Crystal Models
The Solid-State Structure of Metals and Ionic Compounds
Objectives
Understand the concept of the unit cell in crystalline solids.
Construct models of unit cells for several metallic and ionic substances.
Calculate various quantities such as atomic radii and theoretical density from unit cell data.
Introduction
In a piece of crystalline material, either metallic or ionic, the particles show an endless repetition of an orderly
arrangement to generate a three-dimensional structure known as a crystal lattice. Part of a simple lattice is shown in
Figure 1.1. The unit cell is the most convenient small part of the lattice that, if repeated in three dimensions, generates the
entire lattice. An acceptable unit cell for the lattice shown in Figure 0.3(a).
Depending on the relative lengths of the sides of the unit cell and the angles that the sides make with each other, a unit cell
will belong to one of several crystal systems. In this study we consider unit cells in only the cubic system (where all three
sides are equal, designated by the symbol a, and all angles are 90°) and the hexagonal system (where two sides are equal,
designated by a, and one side is longer, designated by c, and the angles are 90°, 90°, and 120°); see Figure 0.4b.
A primitive unit cell is a unit cell in which only the corners are occupied. In some cases unit cells contain other lattice
points in addition to those at the corners; these are called multiple unit cells. A primitive cubic and two multiple cubic
unit cells are shown in Experiment Figure 3, along with a multiple hexagonal unit cell.
The “exploded” unit cell diagrams shown in Figure 3 are somewhat misleading because they show that each cell contains
mostly empty space. The packing of atoms in a crystal is more realistically shown by the sketches given in Figure 0.2
Figure 0.5for these same unit cells.
Figure 1.1 Section of a three-dimensional lattice.
a
b
Figure 0.2 Two common crystal systems, (a) is called primitive cubic, and (b) is called hexagonal.
Although the spheres shown in Figure 0.3 and Figure 0.4 touch each other, there is considerable empty space in each unit
cell. There are three types of open spaces (or “holes”) commonly found in unit cells: tetrahedral holes, octahedral holes,
and cubic holes (see Figure 1.7), based on the geometric figure describing the atoms around the hole.
Part I: Metals
The particles making up the crystal structure of a pure metal consist of identical atoms. Most metals crystallize in the unit
cells shown in Figure 0.3 and Figure 0.4. From a careful analysis of these sketches, several properties of the metals can be
determined.
a
b
Figure 0.3 Two unit cells, (a) the primitive cubic, and (b) the body-centered cubic.
a
b
c
Figure 0.4 Two more unit cells, (a) is the face-centered cubic, also called the cubic close-packed, and (b) the hexagonal closepacked. The hexagonal close-packed unit cell is shown extended in (c) to emphasize the hexagonal shape this lattice has.
(a) Primitive Cubic
(c) Body-Centered Cubic
(b) Hexagonal Close-Packed
(d) Face-Centered Cubic or Cubic Close-Packed
Figure 0.5 Packing of atoms or anions in a crystal. Each unit cell shows two views of the packing of spheres. The left view shows
all of the atoms that sit on the outline of the unit cell. The right view shows the same unit cell with all of the atom parts outside
the unit cell taken away. What remains is the actual content of the unit cell.
Consider the element polonium, which will crystallize in the primitive cubic unit cell, shown in Figure 0.3a, with a =
0.336 nm. The length of a side of the unit cell equals twice the radius of an atom (a = 2R); so we can calculate the radius
of a Po atom as
R
a 0.336 nm

 0.168 nm
2
2
0.1
Similar relationships between the unit cell length and the radius of the atom are summarized in Table 0.1 for the other unit
cells.

