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Page I-45 / The Crystal Structures of Solids Lab
The Crystal Structures of Solids Crystals of pure substances can
be analyzed using X-ray diffraction methods to provide valuable
information. The type and strength of intramolecular forces,
density, molar mass, and more can be extracted from a thorough
crystal lattice investigation. No formal lab write-up is required
for this lab; simply complete the handout found at the end of this
lab and turn it in (with all relevant work displayed on adjacent
pages) to your instructor.
Observing the crystals of an ordinary substance (such as table
salt) using a magnifying glass, one sees many planes at right
angles within the solid. This occurs in many common solids, and the
regularity we see implies a deeper regularity in the arrangement of
atoms or ions in the solid. Indeed, when we study crystals by x-ray
diffraction we find that the atomic nuclei are present in
remarkably symmetrical arrays that continue for thousands or
millions of units in three dimensions. Substances having a regular
arrangement of atom-size particles in the solid are called
crystalline, and the solid material consists of crystals. This lab
deals with some of the simpler arrays in which atoms or ions occur
in crystals and what these arrays can tell us about such properties
as atomic sizes, densities of solids, and the efficiency of packing
of particles. Many crystals are unbelievably complex, and we will
limit ourselves to the simplest crystals that have cubic
structures. Cubic structures imply 90° angles and sides of equal
length (hence, a cube.) We will also limit ourselves to the study
of only one or two kinds of systems (namely elements or binary
compounds), yet they will exhibit many of the interesting
properties of more complicated structures. The Simple Cubic (SC)
Crystal The simple cubic unit cell is a cube with an edge length,
d0, equal to the distance from the center of one atom to the center
of the next (see Figure One). The volume of the cube is equal to
(d0)3, expressed as
V = (d0)3 and is very small since d0 is on the order of 0.5 nm.
Using x-ray diffraction we can measure the value of d0 easily to
four significant figures. The number of atoms in a simple cubic
unit cell is equal to one, for only 1/8 of each corner atom is
actually inside the cell. Each atom in the simple cubic unit cell
is actually connected to six other atoms in the cubic lattice;
hence, we say that the coordination number of the atoms in this
structure is equal to six. Many diagrams displaying the simple
cubic unit cell show a gap between adjoining atoms. In an actual
crystal, we consider that the atoms that are closest are touching.
It is on this assumption that we determine atomic radii, r. In the
SC crystal, if we know d0, we can find the radius r of the atoms,
since one side contains 2 atomic radii, or
d0 = 2r
Figure One: The Simple Cubic Crystal
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for simple cubic crystals. Knowing the radius, we can calculate
d0, and then we can calculate the volume of the unit cell. Knowing
that one atom occupies the simple cubic cell, we can calculate the
mass of the unit cell (using the molar mass and Avogadro’s number),
and from this we can determine the density using the volume of the
cell. Essentially no elements crystallize in the simple cubic
structure, however, due to the inefficiency of the packing. The
atoms in the simple cubic crystal are farther apart then they need
to be, and inspection of the SC lattice will reveal a large hole in
the center of the unit cell. Only about 52% of the cell volume is
occupied by atoms, and more “empty space” means less stabilization
for the crystal structure. The Body Centered Cubic (BCC) Crystal In
a body centered cubic crystal, the unit cell still contains the
corner atoms present in the SC structure, but the center of the
cell now contains an additional atom. This means that every BCC
crystal structure holds two net atoms (eight atoms are 1/8 within
the cell, and one whole atom within the center of the cell for two
net atoms). The edge length, d0, can be determined using simple
geometry from the cube diagonal (see Figure Two). The cube diagonal
reaches across the cube, from an atom in the lower left front to an
atom in the upper right back, or from any other appropriate
combination. Geometry dictates the following relationship between
the cube body diagonal and the edge length, d0:
cube diagonal = d0
The cube diagonal encompasses 4 radii lengths, and d0 can be
expressed in terms of the radius of the atom: d0 =
The quantity d0 can be used to find the volume of the cube;
hence, this relationship is important for BCC cubic systems. In a
BCC lattice, each atom touches eight other atoms, and the
coordination number is eight. The BCC lattice is much more stable
than the SC structure, in part due to the higher coordination
number. Many metals at room temperature display the BCC lattice,
including sodium, chromium, tungsten and iron. Note that there are
two atoms per unit cell in the BCC crystal. BCC crystals are more
efficient than SC crystals, occupying approximately 68% of the
total available volume.
