Crystalline Lattices
Introduction
8
Crystalline Lattices
Solids generally fall into two categories: amorphous solids and crystalline solids. The atoms, ions, or
molecules in amorphous solids are not arranged in any particular order and thus have no definite
shape. Crystalline solids are made up of repeating units of atoms, ions, or molecules. Often this
repeating pattern can be observed by the naked eye. For example, the hexagonal shape of a snowflake
ultimately originates from the hexagonal arrangement of water molecules (see your textbook for the
structure of ice). The overall pattern exhibited by these repeating units is known as the crystal lattice.
The simplest group of constituent particles that is repeated is known as the unit cell. Figure 1 shows
a hypothetical crystalline lattice in which the unit cell is shaded gray. To simplify the math, we will
assume that all atoms and ions are spherical (a good assumption for any atom and most small ions).
Figure 1. A crystalline lattice
with its unit cell shaded in
gray.
Lattices in which the constituent particles are all the same (atoms or molecules) are referred to as Atype lattices. Ionic compounds typically crystallize in either the AB or AB2-type lattices depending on
the formula. For example, NaCl and ZnO have 1:1 formulas and would thus have AB lattices while
CaF2 or NaS2 have 1:2 (or 2:1) formulas, thus putting them into the AB2 lattice. Regardless of the
lattice type, there are sub categories within each type of lattice. In the case of the A-type lattices, the
names are based upon the unit cells (e.g. cubic and hexagonal), whereas the AB and AB2-type lattices
are named after representative examples (e.g. the NaCl and CsCl lattice). We will focus on the lesscomplicated cubic and NaCl/CsCl systems for the A and AB-type lattices, respectively.
A-Type Lattices: Cubic Lattices
Simple (primitive) cubic (sc) is the simplest lattice type. It consists of eight spheres arranged so that
the center of each sphere defines one of the eight vertices of a cube (Fig. 2a). The length of the side of
this cube is designated a0. Knowing that the two spheres are touching, we know that a0 = 2r, where r
is the radius of the sphere. Because the center of the sphere sits on a vertex, only 1/8 of the volume of
one of the spheres is actually inside the cube (Fig. 2b). However there are eight spheres defining a
cube, thus the total volume of spheres occupying the cube is 8(1/8) = 1. This value is the unit cell
content, Z. Another important term is the coordination number, which refers to how many atoms are
in direct contact with each other. In the case of the simple cubic system, the unit cell must be extended
to include the other 7/8 of the sphere. Upon doing this, we see that the sphere in the middle is touching
six other spheres (Fig 2c).
Crystalline Lattices
9
Figure 2.
A) Simple cubic packing.
B) Fraction of sphere inside the
cube.
C) Coordination in sc packing.
A
B
C
If we wish to assess the efficiency of how the atoms are packed in the unit cell, we can calculate the
volume of the cube taken up by the spheres. The volume of the cube = a03 = 8r3. Since Z = 1, the total
volume of spheres inside the cube is 4/3r3. Thus the packing efficiency is (4/3r3)/8r3 = ~0.52 or
~52%. This is not a very efficient way to pack spheres as you can imagine. In fact, only oxygen and
fluorine pack in this manner, possibly to minimize lone pair-lone pair interactions.
Body-centered cubic (bcc) is similar to the simple cubic system except that there is now a sphere in
the middle of the cube (Fig 3). Simple inspection tells us that the coordination number in this case is
8, as the central sphere touches all eight of the spheres defining the corners of the cube.
Figure 3.
Body-centered cubic
lattice.
Notice however, that the eight spheres defining the vertices of the cube no longer touch each other,
only the central sphere. While the length of the cube is still given by a0, its relationship to the radius
of the spheres becomes much more complicated (Eq. 1).
𝑎0 =
4𝑟
√3
(Eq. 1)
It should be obvious that Z, the unit cell content, has changed. We still have 8(1/8) = 1 corner spheres
inside the unit cell plus the central sphere; therefore Z is equal to 1 + 1 = 2. Again, we can calculate
the packing efficiency, but must multiply the volume of the sphere by two to account for the unit cell
content (Eq. 2). Thus, we see that slightly more than 2/3 of the cube’s volume is now occupied. Several
elements pack in this way including Na, V, Fe, and W.
packing efficiency =
volume of spheres
volume of cube
=
4
3
4𝑟 3
( )
√3
2( 𝜋𝑟 3 )
=
8𝜋
3
64
3√3
= 0.68 = ~68%
(Eq. 2)
Crystalline Lattices
10
Face-centered cubic (fcc) is also known as cubic close-packed (ccc). There are still the eight spheres
on the vertices, but now there are six spheres on all of the faces and no sphere in the center (Fig. 4a);
note also that the center of the facial spheres are coplanar with those of the spheres on the vertices.
So what is Z for fcc? Considering that each sphere is bisected by the face of the cube (Fig. 4b), each
facial sphere contributes ½ of its volume to the unit cell. There are six faces to a cube and 6(½) = 3.
Therefore the total Z is 1 (from the corners) + 3 (from the faces) = 4. The Pythagorean theorem (a2 +
b2 = c2) can be used for one of the faces to fairly easily determine that for fcc, 𝑎0 = 𝑟√8. You are
encouraged to verify mathematically that the packing efficiency in this case is ~74%. This is the most
efficient way to pack spheres. As for the coordination number, consider the front center sphere in
Figure 4a. It has four other spheres touching it in the same layer. Immediately behind its layer is a
layer of four atoms (shaded darker) which also touch that central sphere; in addition there will be an
identical layer of four sphere in front of it. Thus, the coordination number in fcc is 4 + 4 + 4 = 12.
