AP CUBIC CRYSTAL LATTICES
A. INTRODUCTION
Definitions: A CRYSTAL LATTICE is the repeating arrangement of atoms, molecules and/or ions which
occurs in a crystal.
A UNIT CELL is the smallest repeating unit that exists in a crystal.
A CUBIC CRYSTAL LATTICE is a crystal lattice in which the unit cell is a cube.
Examine the following 2–dimensional set of "lattice points" arranged in the form of a square.
Assume each lattice point is occupied by an atom, such that the nucleus of an atom is centred on the
lattice point.
Notice that the atoms in the above example just touch each other. If the atoms are smaller, they simply
move closer together until they DO touch each other, producing a smaller square.
The interatomic attractions (an atom's nucleus to its neighbour's electrons) and repulsions (the atom's
electrons to neighbouring electrons, and the atom's nucleus to neighbouring nuclei) "trade off" against
each other until the net attractive force felt by each atom just balances the net repulsive force and the
interatomic distances are optimized.
EXERCISE:
1. Look at the diagram below which represents 4 circles (A, B, C and D) forming a square unit cell
(indicated by the dashed lines.)
A
B
C
D
(a) What fraction of circle A is inside the boundaries of the unit cell?
(b) If all 4 circles are identical, what is the combined contribution of all the parts of circles lying
inside the boundaries of the unit cell?
(c) If each circle is actually a sphere, and the square is now a long square tube as shown below,
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A
B
C
D
What fraction of each sphere lies
i) inside the tube?
ii) inside the tube and ABOVE the plane defined by the points ABCD?
(d) What is the combined contribution of all the parts of spheres which lie BOTH inside the tube
and above the ABCD plane?
B. THE SIMPLE CUBIC LATTICE
The simple cubic lattice is a very rare crystal lattice: polonium (Po) is the only element known to
crystallize in this manner. Note: the atoms touch each other along each edge.
DIAGRAM A
DIAGRAM B
The repe atin g un it
A coll ection of re peating units
and
are Po atoms
EXERCISE:
2. Examine Diagram A, above. The imaginary lines outlining each cube go from the CENTRE of
one Po to the CENTRE of the next, with one Po atom at each corner of the cube. The Po atoms
actually touch each other along each edge.
(a) According to Diagram B, how many nearest neighbours does each Po have?
(b) According to Diagram A, what fraction of each Po atom actually lies INSIDE a given cube?
(c) What is the total number of Po atoms, altogether, which actually exist within each unit cell?
(d) If the length of a unit cell in a Po crystal is 3.345 x 10–8 cm, what is the atomic radius of Po?
(e) What is the distance between nearest neighbours in the Po unit cell?
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If we know the length of a cube side, and the atomic mass of the atoms involved, we can calculate the
density of a crystal made up of simple cubic unit cells.
EXAMPLE: Po–210 forms a simple cubic unit cell having a length of 3.345 x 10–8 cm. Given that
Avogadro's Number has a value of 6.0225 x 1023, calculate the density of a Po–210 crystal.
A simple cubic lattice has 8 atoms, each of which is
corners. This implies we have ONE atom per unit cell.
1
8
inside the lattice, at the 8 cube
The volume of a cubic unit cell having an edge length of 3.345 x 10–8 cm is

V = (3.345 x 10–8 cm)3 = 3.7427 x 10–23 cm3 = 3.7427 x 10–23 mL.
The mass of one atom of Po–210 is
m=
1mol
210 g
x
= 3.4869 x 10–22 g.
1mol
6.0225 x 1023
Based on one Po atom per unit cell, the density of the material in the unit cell is

3.4869 x 10 22 g
m
d =  =
= 9.317 g/mL.
V
3.7427 x 10 23 mL
Note: If there were TWO Po atoms in the unit cell the mass calculation would be adjusted in the
following manner before dividing by the unit cell volume.
