Crystal Models The Solid-State Structure of Metals and Ionic Compounds Objectives Understand the concept of the unit cell in crystalline solids. Construct models of unit cells for several metallic and ionic substances. Calculate various quantities such as atomic radii and theoretical density from unit cell data. Introduction In a piece of crystalline material, either metallic or ionic, the particles show an endless repetition of an orderly arrangement to generate a three-dimensional structure known as a crystal lattice. Part of a simple lattice is shown in Figure 1.1. The unit cell is the most convenient small part of the lattice that, if repeated in three dimensions, generates the entire lattice. An acceptable unit cell for the lattice shown in Figure 0.3(a). Depending on the relative lengths of the sides of the unit cell and the angles that the sides make with each other, a unit cell will belong to one of several crystal systems. In this study we consider unit cells in only the cubic system (where all three sides are equal, designated by the symbol a, and all angles are 90 ) and the hexagonal system (where two sides are equal, designated by a, and one side is longer, designated by c, and the angles are 90, 90, and 120 ); see Figure 0.4b. A primitive unit cell is a unit cell in which only the corners are occupied. In some cases unit cells contain other lattice points in addition to those at the corners; these are called multiple unit cells. A primitive cubic and two multiple cubic unit cells are shown in Experiment Figure 3, along with a multiple hexagonal unit cell. The exploded unit cell diagrams shown in Figure 3 are somewhat misleading because they show that each cell contains mostly empty space. The packing of atoms in a crystal is more realistically shown by the sketches given in Figure 0.2 Figure 0.5for these same unit cells. Figure 1.1 Section of a three-dimensional lattice. a Figure 0.2 Two common crystal systems, (a) is called primitive cubic, and (b) is called hexagonal. Although the spheres shown in Figure 0.3 and Figure 0.4 touch each other, there is considerable empty space in each unit cell. There are three types of open spaces (or holes ) commonly found in unit cells: tetrahedral holes, octahedral holes, and cubic holes (see Figure 1.7), based on the geometric figure describing the atoms around the hole. b
Part I: Metals The particles making up the crystal structure of a pure metal consist of identical atoms. Most metals crystallize in the unit cells shown in Figure 0.3 and Figure 0.4. From a careful analysis of these sketches, several properties of the metals can be determined. a Figure 0.3 Two unit cells, (a) the primitive cubic, and (b) the body-centered cubic. b a b c Figure 0.4 Two more unit cells, (a) is the face-centered cubic, also called the cubic close-packed, and (b) the hexagonal closepacked. The hexagonal close-packed unit cell is shown extended in (c) to emphasize the hexagonal shape this lattice has. (a) Primitive Cubic (b) Hexagonal Close-Packed (c) Body-Centered Cubic (d) Face-Centered Cubic or Cubic Close-Packed Figure 0.5 Packing of atoms or anions in a crystal. Each unit cell shows two views of the packing of spheres. The left view shows all of the atoms that sit on the outline of the unit cell. The right view shows the same unit cell with all of the atom parts outside the unit cell taken away. What remains is the actual content of the unit cell. Consider the element polonium, which will crystallize in the primitive cubic unit cell, shown in Figure 0.3a, with a = 0.336 nm. The length of a side of the unit cell equals twice the radius of an atom (a = 2R); so we can calculate the radius of a Po atom as R a 2 0.336 nm 2 0.168 nm 0.1 Similar relationships between the unit cell length and the radius of the atom are summarized in Table 0.1 for the other unit cells.
The total number of atoms contained within the unit cell is known as the unit cell contents (Z). Again, consider the primitive cubic unit cell for polonium as shown in Figure 0.5a. At first glance you might think there are eight atoms occupying this unit cell, but this is not the case. Although eight atoms are used to outline the unit cell, the actual unit cell is defined in terms of the centers of these atoms; consequently, a given corner atom is really part of eight adjacent unit cells (see Figure 0.5a). Thus only l/8 of a given atom belongs to the unit cell under consideration. Because there are eight such atoms, for the primitive cubic unit cell Z is given as follows: 1/ 8corner Z 8corners 1 atom 0.2 atom All body-centered atoms are completely within the unit cell; any face-centered atom is shared by two unit cells; any edgecentered atom is shared by the number of unit cells around that edge (often 3, 4, or 6), the unit cell content for the various unit cells that we are considering is summarized in Table 1 and Figure 7. a b c Figure 0.6 Diagrams of holes between anions in a crystal. Three kinds of holes are shown in two views. (a) shows a tetrahedral hole, (b) showns an octahedral hole, and (c) shows a cubic hole. The top view shows the anions separated along an axis marked by the dashed line. the lower figure shows the top view of the cluster of anions that surround the hole, looking down that axis. a b c Figure 1.7 The positions of holes in a unit cell. (a)shows the tetrahedral hole, the topmost view is exploded on one axis. You can see the shape of a tetrahedron, with its triangular base and peak. the second picture in the column shows the tetrahedral hole all together. The third picture shows the hole exploded on all axes, and the fourth picture shows the hole re-assembled in the same orientation as in picture three. (b) Shows the same set of views of an octahedral hole and (c) shows the same set of views of a cubic hole. The crystal coordination number (CN) of an atom is the number of nearest neighbor atoms in the crystal structure, Figure 8 shows that CN = 6 for a given corner atom in the primitive cubic unit cell. (This is the value for any atom in the unit cell.) Table 0.1 also summarizes the values of CN for the cubic unit cells to be studied in this experiment.
