DISCUSSION
Unlike molecular substances, the strong bonding interactions in ionic and metallic substances
are not confined to a discrete group of atoms (the molecule). Rather these substances form
extended networks of atoms. This lab examines the structures of extended solids by considering
some of the more common and symmetric crystal structures of ionic and metallic substances. It
also serves as an introduction to X-ray diffraction, which is the technique used to determine the
arrangement of atoms in crystals.
1. Translational Symmetry and Unit Cells
Structures of molecules are defined by the positions of all atoms that make up the molecule.
It would be problematic to describe an extended solid in the same manner. First of all there
would be a huge number of atoms to consider, and secondly the number of atoms in the crystal
varies depending upon the macroscopic size of the crystal. Fortunately, there is a simpler way to
describe the structures of extended solids. In all crystals there is a small unit which contains a
unique arrangement of atoms and the crystal is built up by repeating that unit over and over again
in all three directions. This is similar to the way a wall is constructed by stacking identical
bricks together. The repeating unit (or brick) is called a unit cell. The geometrical pattern of
points on which the unit cells are arranged is called a lattice. The structure of an extended solid
is defined by two things: (a) the dimensions and symmetry of the unit cell, and (b) the locations
of atoms within the unit cell.
Before moving to three dimensions it is useful to consider two dimensional unit cells,
because they are simpler to describe and visualize. Figure 1 shows a two dimensional array of
lattice points. All lattice points have an identical environment. That is to say that if you sat on a
lattice point and looked out at your surroundings the view would be identical,
b
a
Figure 1: An arbitrary two dimensional lattice. The lattice vectors, a and b, are denoted with bold
arrows, and the unit cell is shaded in gray.
regardless of which lattice point you were on or how far you could see. In Figure 1 the lattice
points are denoted by the circles. Their positions are defined by the lattice vectors, a and b.
Beginning from any lattice point it is possible to move to any other lattice point through
translation of an integer number of the two lattice vectors. This property, called translational
symmetry, is the defining characteristic of a crystal. The parallelepiped formed by the lattice
vectors (shaded gray in Figure 1) defines the unit cell. In two dimensions the unit cells can be
tiled so that they completely cover the area of the lattice, with no gaps. In three dimensions the
unit cells can be stacked together to fill all space.
You might think that unit cells could adopt any number of different shapes. However, in two
dimensions there are only five possible lattices. In crystallography these are called Bravais
Lattices. The oblique lattice, which represents the most general case is represented in Figure 1.
The remaining four 2D Bravais Lattices are shown in Figure 2. Other more complicated shapes
you might imagine, such as pentagons and octagons are not valid lattices, because they do not
tile together in a way that fills all space. Note that while hexagons do tile all space, as shown in
Figure 2d the unit cell does not have the shape of a hexagon. That’s because the unit cell must
be defined with two and only two vectors, and it would take three vectors to define a hexagon.
Nonetheless, the hexagonal unit cell shown in Figure 2d has a defined shape derived from a
hexagon, and it tiles together to fill all space.
b
γ
a
b
a
(a)
γ
(b)
b
a
b
γ
a
(c)
(d)
γ
Figure 2: The two dimensional Bravais lattices: (a) primitive rectangular (a ≠ b, γ = 90°), (b) centered
rectangular (a ≠ b, γ = 90°), (c) square (a = b, γ = 90°), and (d) hexagonal (a = b, γ = 120°). The fifth
2D Bravais Lattice, oblique (a ≠ b, γ ≠ 90°) is illustrated in Figure 1. In (b) the primitive unit cell is
shown with dashed lines.
The distinction between a primitive rectangular lattice and a centered rectangular lattice
merits further discussion. The centered rectangular lattice (Figure 2b) differs from the primitive
rectangular lattice (Figure 2a) by the addition of a lattice point at the center of the unit cell.
While it is possible to draw a smaller primitive unit cell that only has lattice points at the corners
of the unit cell (as denoted by a dashed line in Figure 2b), the lattice vectors of the primitive unit
cell are no longer perpendicular to each other. Hence the centered rectangular unit cell shown in
Figure 2b is normally favored. It is important to note that the lattice points at the corners of the
unit cell and the lattice points at the centers of the unit cell have identical surroundings. We will
see when we move to three dimensions that centered lattices are fairly common.
2. Structures of Metallic Substances – Cubic and Hexagonal Close Packing
Metals tend to form crystal structures where the atoms are packed together as closely as
possible. If we approximate the atom as a sphere and construct a close packed layer of spheres
each atom would have six neighbors as shown by the large grey spheres in Figure 3a.
(b) 2-Layer AB Stacking
(a) close packed
layer
(c) Alternate 2-Layer AB Stacking
Figure 3: (a) A close packed layer of spheres. The gray dots represent one set of depressions where
the next layer could sit, as shown in (b). The black dots represent an alternate set of depressions
where the next layer could sit as shown in (c).
To form a three dimensional structure the next step would be to stack layers together. The
spheres will be most closely packed if the next layer of spheres sits in the depressions marked
with either the red or blue dots in Figure 3a. Two different colors are used because the
depressions are too close together for the next layer of spheres to sit on all of the depressions.
The next layer must either sit in the depressions marked with gray dots, as shown in Figure 3b
or in the depressions marked with black dots, as shown in Figure 3c.
If we only have two layers then it doesn’t matter which of the two types of stacking takes
place. They are indistinguishable. However, when we stack a large number of layers together to
form a three dimensional crystal the stacking pattern makes a difference. There are many
different ways you might imagine stacking together close packed layers of spheres. Fortunately,
the structures of most metals can be described by two simple stacking sequences. Those two
sequences are shown in Figures 4 and 5 below. If third layer lies directly over the positions
where the spheres in the first layer sit, and the fourth layer lies directly over the positions where
the spheres in the second layer sit, the stacking pattern repeats every other layer and is
designated as ABAB… stacking. This type of stacking leads to a 3D structure called hexagonal
close packing (hcp) (shown in Figure 4). If the third layer is offset in such a way that it lies
over the blue dots in Figure 3a it will not sit directly above either layers 1 or 2. The three layer
sequence repeats beginning with the fourth layer (which lies directly over the spheres in the first
layer) to give an ABCABC… stacking pattern. This type of structure is called cubic close
packing (ccp) (shown in Figure 5).
