Title: Molecular Geometry and Chemical Bonding Theory
1RVCC Fall 2008 CHEM 103 General Chemistry I
Chapter 9Molecular Structures
Chemistry The Molecular Science, 3rd Ed. by
Moore, Stanitski, and Jurs
2Molecular Structure
- Molecular geometry is the general shape of a
molecule or the arrangement of atoms in three
dimensional space.
Physical and chemical properties depend on the
geometry of a molecule.
3Molecular Structures
3-D Model
3-D Drawing
4Does it matter?The Thalidomide Story
?
The chemical structure of thalidomide.
models enantiomers (mirror image)
The Same and Not the Same, by Roald
Hoffmann 1995, Columbia University Press
5Does it matter?Fatty Acids
trans fatty acid
cis fatty acid
6VSEPR Model
- The Valence Shell Electron Pair Repulsion model
predicts the shapes of molecules and ions by
assuming that the valence shell electron pairs
are arranged as far from one another as possible
to minimize the repulsion between them.
7VSEPR Model
H
- Electron Pair Geometry
- is determined by the number and arrangement of
all electron pairs (bonding and lone) around the
central atom. - Molecular geometry
- is determined by the arrangement of atoms (or
bonding electron pairs only) around the central
atom.
H
H
N
In molecules with no lone pairs, Electron Pair
Geometry Molecular Geometry
8Fig. 9-4, p.383
- AXE shorthand notation
-
- A - central atom
- X - terminal atoms
- E - lone pair electrons
AX3E0
9Predicting Molecular Geometry VSEPR
- Only five basic shapes.
- When a lone pair replaces an atom, the molecular
geometry changes as well as the angles.
e- pairs 2 3 4 5 6
Fig. 9-4, p.383
10Predicting Molecular Geometry VSEPR
- Draw the Lewis structure.
- Determine how many electron pairs (bonded and
non-bonded) are around the central atom. Treat
a multiple bond like a single bond when
determining a shape. - Write the AXE shorthand notation.
- Determine the electron pair geometry (one of
the five basic shapes). - If the molecule has lone pairs around the central
atom, then determine the molecular geometry.
(This is a subset of the electron geometry.)
11Linear (Electron Geometry)Two e- pairs about
central atom
bond lone Molecular pairs
pairs Geometry 2 0 linear
1 1-3 linear
The molecular geometry here is the same as the
electronic geometry even though there is a lone
pair. Two points make a line.
12Predicting Molecular Geometry
Example 1 BeCl2
AX2E0
1. Draw the Lewis structure
2. Two electron pairs around the central atom.
Two bonded and Zero lone pairs.
electron pair geometry molecular geometry
Geometry is Linear. Bond angle is 180o.
13Trigonal Planar (Electron Geometry)Three e-
pairs about central atom
bond lone Molecular pairs
pairs Geometry
Model
3 0 triangular planar
2 1 angular (bent)
1 2 linear
14Predicting Molecular Geometry
Example 2 BF3
..
AX3E0
Three electron pairs around the central atom.
Three bonded and Zero lone pairs.
triangular planar (or trigonal planar)
15Predicting Molecular Geometry
Example 3
SO2
AX3E0 AX2E1
Three electron pairs around the central atom.
Two bonded and One lone pairs.
electron geometry triangular planar. molecular
geometry bent or angular.
16Tetrahedral (Electron Geometry)Four e- pairs
about central atom
bond lone pairs pairs 4 0
tetrahedral
Model
3 1 triangular
pyramidal
2 2 angular
(bent)
17Predicting Molecular Geometry
Example 4
AX4E0
Four electron pairs around the central atom.
Zero lone pairs.
tetrahedral
electron pair geometry molecular geometry
18Predicting Molecular Geometry
Example 5
AX4E0 AX3E1
Four electron pairs around the central atom.
Three bonded and One lone pair.
electron geometry tetrahedral. molecular
geometry triangular pyramidal
19Predicting Molecular Geometry
Example 6
AX4E0 AX2E2
Four electron pairs around the central atom.
Two bonded and Two lone pairs.
O H H
electron geometry tetrahedral molecular
geometry angular or bent
20Predicting Molecular Geometry
- Tetrahedral - bond angles
Order of increasing repulsion bonding
pair-bonding pair lt bonding pair-lone pair lt
lone pair-lone pair
21Trigonal Bipyramidal (Electron Geometry)Five e-
pairs about central atom
Put lone pairs in the equatorial positions.
- The atoms are non-equivalent.
- Green atoms are axial
- blue atoms are equatorial.
22Predicting Molecular Geometry
Example 7 PF5
AX5E0
Five electron pairs around the central atom.
Zero lone pairs.
electron and molecular geometry trigonal
bipyramidal
23Predicting Molecular Geometry
Example 8 SF4
AX5E0 AX4E1
Five electron pairs around the central atom.