3
4r3
Figure Two: Body Centered Cubic Crystal
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Close Packed Structures Although many elements prefer the BCC
crystal arrangement, still more prefer structures in which the
atoms are close packed. In close packed structures there are layers
of atoms in which each atom is in contact with six others, as in
the sketch below:
This is the way in which billiard balls lie in a rack or the
honeycomb cells are arranged in a bees' nest. It is the most
efficient way one can pack spheres, with about 74% of the volume in
a close packed structure filled with atoms. There is more than one
way whereby close packed crystal structures can be stacked. One of
the stacking methods is cubic and is called the Face Centered Cubic
(FCC). The other is called hexagonal packing. We shall look at both
close packed structures. The Face Centered Cubic (FCC) Crystal In
the face centered cubic crystal unit cell there are atoms in each
corner of the cell (as in the SC cell discussed earlier) and there
is another atom at the center of each of the six faces. This means
that FCC cubic systems consist of four net atoms per unit cell
(eight atoms are 1/8 within the cell, and six faces hold an atom
which is 1/2 within the cell for four net atoms). See Figure Three.
The edge length d0 can be determined in an FCC crystal from the
face diagonal which is defined as the distance across one face of
the cube. Using geometry, we can find the edge length from the face
diagonal using the following equation:
face diagonal = d0
The face diagonal encompasses 4 radii lengths, and d0 can be
expressed in terms of the radius r:
d0 =
This expression can be used to find the volume of the cube;
hence, this relationship is important for FCC cubic systems. The
coordination number in an FCC lattice is 12, implying that FCC
lattices are quite stable.
2
4r2
Figure Three: Face Centered Cubic Crystal
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It is worthwhile to look at the types of holes within the FCC
lattice for they will be helpful when discussing ionic solids
later. One type of hole is in the center of the FCC lattice; this
type of hole is called an octahedral hole (Figure Four) because the
six atoms around it define an octahedron. There are twelve
octahedral holes in an FCC lattice. In an ionic crystal, the anions
often occupy the sites of the atoms in your cell, and there is a
cation in the center of the octahedral hole. The ratio of the
maximum radius of the cation, r+, compared with the maximum radius
of the anion, r-, is often calculated to determine what type of
hole to use for the cations. If the ratio r+/r- is greater than
0.414 but less than 0.732, the cations will be placed in octahedral
holes. There is another type of hole in the FCC lattice called
tetrahedral holes. There are eight tetrahedral holes in the FCC
lattice, and each atom in a tetrahedral hole is surrounded by four
complimentary atoms (hence the tetrahedron reference). If the ratio
r+/r- is between 0.225 and 0.414, the cations will go into
tetrahedral holes. The close-packed layers of atoms in the FCC
lattice are not parallel to the unit cell faces, but rather are
perpendicular to the cell diagonal. If you look down the cell
diagonal, you see six atoms in a close-packed triangle in the layer
immediately behind the corner atom, and another layer of
close-packed atoms below that, followed by another corner atom. The
layers are indeed closely packed, and as one goes down the diagonal
of this and succeeding cells, the layers repeat their positions in
the order ABCABC…. This implies that atoms in every fourth layer
lie below one another (see Figure Five (b)). Hexagonal Packing
There is another way to stack the layers as in the FCC lattice,
above. The first and second layers will always be in the same
relative positions, but the third layer could be below the first
one if it were shifted properly. This results in a close-packed
structure in which the order of the layers is ABABAB… (see Figure
Five (a)) The crystal obtained from this arrangement of layers is
not cubic but hexagonal. It is another common structure for metals.
Cadmium, zinc and manganese have this structure. As you might
expect, the stability of this structure is very
Figure Four: Octahedral Holes
Figure Five: Hexagonal Close Packing (left) and Cubic Close
Packing (right)
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similar to that of FCC crystals. We find that simply changing
the temperature often converts a metal from one form to another.