Layer 1
Layers 1 & 2
Layers 1,2 & 3
Layers 1,2,3 & 4
Figure 4: An ABAB… stacking sequence giving rise to hexagonal close packing (hcp) of spheres.
Layer 1
Layers 1 & 2
Layers 1,2 & 3
Layers 1,2,3 & 4
Figure 5: An ABCABC… stacking sequence giving rise to cubic close packing (ccp) of spheres.
It should be noted that hexagonal and cubic close packing are equally efficient ways of
packing spheres. In both cases is 74% of the total volume is occupied. To demonstrate the
similarity of these two structures consider the nearest neighbor coordination environment
(the geometry of the atoms which neighbor an arbitrarily chosen central atom) for each
structure. Each sphere has six neighbors within the same layer, plus three nearest neighbors
in the layer above and another three in the layer below. Therefore, we would say that each
atom has a coordination number of 12, as shown in Figure 6. Note that all 12 neighbors are
the same distance away from the central atom (equidistant) in both types of stacking.
(a)
(b)
Figure 6: The nearest neighbor coordination environment of an atom in (a) a cubic close packed
metal and (b) a hexagonal close packed metal. The shading of the atoms illustrates the different close
packed layers.
3. Three Dimensional Translational Symmetry - Close Packed Metals
In this section we will illustrate the relationship between a lattice and a crystal structure by
considering the structures of metals in greater detail. In 3D three lattice vectors are needed to
define the unit cell. This leads to unit cells that come in seven different shapes, or crystal
systems: cubic, tetragonal, orthorhombic, hexagonal, rhombohedral, monoclinic and triclinic.
When these seven different unit cells are combined with three different types of lattice centering
(body centering, face centering and base centering) 14 different Bravais Lattices result. For a
closer look at the 3D Bravais Lattices see http://undergrad-ed.chemistry.ohiostate.edu/xtal/lattice.html. In the discussion that follows we will only consider the crystal
systems needed to describe close packed metals: hexagonal and cubic.
The 3D hexagonal lattice can be derived from the 2D hexagonal lattice by adding a third
lattice vector, c, perpendicular to the a and b lattice vectors of the 2D hexagonal lattice (Figure
2d). The unit cell of an hcp metal is shown in Figure 7. In this structure the unit cell contains
two atoms, one from each layer as can be seen clearly in Figure 7c. The atoms are displaced
from the lattice point by the following vectors:
Atom 1:
1/3a + 2/3b + 1/4c
Atom 2:
2/3a + 1/3b + 3/4c
The crystal structure is built up by placing two atoms per lattice point, displaced from the
lattice point by the above two vectors. This structure illustrates two important concepts in
crystallography. First of all, it is not necessary for the atoms to sit on the lattice points.
Secondly, we are not limited to one atom per lattice point. In principle there can be any number
of atoms per lattice point. To see how a crystal is built up from an initial group of atoms
followed by application of the translational symmetry associated with various Bravais Lattices
try out the exercises on the web at http://undergrad-ed.chemistry.ohio-state.edu/xtal/xtal.html.
(a)
(b)
(c)
Figure 7: The crystal structure of a hexagonal close packed metal: (a) as seen looking down the c
axis, (b) as seen looking roughly perpendicular to the c axis, (c) showing only the unit cell and the
atoms contained in the unit cell. The close packed layers are artificially colored differently to
emphasize the layer stacking.
There are three Bravais Lattices that have cubic unit cells: primitive cubic, body centered
cubic and face centered cubic. Figure 8a illustrates a metal that adopts the primitive cubic
structure with a single atom sitting on each lattice point. Metals with a primitive cubic structure
are very rare, the only example being the radioactive element polonium, which was much in the
news in late 2006 because of its use to poison a former Russian spy. Figure 8b illustrates a body
centered Bravais lattice. Body centered cubic metals are much more common, some examples
include iron, chromium, molybdenum and tungsten. Figure 8c illustrates a face centered Bravais
lattice. Face centered cubic metals are also common, some of the better known examples include
aluminum, lead, copper, silver, gold and platinum.
(a)
(b)
(c)
Figure 8: The three cubic Bravais Lattices: (a) primitive cubic, (b) body centered cubic, (c) face
centered cubic.
At first glance it may seem as though there are 8 atoms in the primitive cubic unit cell
(Figure 8a) and 9 atoms in the body centered cubic unit cell (Figure 8b), and 14 atoms in the
face centered cubic unit cell (Figure 8c). That counting neglects the fact that atoms which reside
on a boundary (corner, edge, face) of a unit cell are shared by more than one unit cell. Hence
only a fraction of the atom truly resides in the unit cell, as shown in Table 1.1 Atoms that reside
on the corner of a unit cell are shared by eight unit cells so that only 1/8 of each corner atom is
1
It is only the position of the center of the atom that matters. Atoms that reside near the boundary of the unit cell,
but not on a corner, edge or face are counted as residing 100% within the unit cell.
actually within the unit cell. Therefore, the primitive cubic unit cell contains only 1/8 × 8 atoms
= 1 atom per unit cell. In the body centered unit cell there are also 8 atoms at the corners plus an
additional atom at the body center raising the atom count to (1/8 × 8) + 1 = 2 atoms per unit cell.
In the face centered unit cell there are 6 atoms on the faces of the unit cell in addition to the
atoms at the corners. Atoms at the face are shared by two unit cells and thus only ½ of each
atom is actually within the unit cell.
Atom Location
# of Unit Cells
Sharing the Atom
Fraction of atom
within unit cell
Corner
8
1/8 or 12.5%
Edge
4
1/4 or 25%
Face
2
1/2 or 50%
Anywhere else
1
1 or 100%
Table 1: The fraction of an atom sitting on the unit cell boundary that resides within the unit cell.
