Molecular Modeling
Theory (adapted from Dr. Christopher Grant at University of Oregon)
Valence Shell Electron Pair Repulsion theory (VSEPR) is a set of rules whereby the chemist may
predict the shape of an isolated molecule. It is based on the premise that groups of electrons
surrounding a central atom repel each other, and that to minimize the overall energy of the
molecule, these groups of electrons try to get as far apart as possible. Groups of electrons can refer
to electrons that participate in a bond (single, double, or triple) to another atom, or to non-bonding
electrons (e.g. lone pair electrons).
The ideal electronic symmetry of a molecule consisting of a central atom surrounded by a number
of substituents (bonded atoms and non-bonding electrons) is characteristic of the total number of
substituents, and is determined solely by geometric considerations -- the substituents are arranged
so as to maximize the distances amongst them. VSEPR is useful for predicting the shape of a
molecule when there are between 2 and 6 substituents around the central atom (the case of one
substituent is not discussed because it is trivial -- the only possible shape for such a molecule is
linear). That means that there are only five unique electronic geometries to remember. For each
electronic geometry, there may be a number of different molecular geometries (the shape of the
molecule when only bonded atoms, not non-bonding electrons are considered). Molecular
geometries are really just special cases of the parent electronic geometry -- this will hopefully be
evident from the models that you build below.
Background (adapted from Dr. Paul Kiprof at University of Minnesota Duluth)
Most of the time it is impossible to predict the molecular geometry of a compound based solely on
the number of atoms that surround a central atom. For example BF3 has a trigonal planar structure,
ammonia NH3 has a trigonal pyramidal geometry and iodine trichloride ICl3 has a T-shaped
molecular shape.
BF3
NH3
ICl3
The VSEPR theory is based on the principle that electron pairs around an atom repel each other.
This is true for bonding electron pairs and lone pairs, which are not involved in bonding. It gives us
the possibility to predict and understand the structure of molecules and ions that consist of a central
atom and a different number of it surrounding atoms.
The starting point is the Lewis dot structure of the compound. The number of electron pairs around
the central atom determines the electron pair geometry. Lone pairs and bonding electron pairs are
placed around the central atom.
The final step is to leave out lone pairs and determine the molecular geometry or molecular shape.
How to draw Lewis structures for molecules that contain no charged atoms
1) Count the total valence electrons for the molecule: To do this, find the number of valence
electrons for each atom in the molecule, and add them up.
2) Figure out how many octet electrons the molecule should have, using the octet rule: The
octet rule tells us that all atoms want eight valence electrons (except for hydrogen, which
wants only two), so they can be like the nearest noble gas. Use the octet rule to figure out
how many electrons each atom in the molecule should have, and add them up. The only
weird element is boron - it wants six electrons.
3) Subtract the valence electrons from octet electrons: Or, in other words, subtract the number
you found in #1 above from the number you found in #2 above. The answer you get will be
equal to the number of bonding electrons in the molecule.
4) Divide the number of bonding electrons by two: Remember, because every bond has two
electrons, the number of bonds in the molecule will be equal to the number of bonding
electrons divided by two.
5) Draw an arrangement of the atoms for the molecule that contains the number of bonds you
found in #4 above: Some handy rules to remember are these:
a.
b.
c.
d.
Hydrogen and the halogens bond once.
The family oxygen is in bonds twice.
The family nitrogen is in bonds three times. So does boron.
The family carbon is in bonds four times.
A good thing to do is to bond all the atoms together by single bonds, and then add the
multiple bonds until the rules above are followed.
6) Find the number of lone pair (nonbonding) electrons by subtracting the bonding electrons
(#3 above) from the valence electrons (#1 above). Arrange these around the atoms until all
of them satisfy the octet rule: Remember, ALL elements EXCEPT hydrogen want eight
electrons around them, total. Hydrogen only wants two electrons.
