A laboratory experiment in crystals and crystal model building
Each group should have 14 orange foam balls of diameter about 3.6 cm
Objectives
When you have completed this experiment, you should understand
1. Hexagonal closet packing (h.c.p.) and cubic closet packing
(c.c.p.) of sphere;
2. What a tetrahedral hole is;
3. What a octahedral hole is;
4. What a face-centered cubic (f.c.c.) lattice is;
5. The equivalence of a c.c.p. lattic and a f.c.c. lattice;
6. The meaning of the term, unit cell
7. Where the tetrahedral and octahedral holes are in a f.c.c.
unit cell.
8. The crystal structures for which the following compounds are
prototypes. These are all based on a face-centered cubic unit
cell:
a) NaCl
b) Zinc blende (ZnS)
c) Fluorite (CaF2)
d) Anti-fluorite
e) Diamond
9. The body-centered cubic (b.c.c.) lattices (e.g. CsCl)
10. The relationship between ionic radius and structures;
11. The techniques for growing a crystal from solution.
Background
Sir W.H. Bragg, one of the first scientists to use X-rays for the study of crystal structures,
was fascinated by the intricate and orderly atomic patterns revealed by his investigations. He
expressed this idea in 1901 "The molecules are locked into crystalline structure when attachments
are made at sufficient points, and the whole has the stabiity of an engineering structure."
Natural crystals have always held a fascination for man, whether they were precious stones,
such as diamonds or rubies, or just a beautifully formed snowflake. Today we recognize, as did
Bragg, that the regular geometric shapes of well-formed crystals reflect orderly three dimensional
network of the atoms, ions, or molecules that constitute the crystal lattice.
Although the arrangements of atoms and ions which lead to regular crystal lattice structures
are quite simple, it is difficult for people to visualize the arrangements in 3-dimensions.
2-dimensional drawings of 3-dimensional objects often do not help much! One of the purposes
some of the common crystal structures so that you may clearly see how the atoms and ions are
arranged. Having built and examined such models, it will be easier for you to make some sense of
drawings in your text and lectures.
An additional purpose of the experiment is to give you a chance to grow crystals of your
own. From this experience, you will not only learn how one goes about growing a crystal of
macroscopic size, but you will clearly see that orderly arrangements of particles on the microscopic
level lead to orderly structures on the macroscopic level.
Perfect crystals, in which the repetitive lattice pattern is completely uniform throughout the
solid, are extremely rare. This is not surprising when you consider that a crystal is typically formed
from a liquid (a solution or the molten solid) or a gas, in which the particles are quite randomly
oriented. To initiate crystal growth, a tiny collection of particles must form in a highly ordered
pattern. This process is called nucleation and it may occur spontaneously when particles of a pure
substance collide with appropriate orientations and with low enough kinetic energies to permit the
establishment of "permanent" inter-particle attractions rather than "rebound collisions". More often,
some foreign material provides a surface on which particles may be adsorbed in orientations
suitable for continued crystal growth. The nucleation of snowflakes, for example, often, involves a
"dust particle" as the center of adsorption for water molecules, and a string suspended in a
supersaturated solution of a salt may provide nucleation sites as ions are adsorbed onto the fibers.
The tiny micro-crystals initially formed may be perfectly ordered, but continued growth
generally introduces imperfections. As the crystal grows, the surface areas for addition of more
particles become continuously larger, thus requiring an ever-increasing availability of the proper
particles, with appropriate orientations. When molecular crystals grow from the pure liquid (e.g., as
pure water freezes) or gas (e.g. in the formation of iodine crystals from the vapour.), there are no
problems with imperfections from foreign particles, but it is most unlikely that uniform growth
rates over all surface areas will continue. Non-uniform growth may result in crystal imperfections.
Crystals grown from mixtures, such as salt solutions or impure liquids, may contain chemical
imperfections from foreign particles trapped in crystal lattices in addition to physical imperfections
resulting form irregular growth patterns.
