E11 THE USE OF MOLECULAR MODELS
THE TASK
To use molecular models to develop a better understanding of molecular size, shape and
bonding.
THE SKILLS
By the end of the experiment you should be able to:
• use molecular modelling kits and reasoning to study some of the properties of
molecules,
• draw Lewis structures and to use these and the valence shell electron pair repulsion
model to predict molecular shapes.
OTHER OUTCOMES
•
You will develop generic scientific skills including the use of models in scientific
theories and an appreciation of their limitations.
INTRODUCTION
Models versus Reality
In this experiment you will use molecular models to explore a number of aspects of molecular
structure and an important molecular property, polarity. You will also look at some cases in
which the models prove too simple to represent the reality at the molecular level successfully.
In order to be able to make proper use of models it is important to understand the relationship
between a model and the reality it tries to represent.
Models are imperfect things, often very useful when applied to the problems for which they
were designed, frequently misleading when extended beyond their range of applicability. Let’s
consider the model kit used in this experiment. It is designed to represent some basic geometric
features of bonding, specifically, bond angles and lengths. These models are at their most
valuable in showing how the structure of a molecule arises as a consequence of the geometry of
bonds about the individual atoms. Pauling and Corey’s determination of the structure of simple
proteins and Watson and Crick’s discovery of the base pairing in DNA were some early
triumphs of the use of simple model kits like the one you will use in this experiment.
The model kits do not convey information about the energetics of bonding, i.e., why a particular
bond forms between two atoms. The kits do not allow for the distortions of the geometry away
from the perfect tetrahedral, octahedral, etc. The kits do not really attempt to depict the
distribution of electrons around the atoms and in the bonds with any accuracy. You need to be
aware of these limitations when you draw conclusions from your model building. (You will see
what we mean in this experiment.) These limitations, however, are also the reason why the
models are as simple to use (and, hence, as useful) as they are.
E11-1
E11-2
20
LAB-WORK
(1) The Framework Molecular Model Kit
(2) Bonding and Valence Shell Electrons
(3) Building Bonds with the Model Kit
(4) Measuring the Distance between Electron Pairs
(5) Building Models of Molecules with Nitrogen and Hydrogen
(6) Modelling Double and Triple Bonds
(7) Molecular Shape and Polarity
(8) A Puzzle: Which Isomer of ClF3 is Correct?
(9) NO3– : Resonance and the Limitations of Models
(1) The Framework Molecular Model Kit
Acknowledgment is made to Prentice-Hall Inc. for permission to reprint sections of the
instruction manual that accompanies their “Framework Molecular Models” kits.
The FMM kit is composed of two basic construction units, metal ‘valence clusters’ and plastic
tubing, by means of which atoms of all elements can be represented. Each type of atom is
colour coded. For example, red tubing represents oxygen, white is hydrogen, blue is nitrogen,
black is carbon, and so on. Valence clusters are coloured silver, brass and copper for easy
identification.
Each valence cluster has metal prongs joined at a common point representing the centre of the
atom. These prongs, which fit snugly into the plastic tubing, point along the symmetry axes of
the valence orbitals, and the angles formed determine bond angles in the assembled models.
Models built from the FMM kit show to scale the mutual relations of atoms of a given structure.
The scale is 1 inch = 1 angstrom (1 Å) = 10–1 nanometre = 10–10 metre = 100 picometre. Note
that one inch is approximately 25 millimetre on your metric ruler.
The molecular framework is specified by
* the internuclear distances, which measure the separation between the centres of two
neighbouring atoms.
* the bond angles which measure whether two atoms both bonded to a third are collinear or at
an angle.
W
COVALENT
RADIUS OF
CARBON
H
X
+
TETRAHEDRON
FOUR
C–H BONDS
H
C
H
X
W
X
X
H
W
Figure E11-1
W
E11-3
For example, take methane (see Figure E11-1). The points H and C are the centres of hydrogen
and carbon atoms, and the distance between them is called the internuclear (or interatomic)
distance. This distance can be conveniently thought of as composed of two parts:
* the C to X or C–X distance (0.77 Å), measuring the size of carbon in the direction of the
bond, and known as the covalent radius of carbon.