Although it is not immediately obvious a face centered cubic metal is the structure that
results from cubic close packing, as described in the previous section (explaining why packing of
hexagonal layers was called cubic close packing). Figure 9 shows the relationship between the
ABC… layer stacking and the face centered cubic unit cell. As with the hcp example the atoms
have been artificially colored to emphasize the close packed layer from which they come. You
can see that the direction perpendicular to the layer stacking corresponds to the body diagonal of
the cubic unit cell.
C
A
B
A
C
A
B
A
(a)
(b)
Figure 9: The crystal structure of a cubic close packed metal: (a) emphasizing the ABC… layer
stacking, (b) emphasizing the face centered cubic unit cell. The close packed layers are artificially
colored differently to emphasize the layers.
4. Structures of Ionic Substances – Octahedral, Tetrahedral and Cubic Holes
Like metallic substances, ionic substances tend to prefer structures with symmetric, closely
packed arrangements of atoms. However, the different bonding interactions lead to different
structures. In metals the bonding preferences favor structures where each atom has as many
neighbors as possible. In ionic compounds we want structures where the cation-anion contacts
are as close as permitted by ionic radii, while at the same time maximizing the anion-anion and
cation-cation distances. Interestingly the crystal structures of many ionic compounds can be
described as a close packed arrangement of anions with cations occupying the cavities in the
close packed anion sublattice. To better understand these structures let’s take a closer look at the
cavities between two close packed layers of anions.
In many ionic structures we can retain layer stacking analogous to those already discussed for
metals, with the caveat that the layers alternate between anions and cations. Remember back to
our discussion of layer stacking illustrated in Figure 3. There were two sets of depressions in
the close packed layer of spheres, one marked with black dots and the other with gray dots.
Imagine an ionic compound where the initial close packed layer of anions (Figure 10a) is
followed by a cation layer (Figure 10b) that occupies the depressions marked by black dots.
This is followed by a second anion layer (Figure 10c) that resides on positions marked by the
gray dots. This leads to a cation surrounded by three anions from each layer (Figure 11a),
which results in a cation coordination number of 6, and an octahedral coordination
environment for the cation.
Alternatively, the cation layer could occupy the depressions marked by gray dots (Figure
10d). In this arrangement the next anion layer would reside directly over the cations (Figure
10e). This kind of packing leads to a cation with three neighbors in the layer below and one
neighbor from the layer above (Figure 11b), which leads to a cation coordination number of 4,
and a tetrahedral coordination environment for the cation. In this case the cation does not sit
at a height directly between the layers, but shifts toward the lower layer in order to attain four
equidistant bonds to the anions. Thus we can say that there are both octahedral and tetrahedral
holes for cations in a close packed array of anions.
If you are paying attention you may be wondering what would make anions (or cations for
that matter) close pack with each other. Electrostatic forces favor the anions remaining far apart
as possible from each other. The answer to this seeming paradox stems from the fact that the
geometric pattern which maximizes the separations of atoms (in a box of fixed size) is the same
as the pattern that describes close packing. In ionic compounds the size of the unit cell is
dictated by the cation-anion distance, and while the anions might look like they are trying to
close pack the anion-anion distances are in fact large enough that the anions are not touching
each other.
(a)
(b)
(d)
(c)
(e)
Figure 10: (a) A close packed layer of anions (in dark gray), (b) cations (in light gray) fill the
octahedral holes, (c) a second layer of anions is placed on top of the cations, (d) cations fill the
tetrahedral holes, (e) a second layer of anions is placed on top of the cations.
(a)
(b)
(c)
Figure 11: Illustrations of (a) octahedral coordination, (b) tetrahedral coordination, and (c) cubic
coordination. In these figures the anions are represented by gray spheres and the cations by black
spheres.
5. Structures of Ionic Substances – Rock Salt, Fluorite & Zinc Blende
We can describe the structures of many ionic compounds using the description of the
previous section. If we limit ourselves to a cubic close packing of anions let’s consider what
structures can be formed by filling octahedral or tetrahedral holes. The unit cell and anion
positions were shown previously in Figure 9b. The octahedral holes are located on the edges
and in the center of the unit cell. If we fill all of the octahedral holes with cations the structure
that results is the rock salt structure, shown in Figure 12a. The octahedral coordination of the
cation at the center of the unit cell is emphasized in this figure, but if we consider neighboring
unit cells we would see that all of the cations are surrounded by six anions. The anions are also
surrounded by six cations so that if we were to interchange the anions and cations the structure
would be the same.
The easiest way to see the tetrahedral holes is divide the unit cell up into 8 smaller cubes of
equal size, one for each quadrant. The tetrahedral holes are located at the center of each of these
smaller cubes. If we fill all of the tetrahedral holes with cations the structure that results is the
antifluorite structure. There are relatively few examples of compounds with the antifluorite
structure, Na2O being one example. If we interchange the anions and cations in antifluorite, the
fluorite structure results, as illustrated in Figure 12b. This arrangement, which has a close
packed array of cations with anions filling the tetrahedral holes, is more common. Perhaps the
best known example is the mineral fluorite, CaF2.
(a)
(b)
(c)
Figure 12: The crystal structures of the (a) rock salt, (b) fluorite, and (c) zinc blende structures. In
these figures the anions are represented by gray spheres and the cations by black spheres.
Another example does not involve close packed layers of atoms, but rather simple cubic
arrays. The anions are arranged in a “squared” formation instead of the previous alternating
configuration. The cation layer occupies the depressions in between the anions. In this
arrangement the next anion layer would reside directly over the previous anion layer. This
packing leads to a cation with four neighbors in the layer below and four neighbors from the
layer above (Figure 11c), which results in a cation coordination number of 8, and a cubic
coordination environment for the cation. We can see this type of coordination environment for
the cation in fluorite as illustrated in Figure 12b.
There is no particular reason why all of the holes have to be filled. There are a number of
important structures that result when a fraction of the octahedral or tetrahedral holes are filled.
One of the most important of these is the zinc blende structure, shown in Figure 12c. The zinc
blende structure is closely related to antifluorite structure. The difference between the two is that
now only ½ of the tetrahedral holes have been filled by cations. You will explore these three
structures in more detail in the laboratory exercises.