 
210 g
1mol Po
2 Po atoms
m=
x
x
= 6.9738 x 10–22 g / unit cell
23
1mol Po
unit
cell
6.0225 x 10 Po atoms
EXERCISES:
3. A molecule
crystallizes
in the simple cubic system.
If the molecule’s molar mass is 174.2 g/mol



and the unit cell edge is 5.357 x 10–8 cm, what is the density of a crystal made up of such
molecules?
4. Bornite (Cu5FeS4) is a mineral which crystallizes in the simple cubic system. If bornite has a
density of 5.008 g/mL, what is the length of the unit cell edge?
C. BODY–CENTRED CUBIC (BCC)
This cubic lattice involves one atom at each corner of the cube and one atom in the centre of the cube
(hence the name "body–centred cubic"). Each of the 8 corner atoms touch the central atom BUT
adjacent corner atoms DO NOT touch each other, as shown in Diagram C, below.
DIAGRAM C
or
#1
#2
The body–centred cubic unit cell
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In Diagram D, below, notice that the black atom is a “central” atom with respect to the 8 shaded bodycentre atoms. Therefore, the role of central atoms and corner atoms is interchangeable. (The black atom
has been “blackened” simply to show its position at the centre and is not actually different from any of the
white atoms.)
DIAGRAM D
EXERCISES:
5. How many atoms altogether are there in one unit cell of the body–centred cubic lattice shown in
either of the representations in Diagram C?
6. If the central atom in Diagram C #1 is a cesium ion, Cs+, and the corner atoms are chloride ions,
Cl–, how many of each ion are present in the unit cell? Does this agree with the expected formula
for cesium chloride? Repeat your arguments for Diagram C #2.
7. Carefully examine Diagram D, above.
(a) How many nearest neighbours does each white (or black) Cl– atom have? What type of atom
is the "nearest neighbour"?
(b) How many nearest neighbours does each grey Cs+ atom have? What type of atom is the
"nearest neighbour"?
8. (Required, but nasty little beast of a problem!) Cesium chloride, CsCl, forms a body–centred
cubic unit cell. In a body–centred arrangement of spheres, the spheres touch each other
across the body diagonal, as shown below (the diagram represents a unit cell cut vertically
across a top diagonal, exposing the shaded central atom). Notice that the spheres DO NOT
touch along the unit cell edge and DO NOT touch along the diagonal at the top and bottom.
cut a long dia gona l
If the atomic radii of Cs+ and Cl– are 1.67 x 10–8 cm and 1.81 x 10–8 cm, respectively, what is the
density of a crystal of CsCl?
9. (Another nasty little brute.) Titanium metal has a body–centred cubic unit cell and a density of
4.50 g/cm3. Calculate the edge length of the unit cell and a value for the atomic radius of
titanium.
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D. FACE–CENTRED CUBIC (FCC) or CUBIC CLOSE–PACKED (CCP)
The face–centred cubic lattice has one atom at each corner of a cube and one atom at the centre of each
cube face, as shown below.
or
Note that the identification of a particular atom as being at the "corner" or the "face–centred" position is
interchangeable, as shown below.
EXERCISES:
10. How many corner atoms altogether (adding up fractional bits of atoms), are there in the face–
centred cubic unit cell? How many face–centred atoms altogether are there in the unit cell? How
many atoms altogether are there in the unit cell? How many nearest neighbours does a face–
centred atom possess? How many nearest neighbours does a corner atom possess?
11. A compound has a face–centred cubic unit cell containing ions of X at the corners and ions of Y at
the faces. Suggest an empirical formula for the compound formed by ions of X and Y.
12. The unit cell for nickel arsenide is shown below. What is the formula of this compound?
Ni
Ni
Ni
Ni
As
Ni
Ni
Ni
Ni
Ni
Ni
As
Ni
Ni
13. Perovskite is a mineral containing calcium, titanium and oxygen. If the calcium atoms occupy
corner positions of a cubic lattice, the oxygen atoms occupy face–centred positions and the
titanium occupies a body centred position, suggest the formula for perovskite.