* The X–H distance (0.30 Å), the covalent radius of hydrogen.
Values of covalent radii are listed in Table E11-2 (page E11-4). The bond length is the distance
between the two nuclei and is the sum of the covalent radii, in this case 1.07 Å.
Use Table E11-2 to calculate the following bond lengths.
BOND
LENGTH (Å)
BOND
C–F
C–N
C–C1
O–H
C–Br
O–O
C–I
S–S
LENGTH (Å)
In CH4 the H–C–H bond angle is 109.47° (109° 28'), and is called the tetrahedral angle. In less
symmetric molecules, such as CH3Cl or NH3, the bond angles vary slightly from the tetrahedral
angle.
When molecules aggregate to form a crystal, or when they collide at moderate speed, their
closest contact occurs at a distance corresponding to the van der Waals envelope. This envelope
denotes how far the atoms within a molecule extend outwards. The sizes of van der Waals radii
are given in Table E11-3. Note that such radii are not measurable precisely, but estimates, as
given here, are nevertheless very useful.
With the axis of the C–H bond as example (see Figure E11-1), the distance C–X denotes the
covalent radius of carbon (black tubing), the distance X–H denotes the covalent radius of
hydrogen (0.30 Å), and the section H to W extending in the nonbonded direction corresponds to
the van der Waals radius of hydrogen (1.20 Å). In Figure E11-1, one must imagine that the
volume occupied by each hydrogen atom is a portion of a sphere with centre H and radius
H–W. Space-filling models, which show the outer surface of molecules, are available for
comparison. With experience it is not difficult to visualise the volume of a molecule, given the
framework of the molecule in an FMM model.
Calculate the distance from the C nucleus to the farthest boundary of the CH4 molecule.
E11-4
Calculate the distance from the C nucleus to the farthest boundary of each of the following
molecules, which have the same shape as CH4.
MOLECULE
DISTANCE (Å)
CF4
CC14
CBr4
For the preparation of bonds, coloured tubing is supplied, which can be used either by itself for
a bond between like atoms, or joined by a linear fastener (see Figure E11-3 on page E11-8) to
tubing of another colour to represent a bond joining different atoms.
Tubing in various colours is provided (Table E11-1), and sections are cut on the chosen scale
(1 inch = 1 angstrom), as specified by Tables E11-2 and E11-3.
Tables E11-1, E11-2 and E11-3 give the data needed for construction of the structures dealt
with in this exercise.
Table E11-1 Colour Coding of Atoms
H
white
O
red
C
black
F
light green
N
blue
Cl
dark green
Table E11-2 Single Bond Covalent Radii (Å)
H
Be
0.89
B
0.80
0.30
C
0.77
N
0.74
O
0.74
F
0.64
Si
1.17
P
1.10
S
1.04
Cl
1.00
3d transition metals
≈ 1.25
Br 1.14
I
1.33
Xe
1.31
Xe
2.1
Table E11-3 Van der Waals Radii (Å)
H
1.2
C
1.8
N
1.5
O
1.4
F
1.4
P
1.9
S
1.8
Cl
1.8
Br
2.0
I
2.2
E11-5
(2) Bonding and Valence Shell Electrons
When an atom forms a single covalent bond it in effect gains one electron into its valence shell.
An isolated F atom has 7 valence electrons (F is in Periodic Group 17). An HF molecule can be
represented by an electron dot structure as
o o
H .F
o
o
o
o o
In the formation of the one covalent bond, both the H and the F atoms have gained one electron
each and now have the stable electronic configuration of the next noble gas atom in the periodic
table.
An isolated O atom has 6 valence electrons (O is in Periodic Group 16). The H2O molecule can
be represented as
o o
H . O.
H
o
o
o
o
In the formation of the two covalent bonds, the O atom has gained two electrons and now has a
stable octet of valence electrons.