The total number of atoms contained within the unit cell is known as the unit cell contents (Z). Again, consider the
primitive cubic unit cell for polonium as shown in Figure 0.5a. At first glance you might think there are eight atoms
occupying this unit cell, but this is not the case. Although eight atoms are used to outline the unit cell, the actual unit cell is
defined in terms of the centers of these atoms; consequently, a given corner atom is really part of eight adjacent unit cells
(see Figure 0.5a). Thus only l/8 of a given atom “belongs” to the unit cell under consideration. Because there are eight
such atoms, for the primitive cubic unit cell Z is given as follows:
 1/ 8corner 
Z  8corners  
  1 atom
 atom 
0.2
All body-centered atoms are completely within the unit cell; any face-centered atom is shared by two unit cells; any edgecentered atom is shared by the number of unit cells around that edge (often 3, 4, or 6), the unit cell content for the various
unit cells that we are considering is summarized in Table 1 and Figure 7.
a
b
c
Figure 0.6 Diagrams of holes between anions in a crystal. Three kinds of holes are shown in two views. (a) shows a tetrahedral
hole, (b) showns an octahedral hole, and (c) shows a cubic hole. The top view shows the anions separated along an axis marked
by the dashed line. the lower figure shows the top view of the cluster of anions that surround the hole, looking down that axis.
a
b
c
Figure 1.7 The positions of holes in a unit cell. (a)Shows the tetrahedral hole, the topmost view is exploded on one axis. You can
see the shape of a tetrahedron, with its triangular base and peak. the second picture in the column shows the tetrahedral hole all
together. The third picture shows the hole exploded on all axes, and the fourth picture shows the hole re-assembled in the same
orientation as in picture three. (b) Shows the same set of views of an octahedral hole and (c) shows the same set of views of a
cubic hole.
The crystal coordination number (CN) of an atom is the number of nearest neighbor atoms in the crystal structure,
Figure 8 shows that CN = 6 for a given corner atom in the primitive cubic unit cell. (This is the value for any atom in the
unit cell.) Table 0.1 also summarizes the values of CN for the cubic unit cells to be studied in this experiment.
Length–Radius
Relationship
Unit cell
Primitive cubic
Body-centered cubic
Face-centered cubic
Hexagonal closest packed
Diamond
Z=
V=
CN =
a=
R=
?=
a = 2R


a 3 = 4R
a 2 = 4R
a = 2R
a 3 = 8R
Z
V
CN
1
2
a3
a3
6
8
4
2
a3
a2c sin 60o
12
12
?
?
?
number of atoms per unit cell

unit cell volume
coordination number
unit cell length
radius of atom or anion
Figure it out yourself!
Table 0.1 Summary of properties of various unit cells.
If the unit cell dimensions and cell content are known for a substance, the volume of the unit cell and the mass of the
atoms in the unit cell can be calculated. For the cell of polonium, the volume is:
V  a3  (0.33610-7 cm) 3  3.79  10-23 cm 3
0.3
and the mass is:

m
Z  M 1 atom 209 g/mol 

 3.47 1022 g
23
NA
6.022 10 atoms/mole
0.4
Knowing the mass and volume of the unit cell allows the calculation of the theoretical (or crystallographic) density (d) of
the substance
m
3.47 1022 g
d 
 9.16 g/cm 3
23
3
v 3.79 10 cm

0.5
Part II: Ionic Crystals
The crystal structures of ionic compounds may be pictured as arrays of closely stacked larger ions, creating holes into
which the smaller counter-ions can fit. Because anions are usually larger than cations, we consider the anions to define the
 the cations to occupy the holes. For example, in the sodium chloride unit cell shown in Figure 0.8,
original unit cell and
the Na+ cations can be considered to occupy the octahedral holes in a slightly expanded Cl –-anion face-centered cubic unit
cell.
The unit cells of ionic compounds can be analyzed to obtain various properties just as were the unit cells of metals. For
example, the cell content for each type of ion in the sodium chloride unit cell is
1/4 cation 
Z   12 edges  
 1 cation  4 cations
 edge 
1/8 anion 
1/2 anions 
Z  8 corners  
 6 sides  
 4 anions
 corner 
 side 

