Chapter 7: The Solid and Liquid Phases

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7-1: Macroscopic Properties of Solids
7-2: The Lattice Concept
7-3: Close Packing
7-4: Ionic Compounds
7-5: X-Ray Crystallography
7-6: Properties of Liquids
7-7: Model of the Liquid Phase
7-8: Interesting Physical Properties
7-9: The Dissolving Power of Liquids
7-10: Phase Transformations
Applications

Major Concept Area: Electrical Forces and Bonding

Specific Concepts in this Chapter:

The vast majority of known substances are solids under ordinary conditions of pressure and temperature. The purpose of this chapter is to provide an understanding of the structures of crystalline solids at the atomic/molecular level. Be forewarned that a study of such structures requires a sound basis in geometry and an ability to visualize in 3-dimensions. Further, it is impossible to obtain an adequate understanding of the important structural ideas from text and 2-dimensional figures. Work with 3-dimensional models is essential. Read through the chapter, understand it as best you can, then refine your understanding with models in the laboratory. The ideas discussed in this chapter provide a solid base for the understanding of two major classes of materials: metals, in which all atoms in the structure are the same; and ionic compounds, composed of cations of one substance and anions of another.

7-1 Macroscopic Properties of Solids. Solids have a number of observable properties in common:

To rationalize these properties we need a simple model of the solid phase. In designing such a model, we make use of two facts: 1) Since the molar volume of a solid is much less than the molar volume of the gas phase of a substance, the molecules must be crowded together in the solid; 2) the orderly shapes of crystals suggests an orderly arrangement of particles. A simple model, then, is that solids consist of tightly-packed orderly arrangements of particles. Figure 7-1 shows a 2-dimensional representation of a solid. Note first how close together the particles are compared with the gas phase (Figure 5-6). This crowding allows very little free volume to the molecules, which strongly resist further compression. The solid is incompressible. It is also clear why diffusion is slow in solids, because each particle is locked in place by the surrounding ones. Only at the surface of the solid, where a particle has neighbors on only 3 sides, can diffusion occur. Although it is not conveyed by Figure 7-1, the particles are in constant motion, just as in the gas phase. A molecule is constantly jiggling in place, bumping into neighbors that hold it in a cage. Just as for gases, the kinetic energies of the particles (atoms, molecules, or ions) of the solid are distributed according to the Maxwell-Boltzmann distribution, with the kinetic energy of an average particle having the value 3kT/2. The major difference between the motion in gas and solid is that in the solid, a particle is hemmed in and undergoes many more collisions per unit time than in the gas. The movement of a particle in the solid is much like a vibration, because the molecule stays in the same location. If the temperature of the solid is increased, the particles vibrate faster, but are still not free to move about because intermolecular forces from the neighbors hold them in place. Finally, the arrangement of particles is very orderly, not at all chaotic like the gas. Each particle is surrounded by a perfect hexagon of other particles; and particles are arranged in rows that remain straight over long distances. It is easy to imagine that this orderly arrangement at the molecular level is manifested in beautiful crystalline shapes at the macroscopic level.

We now begin to look in detail at the orderly arrangements of particles in solids.

7-2 The Lattice Concept. We will describe structures of crystalline solids in terms of the lattice. A lattice is a regular, repeating pattern of points in space. Wallpaper patterns are examples of 2-dimensional lattices. A wallpaper pattern and the equivalent lattice are shown in Figure 7-2a. The lattice is more fundamental than the particular wallpaper pattern shown, because wallpaper patterns using many different design units could be generated from the same lattice. The key feature of the lattice is the arrangement of points. A small collection of lattice points that contains all of the spatial information in the lattice pattern is called a unit cell. A unit cell is shown outlined in the lattice in Figure 7-2a. The unit cell represents the lattice, because the whole lattice can be reproduced from the unit cell by moving (translating) the unit cell in directions parallel to its sides by distances equal to the lengths of its sides.

Figure 7-2b shows a second small collection of points that might be considered as a unit cell. Movement of this collection in directions parallel to its sides does indeed reproduce the entire lattice. However, the unit cell in Figure 7-2a is simpler than that in Figure 7-2b because it involves fewer points -- one point at each of its 4 corners. A unit cell having lattice points at its corners only is called a primitive unit cell. The unit cell in Figure 7-2b is a non-primitive unit cell because it has an additional point at its center. If it is possible to represent a lattice with a primitive unit cell, we will do so. Crystalline solids are represented by 3-dimensional lattices, one example of which is the simple cubic lattice. The unit cell of the simple cubic lattice is shown in Figure 7-3. This is a primitive unit cell because it has points only at the corners. Movement of the unit cell by the distance, d, in directions parallel to its 3 dimensions reproduces the entire simple cubic lattice. The simple cubic lattice is analogous to an orderly arrangement of building blocks stacked face to face in 3-dimensions.