6. Calculations with Extended Solids
In the laboratory exercises you will be asked to analyze the ionic structures in a variety of
ways. You will be asked to determine the empirical formula, the density, the coordination
environments of the atoms, and the relationship between the unit cell dimensions and the ionic
radii of the ions involved. In order to illustrate these calculations we will go through a similar
analysis on strontium titanate, which has the cubic perovskite structure as shown in Figure 13.
Figure 13: The crystal structure of strontium titanate. The titanium atoms are represented by black
spheres, the oxygen atoms by dark gray spheres and the strontium atom by light gray spheres.
We’ll begin by calculating the empirical formula. To do this we simply need to count the
number of atoms contained in the unit cell. Since the unit cell is repeated over and over again in
any crystal the overall stoichiometry of the crystal must equal the stoichiometry of the unit cell.
Remembering that atoms located at the boundaries of the unit cell (corners, edges and faces) only
count as a fraction because they are shared by other unit cells, the number of atoms in the unit
cell is one strontium (1 × 1), one titanium (8 × 1/8) and three oxygens (12 × 1/4). This gives an
empirical formula of SrTiO3.
The density of an object is its mass divided by its volume. We can calculate the macroscopic
density of a substance by determining the mass of the atoms in the unit cell and dividing by the
volume of the unit cell. In this case we just determined that there is 1 strontium atom, 1 titanium
atom and 3 oxygen atoms in the unit cell. The masses of each of these atoms can be obtained by
dividing their atomic weight by Avogadro’s number.
Mass = [{1 × 87.62 g/mol} + {1 × 47.87 g/mol} + {3 × 16.00 g/mol}] × (1 mol/6.023 × 1023)
Mass = 3.046 × 10-22 g
NOTE: The mass is calculated using ALL of the atoms in the unit cell. In this case of SrTiO3,
the number of each atom in the unit cell is the same as the empirical formula. In the case of LiCl
for example, there are 4 Li atoms and 4 Cl atoms in a unit cell (giving an empirical formula of
LiCl). The mass calculation for LiCl must include 4 Li atoms and 4 Cl atoms like so:
Mass = [{4 × 6.94 g/mol} + {4 × 35.45 g/mol}] × (1 mol/6.023 × 1023) = 2.82 × 10-22 g
The volume of a cube is simply the length of the edge cubed. As we will explore in the
sections that follow X-ray diffraction can be used to determine the dimensions of the unit cell.
For the purposes of this exercise it is sufficient to know that the unit cell edge is 3.91 Å (or 3.91
× 10-8 cm). Given this information the density of strontium titanate can be calculated:
Density = Mass/Volume = (3.046 × 10-22 g)/(3.91 × 10-8 cm)3
Density = 5.10 g/cm3
The next question to consider is the coordination number and environment of each atom. For
brevity we will limit the discussion to the cations, titanium and strontium. The coordination of
the strontium ion can be readily seen from the unit cell diagram in Figure 13. There are twelve
edges in a cube and one oxygen atom at every edge. Therefore, strontium is coordinated by
twelve equidistant oxygen atoms (much like the atoms in a close packed metal). The titanium
coordination cannot easily be seen from the unit cell diagram of Figure 13. If we look at the
titanium atom at the lower front corner of the unit cell you can see three neighboring oxygen
atoms that are located on the edges of the unit cell. However, if we consider neighboring unit
cells we find that there are three additional oxygen atoms coordinated to titanium. This gives six
oxygen neighbors in an octahedral geometry. Since the environment at all lattice points must be
identical the same analysis applies to all titanium atoms in the crystal.
Figure 14: The structure of strontium titanate showing three additional unit cells. In the drawing on
the left all of the atoms the three neighboring unit cells are shown. In the drawing on the right all of
the atoms have been stripped away except for those coordinated to the central titanium atom.
Finally we consider the relationship between the size of the unit cell and the ionic radii of the
atoms involved. This analysis is based on two assumptions. First of all, we will treat the ions as
hard spheres. Secondly, we will assume that the ionic attraction between anions and cations will
draw them together until the spheres are touching. For ionic compounds these assumptions are
reasonably valid. With SrTiO3 the situation is slightly more complicated because there are two
possible cation-anion contacts (Ti-O and Sr-O) that could determine the size of the unit cell. We
will consider each in turn.
Figure 15a shows the ionic contacts in the bottom face of the unit cell. The faces of the unit
cell (all faces are equivalent in a cubic unit cell) contain only titanium and oxygen atoms. We
want to define the length of the unit cell edge in terms of the radii of the Ti4+ and O2- ions. The
ions are shown in full size in the lower half of Figure 15a, illustrating how the close contacts
define the unit cell edge. The relationship between the unit cell edge, a, and the ionic radii of the
ions, r(Ti4+) and r(O2-), is straightforward in this case:
a = r(Ti4+) + 2r(O2-) + r(Ti4+) = 2[r(Ti4+) + r(O2-)]
a = 2[0.60 Å + 1.35 Å] = 3.90 Å
(a)
(b)
Figure 15: The ionic contacts, (a) Ti-O and (b) Sr-O, which determine the length of the unit cell edge
in SrTiO3. The upper figures show the planes of atoms with respect to the entire unit cell. The lower
figures show only these planes, where the atoms are shown with their full ionic radius. The white
arrows denote the directions in which the atoms are touching.
Figure 15b shows the ionic contacts in a parallel plane that cuts through the center of the
unit cell. This plane contains only strontium and oxygen atoms. Here the ions touch along the
face diagonal of the unit cell as illustrated in the lower half of Figure 15b. To determine the
length of the unit cell edge in this case we need to apply some simple geometry. The length of
the diagonal of a square is √2 times the length of the edge of the square. This leads to the
following equation:
√2 × a = r(O2-) + 2r(Sr2+) + r(O2-)
a = (2/√2)[r(Sr2+) + r(O2-)] = √2[r(Sr2+) + r(O2-)]
a = √2[1.44 Å + 1.36 Å] = 3.94 Å
While the two values do not agree exactly they are not far apart. If they were this structure
would not be stable. Furthermore, they are in good agreement with the observed unit cell edge
of 3.91 Å, suggesting the assumptions we made were reasonable. Remember that ionic radii
represent the average size of an ion over many different structures. Hence this analysis should
give a good estimate of the unit cell size. In general it will not correspond exactly to the size of
the unit cell.