0.6
The ratio of Z+/Z– is the same as the ratio of ions in the empirical formula. Each ion has a crystal coordination number, but
this now refers to the number of oppositely charged ions that are nearest neighbors. For example, in NaCl, CN+ = 6 and
CN– = 6. The ratio of CN+/CN- is the inverse of the ratio of Z+/Z–. When calculating the theoretical density, it is helpful to
begin by finding the separate masses of the cations and anions in the unit cell. For example, for NaCl having a unit cell
length of a = 0.56402 nm, the masses of the cations and anions in the cell as well as the cell volume are calculated
according to
m 
m 

Z Atomic Mass of cation 4 cations22.99 g/mol 

6.022 10


cations/mol
6.022 10
23
cations/mol
 1.527 10
-22
Z Atomic mass of anion   4 anions 35.45 g/mol   2.355 10

6.022 10 anions/mol 
m  m  2.35510
V  0.56402 nm10
6.022 10 anions/mol
22

7

d
g
22
23
23


23
0.7
g
0.8
g 1.527 1022 g  3.882 1022 g
cm/1 nm

3
1.9
 1.7943 1022 cm3
0.10
m
3.882 1022 g

 2.164 g/cm 3
22
3
V 1.7943 10 cm
0.11

Crystal structures
of ionic substances are influenced both by the relative sizes of the anions and cations, and by the ionic
charges. Depending on the relative sizes of the ions, a cation may fit into a tetrahedral hole (the smallest) or might fit into
a cubic hole (the largest).