The growing of nearly perfect crystals, or of crystals containing "planned imperfections", is
an important aspect of modern solid state technology. Synthetic diamonds are grown for such uses
as diamond tipped drills; sodium chloride crystals are produced for use in the optics systems of
infrared spectrophotometers; and "doped" crystals of silicon or germanium (containing planned
substitutional lattice defects) are prepared for solid state electronic equipment.
Crystal growing is fun. To make it more so, we would like to make it into a contest, which
will run, for the duration of A term. The object is to grow the biggest and most perfectly shaped
crystal of alum (potassium aluminum sulfate) that you can.
Procedure
You will grow your crystals on an individual basis, but will work in pairs on the model
building. The best thing to do, then, is to get your crystal growing started during the first 30
minutes of the period, then pair off and build.
A. Crystal growing
1. Weigh out 10 g of alum and place it in a clean 250-mL beaker. For each gram of alum
used, add 7 mL of water. Heat the mixture and stir it with a stirring rod until a clear
solution is obtained. (If the mixture remains cloudy, allow the mixture to stand for a few
minutes until the sediment has settled out, then carefully decant (i.e. pour off) the clear
solution into another clean 250-mL beaker.) Allow the solution to cool to room
temperature. Pair off and begin model building at this point. Return to step 2 after you
have completed the model building part of the experiment.
2. Tie a piece of thread to a glass rod so that when suspended in the solution the thread will
extend no more than 1 cm below the surface of the solution. Smear grease on the part of
the thread that will be above the solution to keep the solution from creeping up the
thread. Cover the top of the beaker with a piece of filter paper, and use a rubber band to
hold the paper in place. Punch a hole in the center of the paper, lower the thread through
it until the thread is submerged and the stirring rod rests on top of the beaker. Keep the
beaker in a safe storage place. Crystals should have formed on the string and/or at the
bottom of the solution by the time you return for your next lab period.
3. If crystals have formed on the string, crush off all the best one. If not, decant the liquid
from the crystal or crystals at the bottom of the container into another clean container. If
more than one crystal grew, inspect them and pick one that shows good faces. Tie a fine
thread to this "seed" crystal so that it may be suspended from the glass rod.
4. Suspend the crystal in a freshly prepared solution of 10 g of alum in 70 mL of water.
This solution must be allowed to cool to room temperature before the crystal is
introduced. If you would rather have a coloured crystal, you may
5. Substitute a growing solution of "chrome alum", KCr(SO4)4. 12H2O, in place of the
second alum growing solution. Prepare your seed crystal as described above using 8 g of
alum in 56 mL of distilled water repar?? second solution of chrom alum. Using 15 g per
25 mL of H2O. The colour of this solution will be a dark blue-green. Pour the coloured
solution slowly into the growing colourless solution of ordinary alum, while holding the
growing solution up to the light. Stop adding the chrome alum when you almost can't
see through the solution. Suspend your seed crystal in the solution for 1 week. For
subsequent growing stages (the crystal grows in different solutions, each for about one
week), using ordinary alum solutions. This will give you a purple crystal inside a
colourless one. [N.B. It is important to cool your growing solution sufficiently] before
suspending your seed crystal in it, particularly if it was necessary to tie your seed crystal.
If the solution has not been allowed to cool enough it will not be saturated It will start to
dissolve the seed crystal and the seed crystal may slip out of the thread loop. You can
get an indication if the solution is cool enough by carefully observing the portion of the
solution near the suspended crystal. if the solution is less than saturated, the portion of
the solution in contact with the crystal will dissolve some of the crystal. This will
increase the density of that portion of the solution and it will flow to the bottom of the
beaker. On careful observation this density current can be seen. If this happens, remove
the seed crystal and allow the solution to cool further.
B. MODEL BUILDING
Required materials
1. 20 white foam balls, 12 orange foam balls and 8 pink foam balls; all of diameter 3.6 cm.
