# Crystal Structure

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```					Chem 106: Geochemistry
Quantitative Project

Crystal Structure
INTRODUCTION AND PURPOSE

The solid state of matter generally consists of a regular arrangement of atoms, molecules, or ions.
A definite geometric form, called a crystal, is distinctive for the substance in question. The
configuration of particles extending in all directions throughout the crystal is called the space
lattice. The smallest portion of the space lattice which shows the pattern for the whole lattice is
called a unit cell. It may be defined as the unit structure which, when repeated indefinitely, gives the
crystal. The number of nearest neighbors of a particle in a crystal lattice is called its coordination
number. Several of the more basic and frequently found crystal structures will be constructed and
studied. Styrofoam balls will be used to represent the atoms. The structures of monatomic metals
are explored in this experiment.

Cubic structures

Three of the more simple crystal geometries are shown in detail in the attached figures (simple
cubic, body-centered cubic, and face-centered cubic structures). The arrangement of spheres as
whole atoms is shown first, followed by figures which represent more clearly the cubic
configuration of these atoms, where the spheres represent the centers of the atoms.

One important distinction among types of crystal structure is the efficiency of packing in the given
arrangement. The amount of occupied or filled space as contrasted to the amount of empty space is
a measure of packing efficiency. Occupied volume can readily be measured by calculating the
volume of the spheres that represent the packed atoms. Total volume, the sum of the filled and
empty volumes, is determined by calculating the volume of the unit cell. The empty volume is the
difference between the total and the occupied volumes. Note that the unit cell extends only to the
center of the atoms which comprise the edge of each unit cell (see figures).

Empty spaces can play an important part in determining the properties of a substance. For example,
if carbon atoms are intentionally inserted into some of the empty spaces in the body-centered cubic
(bcc) structure of iron, the resulting distorted iron is “hardened” and is called steel.

Closest packing

Among the cubic lattices, the efficiency of packing increases going from simple through body-
centered to face-centered cubic. The term closest packing refers to a way of arranging identical
spheres such that the space is filled most efficiently, and thus each object is in contact with a
maximum number of like objects. A honeycomb is an example of nature’s closest packing.

There are a number of possible orderly close-packed arrangements obtained by stacking layers of
atoms in a perfectly repetitive manner. The two simplest close-packed arrangements are hexagonal
closest packing (hcp) and cubic closest packing (ccp), achieved by stacking layers A and B or
layers A, B, and C as shown in the attached figures. Note that layers A, B, and C are composed of
identical spheres, and only the stacking pattern varies. In ccp, layer B goes above one set of holes
on layer A, and layer C goes above a different set of holes on layer A. The unit cells of both hcp
and ccp arrangements are shown, and illustrate graphically that ccp is the structure previously called
face-centered cubic (fcc). It is possible to rearrange the hcp structure to produce the ccp structure.
EQUIPMENT

Styrofoam balls, toothpicks, calculator.

PROCEDURE 1: Simple cubic lattice

Use assemblies in which balls are in contact with each other (use toothpicks to link the styrofoam
balls together).

Place one 4-ball unit on top of an identical 4-ball unit. Using a third 4-ball unit, extend the model in
the x direction (it need not be pinned in position) so that you have two unit cells. Extend the model
in the y direction by using a 6-ball unit and in the z direction by using a 9-ball unit. You now have a
cube containing eight unit cells. Verify that you can visualize all eight.

1. Focus on one unit cell. Record on the Report Sheet the edge length a of the unit cell in terms of
Note the following:
a. In all calculations, distances to be measured are between the centers of the spheres.
b. Since many atoms of different atomic radius crystallize in a few basic cell types,
calculations will be left in terms of r. Thus, the specific radius for any
particular atom can be inserted conveniently into the general formulas developed for the
basic cell types.

2. Calculate and record the total volume of the unit cell in terms of r.

3. Focus on the sphere hidden in the center of the 27-ball model. Record the fraction of this
sphere that lies in each of the eight unit cells.

4. Note that any sphere lying at the corner of a cubic unit cell contributed this same fraction of a
sphere to the unit cell. There is the equivalent of how many spheres within a simple cubic unit cell?

5. Calculate and record the volume of the sphere(s) within the unit cell in terms of r. [The formula
for the volume of a sphere is V = (4/3)πr3.]

6. Using your values for the total volume and the occupied volume of this cell, calculate and record
the empty volume in terms of r.

7. What percentages of the unit cell are occupied space and empty space? Record, showing
calculations.

8. What is the coordination number of an element arranged in a simple cubic lattice?
PROCEDURE 2: Body-centered cubic lattice
Take a single ball and the two 4-ball units illustrated below (extend the links by taping them
overlapping toothpicks together in series). These units are constructed leaving a space between
spheres.

First layer            Second layer             Third layer

Assemble the layers by placing the single sphere in the depression formed in the middle of the first
layer, then laying the third layer on top so that its spheres are directly over those of the first layer.
Note that the spheres are in contact along the body diagonal, not along the cube edge or face
diagonal. Although there is a space along the cube edge, the edge length can be calculated using
geometry and data from the body diagonal. In an actual crystal, there would be an indefinite set of
these unit cubes with common corners.

b
a      f

a          a

From a study of the right triangles involved in a cube, as illustrated above, where

a = edge length of cube
f = face diagonal length of cube
b = body diagonal length of cube

the following relationships are valid by the Pythagorean Theorem.

Equation 1:   b2 = f2 + a 2 (triangle bfa)
Equation 2:   f2 = a2 + a 2 (triangle faa)
Substituting Equation 2 into Equation 1:

b2 = (a2 + a 2) + a2 = 3a2

1. Calculate and record the body diagonal length b in terms of sphere radius r.

2. Knowing that b2 = 3a2, calculate and record the value of the edge length a in terms of r.

3. Calculate and record the total volume of the unit cell in terms of r.
4. Calculate and record the equivalent number of spheres comprising this unit cell.

5. Calculate and record the volume of the spheres comprising the unit cell.

6. Calculate and record the empty volume in terms of r.

7. What percentages of the unit cell are occupied and empty space? Record.

8. What is the coordination number of an element in a body-centered cubic lattice?

PROCEDURE 3: Face-centered cubic lattice

Take the layers illustrated below.

First layer               Second layer             Third layer

Assemble the three layers. The second layer should be placed so that the four spheres fit between
the outer spheres of the first layer. The spheres of the third layer should be directly over those of
the first layer.

1. Calculate and record the total volume of the unit cell in terms of sphere radius r.

2. Calculate and record the equivalent number of spheres comprising the unit cell.

3. Calculate and record the volume of the equivalent number of spheres comprising the unit cell.

4. Calculate and record the empty volume of the unit cell in terms of r.

5. What percentages of the unit cell are occupied and empty space? Record.

6. What is the coordination number of an atom when packed in a crystal of this type?

PROCEDURE 4: Empty volume determination of aluminum

A sample of aluminum weighs 27.0 g and has a volume of 10.0 cm3. The atomic radius of metallic
aluminum has been determined by x-ray diffraction to be 0.143 nm.

1. Using the mass and atomic weight, calculate the total number of atoms in the piece of metal.
2. Using the total number of atoms in the sample and the volume of an individual atom, calculate
the actual volume occupied by the atoms in cm3. Compare this with the theoretical value for the
face-centered cubic structure exhibited by aluminum.

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