Calcium, for example, is FCC at room temperature, but if heated to
450 °C it converts to close-packed hexagonal. Crystal Structures of
Some Common Binary Compounds We have now dealt with all of the
possible cubic crystal structures for metals. It turns out that the
structures of binary ionic compounds (i.e. MX where M is the cation
and X is the anion in 1:1 stoichiometry) are often related to these
metal structures in a relatively simple way. In many ionic
crystals, the anions (which are large compared to cations – recall
the periodic properties of elements and ions) are essentially in
contact with each other in either an SC or FCC structure. The
cations go into the cubic, octahedral or tetrahedral holes
depending on the cation-anion radius ratios r+/r-. The idea to
remember is that cations tend to go in holes in which it will not
quite fit. This increases the unit cell size from the value it
would have if the anions were touching, which reduces the Coulombic
repulsion energy. According to the radius-ratio rule, large cations
go into cubic holes, smaller cations go into octahedral holes, and
the smallest cations go into tetrahedral holes. One can calculate
the radius ratio and then determine the location of the
cations:
If: r+/r- > 0.732 cations go into cubic holes 0.732 >
r+/r- > 0.414 cations go into octahedral holes 0.414 > r+/r-
> 0.225 cations go into tetrahedral holes
These are the three common cubic structures of binary compounds.
The radius-ratio rule allows us to predict the structure a given
compound will have. It does not always work, but it is correct more
often than not. Examples follow for each of the common binary
compounds. The NaCl Crystal (Octahedral Cation Holes) To apply the
radius rule to NaCl, we need to calculate the r+/r- ratio. From
tables we find r+ for Na+ equals 0.0950 nm and r- for Cl- equals
0.181 nm. The radius ratio is 0.095/0.181 = 0.525, and the ratio
implies that the Na+ ions will be placed in octahedral holes since
this value is less than 0.732 and greater than 0.414 (See Figure
Six.) To build a NaCl lattice, we place a Cl- ion in each FCC
position. The Na+ ions go into each of the octahedral holes in the
FCC lattice. There are twelve octahedral holes in the FCC lattice
(one on each edge), and each Na+ ion will be 1/4 in the unit cell.
In addition, one sodium atom fits in the center of the FCC lattice
of Cl- atoms. Therefore, the number of Na+ ions per NaCl unit cell
will be (12 octahedral atoms * 1/4) + (1 Na+ in center) = 4 net Na+
atoms per unit cell. We have already seen that a FCC lattice holds
four net atoms, which means each NaCl unit cell contains four net
Cl- ions. Since there are four net Na+ cations and four net Cl-
anions, the resulting structure is Na4Cl4 or, as is more commonly
expressed in empirical form, NaCl. The edge of an octahedral ionic
solid can be related to the cation and anion radii using
d0 = 2r+ + 2r-
Figure Six: The NaCl crystal
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The CsCl Crystal (Cubic Cation Holes) Cesium chloride is a
binary ionic compound just like NaCl, but due to the larger r+
value of the Cs+ ion (0.169 nm), the r+/r- radius ratio is 0.933.
Using the radius-ratio rule, we find that the Cs+ cations should be
placed in cubic holes (see Figure Seven). The structure of CsCl
will look like the BCC lattice except that the atom in the center
will be Cs+ while the corner atoms will be Cl- ions. Note that this
structure is not BCC because the corner and center atoms have
different identities. The resulting empirical formula for cesium
chloride is (1 net Cs+ atom in center) + (8 Cl- atoms * 1/8 each
atom in unit cell) = CsCl. The edge of a cubic ionic solid can be
related to the cation and anion radii using
The ZnS Crystal (Tetrahedral Cation Holes) To determine the
structure of ZnS (another binary compound), we first determine the
r+/r- radius ratio. From tables we find r+ for Zn2+ is 0.0740 nm
while r- for S2- is 0.184 nm. The r+/r- radius ratio is, therefore,
0.402, and this implies that the Zn2+ cations will be placed in
tetrahedral holes. (See Figure Eight) To build the ZnS crystal,
first construct a FCC crystal lattice using S2- ions. The Zn2+
cations occupy four of a possible eight tetrahedral holes. The
resulting formula for ZnS will consist of four Zn atoms (each
tetrahedral cation is completely within the unit cell) and four S
atoms (a FCC lattice has four net atoms), and the resulting
empirical formula is ZnS (zinc(II) sulfide).