(2) A transparent square plate of side length about 20 cm. (3) A piece of transparent
plastic strip of length about 60 cm and of width about 6 cm. (4) White glue or thermal
molten plastic glue for adhering foam balls together (5) A paper clip
1. Closest packing
a) On a 20 cm square transparent plate use a paper clip to fix the two ends of a transparent
plastic strip such that the strip forms a boundary for the foam balls.
b) Put 19 white foam balls into the space
surrounded by the plastic strip. Arrange the
balls such that a hexagon of 3 balls on each side
is formed. Arrange the balls in the manner of
closest packing. Such an arrangement allows
the largest possible number of spheres to fit into
a given area. That is the density of spheres is a
maximum. Note that each sphere in the layer is
surrounded by a perfect hexagon of spheres, all
touching it.
c) Look directly into your close-packed layer
from above. You should see two types of
triangularly-shaped depressions between
spheres. These depressions are shown in the
right figure. One type has a vertex pointing
"up", the other type has a "up" vertex pointing
"down". We will call these "u" and "d"
depressions, respectively. Note that a "down"
depression of one type has vertices in common
with 3 holes of the second type, and vice
versa.
d) Place an orange sphere on the white layer of atoms. If it is placed directly on top of one of
the red spheres, it will spontaneously roll into one of the depressions, or holes, formed by
three adjacent white spheres. This leads to the closest packing arrangement of spheres in
two layers. Now place the orange sphere in the first "d" hole in the upper left hand corner of
the white layer.
Focus your attention on the orange sphere and the three
white spheres that it touches. They should look as
shown in the following diagram. These 4 touching
spheres lie at the corners of a regular tetrahedron (a
polyhedron with 4 vertices and 4 faces, each an
equilateral triangle). The hole at the center of the 4
spheres is therefore called a tetrahedron hole.
e) Add 12 orange spheres to your second layer, using only "d" type depressions. In doing this,
you find that none of the "u" type depressions in the first (white) layer are used. If you look
down through the "d" type holes in the top layer, you can see the "u" holes in the bottom
layer staring you in the face.
f) Now remove 8 spheres away and remain four
spheres at the center part. Arrange these spheres
in the shape of a parallelogram. Focus your
attention on the two equilateral triangles of
orange spheres of the parallelogram. One
triangle, together with a white sphere in the first
layer forms a tetrahedron hole. Another triangle,
together with an inverted triangle (formed by
three white spheres) forms another kind of hole.
The hole lies at the corners of a regular
octahedron which looks as in the right diagram.
The hole at the center of the 6 spheres is
therefore called an octahedral hole. After you
have recognized these holes, fill in other orange
spheres to construct the whole second layer.
g) In constructing a third layer of spheres above
the first two, we are confronted with a choice.
The layer can be constructed using "u" type
depressions in the orange layer so that the atom
lie in rows directly above the white spheres in
the first layer; or the spheres in the third layer
can be placed in "d" type depressions in the
orange layer so that neither orange nor white
spheres lie directly below them.
h) Using 7 white spheres construct the third layer
so that each sphere on the third layer is directly
above a sphere give you a central sphere
surrounded by a hexagon. The hexagon lies
directly over an identical hexagon in the first
layer (find it!). These two hexagons are
separated by an equilateral triangle of orange
spheres in the second layer. We summarize this
diagrammatically as in the right diagram.
We might characterize this packing arrangement as w,o,w,o,..., the letters indicating that the
colours and locations of spheres repeat every other layer. This type of packing arrangement is
referred to as hexagonal closest packed (h.c.p.). Its more general designation is 1212.... . Note
that this structure has hexagonal-shaped channels running through the structure in the z-direction.