a
b
Figure 0.8 The Sodium chloride unit cell
Radius ratio (r+/r- )
Coordination number (CN+)
Hole geometry
Holes per anion
0.225 – 0.414
0.414 – 0.732
> 0.732
4
6
8
tetrahedral
octahedral
cubic
2
1
1
Table 0.2 The radius ratio rule
Table 0.2 indicates the values of the radius ratio (r+/r-) needed by a cation so that it will fit into a certain type of hole.
Using r+ = 0.097 nm for the Na+ cation and 0.181 nm for the Cl– anion in NaCl gives
r+/r- = 0.097 nm/0.181 nm = 0.54
0.12
This radius ratio value corresponds to the “octahedral hole geometry” entry in the table. Because there is one octahedral
hole per Cl– anion, all these holes must be occupied by Na+ cations in order to achieve the one-to-one ratio of cation to
anion required by the formula NaCl. The empirical relationships expressed in Table 0.2 are actually violated by many
crystalline species and should be used only as general guidelines.
Experimental Procedure
Your instructor will familiarize you with the kit of spheres and connectors you will be using to build the models of the unit
cells. The kit will contain four different sizes of spheres. For each structure, construct the model and analyze it to
determine the unit cell parameters requested on the table at the end of this experiment.
Caution! Please stuff wet paper towels into the drains of the sinks at your lab station. Otherwise, you may lose a few
beads into the drains.
Part I: Metals
(Use spheres of uniform size for each structure in this part.)
A. -Ni Structure
C. -Fe Structure
Prepare the three layers shown in the Figure. Place layer
2 on layer 1; then place layer 3 on layer 2. Keep the
relative orientations the same as given in the figure and
a cubic structure will result.
Prepare the three layers shown. Place layer 2 on layer 1
so that this single sphere nestles down in the hole
between the four atoms in layer 1. Place layer 3 on layer
2. Keep the relative orientations of layer 1 and 3 the
same as given in the figure to generate a cubic structure.
D. Silicon Structure
B. -Ni Structure
Prepare the three layers shown. Place layer 2 on layer 1
so that this single sphere nestles down in the hole
between the three left-most atoms of layer 1. Place layer
3 on layer 2. Keep the relative orientations of layers 1
and 3 the same as given in the figure, and a hexagonal
closest-packed structure will result. It may be necessary
to support the right-hand portion of the top layer to keep
it from falling.
Prepare the five layers shown. Assemble the layers,
keeping the relative orientations of the layers the same
as given. It may be necessary to support two of the
corners of the model. This structure is called a diamondlike structure.
Part II: Ionic Crystals
A. CsCl Structure
B. AgCl Structure
Prepare the three layers shown above, using the large
and medium spheres. The large spheres in the first and
third layers should not quite touch each other in order to
make room for the medium sphere of the second layer.
Assemble the layers, keeping the relative orientations of
the layers the same as given in the figure. A cubic
structure will result.
Prepare the three layers shown, using the large and
small spheres, Assemble the layers, keeping the relative
orientations of the layers the same as given in the figure.
A cubic structure will result.
C. CuI Structure
D. Li2S Structure
Prepare the five layers shown, using the large and small
spheres. Assemble the layers, keeping the relative
orientations of the layers the same as given in the figure
and, again, slightly expand the structures in layers 1, 3,
and 5. The small spheres should nestle down in the
holes created by the large spheres. A cubic structure will
result.
Prepare the five layers shown, using the large and small
spheres. Assemble the layers, keeping the relative
orientations of the layers the same as given in the figure.
A cubic structure will result.
Enrichment links: http://www.umanitoba.ca/chemistry/courses/chem130/molecules/crystals/130_crystal_models.html
Calculations
Part I: Metals
In each case, use your model to identify the type of unit cell.
A.
-Ni Structure
Using a = 0.35238 nm, calculate the atomic radius from the appropriate unit cell length-radius relation given in Table
1. Using the unit cell content (Z), calculate the mass of the atoms in the unit cell. Use the value of a to calculate the
volume of the unit cell. Now find the theoretical density of metallic nickel in this crystal form.
B.
-Ni Structure
Identify the unit cell (a = 0.266 nm, c = 0.429 nm), and calculate the atomic radius and the theoretical density of
metallic nickel in this second crystal form.
C.
-Fe Structure
Identify the unit cell (a = 0.28664 nm), and calculate the atomic radius and the theoretical density of metallic iron in
this crystal form.
Silicon Structure
Identify the unit cell (a = 0.54305 nm), and calculate the atomic radius and theoretical density of silicon.
D.
Part II: Ionic Compounds
A.
CsCl Structure
Using r+ = 0.167 nm and r– = 0.18 nm, calculate r+/r-. From Table 2, ascertain the hole geometry and the cation
coordination number (CN+). Determine if all these holes are occupied by the cations in order to achieve the cation-toanion ratio expressed by the empirical formula of the compound.
Identify each ion in your model as being corner, body-centered, face-centered, or edge-centered. Calculate Z+ and Z–,
using the fractional contribution. Use Z+ and Z– to calculate the masses of all ions (m+, m–, m) in the unit cell as
described previously. Find the volume of the cubic unit cell given that a = 0.4110 nm. Calculate the theoretical
density of CsCl. Finally, describe the type of cubic unit cell defined by the Cl anions, and use Table 0.1 to determine
the coordination number (CN–) of these anions.
Note: The entry for Z in Table 0.1 should be the same as your
calculated value for Z.
B. AgCl Structure
Using r+ = 0.126 nm, r– = 0.181 nm, and a = 0.55491 nm, repeat the calculations for AgCl.
C. CuI Structure
Using r+ = 0.096 nm, r– = 0.220 nm, and a = 0.6047 nm, repeat the calculations for CuI.
D. Li2S Structure
Using r+ = 0.068 nm, r– = 0.220 nm, and a = 0.570 nm, repeat the calculations for Li2S.
Part I: Metals
Structure
Arrangement of
Radius (cm)
Atoms
CN
Z
Mass (g)
Volume (cm) Density (g/cm3)
Ni
-Ni
-Fe
Si
Part II: Ionic Crystals
Structure
Arrangement of
Fraction of
Hole Geometry
Anions
Holes Filled
CN+
CN-
Z+
Z-
CsCl
AgCl
CuI
Li2S
Structure
CsCl
AgCl
CuI
Li2S
m+(g)
m-(g)
Mass (g)
Volume (cm3)
Density (g/cm3)