€
d0 = 2r+ + 2r-
3
Figure Seven: The CsCl Crystal
Figure Eight: The ZnS Crystal
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Summary of Crystal Lattice Types Figure Nine shows the three
main cubic unit crystal types that we have explored. Recall that
binary ionic compounds utilize these shapes – in their own way – by
placing the cations in the octahedral, tetrahedral, or cubic
"holes" within the lattice; the anions occupy the "normal" cubic
positions. The r+/r- radius ratio determines which of the three
possible holes in which the cations are placed. Relevant
relationships between the edge length, d0, and the cation and anion
radii are given in Figure Ten.
Lattice Type Simple Cubic Body Centered Cubic Face Centered
Cubic # net atoms per cell 1 2 4 d0 (edge) in relation
to r d0 = 2r d0 = d0 =
Figure Nine: Summary of the Three Cubic Unit Cell Types
Binary Cell Type Cubic Octahedral Tetrahedral # net molecules
per cell 1 4 4 radius ratio qualifying
values r+/r- > 0.732 0.732 > r+/r- > 0.414 0.414 >
r+/r- > 0.225
d0 (edge) in relation to r
d0 = 2r+ + 2r-
Figure Ten: Summary of the Three Common Binary Cell Types
4r3
4r2
€
d0 = 2r+ + 2r-
3
€
d0 = 2r+ + 2r-
2
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Page I-52 / The Crystal Structures of Solids Lab
The Crystal Structures of Solids Complete the following problems
using the table of radii at the bottom of this page. All answers
must be provided on the worksheet, and relevant work must be
stapled to the back of the worksheet.
1. Experimentally determine the density of an unknown metal
solid to at least three significant figures using any
equipment found in your lab drawer. Explain the process (and
show calculations) used to determine the density in three sentences
or less on both this sheet and in your lab notebook.
2. What element forms a face centered cubic cell, has a density
of 8.92 g/cm3, and a radius of 128 pm? 3. Chromium forms a body
centered cubic crystal. If the length of an edge is 2.884
angstroms, calculate the
density (g/cm3) and the radius of a chromium atom in angstroms.
4. Sodium forms a body centered cubic crystal. Calculate the
density of sodium metal. Propose a simple
experiment to confirm your calculated density of sodium in the
lab. (Note: use the table of radii below!) 5. Calculate the radius
ratio for TlBr. What kind of cation holes will be filled in a
crystal of TlBr? 6. Calculate the radius ratio for KI. What kind of
cation holes will be filled in a crystal of KI? 7. Consider a cubic
array of bromide ions in which the anions are touching along the
face of the cube. What is
the maximum radius (in angstroms) that a cation could have and
still fit in the octahedral holes on the edges of the FCC unit
cell?
8. Clausthalite is a mineral composed of lead(II) selenide. The
mineral adopts a NaCl octahedral-type structure.
If the density of PbSe is 8.27 g/cm3, calculate the radius of
the lead(II) ion. (The radius of selenide ion is given below.)
9. NaBr forms a crystal lattice similar to octahedral NaCl.
Calculate the density of NaBr. Propose a simple
experiment to confirm your calculated density in the lab.
Atom Atomic Radius (pm) Ionic Radius (pm)
Na 186 95.0 Se 117 191 (Se2-) Cs 265 169 Cl 99.0 181 Br 114 196
Zn 135 74.0 Tl 170. 147 (Tl+) K 227 133 S 104 184 I 133 220.
conversion V = edge3 density molar mass (g/mol) Avogadro (6.022
x 1023)
1 pm = 10-12 m / 1 Å = 10-10 m / 1 cm = 10-2 m 4 atoms = 1 fcc
cell, etc.
radius↔ edge↔ volume↔mass (g)↔moles↔ atoms / molecules
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Worksheet: The Crystal Structures of Solids Name: All final
answers to the questions (which appear on the previous page) must
be provided on this worksheet, and relevant work must be stapled to
the back on separate paper to receive full credit.
1. Density of unknown metal (g cm-3): ________ Unknown letter:
________ Brief description (with calculations) of method used to
calculate density: 2. Element: ______ 3. Density of Cr (g cm-3):
_________ Radius of Cr (angstroms): ________ 4. Density of sodium
(g cm-3) = ________ Proposed Experiment: 5. Radius ratio for TlBr =
_________ Type of cation holes = _________ 6. Radius ratio for KI =
_________ Type of cation holes = _________ 7. Maximum radius of
cation (angstroms): ________ 8. Radius of the lead(II) ion
(angstroms): ________ 9. Density of NaBr (g cm-3): ________
Proposed Experiment:
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