An isolated N atom has
can be represented as
In the formation of the
and has a stable
valence electrons (N is in Group 15). An NH3 molecule
.
covalent bonds, N has gained
of valence electrons.
electrons
E11-6
An isolated C atom has
valence electrons (C is in Group 14). A CH4 molecule
.
can be represented as
In the formation of the
and has a stable
covalent bonds, C has gained
electrons
of valence electrons.
Demonstrator's
Initials
Where one of the electron pairs is a bonding pair, the coloured tubing representing it must be
joined at its outer end to a length of the appropriate tubing representing the other atom. Where
one of the electron pairs is a nonbonding pair, the coloured tubing (cut to the appropriate van
der Waals radius length) represents the electron pair. A nonbonding electron pair is often called
a lone pair.
Complete the following table, showing the number of bonding pairs and nonbonding pairs in the
valence shell of the atom indicated.
Atom
Number of
bonding pairs
Number of
nonbonding pairs
Total number of
valence shell
electron pairs
F in HF
O in H2O
N in NH3
B in BF3
Cl in ClF3
Demonstrator's
Initials
E11-7
A prediction of the geometry of a molecule can be made based on the simple idea that the
electron pairs in the valence shell (non-bonding pairs as well as bonding pairs) seek to maximise
their distance from one another in order to minimise the Coulombic repulsions. This idea is
referred to as Valence Shell Electron Pair Repulsion (or VSEPR, for short). For example,
consider an atom which, in a molecule, has 4 electron pairs in its valence shell. The VSEPR
model would predict that those 4 pairs would arrange themselves tetrahedrally about the atom,
as this is the arrangement which maximises the distance of all the pairs from each other.
However, the shape of the molecule or ion is that figure defined by the atoms themselves and
disregarding any lone pairs which may be present, even though those lone pairs have
participated in determining the shape. Thus if the tetrahedral arrangement of valence shell
electron pairs in the example above were all bonding pairs, then the shape of that molecule or
ion would also be tetrahedral, but if for example two of those electron pairs were lone pairs,
then the shape of the molecule or ion would be angular, the result of disregarding two of the
corners of the tetrahedrally disposed valence shell pairs.
Complete the following table.
Central
atom
Electron dot
structure
Total number of
valence electron
pairs about
central atom
Geometry of
electron pairs
about central
atom
Number
of
bonding
pairs
O in H3O+
B in BF3
Cl in ClF5
Demonstrator's
Initials
Shape of
molecule or ion
E11-8
(3) Building Bonds with the Model Kit
C–C Bond
A bond between a pair of like atoms is cut from tubing of one colour. The decimal-inch scale
box can be used.
Calculate the length of a C–C single bond.
C–H Bond
As shown in figure E11-2, a C–H bond could be constructed by joining in sequence:
•
a 0.77 inch length of black tubing representing the covalent radius of carbon.
•
a 0.30 inch length of white tubing representing the covalent radius of hydrogen, and which
has a permanently attached linear fastener at both ends.
•
a 1.20 inch length of white tubing representing the van der Waals radius of hydrogen.
CARBON
NUCLEUS
COVALENT RADIUS
OF CARBON (0.77 Å)
LINEAR
FASTENER
COVALENT RADIUS
OF HYDROGEN (0.30 Å)
HYDROGEN
NUCLEUS
VAN DER WAALS RADIUS
OF HYDROGEN (1.20 Å)
Figure E11-2
The hydrogen atom occupies a sphere centred at the nucleus. You should mentally fill in the
volume of the atom on this framework.
C–F Bond
A 0.77 inch length of black tubing (covalent radius of carbon 0.77 Å) is attached to a 0.64 inch
length of light-green tubing (covalent radius of fluorine 0.64 Å) by means of a linear fastener to
form the carbon–fluorine bond, as shown in Figure E11-3.