Another example to demonstrate the relationship of the unit cell edge, a, and the ionic radii
would be cesium chloride, shown in Figure 16a. In this case the ionic contact is not along an
edge or a face, but along the body diagonal of the unit cell, illustrated in Figure 16b.
Figure 16: (a) The unit cell of CsCl, and (b) the ionic contact of Cs-Cl across the body diagonal
which determines the length of the unit cell edge in CsCl. The white arrows denote the direction in
which the atoms are touching.
Again, we will need to use simple geometry to determine the length of the unit cell edge.
The length of the body diagonal of a cube is √3 times the length of the edge of the cube. This
leads to the following equation:
√3 × a = r(Cl-) + 2r(Cs+) + r(Cl-) = 2[r(Cs+) + r(Cl-)]
a = (2/√3)[r(Cs+) + r(Cl-)]
a = (2/√3)[1.69 Å + 1.81 Å] = 4.04 Å
7. Diffraction
How do we know the arrangements and spacing of atoms in crystals? Over the last few
decades electron microscopy techniques have advanced to the point where individual atoms can
be seen on surfaces. However, this technique is not trivial to apply and even when it can be used
it only gives information about atoms at the surface. The much older and more common way of
determining crystal structures is to use a technique called X-ray diffraction. We are going to
look at this technique in more detail, but before we do that let’s review the concept of diffraction.
Destructive Interference
Constructive Interference
Figure 17: Examples of two waves adding together either destructively or constructively. This image
was taken from the following web site:
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html.
When two waves of the same wavelength and amplitude interact they can add constructively
or destructively, as shown in Figure 17, or any combination in between. The difference between
constructive and destructive interference arises because of a phase shift of the second wave. If
the two waves start from the same point the wave crests and troughs are matched to each other
and the waves add constructively to give a resultant wave with the same wavelength, but twice
the amplitude. If the second wave starts 180° out of phase the crest of wave 1 is matched to the
trough of wave 2 and the two waves completely cancel out. To explore how changes in phase,
amplitude and wavelength lead to different interference patterns try the exercises at:
http://undergrad-ed.chemistry.ohio-state.edu/xtal/x-ray.html.
The classic example of a diffraction experiment is to pass monochromatic (single
wavelength) light through a pair of slits. Once the light waves pass through the slits they spread
out. In terms of Figure 17 we can think of the slits as the starting point for each wave. When
the waves come together they can have different phases depending upon how far each wave has
traveled. If you project the waves onto a screen you will see a series of bright and dark spots.
These correspond to places where the waves are interacting constructively and destructively,
respectively. We will not go into more detail on the double-slit experiment, except to say that by
measuring the spacing between bright spots and the distance from the slits to the screen it is
possible to determine the spacing between slits. A more complete description of this experiment
can be found on the web at:
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/DoubleSlit/DoubleSlit.html.
In Part B of the laboratory exercises you are going to do an optical diffraction experiment of
your own. The source of monochromatic light is a laser. The laser will pass through a slide with
a pattern such as the one shown in Figure 18. The dark lines absorb the light, but the white lines
pass the light and act as grid of regularly spaced horizontal slits. By measuring the distance
between bright “diffraction” spots, X, where constructive interference occurs, as well as the
distance between the slide and the screen, L, you will determine the distance between the “slits”,
d, which is too small to be deduced from a visual inspection of the slide.
d
Diffraction
Spots
Diffraction
Slide
X
Laser
L
Figure 18: The optical diffraction occurring in Part B of the experiment.
Let’s look more closely at the how the interference occurs in this experiment. Figure 19
gives a close up view of the wave paths. Constructive interference will be maximized for light
waves traveling parallel to each other, so we will look at parallel waves that pass through two
different slits. The waves are in phase and parallel upon passing through the slit, but wave 1 has
to travel farther than wave 2. This will introduce a phase shift between the two waves. If the
extra distance that wave 1 has to travel is an integer multiple of the wavelength, the two waves
will still be in phase and constructive interference will result. If the extra path length is not an
integer multiple of the wavelength then some degree of destructive interference will take place.
φ
Wave 2
Wave 1
d
φ
u
Figure 19: Fraunhofer diffraction of light waves passing through parallel slits.
Using trigonometry we can easily calculate the relationship between the difference in path
length, u, the spacing of the slits, d, and the angle of the diffracted light beams, φ:
u = d sin φ
φ = tan-1 (X/L)
For constructive interference to occur the extra path length has to be an integer multiple of the
wavelength:
nλ = u
nλ = d sin φ
where n is an integer. If we take n to be 1 and use the wavelength of the light (λ = 635 nm) we
can calculate the slit spacing, d, from the values of X and L measured in Part B of the
experiment.
8. X-ray Diffraction and Bragg’s Law
Atoms scatter electromagnetic radiation and in principle the spacing between atoms can be
determined by analyzing the diffraction patterns of light scattered from crystals. However, if you
were to replace the slide in the previous experiment with a crystal you would not see a
diffraction pattern. Diffraction effects are most pronounced when the wavelength of the
radiation, λ, and the spacing between slits, d, are similar in magnitude. The optical diffraction
experiment discussed above works best when the spacing between slits is comparable to visible
light, on the order of hundreds of nanometers. The spacing between atoms in crystals is on the
order of 1-4 Å (0.1 to 0.4 nm). To create a diffraction pattern from a crystal we need light with a
much smaller wavelength. The type of electromagnetic radiation which possesses a wavelength
comparable to interatomic distances is X-ray radiation.