It was shown in the 1800s by Bravais that there are only 7 shapes (called crystal types) that a unit cell can have and only 14 possible distributions of points (lattice types) within these 7 shapes. The 7 shapes, characterized by the relative magnitudes of the repeat distances in three directions and the angles between these directions, are summarized in Table 7-1.

Table 7-1: The Bravais Crystal Types
System Edge Lengths Angles
Cubic a = b = c all 90o
Tetragonal a = b not equal to c all 90o
Orthorhombic a not equal to b not equal to c all 90o
Monoclinic (once-angled) a not equal to b not equal to c a = b = 90o, g not equal to 90o
triclinic (thrice-angled) a not equal to b not equal to c a not equal to b not equal to g not equal to 90o
Rhombohedral a = b = c = 90o
Hexagonal a = b not equal to c a = b = 90o; g = 120o

We will discuss cubic and hexagonal lattices in the context of a discussion of close-packing ideas.

7-3 Close Packing. What is the most efficient way of arranging spheres in 3-dimensional space? We take "efficient" to mean economical in terms of space. We begin by building one layer of spheres that are arranged as closely to one another as possible. Such a layer is shown in Figure 7-4a. Note that this arrangement is identical to that pictured earlier in Figure 7-1. Spheres are lined up in straight rows, with the spheres in one row fitting snugly into the indentations in adjacent rows. Each sphere is surrounded by a perfect hexagon of spheres, all touching it. There are numerous gaps, or holes, in the layer, of triangular shape. For descriptive purposes here, we label these holes "u" if the triangle has a single vertex pointing up, or "d" if the triangle has vertex down. The spheres in Figure 7-4a form a close-packed layer.

We now add a single sphere of a second layer on top of the first (the black-shaded sphere in Figure 7-4b). The added sphere rests in a low point of the first layer, corresponding with a triangular hole, in contact with 3 spheres of the first layer. The 4 touching spheres lie at the vertices of a regular tetrahedron. There remains some unfilled space at the center of this 4-sphere group that is called a tetrahedral hole. Each time we add a sphere to the second layer, we create a tetrahedral hole between the layers. When additional spheres are added to the second layer, it turns out that they must all go in either "u" holes or "d" holes, but not both. Thus only half of the triangular depressions in the first layer can be used for placement of 2nd-layer spheres. Triangular holes in the 2nd layer lie directly over the unused holes of the first layer, so that channels pass right through both layers. The point at the center of such a channel is surrounded by 6 spheres (gray-shaded in Figure 7-4b), arranged as 2 equilateral triangles, one triangle in the 1st and one in the 2nd layer. These 6 spheres form a regular octahedron, a geometrical solid with 6 vertices and 8 equilateral triangular faces. A regular octahedron is shown in Figure 7-4c. The space surrounded by the 6 spheres is called an octahedral hole. An octahedral hole is larger than a tetrahedral hole because 6 spheres cannot be crowded as closely together as 4 spheres.

There are 2 ways in which a third layer of spheres can be added to the existing 2 layers in close-packed fashion. In the first way, spheres are place directly over spheres of the first layer. The third and first layers then have spheres in the same locations. The 4th layer then goes over the second, the 5th over the 3rd, and so on, so a pattern that repeats every other layer is obtained. This is characterized as a 1212... pattern. Figure 7-5a shows the unit cell of this 3-dimensional lattice. The unit cell is hexagonal in shape; the 1212... close-packing arrangement is therefore called hexagonal close packing (hcp). Note that the coordination number (the number of nearest neighbors) of any sphere in hcp is 12.

The second way of adding a third layer is to place spheres directly over the octahedral holes between the first two layers. In this case, the third-layer spheres do not lie over spheres in layer 1. This is still a close-packing arrangement because the coordination number of any sphere is 12, just as in hcp. The 4th layer of spheres is then placed directly over the spheres in the first layer. This type of close packing is designated 123123... to indicate that the pattern repeats after 3 layers. The unit cell for 123... close packing is shown in Figure 7-5b. Although it may not be obvious from the figure, this unit cell is a cube (this is a situation in which a 2-dimensional drawing does not adequately convey what is true). This type of close packing is therefore called cubic close packing (ccp).

