cula
r G
eom
etr
yp.
xx
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular GeometryLINEAR
Example formula: BeF2
Type of molecule: AB2 Bond angle: 180° Shared pairs on the central atom: 2Unshared pairs on the central atom:
0
- Be -
F::F: :
: :
Chapter 6 Section 5 Molecular Geometry pages 197-207
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TRIGONAL PLANAR
Example formula: BF3
Type of molecule: AB3 Bond angle: 120° Shared pairs on the central atom: 3Unshared pairs on the central atom:
0
Molecular Geometry
F:::
:F::
:F::
B
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular GeometryTETRAHEDRAL
Example formula: CH4
Type of molecule: AB4 Bond angle: 109.5° Shared pairs on the central atom: 4Unshared pairs on the central atom:
0
CHH
H
H
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Molecular GeometryANGULAR
Example formula: H2O
Type of molecule: AB2E2
Bond angle: 104.5° Shared pairs on the central atom: 2Unshared pairs on the central atom:
2
HO
H
::
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular GeometryTRIGONAL PYRAMIDAL
Example formula: NH3
Type of molecule: AB3EBond angle: 107.5° Shared pairs on the central atom: 3Unshared pairs on the central atom:
1
H H
N
:
H
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular GeometryTrigonal Bipyramidal
Example formula: PCl5Type of molecule: AB5
Bond angle: 90°, 120° Shared pairs on the central atom: 5Unshared pairs on the central atom: 0
H
:
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular GeometryOctahedral
Example formula: SF6
Type of molecule: AB6
Bond angle: 90° Shared pairs on the central atom: 6Unshared pairs on the central atom: 0
H
:
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular Geometry• Unshared pairs occupies more
space around the central atom than shared pairs
• Unshared pairs repel other electrons more strongly than shared pairs
• Multiple bonds are treated the same as single bonds
• Polyatomic ions are treated like molecules.
Chapter 6 Section 5 Molecular Geometry pages 197-207
15
Hybridization• The mixing of two or more atomic
orbitals of similar energies on the same atom to produce new hybrid atomic orbitals of equal energy
• Example CH4
C = _ _ __ 1s 2s 2p _ _ _ _ 1s sp3
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Hybridization• s and p orbitals have different
shapes• The 2s & 2p hybridize to make
four identical orbitals– named sp3
– The 3 is from the three p orbitals used
– But the 1 is not written for the s
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Hybridization• All sp3 orbitals have the same
energy– Higher than 2s but– Lower than 2p
• Hybrid orbitals – orbitals of equal energy produced by the combination of two or more orbitals.
Chapter 6 Section 5 Molecular Geometry pages 197-207
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HybridizationN = _ _ _ 1s 2s 2p _ _ _ 1s sp3
O = _ _ 1s 2s 2p _ _ 1s sp3
Chapter 6 Section 5 Molecular Geometry pages 197-207
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HybridizationBe = 1s 2s _ _ __ 1s spB = _ __ __ 1s 2s 2p _ _ _ __ 1s sp2
Uses one p orbital
Uses two p orbitals
Chapter 6 Section 5 Molecular Geometry pages 197-207
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H
yb
rid
izati
on
p.
xx
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Comparing Molecular & Ionic Compoundsp
. xx
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular Polarity
• Dipole: created by equal but opposite charges that are separated by a short distance
H - ClLower EN
Higher ENpolar bond =
dipole
2.1
3.0
δ+ δ-
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity• Molecule polarity for
compounds with more than one bond depends on …
bond polarity and
molecule geometry.
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity1. Draw the Lewis Structure true
to shape. Example NH3
NH
HH
:
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity2. Find all the partial positive and
negatives for each atom in the molecule
HH
H
Look at each bond.High EN = δ- Low EN = δ+
δ-
δ+
δ+
δ+
2.1
3.0N
:
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity3. Look at around the “outside” of
the molecule.
NH
HH
:
All the same δ = NP; Different δ = P
δ-
δ+
δ+
δ+
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity1. Draw the Lewis Structure true
to shape. Example CH4
CH
HH
H
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity2. Find all the partial positive and
negatives for each atom in the molecule
Look at each bond.High EN = δ- Low EN = δ+
δ+
δ+
δ+
δ+
2.1
2.5C
HH
H
H
2.12.
1
2.1
δ-
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecule Polarity3. Look at around the “outside” of
the molecule.
δ+
δ+
δ+
δ+
CH
HH
H
δ-
All the same δ = NP; Different δ = PCarbon is not on the “outside”.
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Intramolecular Forces• In covalent bonding, atoms form
stable units called molecules by sharing electrons.