At the beginning of the 20th century Maxwell von Laue realized that X-rays could give a
diffraction pattern when scattered from crystals. Not long after that Lawrence Bragg worked out
way to analyze crystal structures by collecting X-ray diffraction patterns. The development of Xray diffraction allowed scientists to determine the arrangement of atoms in ionic, metallic and
molecular crystals. Much of the material that has been presented in this course (atomic and ionic
radii, molecular shape, bonding, etc.) could only be understood once X-ray diffraction techniques
were developed.
The diffraction of X-rays is very similar to the Fraunhofer diffraction illustrated in Figure
17. Rather than visible light passing through slits the X-rays are directly scattered by the atoms.
Thus the atoms act like the slits in an optical diffraction experiment. Another difference is that
we will be looking at X-rays that are reflected off parallel planes of atoms, rather than
transmitted through slits. The geometry of an X-ray diffraction experiment is illustrated in
Figure 20. The incoming beams of X-rays are in phase with each other. However, the upper
(red) beam travels a shorter distance than the lower (green) beam. This is because the lower
beam must travel the extra distances from points B to C and C to D (both equal to d sinθ where d
is the distance between planes of atoms). For the two beams of X-rays to be in phase again, the
extra distance traveled (2d sin θ) must be an integral multiple of the X-ray wavelength (λ), which
gives rise to one of the most famous equations in science: Bragg’s Law. Bragg’s Law identifies
the conditions that must be met for constructive interference to occur:
BC + CD = 2d sinθ = nλ
n = 1, 2, 3, ...
Figure 20: Bragg diffraction of X-rays reflecting from parallel planes of atoms in a crystal.
To see a 3D visualization of Bragg diffraction visit the following web site: http://undergraded.chemistry.ohio-state.edu/xtal/x-ray.html.
9. Lattice Planes and Miller Indices
In the introduction to X-ray diffraction given in the previous section the derivation of
Bragg’s Law is given for scattering of X-rays off of parallel planes of atoms. As you can
imagine it’s possible to define many different parallel planes of atoms in a crystal. In general
each set of parallel planes will produce constructive interference at the correct angle giving rise
to a bright spot in the diffraction pattern. Before going any further we need a nomenclature for
describing these planes and an equation for determining the distance between planes, the
interplanar spacing, d.
The nomenclature for labeling planes in crystals is based on where the plane intercepts the
crystal axes. For a cubic crystal system the lattice vectors are parallel to the three-dimensional
Cartesian coordinate system. The a lattice vector points in the x-direction, the b lattice vector
points in the y-direction and the c lattice vector points in the z-direction. (See the relationship
between x-y-z & a-b-c in Figures 21a and 21b.) Consider the plane shown in Figure 21. This
plane never intersects the x-axis (it is parallel to the x-axis) so we can say that its intercept with
the x-axis is ∞. It intercepts the y-axis at the unit cell boundary, so we can say that its intercept
is one lattice vector from the origin or 1 b. It intersects the z-axis at one-half lattice vector from
the origin, or ½ c. If we take the inverse of the coefficients of the intercepts (∞, 1, ½) we get
three integers, h, k and l respectively, which collectively are called the Miller Indices. In this
example: h = 1/∞ = 0, k = 1/1 and l = 1/(½) = 2.2 To denote a plane the Miller Indices are
surrounded parentheses (hkl). For example the plane in Figure 21c is the (012) plane.
z
x
c
y
(a)
a
b
(b)
(c)
Figure 21: (a) The Cartesian coordinate system, and (b) the lattice vectors that make up the crystal axes.
(c) A cubic unit cell where the lattice vectors are shown with arrows. The (012) lattice plane is shown in
black.
A few examples of parallel planes that exist in a cubic close packed metal are shown in
Figure 22. One thing that can be confusing when you first start assigning planes is to know
which of the parallel planes to use for finding the intercepts with the x, y and z-axes. The correct
plane to use is the one that immediately neighbors the plane that intersects the origin. In Figure
22 this plane is shaded in blue to distinguish it from the other parallel planes. Once you have
found this plane the next step is to determine the intercepts with the axes and from those
intercepts to calculate the Miller Indices as outlined in the previous paragraph.
22
The Miller Indices, h, k and l, are always integers. If this is not the case after dividing by the intercepts all three
are multiplied by a common factor to convert to them to integers.
(001)
(010)
(110)
(002)
(012)
(111)
Figure 22: Various parallel planes and their Miller Indices (hkl) for a cubic close packed metal. Note
the orientation of the lattice vectors above each drawing. In each case 2-4 parallel lattice planes are
shown to illustrate the spacing between planes, but the lattice shown in black is the one used to
determine the Miller Indices. The atom shaded in black represents the origin.
For the purposes of X-ray diffraction it is the spacing between planes that is important.
While in principle the spacing between planes, d, may be different for all of the examples in
Figure 22, they all depend upon a single variable, the length of the unit cell edge, a. It should be
obvious that for a cubic unit cell the spacing between (001) planes, d001, must be equal to the
length of the unit cell edge, a, as must be the spacing of the (010) planes, d010. It is also
relatively easy to see that the spacing of the (002) planes, d002, is one-half the value of d001. The
relationship between the d001 and the other examples in Figure 22, such as d012 or d111, is not so
immediately obvious. Fortunately, there is simple equation that relates the interplanar spacing,
dhkl, the Miller Indices, and the cell edge, a, for any set of lattice planes:
(1/dhkl)2 = (h2 + k2 + l2)/a2
This equation is valid for any cubic crystal. There are similar equations for the other crystal
systems (hexagonal, tetragonal, orthorhombic, etc.). Let’s do a quick example to illustrate how
this equation is used. Platinum metal adopts a cubic close packed structure with a unit cell edge
of a = 3.92 Å. If we want to know the spacing between close packed layers, which correspond to
(111) planes, then we need to find d111 interplanar spacing:
(1/d111)2 = (12 + 12 + 12)/(3.92 Å)2
d111 = 3.92/√3 = 2.26 Å
10. Single Crystal and Powder X-ray Diffraction
Finally we’re at the point where we can start to understand what an X-ray diffraction pattern
tells us about the crystal structure. Before we discuss how to interpret the data, let’s take a brief
overview of how X-ray diffraction data is collected. There are two varieties of X-ray diffraction
experiments: single crystal and powder diffraction. These are illustrated in Figure 23.