Unit cell Length Radius Relationship Z V CN Primitive cubic a = 2R 1 a 3 6 Body-centered cubic a 3 = 4R 2 a 3 8 Face-centered cubic a 2 = 4R 4 a 3 12 Hexagonal closest packed a = 2R 2 a 2 c sin 60 o 12 Diamond a 3 = 8R??? Z = number of atoms per unit cell V = unit cell volume CN = coordination number a = unit cell length R = radius of atom or anion? = Figure it out yourself! Table 0.1 Summary of properties of various unit cells. If the unit cell dimensions and cell content are known for a substance, the volume of the unit cell and the mass of the atoms in the unit cell can be calculated. For the cell of polonium, the volume is: and the mass is: V a 3 (0.33610-7 cm) 3 3.79 10-23 cm 3 0.3 m Z M N A 1 atom 209 g/mol 6.022 10 23 atoms/mole 3.47 1022 g 0.4 Knowing the mass and volume of the unit cell allows the calculation of the theoretical (or crystallographic) density (d) of the substance d m v 3.47 1022 g 3.79 10 23 cm 3 9.16 g/cm 3 0.5 Part II: Ionic Crystals The crystal structures of ionic compounds may be pictured as arrays of closely stacked larger ions, creating holes into which the smaller counter-ions can fit. Because anions are usually larger than cations, we consider the anions to define the original unit cell and the cations to occupy the holes. For example, in the sodium chloride unit cell shown in Figure 0.8, the Na + cations can be considered to occupy the octahedral holes in a slightly expanded Cl -anion face-centered cubic unit cell. The unit cells of ionic compounds can be analyzed to obtain various properties just as were the unit cells of metals. For example, the cell content for each type of ion in the sodium chloride unit cell is Z 12 edges Z 8 corners 1/4 cation 1 cation 4 cations edge 1/8 anion 1/2 anions 6 sides 4 anions corner side 0.6
The ratio of Z + /Z is the same as the ratio of ions in the empirical formula. Each ion has a crystal coordination number, but this now refers to the number of oppositely charged ions that are nearest neighbors. For example, in NaCl, CN + = 6 and CN = 6. The ratio of CN + /CN - is the inverse of the ratio of Z + /Z. When calculating the theoretical density, it is helpful to begin by finding the separate masses of the cations and anions in the unit cell. For example, for NaCl having a unit cell length of a = 0.56402 nm, the masses of the cations and anions in the cell as well as the cell volume are calculated according to m Z Atomic Mass of cation 6.02210 23 cations/mol 4 cations 22.99 g/mol 6.02210 23 1.52710-22 g 0.7 cations/mol m Z Atomic mass of anion 6.022 10 23 anions/mol 4 anions 35.45 g/mol 6.022 10 23 anions/mol 2.3551022 g 0.8 m m 2.35510 22 g 1.527 10 22 g 3.882 10 22 g 1.9 V 0.56402 nm10 7 cm/1 nm 3 1.794310 22 cm 3 0.10 d m V 3.882 1022 g 1.794310 22 cm 2.164 g/cm 3 0.11 3 Crystal structures of ionic substances are influenced both by the relative sizes of the anions and cations, and by the ionic charges. Depending on the relative sizes of the ions, a cation may fit into a tetrahedral hole (the smallest) or might fit into a cubic hole (the largest). a Figure 0.8 The Sodium chloride unit cell b Radius ratio (r + /r - ) Coordination number (CN+) Hole geometry Holes per anion 0.225 0.414 4 tetrahedral 2 0.414 0.732 6 octahedral 1 > 0.732 8 cubic 1 Table 0.2 The radius ratio rule Table 0.2 indicates the values of the radius ratio (r + /r - ) needed by a cation so that it will fit into a certain type of hole. Using r + = 0.097 nm for the Na + cation and 0.181 nm for the Cl anion in NaCl gives r + /r - = 0.097 nm/0.181 nm = 0.54 0.12
This radius ratio value corresponds to the octahedral hole geometry entry in the table. Because there is one octahedral hole per Cl anion, all these holes must be occupied by Na + cations in order to achieve the one-to-one ratio of cation to anion required by the formula NaCl. The empirical relationships expressed in Table 0.2 are actually violated by many crystalline species and should be used only as general guidelines. Experimental Procedure Your instructor will familiarize you with the kit of spheres and connectors you will be using to build the models of the unit cells. The kit will contain four different sizes of spheres. For each structure, construct the model and analyze it to determine the unit cell parameters requested on the table at the end of this experiment. Caution! Please stuff wet paper towels into the drains of the sinks at your lab station. Otherwise, you may lose a few beads into the drains. Part I: Metals (Use spheres of uniform size for each structure in this part.) A. -Ni Structure C. -Fe Structure Prepare the three layers shown in the Figure. Place layer 2 on layer 1; then place layer 3 on layer 2. Keep the relative orientations the same as given in the figure and a cubic structure will