LINEAR
FASTENER
COVALENT RADIUS
OF CARBON (0.77 Å)
COVALENT RADIUS
OF FLUORINE (0.64 Å)
Figure E11-3
CARBON
NUCLEUS
FLUORINE
NUCLEUS
ASSEMBLED C–F BOND (1.41 Å)
E11-9
Single bonds have axial symmetry (like the tubing which represents them). Bonds with axial
symmetry are termed σ bonds (sigma bonds). In this exercise, all tubing connecting two
clusters, and all tubing connecting one cluster to an H nucleus, represents σ bonds.
(4) Measuring the Distance between Electron Pairs
In this exercise you will examine the properties of three of the five basic geometries adopted by
valence shell electron pairs. (The two missing geometries are the linear and trigonal shapes.)
From the model kit select metal clusters with 4, 5 and 6 prongs, respectively. Each cluster is
associated with a solid figure in the following way. Take, for example, the 4 prong cluster and
put it on the bench. The 3 points of contact with the bench make up the corners of a triangular
face. The 4 prong cluster has 4 such equivalent triangular faces. The solid figure made up by
these 4 faces is called a tetrahedron. By a similar sort of reasoning, the 5 prong cluster is
described as trigonal bipyramidal (meaning two triangular pyramids joined by their bases) and
the 6 prong cluster is described as octahedral. Diagrams of these three solids are shown in
Figure E11-4 below.
f ace
prong
edge
corner
tetrahedron
trigonal bipyramid
octahedron
Figure E11-4
By examining these three clusters, complete Table E11-4.
Table E11-4 Clusters and Their Associated Solid Figures
SHAPES
GEOMETRIC PROPERTIES
Number of ‘prongs’ in associated
model cluster
Number of faces of this figure
Number of corners of this figure
Number of edges of this figure
Tetrahedron
Trigonal
bipyramid
Octahedron
E11-10
By inspection only, identify which of the clusters have all prongs equivalent?
Sketch the cluster in which the prongs are not equivalent and identify the “axial” and
“equatorial” prongs.
Verify by measurement that the prongs on the tetrahedral cluster are farther apart than the
neighbouring prongs in a cluster of four prongs in a square arrangement. Such an arrangement
can be found in the octahedral cluster. Record here these distances for comparison.
Tetrahedral:
Square:
Indicate on the diagram below all the different bond angles and give the size of each different
angle.
Demonstrator's
Initials
E11-11
(5) Building Models of Molecules with Nitrogen and Hydrogen
In this exercise you will build the following molecules and ions: NH3, NH4+ and NH2–. These
species all have the same number of electrons and, hence, are referred to as isoelectronic with
one another. This exercise demonstrates the important role of the nonbonding electron pairs in
determining molecular shape.
Complete the following table. Do not forget to take into account the charge on the ions when
determining the number of valence electrons.
Species
Electron dot
diagram
Number of valence electron pairs
non-bonding
Total
σ-bonding
NH3
ammonia
NH4+
ammonium ion
NH2–
amide ion
Each molecule has
nitrogen in a
valence electron pairs which are arranged about the
geometry.
Using blue tubes for nitrogen and white tubes for hydrogen, build models of each of the three
species, including the lone pairs. Indicate the shape of each species in the table below.
Species
Geometrical arrangement of
valence-shell electron pairs
Shape of species
NH3
NH4+
NH2–
Demonstrator's
Initials
E11-12
(6) Modelling Double and Triple Bonds
Frequently there are molecules whose structure can only be explained by having more than one
electron pair taking part in a bond. Consider, for example, the case of O2 which has a total
number of valence electron pairs of 6. The only way these pairs can be distributed between the
atoms such that each atom has a complete octet of electrons about it is with the following
arrangement:
...O....O...
. .
The two shared electron pairs are said to make a double bond which is depicted as O=O. When
three pairs are shared between two atoms we call this a triple bond and depict it as N≡N, to use
N2 as an example. For the sake of the VSEPR model, multiple bonds are assumed to act as a
single electron pair when considering Coulombic repulsions.
Draw electron dot diagrams, including all valence electron pairs, for the following molecules
and indicate how many double and triple bonds there are.