Hcp and ccp are the two ways of close packing spheres in 3 dimensions. They have several features in common:

There are a number of substances that crystallize in one or the other of these two packing arrangements. Beryllium, magnesium, and zinc metal all form hexagonal close packed lattices. Nickel, copper, gold, and aluminum form ccp lattices. Cubic close packing is particularly common. We will take a closer look at cubic lattices now.

The three Bravais cubic lattices are shown in Figure 7-6. The simple cubic (sc) lattice (Figure 7-6a) is primitive, with points (spheres or atoms) only at the vertices of the cube. Metallic polonium and non-metallic dioxygen (O2) crystallize with the simple cubic lattice. The coordination number of an atom in the simple cubic lattice is 6. To determine this, pick one atom in the unit cell; find an atom that appears to be closest to the atom you picked; then count the number of other atoms that are at this same distance from the atom you picked. In counting coordination numbers, you may have to go outside of the single unit cell pictured in Figure 7-6a. In the simple cubic lattice, identical unit cells surround the pictured one on all sides. Some of the nearest neighbors of your picked atom lie in these neighboring unit cells. Finally, the simple cubic unit cell contains only 1 atom, despite the fact that you see 8 atoms drawn in the figure. We say that the simple cube has 1 atom per unit cell (1 atom/uc). The procedure for counting atoms in unit cells is described in the paragraph below. The body-centered cubic (bcc) lattice (Figure 7-6b) is non-primitive, with points at the corners and the center of the unit cell (the center of a cube is called the cube body center). Metallic iron and barium crystallize in the bcc lattice. The coordination number is 8 for this lattice (count the nearest neighbors of the body-center atom), and the body-centered cube has 2 atoms per unit cell. Finally, the face-centered cubic (fcc) lattice (Figure 7-6c) is also non-primitive with points at each corner and in the center of each of the 6 faces of the cube. Metallic nickel and copper adopt this structure. The coordination number is 12 for this lattice, which means that it must be close packed! The face-centered cubic lattice and the cubic close packed lattice are identical. It is very difficult to see this in two-dimensional drawings. To see it, you must imagine looking along a body diagonal of the face-centered cube. A body diagonal runs on a straight line from one corner, through the body center of the cube, to the diagonally opposite corner. As you look along the body diagonal, you will see the pattern of spheres shown in Figure 7-5b: a single sphere; a triangle consisting of 6 spheres; a second triangle of 6 spheres, but rotated 60o with respect to the first one; and finally another single sphere immediately below the first one. Figure 7-7 is an attempt to outline the face-centered cube in a ccp arrangement of spheres. The cube is admittedly difficult to visualize from the drawings. The face-centered cubic lattice has 4 atoms per unit cell.

We now address the problem of determining the number of atoms per unit cell. This is quite simple once it is realized that atoms are generally shared among several unit cells and therefore must be partitioned accordingly. An atom at a corner (vertex) of a cube is shared among 8 unit cells. To see this, imagine the simple cube in Figure 7-6a to be a building block, and focus on the upper right front corner of the building block. Imagine that this corner is labelled "a". In your mind, place identical blocks on top, to the right, and in front of the block in the figure. Then place additional blocks in the remaining places adjoining these blocks to generate the 8-block arrangement in Figure 7-8. Point "a" is now at the center of this collection, and is simultaneously a corner for each of the 8 blocks; 8 blocks share a vertex. This point cannot be said to belong entirely to any particular block. Instead, each block gets 1/8 of it. The counting rule for a corner atom, then, is that only 1/8 of it is assigned to a single unit cell. Similar thinking leads to the conclusion that a face-center atom is shared between 2 unit cells, so 1/2 of it is assigned to a particular unit cell; and a body-center atom is not shared at all, so it belongs entirely to the unit cell that it is found in. The atom count for a body-centered unit cell then goes like this:

# atoms/uc = 8 corner atoms x 1/8 + 1 body-center atom x 1

We will have occasion also to partition points located along edges of the cube. Edges are the lines connecting adjacent vertices. Since an edge is common to 4 cubes arranged in a square, 1/4 of any point along an edge is assigned to a particular cube. The point-counting rules are summarized in Table 7-2.

