Forces and Bonding: Solids

To understand several fundamental aspects of the structures of crystalline solids:

Sir W. H. Bragg, one of the first scientists to use X-rays to study crystal structures, was fascinated by the intricate, orderly atomic patterns revealed by his investigations. In 1901 he likened the stability of crystalline structures to that of an engineering structure.

Natural crystals, from gemstones to snowflakes, have always held a fascination for man. Today we recognize that the regular geometric shapes of crystals reflect orderly three-dimensional networks of the atoms, ions, or molecules that form the crystal lattice.

Although the arrangements of atoms and ions that lead to regular crystal lattice structures are quite simple, it is hard to visualize the arrangements in 3-dimensions. Two-dimensional drawings of 3-dimensional objects do not help! The purpose of this workshop is to give you the opportunity to build models of some common crystal structures and clearly see how the atoms and ions are arranged. Having built and examined models, it will be easier for you to make sense of drawings in the text and class discussions.

Perfect crystals, in which the repetitive lattice pattern is completely uniform throughout, are extremely rare. This is not surprising when you consider that a crystal is formed from a liquid (a solution or the molten solid) or a gas, in which particles are randomly oriented. To initiate crystal growth, a tiny collection of particles must form in a highly ordered pattern. This process, called nucleation, may occur spontaneously when particles of a pure substance collide with appropriate orientations and with low enough kinetic energies to allow "permanent" interparticle attractions rather than "rebound collisions." More often, a foreign material provides a surface on which particles may be adsorbed in orientations suitable for continued crystal growth. The nucleation of snowflakes, for example, often involves a "dust particle" as the center of adsorption for water molecules, and a string suspended in a supersaturated solution of a salt may provide nucleation sites as ions are adsorbed onto the fibers.

The tiny microcrystals initially formed may be perfectly ordered, but continued growth introduces imperfections. As the crystal grows, the surface area for addition of more particles becomes larger, requiring an ever-increasing availability of the proper particles with appropriate orientations. When molecular crystals grow from the pure liquid (e.g., as pure water freezes) or gas (e.g., in formation of iodine crystals from vapor), there are no problems with imperfections from foreign particles, but it is unlikely that uniform growth rates over all surface areas will continue. Nonuniform growth results in crystal imperfections. Crystals grown from mixtures, such as salt solutions or impure liquids, may contain foreign particles trapped in the lattice in addition to physical imperfections resulting from irregular growth patterns.

The growing of nearly perfect crystals, or of crystals containing "planned imperfections," is important in modern solid state technology. Synthetic diamonds are grown for such uses as diamond-tipped drills; sodium chloride crystals are produced for optics systems of infrared spectrometers; and "doped" crystals of silicon or germanium (containing planned substitutional lattice defects) are prepared for solid state electronic equipment.

1. An atom (or hole) at a vertex (corner) of a cubic unit cell is shared with 7 other identical unit cells, or a total of 8 unit cells. Thus only 1/8 of that atom can be assigned to the unit cell of interest. Then Contribution from vertices = Number of occupied vertices * 1/8
2. An atom (or hole) anywhere along an edge of a cubic unit cell is shared among a total of 4 unit cells. Thus Contribution from edges = Number of occupied edges * 1/4
3. An atom (or hole) anywhere in a face of a cubic unit cell is shared between 2 unit cells. Thus Contribution from faces = Number of occupied faces * 1/2
4. Finally, an atom (or hole) contained entirely within the body of (within the boundaries of) a cubic unit cell belongs entirely to that unit cell. Thus Contribution from body = Number of atoms (or holes) contained in body

Focus Questions are included in the Experimental Procedure, below.

Crystal model kit, consisting of

Write answers to questions in your notebook. Each group should obtain a model kit from the instructor. As you do the experiment, make simple sketches of the models. Answer questions as you go. Indicate which question you are answering by number.

1. Closest Packing

a. Place the grooved closest-packing shelf on the top shelf of the kit with grooves running left to right.

b. Make a close-packed layer of red spheres using the grooves, consisting of 7 rows of spheres. The number of spheres per row should alternate 9,8. . .9. In a close packed layer, the atoms of one row fit the indentations in the neighboring rows, as shown below.

This arrangement allows the largest possible number of spheres to fit in a given area--i.e., the density of spheres is maximum. Each sphere is surrounded by a hexagon of spheres touching it.

c. View your close-packed layer from above. You will see two types of triangular holes between spheres, shown below.

One type has a vertex pointing "up" (as you view it), the other a vertex pointing "down". We call these "u" and "d" holes, respectively. A hole of one type has vertices in common with 3 holes of the second type.

d. Place a blue sphere in the first "d" hole in the upper left corner of the red layer. This begins the closest packing arrangement of spheres in two layers.

e. The blue sphere and the three red spheres that it touches should look as shown here.

These 4 spheres form a regular tetrahedron (polyhedron with 4 vertices and 4 faces, each an equilateral triangle). The space at the center of the 4 spheres is a tetrahedral hole.

f. Add blue spheres to the second layer, using "d" holes, to give a pattern that looks as shown below (ignore x's).

Construct a second identical blue pattern, separated from the first, in the lower right corner of the red layer. Use only "d" holes. When done, observe that none of the "u" holes in the red layer are used. If you look through the "d" holes in the blue layer, you can see the "u" holes in the red layer.

g. Focus on the equilateral triangle of blue spheres marked by x's in the diagram above. Immediately under this triangle, find a similar triangle of red spheres rotated 60^o with respect to the blue triangle. The two triangles are oriented as shown here

These 6 spheres form a regular octahedron (polyhedron with 6 vertices and 8 equilateral triangle faces), which looks as shown here. The space at the center of the 6 spheres is an octahedral hole.

h. There are two ways to place a third layer of spheres above the first two. Using "u" holes in the blue layer will place atoms in rows directly above the red spheres in the first layer. Using "d" holes in the blue layer, neither blue nor red spheres will lie directly below the spheres in the third layer.