Molecule
Electron dot diagram
Total number of valence
electron pairs (all atoms)
Number and type of
multiple bonds
N2
CO2
CH2O
In order to make models which include double and triple bonds, we need to know some more
about how the electron pairs in multiple bonds are arranged in space. Let us start with the
double bond. The two electron pairs in the double bond are distributed in quite different ways.
The pair that occupies the space along the axis between the two atoms forms the σ-bond. The
second pair occupies the space above and below the axis between the two atoms. This space is
formed by the overlap of the two p-orbitals as shown in Figure E11-5. The resulting bond is
called a π bond (pi bond).
Figure E11-5.
E11-13
A triple bond consists of one σ-bond and two π-bonds, one π-bond above and below the σ-bond
axis and the other π-bond in front and behind this axis, as shown in Figure E11-6.
Two lobes of
one π-bond
Two lobes of
one π-bond
Axis of σ-bond
Figure E11-6.
To see how these two types of bond can be represented with the model kit, consider the case of
ethylene H2C=CH2. The lengths of tubing required are calculated from the appropriate π-bond
covalent radii in Table E11-4 and the π-bond van der Waals radii in Table E11-5. As shown in
Figure E11-7 the π bond is made from two ‘bridges’, each built out of three tubes and two right
angle fasteners, one above and one below the plane of the molecule. (Pre-assembled π-bond
bridges are included in your FMM kit. Do not attempt to pull them apart.) Note that the cluster
used for each atom at the ends of the double bond, in this case carbon, is trigonal bipyramidal.
Table E11-4 π-bond covalent radii
C
N
Table E11-5 π-bond van der waals radii
O
Period-2 atoms (C, N, O)
1.5
Period-2 atoms (Si, P, S)
1.9
Double 0.67
double
0.62
Double 0.55
Triple
triple
0.55
Triple
0.60
0.50
Figure E11-7.
What kind of cluster from the model kit is needed to build models of the following molecules?
Molecule
Number and type of multiple bond
C2H4
1 double
Type of cluster required
C atom:
N2
N atom:
CO2
C atom:
O2
Both O atoms:
O atom:
E11-14
Build models of N2, CO2 and O2 including the nonbonding pairs and have them checked by your
demonstrator. IMPORTANT: When constructing these models, make the σ framework
first and then add the π-bonds.
Demonstrator's
Initials
(7) Molecular Shape and Polarity
The shape of a molecule determines many of its properties and chemical behaviour. Modern
drug design, for example, makes extensive use of molecular models (typically ‘built’ in a
computer instead of by hand) in order to see how a drug will bind to large biological molecules
like proteins or DNA. One very important property, particularly for small molecules, is how the
electrostatic charge is distributed about the molecule; specifically, does the molecule have a
positive ‘side’ and a negative ‘side’? A molecule which does have such a distribution is said to
have a dipole moment and is called polar,
.O.
δ−
eg. H2O
H
H
δ+
A molecule which has no such positive and negative ‘sides’ has no dipole moment and is
described as nonpolar,
eg. N2
N
N
The dipole moment of a molecule determines properties such as molecular solubility and
whether the molecule will absorb microwave radiation.
Examine the CO2 molecule built earlier. Would you expect this molecule to be polar or
nonpolar? Explain your answer.
E11-15
(8) A Puzzle: Which isomer of ClF3 is the most stable?
In this exercise we will see how a combination of modelling and experiment can be used to
predict a molecular structure. The molecule in question is ClF3. The central Cl atom has
valence electron pairs around it which, according to the VSEPR model adopt a
geometry. There are 3 different ways of arranging the three
fluorine atoms about the chlorine resulting in 3 different structural isomers.
Build a model of any one of the isomers of ClF3. Make sure to include the non-bonding pairs
on the chlorine atom.
What is the approximate angle between two equatorially-oriented electron
pairs?
What is the approximate angle between an equatorially-oriented electron
pair and an axially-oriented electron pair?
In which of these two types of interactions will the repulsion between the two electron pairs be
greater?