Table 7-2: Atom Counting Rules for Cubic Unit Cells
Atom location Fraction Assigned to uc
Corner 1/8
Edge 1/4
Face 1/2
Body-center 1

If, as we have said, the face-centered cubic lattice is in fact cubic close packed, then there should be tetrahedral and octahedral holes in the fcc lattice. Further, we should find 4 octahedral and 8 tetrahedral holes in a fcc unit cell based on the the last-listed common feature of close-packed lattices above. To locate these holes it is convenient to place our cube in a coordinate system. The origin of coordinates is located at the lower left front corner of the cube, with the x-axis along the bottom front edge, the y-axis along the left vertical edge, and the z axis along the left bottom edge. This is shown in Figure 7-9a. We specify distances along the 3 axes in terms of fractions of the cube edge length. The set of coordinates (1,0,0) means to proceed 1 edge length along x. These are the coordinates of the lower right front vertex of the cube (labelled 1 in the figure). Similarly, (1/2,1/2,1/2) specifies the body center of the cube, labelled 2 in the figure. Now examine the face-centered cube in Figure 7-9b. We find the most obvious octahedral hole at the body center. This point is surrounded by 6 atoms all at the same distance from it. These are the atoms at the face centers of the cube. Similar octahedral holes are located at the cube edge centers. For example, the point labelled 3 in Figure 7-9b is an octahedral hole. It is surrounded at equal distance by the atoms at (1/2,1/2,1), (1,0,1), (1,1,1), and (1,1/2,1/2). The remaining 2 atoms are not part of the unit cell shown, but can still be located with coordinate sets (1 1/2,1/2,1) and (1,1/2,-1 1/2). (Translating these coordinates is good practice in the use of the unit cell coordinate system.) The cube has 12 edges, so there are 12 holes of this type. The counting rules tell us that only 1/4 of each of these edge-centered octahedral holes belongs to the unit cell of interest. The total number of octahedral holes in the unit cell is then 1(body center) + 12(1/4) = 4. The number of octahedral holes is the same as the number of atoms in the unit cell!

A tetrahedral hole in the fcc unit cell is illustrated in Figure 7-9c. It is defined by the four shaded atoms, and has coordinates (1/4,1/4,1/4). There are 7 other tetrahedral holes in the unit cell, all with coordinate combinations involving 1/4 and 3/4, and all defined by 1 corner and 3 face-center atoms. Try to locate them in the figure, and write down their coordinates. Each tetrahedral hole lies entirely within the cube body, so none of them are shared with other unit cells. The number of tetrahedral holes per fcc unit cell is therefore 8, twice the number of atoms in the unit cell.

Octahedral and tetrahedral holes are important in the discussion of ionic compounds.

7-4 Ionic Compounds. The ideas discussed thus far apply to crystals in which the units -- spheres or atoms -- are all identical. Crystals of pure metals, with close-packed metal atoms, and even crystals of molecular solids (for example, methane, CH4, in which there is a methane molecule at each lattice point) are examples of real situations in which these ideas apply. Ionic compounds are different because there are two distinctly different units -- the cation and the anion -- that must be accommodated in the lattice. The resulting lattices are often called compound lattices because they contain two different components. The cation and the anion differ in charge, of course, and in chemical identity (usually). They also differ in size. Generally speaking, cations are smaller than anions, for easily-understood reasons. Cations contain an excess of protons over electrons. The Coulomb pull of the excess positive charge causes the electron cloud to contract, so that it is smaller than in the corresponding atom (e.g., Na+ is smaller than Na). Anions contain an excess of electrons over protons. The repulsion between the electrons exceeds that in the corresponding neutral atom, causing an expansion of the electron cloud (e.g., Cl- is larger than Cl). Except for a few rare cases, the cation in an ionic compound is smaller than the anion. In forming a lattice, the space requirements are more important for the larger anion, which tends to dictate which lattice is adopted. The (smaller) cations then fit into the holes, or interstices, of the anion lattice. Many common ionic compounds have lattices that consist of cubic close packed anions, with cations in tetrahedral and/or octahedral holes. Stoichiometries of up to 3 cations per anion can be accommodated in this framework. We now discuss some common ionic lattices.

The Sodium Chloride (Rock Salt) Lattice. The NaCl lattice is shown in Figure 7-10a. It can be described as a cubic close packed (face-centered cubic) anion lattice, with Na+ ions in the octahedral holes. The coordination number (number of nearest neighbors of opposite charge) of each Na+ ion is 6, and of each Cl- ion is 6. The cation and anion must have equal coordination numbers because the stoichiometry of the compound is 1:1 (1 cation per anion). The unit cell contains 4 formula units of NaCl (4 Cl- ions, which define the fcc unit cell; 4 Na+ ions because there are 4 octahedral holes per unit cell). Many ionic compounds exhibit the NaCl lattice, including most of the group 1 halides, most of the group 2 oxides, and a number of d-metal oxides.