Four bonded and One lone pair.
electron geometry trigonal bipyramidal
molecular geometry seesaw
24Predicting Molecular Geometry
Example 9 BrF3
AX5E0 AX3E2
Five electron pairs around the central atom.
Three bonded and Two lone pairs.
electron geometry trigonal bipyramidal
molecular geometry T-shaped
25Predicting Molecular Geometry
Example 10 XeF2
AX5E0 AX2E3
Five electron pairs around the central atom.
Two bonded and Three lone pairs.
electron geometry trigonal bipyramidal
molecular geometry linear
26Octahedral (Electron Geometry)Six e- pairs
about central atom
Equivalent atoms
27Predicting Molecular Geometry
Example 11 SF6
AX6E0
Six electron pairs around the central atom. Six
bonded and Zero lone pairs.
electron geometry octahedral molecular
geometry octahedral
28Predicting Molecular Geometry
Example 12 IF5
AX6E0 AX5E1
Six electron pairs around the central atom.
Five bonded and Two lone pairs.
electron geometry octahedral molecular
geometry square pyramidal
29Predicting Molecular Geometry
Example 13 XeF4
AX6E0 AX5E1
Six electron pairs around the central atom.
Four bonded and Two lone pairs.
electron geometry octahedral molecular
geometry square planar
30Predicting Molecular Geometry
Fig. 9-5, p.391
31Predicting Molecular Geometry
Fig. 9-6, p.393
32Practice
- ICl
- ICl3
- ICl5
- GeF4
- SeF4
- XeF4
33Bond Angles
?CHO
Give the approximate values for the indicated
bond angles.
?COH
?OCN
?HNH
34Molecular Geometry Dipole Moment and Polarity
- Electronegativity (EN) values are used to predict
the polarity of covalent bonds. The greater ?EN,
the more polar will be the bond. A polar bond
has a dipole or slight separation of charge (from
the unequal sharing of bond electrons). Chapter
8 - The polarity of a molecule depends on the sum of
all the bond dipoles (vectors). If there is a
net dipole for the molecule, than the molecule is
polar. A molecule that has polar bonds may or
may not be polar. - The dipole moment (µ) is a measure of the degree
of charge separation or the polarity.
35Molecular Geometry Dipole Moment and Polarity
d
d-
d-
nonpolar, bp-79?C
dipole moment, µ 0 D
d-
..
..
Net dipole
d
d
polar, bp100?C
dipole moment, µ 1.85 D
36Molecular Geometry Dipole Moment and Polarity
- In general, a molecule is polar if
- it isnt a basic VSEPR shape (symmetrical)
- Ex H2O, bent (polar)
- or if the terminal atoms/groups in a
- basic VSEPR shape differ.
- Ex CH2Cl2, tetrahedral (polar)
-
-
37Dipole Moment and Molecular Geometry
- Molecules that exhibit any asymmetry in the
distribution of electrons would have a nonzero
net dipole moment. These molecules are considered
polar.
Non polar VSEPR shape identical atoms
Polar VSEPR shape atoms differ
38Dipole Moment and Molecular Geometry
39Molecular Geometry Dipole Moment and Polarity
Non polar Atoms differ. BUT can be divided into
nonpolar VSEPR shapes linear triangular planar
Non polar VSEPR shape identical atoms
Polar Atoms differ. Doesnt divide into nonpolar
VSEPR shapes
Polar VSEPR shape atoms differ
40Dipole Moment and Molecular Geometry
SF4
F
ClF3
SeeSaw No symmetry ? polar
Cl
F
T-shaped No symmetry ? polar
F
F
XeF2
XeF4
Xe
Linear Symmetric ? non polar
F
Square Planar Symmetric ? non polar
41Molecular Geometry Dipole Moment and Polarity
CO, PCl3, BCl3, GeH4, CF4
- Which compound is the most polar?
- Which compounds on the list are non-polar?
42Orbitals Consistent with Molecular Shape
- Lewis dot VSEPR gives the correct shape for a
molecule. BUT
How do atomic orbitals (s, p, d ) produce these
shapes?
Valence bond theory describes a bond as an
overlap of atomic (hybrid) orbitals.
43Valence Bond Model
H2
HF
44Valence Bond Theory
and, why do we draw the Lewis structures like
we do?
- This works for H2 and HF, but
- Why does Be form compounds?
- no unpaired electrons
- Why does C form 4 equivalent bonds at tetrahedral
angles? - only two unpaired electrons
- p orbitals are at 90 to each other (not 109.5)
45Orbitals Consistent with Molecular Shapes
- Atomic orbitals (AOs) can be hybridized (mixed).
- Sets of identical hybrid orbitals form identical
bonds. - AOs that hybridize hybrids orbitals .