14. The collapse of a large star results in a supernova explosion. The outward part of the explosion
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synthesizes a vast quantity of elements, in addition to a huge amount of energy and energetic
light particles. The blast also compresses the core of the star in such a way that all the neutrons,
protons and electrons are compacted into a solid mass of neutrons (a neutron star). Assume the
resulting collection of neutrons possesses a face–centred cubic lattice. If neutrons have a molar
mass of 1.0087 g and a diameter of about 1 x 10–13 cm, what is the density of a neutron star?
15. Iridium (Ir) has a face–centred cubic unit cell with an edge length of 3.833 x 10–8 cm. The density
of iridium is 22.61 g/cm3. Use these data to calculate a value for Avogadro's number.
16. A metallic solid with atoms in a face–centred cubic unit cell with an edge length of 3.92 x 10–8 cm
has a density of 21.45 g/cm3. Calculate the atomic mass and the atomic radius of the metal.
What element might this metal be?
17. Copper forms a face–centred cubic unit cell having an edge length of 3.62 x 10–8 cm.
(a) What is the atomic radius of a copper atom?
(b) Calculate the density of copper metal.
18. Silver atoms have a radius of 1.44 x 10–8 cm and crystallize in a face–centred cubic unit cell.
(a) What is the edge length of the silver unit cell?
(b) Calculate the density of metallic silver.
19. A crystal of atacamite consists of a collection of Cu(OH) 2 molecules in which one–quarter of the
hydroxide ions have been substituted by chloride ions. What is the simplest formula for
atacamite?
20. Magnetite crystals consist of Fe2+, Fe3+ and O2– ions arranged in such a way that 1 3 of the iron
atoms have a 2+ charge and 2 3 have a 3+ charge. What is the simplest formula for magnetite?
21. Dioptase is a beautiful green mineral consisting of SiO 2 groups (where the Si can be thought of
as Si4+) in which 50% of the Si4+ are replaced by Cu2+ ions. Sufficient H+ions are also present
so that the resulting 
molecule is electrically neutral. What is the simplest formula for dioptase?
E. THE SODIUM CHLORIDE STRUCTURE
The sodium chloride unit cell structure is shown below.
or
EXERCISE:
22. Examine the above unit cell for NaCl. Which of the ions are Na+: the smaller or larger ones?
Ignoring the Cl– ions, what type of cubic unit cell structure is occupied by the Na+ ions? Ignoring
the Na+ ions, what type of cubic unit cell structure is occupied by the Cl– ions?
23. Describe the arrangement of the Na+ ions around each Cl– ion. Describe the arrangement of the
Cl– ions around each Na+ ion.
24. How many Na+ ions are there altogether in the NaCl unit cell? How many Cl– ions? What is the
ratio of Na+ : Cl– ions?
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E. HEXAGONAL CLOSE–PACKED (HCP)
Although this is not a "cubic" structure, it is very important and is normally covered when discussing cubic
structures.
One layer of the hexagonal structure is packed together as follows (only part of a layer is shown).
Placing three atoms of another layer on top of the first produces the following.
Note that each atom in the second layer sits on a triangular "hole" in the layer below. By adding a third
layer such that each atom in the top layer lies directly over an atom in the first (lowermost) layer, the
resulting structure is called HEXAGONAL CLOSE–PACKED.
Interestingly, if we leave the lower two layers undisturbed and rotate the top layer by 60 o, the result is a
face–centred cubic structure! (Build a model to prove it to yourself.)
EXERCISE:
25. How many nearest neighbours does a given atom have in a hexagonal close–packed structure?
Compare this to the number of neighbours in face–centred cubic.
26. The radioactive element americium, Am, undergoes a reversible phase change at 1074oC from its
alpha form to its beta form. The alpha form is hexagonal close–packed and the beta form is face–
centred cubic. What is likely true about the density of the alpha versus beta forms at 1074 oC?