As non-bonding electron pairs have, effectively, a stronger Coulombic repulsion than do
bonding pairs, it follows that the lowest energy structure will be that in which the interactions
involving the non-bonding pairs and all the other electron pairs have been minimised. That is,
the most stable isomer is likely to be the one in which the number of 90° interactions between a
non-bonding pair and the other electron pairs is minimised. Consider SF4 as an example. It is
known to have the ‘see-saw’ shape with the non-bonding pair occupying the equatorial position.
F
F
S
F
F
3 × 90° interactions
F
S
F
F
F
2 × 90° interactions
The problem is to predict which of the three possible isomers of ClF3 is the most stable. Sketch
each of the three isomers of ClF3 in the spaces provided in the following table and, for each
isomer, indicate the number of axial and equatorial non-bonding pairs on the central chlorine
atom. Record the total number of 90° interactions between the non-bonding pairs and any other
electron pair. (Change your model to represent each of the other isomers, if necessary.)
E11-16
I
II
III
Number of axial
non-bonding pairs
Number of equatorial
non-bonding pairs
Total number of 90°
interactions between
the Cl non-bonding
pairs and any other
electron pair
Arrange the three isomers above in order of increasing energy. Use the labels I, II and III to
indicate the 3 isomers.
lowest
energy
highest
energy
It has been determined experimentally that ClF3 is a polar molecule. Does this additional
information eliminate any of the isomers? If so, which one(s)?
By selecting the lowest energy isomer, sketch your prediction for the structure of ClF3 in the
space below. Indicate on your diagram the approximate values of the bond angles.
Demonstrator's
Initials
E11-17
(9) NO3– : Resonance and the Limitations of Simple Models
In this exercise we will look at one of the limitations of simple geometrical models.
Consider the nitrate ion, NO3–. This molecular ion has a total of
valence
electron pairs.
Draw an electron dot diagram of NO3– and indicate how these electron pairs are distributed.
Based on this electron dot diagram, build a model of NO3– including any multiple bonds and
nonbonding electron pairs. Before you start, consider the following: are all the NO bonds
equivalent? Give the cluster type that should be used for each of the atoms.
N atom:
cluster
Double bonded O atom:
Both single bonded O atoms:
cluster
clusters
Measurements of the rotation of the nitrate ion (probed using microwaves) indicate that all three
NO bonds are equivalent in length. This is (or should be) in contradiction of your model. The
shortcoming of the model (and the electron dot representation) is that the valence electrons do
not have to be always located either between two atoms (in the case of a bonding pair) or on a
single atom (in the case of a nonbonding pair). Electrons can be distributed between 3 or more
atoms. In the case of NO3– we can represent this as shown in Figure E11-8.
_
O
O N
O
Figure E11-8
As usual each solid line represents a bonding electron pair. The dotted lines represent three
electron pairs distributed between all 4 atoms. Note that each NO bond is now equivalent, in
agreement with experiment. NO3– is an example of the limitation of the bonding model we have
used. Specifically, it shows up the weakness of the simplifying assumption that bonding
electron pairs must be located only between a specific pair of atoms.
E11-18
In order to get around this problem while retaining the simplicity of this model, the idea of
resonance structures was introduced. Examine your model of the nitrate ion. There were three
choices for which NO bond would be the double bond and, hence, three possible structures
(related to each other by a rotation through 120°) as shown in Figure E11-9. In the resonance
picture, the structure of certain species (such as NO3–) cannot be depicted accurately by a single
electron dot formula. Rather, the “true” structure of NO3– is an average (or resonance hybrid)
of all the contributing or resonance forms shown in Figure E11-9.
-
O
O
O N
O N
O
O
O N
O
O
Figure E11-9
The formula of formic acid is usually written HCOOH, whereas the formula of the formate ion
is usually given as HCO2–. Draw the electron dot formulas of these two species and explain
why the two O atoms in formic acid are different, but the two O atoms in the formate ion are
equivalent.
Demonstrator's
Initials