- s p sp sp
- s p p sp2 sp2 sp2
- etc.
46sp Hybrid Orbitals
sp hybridization occurs around the central atom
whenever there are two regions of high e-
density. Two equivalent covalent bonds form
(180 apart) LINEAR.
47sp Hybrid Orbitals
48 sp2 Hybrid Orbitals
AX3E0, Ex BF3
The result is THREE equivalent hybrid orbitals,
in a VSEPR basic shape of trigonal planar.
p. 396
49 sp3 Hybrid Orbitals
AX4E0, Ex CH4
TETRAHEDRAL
50 sp3 Hybrid Orbitals
AX3E1 ( NH3) and AX2E3 ( H2O)
51Orbitals Consistent with Molecular Shapes
- Describe bonding in PCl5 using hybrid orbitals.
AX5E0
trigonal bipyramidal We need 5 orbitals.
52sp3d Hybrid Orbitals
3d
valence shell
3p
hybridization
five equal sp3d hybrid orbitals
X
3s
P atom (ground state)
53sp3d Hybrid Orbitals
54Orbitals Consistent with Molecular Shapes
- Describe the bonding in SF6 using hybrid orbitals.
AX6E0
Octahedral We need 6 orbitals.
55sp3d 2 Hybrid Orbitals
3d
X
hybridization
3p
six equal sp3d2 hybrid orbitals
X
3s
S atom (ground state)
56sp3d 2 Hybrid Orbitals
57Summary - Hybrid Orbitals
58Hybridization
Mixed Hybrids () Remaining Geometry sp
sp (2) pp Linear spp sp2 (3)
p Triangular planar sppp sp3
(4) Tetrahedral
Mixed Hybrids () Remaining Geometry spppd
sp3d (5) dddd Triangular bipyramid spppdd
sp3d2 (6) ddd Octahedral
59Practice
What are the hybridization and approximate bond
angles for each C, N, O in the given molecules?
60What about multiple bonding!
- According to valence bond theory hybrid orbitals
include - single bonds
- lone pairs
- one of the bonds in a multiple bond.
- The electrons in the unhybridized atomic orbitals
are used to form the additional multiple bonds.
61Multiple Bonding
- A s (sigma) bond is an overlap of orbitals
(hybrids) along the bond axis. - A p (pi) bond is a overlap of parallel p
orbitals, creating an electron distribution above
and below the bond axis.
62Multiple Bonding
(unhybridized)
2p
2p
sp2
Energy
(3 sp2 hybrid 1 unhybridized p)
C atom (ground state)
63Multiple Bonding
(unhybridized)
2p
2p
sp2
2s
Energy
(3 sp2 hybrid 1 unhybridized p)
O atom (ground state)
64Multiple Bonding
s p
65Multiple Bonding
66Multiple Bonding
67Practice
s
s p
s
s
- Identify the pi and sigma bonds in the given
molecules.
s p, p
s
68Types of Intermolecular Forces
Intermolecular Interactions
London Forces
Dipole-Dipole Forces
Hydrogen Bonding
(0.05 40 kJ/mol)
(5 25 kJ/mol)
(10 40 kJ/mol)
(Intramolecular Covalent Bond 150 1000 kJ/mol)
69Types of Intermolecular Forces
London Forces (dispersion forces)
When electrons are momentarily unevenly
distributed in the molecule, polarization occurs.
Induced Dipole
All molecules, EVEN nonpolar ones experience
London Forces! (Nonpolar molecules do not
experience any other intermolecular interaction)
70Types of Intermolecular Forces
To boil (l g), molecules must have enough
energy to overcome their intermolecular forces.
The higher the intermolecular force the higher
the boiling point!
Dispersion Forces increase with increased number
of electrons.
increased polarizability
71Types of Intermolecular Forces
A polar molecule is a Permanent Dipole that
creates .
Dipole-Dipole forces
72Types of Intermolecular Forces
The more polar the molecule (at a given size)
the higher the boiling point!
73Types of Intermolecular Forces
Hydrogen bond
- is established by the attraction between
hydrogen and an electron pair on a small, very
electronegative atom.
d
XH - - - Z X N, O, F Z N, O, F
This bond is responsible of determining the
three dimensional structure of large proteins
molecules
74Types of Intermolecular Forces
Water One molecule can participate in four H
bonds with other molecules.
Because of the hydrogen bond, water has a boiling
point 200 C higher than if the bond were not
present.
75Practice
- Explain the following boiling points,
- HF (20C), HCl (-80C), HBr (-60C), HI (-25C)
- Which of the following will form H-bonds
- CH2Br2, CH3OCH2CH3, CH3CH2OH, H2NCH2COOH
-
- What types of forces must be overcome in these
changes? - The sublimation of solid C10H8
- The decomposition of water into H2 and O2
- The evaporation of PCl3