2. The face-centered cubic lattice
a) Using 8 pink spheres to construct the third layer so that neither white nor orange spheres lie
directly below the pink spheres by placing pink spheres in the "d" type depressions of the
orange sphere. Finally, place a white sphere in the center hole of the pink layer. This lies
directly over a white sphere in the first layer. We might characterize this packing arrangement
as w,o,p,w,o,p,... It is called cubic closest packing (c.c.p.) and is denoted generally as
123123...
b) Build up a unit cell of cubic closest packing
(1) 18 white spheres and 1 orange sphere to form a hexagon in the closest structure. the center
sphere of the first layer to be an orange sphere.
(2) Only 6 spheres of the second layer to be orange; these spheres put into the 'u'-type
depressions and they are arranged such that
(i)
they form an equilateral triangle and
(ii) its central depression is just above the orange sphere in the first layer.
(3) 6 orange spheres in the third layers (before doing so, you must add 6 white spheres
surrounding the orange spheres in the second layer.) They are arranged just above the
'd'-type depressions of the first layer.
(4) One orange sphere in the central depression of the third layer. This sphere should be just
above the orange sphere in the first layer. It forms the fourth layer. The orange spheres so
built constitute a unit cell of c.c.p. structure. Now observe the orange spheres carefully
from different positions. The orange spheres have a cubic shape! Can you identify
diagonal on the face of this cube? The unit cell under inspection has the shape of a cube.
Each face of the cube has a centre atom. This structure also known as face centered cubic
(f.c.c.)
3. Lattice holes
There are two kinds of holes (vacancies) in the closest packing, say tetrahedral hole and
octahedral hole.
Questions for discussion
1. What is the coordination number (i.e. the number of atoms touching the atom) of an atom in a
ccp structure? In a hcp structure? Ues your models to help you decide.
2. Draw a diagram of the f.c.c. unit cell you have just constructed. Show the edge of the cube
clearly.
3. This question is about face-centred cubic.
(a) Find the hole, or vacancy, at the center of the fcc unit cell.
4.
5.
6.
7.
(b) By how many spheres the vacancy is surrounded?
(c) What is the shape formed by the above 6 spheres?
(d) Give a name for the hole.
How many octahedral holes are there in your unit cell (the vacancies at the centres of the cube
edges are also octahedral holes which belong only partially to the unit cell considered.)?
How many tetrahedral holes are there in your unit cell?
How many atoms are there in your unit cell? (Remember, some of the atoms belong only
partially to this unit cell!).
Which is larger, a tetrahedral hole or an octahedral hole?
4. Structures based on the f.c.c. lattice
a)
b)
c)
Rocksalt (NaCl) structure : Many other compounds crystallize with the NaCl-type
structure: several alkali halides (e.g. KBr), hydrides (e.g. NaH), and some alkaline earth
oxides (e.g. CaO). Even such "complex" compounds as NH4Cl have this structure.
Study the zinc-blend, ZnS-type and the fluorite-type (CaF2) structure
Study the structures of diamond.
Questions for discussion
8. Draw the structures.
(a) How many Na+ ions per unit cell?
(b) How many Cl- ions per unit cell?
(c) Are these numbers consistent with the 1:1 stoichiometry implied by the formula?
(d) What is the coordination number of each cation in this type structure?
(e) What is the coordination number of each anion?
9.
(a) Draw the Zinc-blend structure.
(b) Draw the Fluorite structure
10. (a) Confirm the 1:1 ion ratio in zinc-blend structure (Reference: see part a).
(b) Confirm the 2:1 ion ratio in fluorite structure.
(c) What is the coordination number of Zn2+?
(d) What is the coordination number of Ca2+?
(e) What is the coordination number of S2-?
(f) What is the coordination number of F-?
11. What is the coordination number of each carbon atom in the diamond lattice (check several
atoms in the model!)
4.Body centered cubic arrangement
The alkali metals, barium, the vanadium and chromium groups of elements all have the bcc
arrangement of atoms. Study the unit cell of body centered cubic lattice (bcc) and Draw the
structure. What is the coordination number? What is the number of particle in a unit cell of b.c.c.?
END