Diffracted
Beams
Diffracted
Beam
Incident
Beam
Incident
Beam
Single Crystal Diffraction
Powder Diffraction
Figure 23: The interaction of X-rays with a single crystal (shown on the left) and a polycrystalline
powder (shown on the right). The lower figures illustrate the kind of data that is collected in a single
crystal diffraction experiment (on the left) and a powder X-ray diffraction experiment (on the right).
In a single crystal diffraction experiment you have to isolate one crystal. The angle between the
X-ray beam and the crystal is then varied and diffraction patterns, like the one shown in Figure
23, are collected at various angles. This experiment is much like the optical diffraction
experiment shown in Figure 18. Typically hundreds of diffraction patterns are collected at
various angles and analyzed by a computer to determine the unit cell dimensions and the
locations of the atoms within the unit cell. In a powder diffraction experiment the powder
contains thousands of randomly oriented tiny crystals. The angle between the powder and the
incoming beam is then varied. Although most of the crystals are misaligned, the large number of
crystals means that there are always some crystals that are properly aligned to diffract the
incoming beam. Of course diffraction only occurs if Bragg’s Law is obeyed. Instead of
hundreds or thousands of diffraction patterns the result is a single diffraction pattern as shown at
the bottom right of Figure 23.
A schematic of an X-ray powder diffractometer is shown in Figure 24. It is not necessary to
describe all of the elements of the instrument here. It is sufficient to know that during the
experiment the angle of the sample, which is a flat plate, and the detector continually change so
that the sample is always at angle where it can act like a mirror refocusing the X-rays that come
from the source back onto the detector. However, because of interference effects an appreciable
amount of X-rays will only reach the detector at angles which satisfy Bragg’s Law. Unlike the
optical diffraction experiment the diffraction angle θ is calculated directly by the diffractometer.
Divergence
Slit
Horizontal
Diffraction
Circle
θ
Sample (Vertical
Flat Plate)
Divergent
X-ray
Source
2θ
Antiscatter
Slit
Receiving Slit
Horizontal
Soller Slits
Detector
Figure 24: A schematic of an X-ray powder diffractometer.
11. Interpreting X-ray Powder Diffraction Patterns
In this experiment you are going to receive the X-ray powder diffraction pattern of an
unknown cubic compound, with the 2θ values of the peaks marked. You will be asked to (a)
determine the length of the unit cell edge, a, and (b) determine the hkl value corresponding to
each peak. This process is called indexing a powder pattern. We will illustrate this process
with an example.
75.322
69.713
63.893
35.576
Intensity
43.944
Let’s take a look at the body centered cubiccompound, Ag2Se. The X-ray powder diffraction
pattern for Ag2Se is shown in Figure 25. Between 20-80° 2θ (or 2-theta) five peaks are
observed in the diffraction pattern. Each peak falls at an angle (marked on the plot) where the
spacing between planes of atoms in the crystal satisfies Bragg’s Law and constructive
interference results.
2-Theta (°)
Figure 25: The X-ray powder diffraction pattern of Ag2Se.
The first step in the indexing process is to use Bragg’s Law:
λ = 2d sin θ
to convert all of the 2θ values to values of interplanar spacing, d. The values of θ are obtained by
dividing the 2θ values by 2. This data was collected with a diffractometer equipped with a
copper X-ray source, as are most powder diffractometers. Copper sources emit X-rays with a
wavelength, λ = 1.5406 Å. This wavelength should be used in all of the calculations.
We can rearrange Bragg’s Law to complete the calculations:
d = λ / (2
θ)
( sinθ)
For the first peak this results in:
d = 1.5406 Å / (2 sin(17.788°°)) = 2.5215 Å
All of the results of this conversion are shown in Table 2:
1
2θ
θ (°°)
35.576
θ (°°)
17.788
2.5215
2
43.944
21.972
2.0588
3
63.893
31.9465
1.4558
4
69.713
34.8565
1.3478
5
75.322
37.661
1.2608
Peak #
d (Å)
Table 2: Values of interplanar spacing, d, for the peaks in the Ag2Se diffraction pattern.
Given the d-values for each peak, the next task is to find a different set of Miller Indices, hkl,
for each peak that satisfy the equation (where a is the same for all peaks in the pattern):
(1/dhkl)2 = (h2 + k2 + l2)/a2
In practice, the most common way to do this is to create a table of 1/d2 values and look for a
common factor, Z, which can be divided into each value to give an integer quotient. For some
cubic structures the first observed peak is the (100) peak. If that is the case, the common factor
is equal to the 1/d2 value of the first peak. In this case that would be Z = 0.15728. In this case,
we do NOT get even integers. So, how do we pick a Z value that will work? There are two
common methods to discover the Z value:
1. Divide the first 1/d2 value (in this case: 0.15728) by successive integers. 0.15728/2 =
0.07864, 0.15728/3 = 0.05243, etc. Test these values as Z values to see if they result in integers.
In this example, if we try 0.07864 we get even integers, so this is a good Z value.
2. Look for the smallest difference between the 1/d2 values. For example, the difference
between the last two 1/d2 values is 0.62908 - 0.55049 or 0.07859. After finding the difference
between all of the other 1/d2 values, we discover that this is the smallest difference. (This value
is very close to the Z value you get from the other method, so either can be used - the difference
would be negligible.)
Now that we know our Z value, we can divide it into our 1/d2 values. In Table 3 all of the
1/d2 values have been divided by this common factor and the validity of choosing this value for Z
is confirmed by the fact that all of the quotients in the last column are integers.
The next task is to find values of h, k and l that when squared and summed (h2 + k2 + l2) will
equal the integers in the (1/d2)/Z column of Tables 3 & 4. The h, k and l values are shown in
Table 4. Note that in a powder diffraction pattern, lattice planes that have the same interplanar
spacing, for example the (001), (010) and (100) in a cubic crystal, will all contribute to the
intensity of the same peak. In terms of indexing it doesn’t matter which one we use.