1. Using red spheres and the upper left blue layer, place the third layer over the first layer. This gives a central sphere surrounded by a hexagon lying directly over an identical hexagon in the first layer (find it). The two hexagons are separated by an equilateral triangle of blue spheres in the second layer: We characterize this packing arrangement as RBRB..., the letters indicating that the colors and locations of spheres repeat every other layer. This packing arrangement is hexagonal closest packed (hcp), shown in a) below. Its more general designation is 1212. . . . Hexagonal channels run vertically through the structure.

2. Using yellow spheres and the other blue layer, place the third layer so that neither red nor blue spheres lie below the yellow spheres. Use "d" holes of the blue layer. This gives a yellow triangle of 6 spheres, vertex down. Finally, put a red sphere in the center hole of the yellow layer, lying directly over a red sphere in the first layer: This packing arrangement, called cubic closest packing (ccp) and shown in b) above, is characterized as RBYRBY. . . or 123123. . . .

1. What is the coordination number (the number of atoms touching an atom) of an atom in a ccp structure? In a hcp structure?

Save these two models for later reference and proceed to the next part.

2. Face-centered cubic lattice. (Steps 2a-2c show that the face centered cubic lattice and cubic closest packing are identical. Be sure you understand everything in a step before proceeding. If you have trouble, get help!)

2. How many nearest neighbors does an atom have in the fcc model? (This should be the same as the coordination number in the ccp structure in section 1.)

3. Draw a diagram of the cube with atoms at the edge centers and body center of the cube. Connect atoms to show the two triangles and hexagon.

3. Lattice Holes.

4. How many octahedral holes are in the unit cell (vacancies at the centers of cube edges are octahedral holes belonging partially to the unit cell).

5. How many tetrahedral holes are there in the unit cell?
6. How many atoms are there in the unit cell? (Some atoms belong only partially to this unit cell!).
7. Do your findings support the statement: there are 2 tetrahedral holes and 1 octahedral hole per atom in a fcc unit cell?
8. Which is larger, a tetrahedral or an octahedral hole?

4. Structures of compounds based on the fcc lattice.

9. How many Na⁺ ions are there per unit cell of the NaCl structure?
10. How many Cl^- ions are there per unit cell?
11. Is this consistent with the 1:1 stoichiometry implied by the formula?
12. What is the coordination number of each cation in the structure?
13. What is the coordination number of each anion?

14. Confirm the 1:1 ion ratio in ZnS.
15. What is the coordination number of Zn²⁺ (yellow spheres)?
16. What is the coordination number of S^2- (red spheres)?

17. Confirm the 1:2 ion ratio in CaF₂.
18. What is the coordination number of Ca²⁺ (red spheres)?
19. What is the coordination number of F^- (yellow spheres)?

20. From the name of the structure and the chemical formulas CaF₂ and Na₂O, describe cation and anion positions in the anti-fluorite structure.
21. What are the coordination numbers?

22. What is the coordination number of a C atom in the diamond lattice (check several atoms in your model!)?

5. The Body Centered Cubic Lattice.

23. How many atoms are there per unit cell in the bcc lattice?
24. What is the coordination number of an atom?
25. Can you find any octahedral or tetrahedral holes by extending the lattice?

6. Simple Cubic Lattice

26. What is the coordination number of Cs⁺? Of Cl^-?
27. Calculate the number of Cs⁺ and Cl^- ions per unit cell and verify the stoichiometry of CsCl.

7. Back to Lattice holes

a. Use the three bottom shelves to make a fcc lattice at the right side of the kit.

b. The fcc and ccp arrangements are equivalent. The unit cell just constructed is not actually close-packed, however. It is expanded for better visibility. Imagine that the spheres uniformly move together until spheres along the face diagonal of the cube are just touching, as shown below.

The corner spheres do not touch, but each corner sphere touches the face center sphere. In terms of the sphere radius, r, the face diagonal has length 4r.

c. Find a tetrahedral hole and an octahedral hole. Which is larger?

Whether a cation occupies a tetrahedral or an octahedral hole in a close-packed anion lattice depends on its size compared to the sizes of the holes. In ZnS, Zn²⁺ is small enough relative to S^2- to fit tetrahedral holes. In NaCl, Na⁺ is too big for tetrahedral holes, but fits octahedral holes. In CsCl, Cs⁺ is so big compared to Cl^- that it cannot fit into either type of hole. The simple cubic anion lattice is adopted.

28. Based on the considerations in b above, and assuming an anion radius r_o, calculate, in terms of r_o, the radius of a cation that will just fit an octahedral hole in a fcc anion lattice. (Hint: Consider a face of the face-centered cube, and use the Pythagorean Theorem.)
29. Calculate the radius of the cation that can just fit the hole in a simple cubic lattice of anions of radius r_o. (Hint: Consider a right triangle made from a body diagonal, a face diagonal, and an edge of the simple cube.)
30. Calculate the just-fit radius for a tetrahedral hole in a fcc anion lattice. (Hint: Focus on one of the 8 small cubes that make up the fcc unit cell).

When finished, return spheres to boxes.

No disposal required.

Preparation Questions