Figure 4.
A) Face-centered cubic lattice.
B) Fraction of sphere inside the cube.
A
B
The AB-Type Lattices: NaCl and CsCl
With two ions, the ionic or AB-type lattices are typically defined by the larger of the two ions. In other
words, the smaller ions “sit” in the holes created by the larger ions. In terms of packing efficiency,
things become a bit more complicated because now things can have different sizes, so this type of
analysis will be best left to the pros. Note: Both the NaCl and CsCl lattices look similar to some of the
A-type lattices. However, since the constituent particles in NaCl and CsCl are NOT the same, by
definition, they do NOT have A-type lattices!
The “rock salt” (NaCl) lattice resembles two interpenetrating,
face-centered cubic A-type lattices in which the Cl− ions are
arranged in a face-centered cubic arrangement and the Na+ ions
occupy the eight edges of the lattice as well as the center (Fig. 5).
The coordination number for this lattice is how many of the
counterions are touching. In this case, the coordination number
for both ions is 6 (this is shown with dashes for Na+ in the center
of the lattice in Figure 5). If you are up for the challenge, show that
the coordination number is the same for Cl−. What about the unit
cell content? Count the Na+ and Cl− separately. Since Cl− is in a
face-centered arrangement, we already know that there should
be 4 Cl−. The Na+ ions are arranged in such a way that one ion
occupies the center of the lattice and twelve additional ions lie on
all 12 edges of the cube. Since a corner has 1/8 of a sphere and a
Figure 5. The “rock salt” (NaCl)
lattice. For clarity, the ions are not
shown to be touching each other.
Crystalline Lattices
11
face has ½ of a sphere, it makes sense that an edge should have ¼ of a sphere. Thus for Na + we have
1 + 12(¼) = 1 + 3 = 4. With four Cl− and four Na+, we would say that Z = 4 because there are four NaCl
“units” in the lattice.
The “cesium chloride” (CsCl) lattice is much easier to visualize because it consists of two
interpenetrating simple cubic systems (Fig. 6). Analysis of the unit cell content is rather easy; there
is 1 Cl− in the center and 8(1/8) = 1 Cs+. Thus, Z for the CsCl lattice is 1 (i.e. 1 CsCl “unit” in the lattice).
Likewise, the coordination number of eight for both ions is straightforward.
Figure 6. The CsCl lattice. For
clarity, the ions are not shown
to be touching each other.
Equations/Sample Calculations
Density:
𝑑=
mass
volume
=
𝑍(𝑀𝑊)
𝑁𝐴 (𝑎𝑜 )3
Z = unit cell number
MW = molecular weight (g/mol)
NA = Avogadro’s number, 6.022 × 1023 mol‒1
ao = unit cell length, convert Å to cm (1Å = 10‒8cm)
Volume of atoms or ions in unit cell: (don’t forget the other ion when applicable)
4
𝑉𝑎 = 𝑍 × 𝜋𝑟 3
3
Volume of unit cell:
𝑉𝑢𝑐 = 𝑎𝑜 3
% Volume or Packing efficiency:
% Volume =
𝑍(4⁄3 𝜋𝑟 3 )
(𝑎𝑜
)3
× 100% =
Va
× 100%
Vuc
Pre-Lab Quiz Review
•
•
•
•
Describe and/or recognize the differences between A and AB-type lattices
Define unit cell content, coordination number
Calculate unit cell content
Calculate density, radius, or unit cell length for the cubic A-type lattice
Crystalline Lattices
12
Materials and Chemicals Required
Inorganic Lattice Modeling Kits
Experimental Procedure
You will be given a crystalline lattice model kit and an instruction manual. Build your assigned
model(s) and provide the information listed below. Details for constructing the models are provided
in the manual. Information regarding the unit cell lengths is given in Table 1.
1. Z, the unit cell content (the number of formula units in a unit cell)
2. CN, the coordination number
3. r, the radius of the atom or ion (for NaCl and CsCl assume rCl– = 1.67 Å)
4. volume of the atoms or ions in the unit cell (using the radius), in Å3
5. packing efficiency of the unit cell, as a %
6. calculated density of the cell in g/cm3
Table 1.
Relationship of a0 to
r
Page # in
model kit
manual
𝑎0 = 2𝑟√2
27
Lattice type
Substance
Unit cell length,
a0 (Å)
fcc
gold
4.07
bcc
niobium
3.35
CsCl
cesium chloride
4.11
𝑎0 =
NaCl (rock
salt)
sodium chloride
5.64
𝑎0 = 2𝑟𝑁𝑎+ + 2𝑟𝐶𝑙−
diamond
*requires the
use of spacers
diamond
3.57
𝑎0 =
4𝑟
√3
2𝑟Cs+ + 2𝑟Cl−
√3
𝑎0 =
8𝑟
√3
18
11
33
23
Crystalline Lattices
13
Waste
No waste for this lab. Return materials back into their containers and give the containers to the TA
or instructor.
Data Required
Fill out the crystal lattice results sheet with all the information from steps 1-6 of the procedure. You
will analyze 5 different crystal lattices: gold, niobium, cesium chloride, sodium chloride, and
diamond.
Items for Discussion and Interpretation of Results
•
•
•
How does coordination number relate to packing efficiency?
Considering the structures of the lattices, does it make sense that we do not refer to them as
molecules and that we only write their empirical formulas?
Why do different elements pack differently? Hint: Think about size of the spheres and the
holes created.
Lab Exam Hints
•
•
•
See Pre-lab quiz items
List/recognize the coordination numbers for all of the models in this experiment
Be able to construct any of the models