θ (°°)
17.788
d (Å)
1/d2
(1/d2)/Z
1
2θ
θ (°°)
35.576
2.5215
0.15728
2
2
43.944
21.972
2.0588
0.23592
3
3
63.893
31.9465
1.4558
0.47184
6
4
69.713
34.8565
1.3478
0.55049
7
5
75.322
37.661
1.2608
0.62908
8
Peak #
Table 3: Values of 1/d2 for Ag2Se and the integer that results when each is divided by Z = 0.07864.
θ (°°)
17.788
d (Å)
1/d2
(1/d2)/Z
h
k
l
1
2θ
θ (°°)
35.576
2.5215
0.15728
2
1
1
0
2
43.944
21.972
2.0588
0.23592
3
1
1
1
3
63.893
31.9465
1.4558
0.47184
6
2
1
1
4
69.713
34.8565
1.3478
0.55049
7
?
?
?
5
75.322
37.661
1.2608
0.62908
8
2
2
0
Peak #
Table 4: Miller Indices for each peak in the Ag2Se diffraction pattern.
Notice that we run into a problem with the next to last peak: there are no combinations of h2
+ k + l2 that will equal 7! This can sometimes be a problem. Even though we chose a correct Z
value to get even integers, our hkl values aren’t adding up. One simple solution would be to
multiply all of the (1/d2)/Z values by successive integers and test the hkl values. Let’s start with
multiplying by two (results shown in Table 5):
2
θ (°°)
17.788
d (Å)
1/d2
(1/d2)/Z
h
k
l
a
1
2θ
θ (°°)
35.576
2.5215
0.15728
4
2
0
0
5.04
2
43.944
21.972
2.0588
0.23592
6
2
1
1
5.04
3
63.893
31.9465
1.4558
0.47184
12
2
2
2
5.04
4
69.713
34.8565
1.3478
0.55049
14
3
2
1
5.04
5
75.322
37.661
1.2608
0.62908
16
4
0
0
5.04
Peak #
Table 5: Miller Indices for each peak in the Ag2Se pattern after multiplying (1/d2)/Z by two.
We can see that multiplying by two was the right choice. All of our final Miller Indices are
correct.
The last step is to calculate the length of the cell edge, a. Given the value of d and the Miller
Indices we could calculate a from any of the peaks in the pattern. For a very good pattern the
value will be the same no matter what peak we choose. For Ag2Se the math is easiest for the
(110) peak, where we can calculate that a = 5.04 Å. By rearranging our above equation, we get:
a2 = (h2 + k2 + l2) / (1/d2)
For the first peak: a2 = (2) / (0.07864) = 25.432
a = √(25.432) = 5.04 Å
12. Systematic Absences (Missing Peaks)
The example used in the previous section to illustrate how powder diffraction patterns are
indexed is very straightforward. In practice there are a few things that can complicate the
process. The mathematics behind the process become more difficult as the symmetry is lowered
and more variables are needed to define the unit cell dimensions. For example, two independent
variables are needed to define the dimensions of a hexagonal unit cell, three are needed for an
orthorhombic unit cell, and six are needed for a triclinic unit cell. For low symmetry structures
computers are typically used to index the diffraction pattern. That process will not be described
here. Another complicating factor arises when some of the diffraction peaks are missing. In that
case our assumption that the first peak corresponds to diffraction from (100) planes is not
necessarily valid.
The intensity of each diffraction peak depends upon the arrangement of atoms within the unit
cell. Sometimes the intensity of a peak is so weak that the sensitivity of the instrument is not
sufficient to observe the peak. This was the case with the 110, 220 and 310 peaks in the Ag2Se
example. Other times a peak is truly absent regardless of the sensitivity of the instrument or how
hard we look. That is to say the intensity of such a peak is zero by definition. Diffraction peaks
that fall into this latter category are called systematic absences. This was the case with the 100,
111, 210, 300, etc. peaks in the Ag2Se example. Systematic absences arise because there is some
element of translational symmetry present in the structure, smaller than the lattice vectors (a, b
and c). For example, centered lattices have vectors that relate the lattice points on the corners of
the unit cell with other lattice points located on either the faces or the body center. Whenever the
Bravais lattice is centered rather than primitive there will be a subset of peaks that are
systematically absent. The rules for systematic absences in face centered and body centered
Bravais lattices are given in Table 6. Base centering will also give rise to absences as does the
presence of certain symmetry elements, such as screw axes and glide planes. However, we will
not encounter these in the laboratory exercise. So they will not be discussed further here.
Bravais
Lattice
Allowed Peaks
Body
Centered
All peaks where h+k+l
is an even number
Face
Centered
All peaks where h,k & l
are either all even or all
odd numbers
Examples
of Allowed
Peaks
(110), (200)
(211), (220)
(200), (111)
(220), (311)
Systematically Absent
Peaks
All peaks where h+k+l
is an odd number
All peaks where h,k & l
are a mixture of even
and odd numbers
Examples
of Absent
Peaks
(100), (111)
(210), (221)
(100), (110)
(210), (211)
Table 6: Systematic absences for body centered and face centered lattices.
To understand the reason why systematic absences exist consider the lattice planes shown in
Figure 22 for a face centered cubic lattice. If we apply the systematic absence rules given in
Table 6 to these six set of planes the (001), (010), (110) and (012) peaks would be systematically
absent. Notice that for these planes there is a plane of atoms located exactly halfway in between
the lattice planes, whereas for the (002) and (111) planes that is not the case. X-rays will also be
scattered from the atoms located between the planes. It so happens that for a face centered
Bravais lattice the X-rays scattered off of the atoms contained in the planes and the atoms located
halfway between the planes will be 180° out of phase. This will lead to perfect destructive
interference.
How do absences, both systematic and accidental, impact the process of indexing a powder
pattern? It simply means that you cannot assume the first peak is the (100) peak. For a body
centered cubic structure it would typically be the (110) peak, whereas for a face centered cubic
structure it would typically be the (111) peak. In some cases even these peaks might be
accidentally missing. The reality of missing peaks simply means that more trial and error may
be needed to find the common factor, Z.