Crystal Structures

Of the five solids we discussed in the last page, three form strong crystal structures: metallic, ionic, and network solids. These solids form regular matrices; seven different crystal patterns are used in nature, with different angles between each component atom or molecule and different edge lengths. However, we'll only look at some of the more common crystal shapes.

As we touched on in the previous page with out discussion of ionic and network solids, enough information must be provided to allow us to draw the entire crystal matrix. The smallest units of a crystal matrix are called the unit cells, and they are the smallest pieces of the crystal pattern that can convey all the information about how the atoms or units fit together.

We will discuss three basic packing schemes, each with its own unit cell. The first is simple cubic, in which all the component atoms line up in rows, forming very simple, regular cubes. Since simple-cubic packing involves only one type of atom, the unit cell is just one atom, which is repeated in every direction. Network solids are the only type of crystals to use simple-cubic packing.

A variation of the simple-cubic structure is the body-centered cubic, in which an additional atom is placed in the center of the cube. Since this center atom takes up space, the other atoms are forces apart and do not touch each other; only the center atom is in contact with other atoms. The unit cell of a body-centered cube consists of the center atom, along with 1/8 of each of the 8 corner atoms, giving two atoms in the cell.

One additional packing scheme is the face-centered cubic pack, in which each side a typical the box has four atoms, with one extra in the center. Repeating this configuration for all 6 sides of a box gives the three-dimensional structure. The unit cell for this crystal takes one-half of each of the six "face-centered" atoms, and one-eighth of each of the 8 corner atoms, giving a total of 4 atoms per unit cell.

For both body-centered and face-centered schemes, the crystal may be consisted of one type of atom (as in a metal) or various kinds (as in ionic solids).

So why are packing schemes and unit cells so important? They allow us to determine the atomic radius of a substance given its density and packing type. For instance, let's take copper, which crystallizes in a face-centered structure and has a density of 8.92 grams per cubic centimeter. First, let's see what the mass of a unit cell is; we can combine this with the density of copper to find the volume of the unit cell, and hence the radius of copper. Copper has a mass of 63.546 grams per mole, with 6.022 x 10²³ atoms per mole, and four atoms per unit cell. Therefore, one unit cell has a mass of:

   (63.546 g / 1 mol) * (1 mol / 6.022 x 10²³ atoms) * (4 atoms / 1 unit cell) = 4.2209 x 10^-22 g / unit cell

We know that one cubic centimeter of copper weighs 8.92 grams, while each unit cell weighs 4.2209 x 10^-22 grams. Multiplying these figures like so gives the volume per unit cell:

   (4.2209 x 10^-22 grams / 1 unit cell) * (1 cm³ / 8.92 g) = 4.732 x 10^-23 cm³ / unit cell

Now, we know that each unit cell has a volume of 4.732 x 10^-23 cm³ cubic centimeters. Since volume is measured by multiplying height by width by depth, and since each unit cell is a cube (with each dimension being the same), we can take the cube root of the volume to find the cell edge length:

   (4.732 x 10^-23 cm³)^1/3 = 3.617 x 10^-8 cm

Now, we must consider the configuration of one side of the face-centered cube unit cell: there is one central atom with one atom touching it on each corner. The cell edge is measured from the center of one corner atom to the center of an adjacent atom. We also know that a diagonal, from the center of one corner atom to the opposite, passes completely through the center atom, meaning that it is composed of 4 atomic radii. Via the Pythagorean Theorem, the cell diagonal (4 radii) squared is equal to twice the square of the cell edge:

   (4 radius)² = 2 * (3.617 x 10^-8 cm)²

Taking the square root of each side gives us:

   4 radius = 2^1/2 * (3.617 x 10^-8 cm)

Therefore, 4 times the atomic radius is the square root of two times the cell edge. Simple division gives an atomic radius of 1.279 x 10^-8 centimeters, or 127.9 picometers. Whew! Now, wasn't that fun!

The same principle applies to the other crystal packing structures, except that the diagonal-edge ratio will be different for each type; a body-centered cube's diagonal (4 atomic radii) is 3^1/2 (the square root of 3) times the edge, while a simple-cube's edge is simply twice the atomic radius (a diagonal is unnecessary).

You may also be asked to find the maximum radius of an atom that could fit inside a given crystal structure. Again, looking at the unit cell is essential in finding the width of the gaps (called the interstitial radius). This type of problem is particularly important in ionic crystals, as two different types of atom are involved, with two different radii. For example, Cl^-, the larger ion in the NaCl crystal, packs in a face-centered cube arrangement. Na⁺ fits into the holes between the corner atoms of each face. Given the radii of the ions (Cl^- is 167 pm, and Na⁺ is 116 pm), we can find the density of the crystal (or vice-versa).

First, let's find the cell edge length: one radius of each corner chlorine ion is counted, plus two for the sodium ion (because the edge passes through the whole atom), for a cell edge of

   2 * 167 pm + 2 * 116 pm = 566 pm

Cubing this gives the volume of the cell:

   (566 pm)³ = 1.81 x 10⁸ pm³, or 1.81 x 10^-22 cm³

Since the molar mass of NaCl is 58.44 grams, we can calculate the mass of one mole of "NaCl" pairs (one sodium and one chlorine atom):

   (58.44 g / 1 mol) * (1 mol / 6.022 x 10²³ NaCl pairs) = 9.7044 x 10^-23 g per NaCl pair

We know that there are 4 NaCl pairs per unit cell of the NaCl crystal, letting us find the mass of one unit cell:

   (9.7044 x 10^-23 g / 1 NaCl pair) * (4 NaCl pairs / 1 unit cell) = 3.8818 x 10^-22 g / unit cell

Finally, since 1 unit cell has a known volume and a known mass, we can calculate the density of crystal in general (the unit cells will cancel):

   (3.8818 x 10^-22 g / 1 unit cell) * (1 unit cell / 1.81 x 10^-22 cm³) = 2.1446 grams per cubic centimeter!

There is the final density, which is quite close to the experimental density.

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