- termed intramolecular (within the
molecule) bonding
- covalent bonding is an
intramolecular force
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Intermolecular Forces• Forces between neutral molecules
also exist which are termed intermolecular forces.
- London Dispersion Forces (LDF)
- Dipole-Dipole Interactions (DD)
- Hydrogen Bonding (HB)
• DD and LDF together are often termed Van der Waals forces.
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Intermolecular Forces
• Intermolecular forces (IMF) are responsible for the ability of substances to exist in condensed states (i.e. liquids and solids)
- all substances have intermolecular forces (or interatomic as in the cases of monatomic gases - i.e. noble gases)
- when intermolecular forces are broken (e.g. substances melt or vaporize), the molecules remain intact - covalent bonds are not broken.
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London Dispersion Forces• London Dispersion Forces (LDF) –
intermolecular forces resulting from the instantaneous dipoles created by the constant motion of electrons.
- present between all atoms and molecules.
- increase with increasing numbers of
electrons
- LDF are the weakest IMF• Consider Helium:
Chapter 6 Section 5 Molecular Geometry pages 197-207
34
London Dispersion Forces• Nonpolar molecules don’t have
dipoles• However at any instance the
electron distribution may be uneven.
• An instantaneous dipole can occur and induce dipoles in other molecules
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Dipole-Dipole Interactions• Dipole-Dipole Interactions (DD) –
attractive forces between polar molecules due to the electrostatic attraction of partial positive and negative charges.
- present only between polar molecules
- stronger than LDF, but weaker than
H-bonding
- explained by Coulomb’s Law
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Molecular Polarity
• Dipole: created by equal but opposite charges that are separated by a short distance
H - ClLower EN
Higher ENpolar bond =
dipole
2.1
3.0
δ+ δ-
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Dipole-Dipole Interactions
• Dipole-Dipole (DD) Interactions are due to the electrostatic attractions between δ+ and δ- in different polar molecules.
H - Clδ+ δ-
H - Clδ+ δ-||||||||||||||||||||
Dipole-Dipole Interaction
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Comparing Dipole Dipole Forcesp
. xx
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Induced Dipole• Polar molecules cause a dipole
in a nonpolar molecule
O
H
H
δ+
δ- O O::
::
::
δ+ δ
+δ-
|||||||||||||||||||||||||
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Hydrogen Bonding
• Hydrogen Bonding (HB) is an extreme form of dipole-dipole interaction between the hydrogen atom in a polar bond, specificallyO-H, N-H or H-F, and the electronegative O, N or F atom of another molecule.- strongest IMF
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Hydrogen Bonding
• Due to the combined effect of the large dipole between the hydrogen and the electronegative element (O, N or F) and the presence of the lone pair of electrons on the O, N or F.
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Hydrogen Bonding• Consider H2O:
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Hydrogen Bonding• H-F, H-O or H-N bonds have a
large electronegativity difference
• These bonds are very polar.
• Molecules with these bonds have very strong dipole-dipole forces
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Hyd
rog
en
Bon
din
g
p.
xx
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Hydrogen Bonding• Compare
PH3 & NH3 H2O & H2S on p. 204
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Hydrogen Bonding• Compare the boiling points of
CH4, NH3, HF and H2O:
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Hydrogen Bonding• The double helix of DNA is
possible due to hydrogen bonding
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Hydrogen Bonding• Quick Reference for Predicting
IMF for a Particular Molecule:
Type of Molecule Intermolecular Forces
Non-Polar London Dispersion
Polar without H-O, H-N or H-FLondon Dispersion
Dipole-Dipole
Polar with H-O, H-N or H-FLondon Dispersion
Dipole-DipoleHydrogen Bonding
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Modern ChemistryChapter 6
Chemical Bonding
Sections 1-5Introduction to Chemical Bonding
Covalent Bonding & Molecular CompoundsIonic Bonding & Ionic Compounds
Metallic BondingMolecular Geometry
Chapter 6 Section 5 Molecular Geometry pages 197-207
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Do Now
Draw the Structures of AlCl3and CH4.
What would the bond angles need to be between the
valence electron pairs in each molecule to minimize electron
repulsion.
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Do Now
For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Determine the hybridization of the central element.
SiCl4 H2S
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Do Now
For the following molecules/ions:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Determine the hybridization of the central element.
CHCl3 IF4+
53
Do Now
For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Discuss the intermolecular forces between each molecule.
SiCl4 NH2Cl COH2
54
Do Now
For the following molecules:- Draw the Lewis Structure- Assign the A,B,E form- Assign the molecular geometry- Determine the polarity- Discuss the intermolecular forces between each molecule.
SbCl5 SeCl4 N2H4