result. B. -Ni Structure Prepare the three layers shown. Place layer 2 on layer 1 so that this single sphere nestles down in the hole between the four atoms in layer 1. Place layer 3 on layer 2. Keep the relative orientations of layer 1 and 3 the same as given in the figure to generate a cubic structure. D. Silicon Structure Prepare the three layers shown. Place layer 2 on layer 1 so that this single sphere nestles down in the hole between the three left-most atoms of layer 1. Place layer 3 on layer 2. Keep the relative orientations of layers 1 and 3 the same as given in the figure, and a hexagonal closest-packed structure will result. It may be necessary to support the right-hand portion of the top layer to keep it from falling. Part II: Ionic Crystals A. CsCl Structure Prepare the five layers shown. Assemble the layers, keeping the relative orientations of the layers the same as given. It may be necessary to support two of the corners of the model. This structure is called a diamondlike structure. B. AgCl Structure Prepare the three layers shown above, using the large and medium spheres. The large spheres in the first and third layers should not quite touch each other in order to make room for the medium sphere of the second layer. Assemble the layers, keeping the relative orientations of the layers the same as given in the figure. A cubic structure will result. Prepare the three layers shown, using the large and small spheres, Assemble the layers, keeping the relative orientations of the layers the same as given in the figure. A cubic structure will result.
C. CuI Structure D. Li 2 S Structure Prepare the five layers shown, using the large and small spheres. Assemble the layers, keeping the relative orientations of the layers the same as given in the figure and, again, slightly expand the structures in layers 1, 3, and 5. The small spheres should nestle down in the holes created by the large spheres. A cubic structure will result. Prepare the five layers shown, using the large and small spheres. Assemble the layers, keeping the relative orientations of the layers the same as given in the figure. A cubic structure will result. Enrichment links: http://www.umanitoba.ca/chemistry/courses/chem130/molecules/crystals/130_crystal_models.html Calculations Part I: Metals In each case, use your model to identify the type of unit cell. A. -Ni Structure Using a = 0.35238 nm, calculate the atomic radius from the appropriate unit cell length-radius relation given in Table 1. Using the unit cell content (Z), calculate the mass of the atoms in the unit cell. Use the value of a to calculate the volume of the unit cell. Now find the theoretical density of metallic nickel in this crystal form. B. -Ni Structure Identify the unit cell (a = 0.266 nm, c = 0.429 nm), and calculate the atomic radius and the theoretical density of metallic nickel in this second crystal form. C. -Fe Structure Identify the unit cell (a = 0.28664 nm), and calculate the atomic radius and the theoretical density of metallic iron in this crystal form. D. Silicon Structure Identify the unit cell (a = 0.54305 nm), and calculate the atomic radius and theoretical density of silicon. Part II: Ionic Compounds A. CsCl Structure Using r + = 0.167 nm and r = 0.18 nm, calculate r + /r -. From Table 2, ascertain the hole geometry and the cation coordination number (CN + ). Determine if all these holes are occupied by the cations in order to achieve the cation-toanion ratio expressed by the empirical formula of the compound. Identify each ion in your model as being corner, body-centered, face-centered, or edge-centered. Calculate Z + and Z, using the fractional contribution. Use Z + and Z to calculate the masses of all ions (m +, m, m) in the unit cell as described previously. Find the volume of the cubic unit cell given that a = 0.4110 nm. Calculate the theoretical density of CsCl. Finally, describe the type of cubic unit cell defined by the Cl anions, and use Table 0.1 to determine the coordination number (CN ) of these anions. Note: The entry for Z in Table 0.1 should be the same as your calculated value for Z. B. AgCl Structure Using r + = 0.126 nm, r = 0.181 nm, and a = 0.55491 nm, repeat the calculations for AgCl. C. CuI Structure Using r + = 0.096 nm, r = 0.220 nm, and a = 0.6047 nm, repeat the calculations for CuI. D. Li 2 S Structure Using r + = 0.068 nm, r = 0.220 nm, and a = 0.570 nm, repeat the calculations for Li 2 S.
Arrangement of Structure Atoms Part I: Metals Radius (cm) CN Z Mass (g) Volume (cm) Density (g/cm 3 ) Ni -Ni -Fe Si Structure Arrangement of Hole Geometry Anions Part II: Ionic Crystals Fraction of Holes Filled CN + CN - Z + Z - CsCl AgCl CuI Li 2 S Structure m + (g) m - (g) Mass (g) Volume (cm 3 ) Density (g/cm 3 ) CsCl AgCl CuI Li 2 S