Pre-Laboratory Exercises
1. Complete the following table:
Directions
of e-
Directions
of Atoms
2
2
Directions
of e-
Directions
of Atoms
3
3
3
2
Directions
of e-
Directions
of Atoms
4
4
4
4
3
2
Directions
of e-
Directions
of Atoms
5
5
5
5
5
4
3
2
Directions
of e-
Directions
of Atoms
6
6
6
6
6
6
5
4
3
2
Geometry Name
~ Bond Angles
Hybridization
Example
Geometry Name
~ Bond Angles
Hybridization
Example
Geometry Name
~ Bond Angles
Hybridization
Example
Geometry Name
~ Bond Angles
Hybridization
Example
Geometry Name
~ Bond Angles
Hybridization
Example
2. Explain why a Td molecule containing four identical bonds could never be a polar molecule.
3. Determine whether the compound N2O is polar (draw the Lewis structure).
4. Draw the Lewis structure and predict the geometry for the polyatomic carbonate ion (CO32-).
Procedure
Using a molecular model kit, build each compound and fill in the data table. Work across the data
table until the entire row is complete. When the row is complete, have your work checked by your
instructor (initials).
Formula
H2O
NH3
BCl3
PF3
PF5
ClF5
CO2
C2H6
C2H4
C2H2
HCHO
CH3OH
H2CCCl2
Lewis Structure
Geometry
Name
~ Bond
Angles
Hybridization
Polar or
Non-Polar
Instructor’s
Initials
Questions (adapted from Purdue University “The Shape of Molecules”)
1. On the basis of the VSEPR model, which of the following molecules is linear?
a) H2O
b) BF3
c) BeF2
d) CH4
e) none of these
2. The molecular shape of H3O+ is:
a) linear.
b) planar.
c) see-saw.
d) trigonal planar.
e) trigonal pyramidal.
3. Which of the following is a planar molecule or ion?
4. The molecular shape of SeF4 is:
a) see-saw.
b) square planar.
c) tetrahedral.
d) square pyramidal.
e) impossible to predict.
5. Predict the geometry for the center atom of each of the following molecules:
6. Predict the approximate bond angle for each of the hybrid orbitals:
a) sp
b) sp2
c) sp3
d) sp3d
e) sp3d2
7. Predict the bond angles between adjacent bonds:
a) CO2
b) CH4
c) H2O
d) NH3
e) PH3
8. What are the approximate values of the bond angles, a and b, in the ion illustrated below?
a) a is ~90o and b is ~90o
b) a is ~109o and b is ~109o
c) a is ~109o and b is ~120o
d) a is ~120o and b is ~109o
e) a is ~90o and b is ~180o
9. For the amino acid alanine, what is the bond angle about the indicated carbon atom?
a) 109.5o
b) 120o
c) 180o
d) 90o
e) 360o
10. Identify the molecular geometry of formaldehyde, CH2O.
a) linear
b) trigonal planar
c) bent
d) tetrahedral
e) trigonal bipyramidal
Link to Organic Chemistry
1. Benzene (C6H6) consists of a six-membered ring of carbon atoms with one hydrogen bonded of
each carbon.
a. Draw the Lewis dot structure, including resonance structures for benzene.
b. Build a framework and a ball and stick model for benzene. The framework model
illustrates a benzene model with the double bond (pi bond) delocalized and the ball and stick
model shows the double bond (pi bond) localized between specific carbon atoms.
c. An important observation supporting the need for resonance structures or delocalized
bonds is the fact that there are only three different structures for the compound
dichlorobenzene (C6H4Cl2). Draw diagrams of the three structures for dichlorobenzene. (Use
the framework model)
d. If benzene did not have resonance, and had three isolated double bonds (ball and stick
model), how many structures would exist for dichlorobenzene? Draw the structures.
2. Build a ball and stick model and draw a projection formula for ethane, C2H6. Indicate the
geometry around each C atom.
a. Replace a hydrogen atom on the ethane model with a chlorine atom. Draw a flat formula
and a projection formula for the ethyl chloride molecule, C2H5Cl, that you have constructed.
Does it matter which hydrogen atom of ethane, C2H6, is replaced to make ethyl chloride?
Are the six hydrogen atoms equivalent? How many isomers of C2H5Cl, ethyl chloride, can
you construct?
b. Replace a hydrogen atom on the ethyl chloride model, C2H5Cl, with another chlorine
atom to form C2H4Cl2. Draw flat structural formula for all possible structures for C2H4Cl2.
Does it matter which hydrogen atom of ethyl chloride is replaced to make C2H4Cl2? Are all
five hydrogen atoms in ethyl chloride, C2H5Cl, equivalent?