The Zinc Blende Lattice. Zinc blende is the common name for one crystal modification of the common mineral, ZnS. The other form is known as wurtzite. Wurtzite consists of an hcp array of S2- ions with Zn2+ ions in one-half of the tetrahedral holes; it will not concern us further here, because we are restricting our attention to cubic lattices. Zinc blende consists of a ccp array of S2-, with Zn2+ in half of the tetrahedral holes. The lattice is shown in Figure 7-10b. Since there are 8 tetrahedral holes in the sulfide ion lattice, but only 4 S2- ions, only half of the holes are used to accommodate 4 Zn2+ ions. This is consistent with the 1:1 stoichiometry of the compound. The zinc ions occupy tetrahedral holes so that they can be as far apart from one another as possible to minimize repulsions between positive ions. If the first Zn2+ ion occupies the hole at (1/4,3/4,1/4), the other three will be in holes at (3/4,1/4,1/4), (1/4,1/4,3/4), and (3/4,3/4,3/4). You should locate these holes and verify that the Zn2+ ions describe a tetrahedron. The cation and anion coordination numbers are both 4 in zincblende. As for NaCl, there are 4 formula units per unit cell. Other ionic compounds that adopt the zinc blende lattice include BeS, CdS, HgS, and AgI.

The Calcium Fluoride (fluorite) Lattice. This compound has formula CaF2, and exhibits the lattice shown in Figure 7-10c. The Ca2+ ion is virtually the same size as the F- ion, one of those rare situations referred to above, and forms a face-centered cubic lattice. The F- ions fill all of the tetrahedral holes in the cation lattice. Since there are 8 such holes per 4 Ca2+ ions, the stoichiometry is nicely accommodated. The coordination number of the Ca2+ ion is 8, and that of the F- ion is 4. (Note that the number of cations per formula unit multiplied by the cation coordination number is equal to the product of the number of anions per formula unit and the anion coordination number.) There are again 4 formula units per unit cell. The fluorite structure is very common for ionic compounds of 1:2 (or 2:1) stoichiometry.

The Cryolite Lattice. Cryolite is the common name for the mineral, Na3AlF6. Though rare, this substance is of extreme commercial value, because it is used in the process of extraction of aluminum from bauxite ore. Cryolite consists of Na+ cations and AlF63- anions. It adopts a lattice in which the anions are cubic close packed (fcc) and the cations occupy all of the octahedral and tetrahedral holes.

The Diamond Lattice. Diamond is one crystal modification of the element carbon. It is, of course, not ionic, nor even a compound. It is dealt with here because it, too, is based on a fcc unit cell. The diamond lattice is just like the zinc blende lattice, with carbon atoms in place of both Zn2+ and S2- ions. We therefore describe it as a fcc array of carbon atoms with carbon atoms in one-half the tetrahedral holes. The coordination number of each carbon atom in diamond is 4. The lattice is pictured in Figure 7-10d.

The Cesium Chloride Lattice. The cesium chloride lattice is shown in Figure 7-10e. It is a more open lattice than those considered thus far. This lattice consists of a simple cubic array of chloride ions, with Cs+ ions in cubic holes. A cubic hole is the space at the body center of a simple cubic unit cell. There is 1 formula unit of CsCl per unit cell, and the coordination number of both cation and anion is 8. Ionic compounds that adopt the CsCl lattice include CsBr, CsI, the halides of thallium(I), and NH4Cl.

Radius Ratio Rules. Compounds of 1:1 stoichiometry commonly adopt one of three lattices discussed above: the zinc blende lattice, the rock salt lattice, or the cesium chloride lattice. Of interest now is whether or not we can predict, or at least understand, why a particular compound favors one lattice over the others. A number of factors are important in choice of lattice, but one of these dominates in importance: the relative sizes of cation and anion. Because we view the cation as occupying holes in the anion lattice, we begin by calculating the sizes of the various hole types in relation to the size of the anion involved. The most straightforward calculation is for the octahedral hole, which is shown in Figure 7-11. The figure shows one face of the fcc unit cell, with the anions just touching along a face diagonal of the cube. There is an octahedral hole at the center of each of the 4 cube edges shown; we focus on the right edge. The size of the octahedral hole will be taken as the distance from the edge center to, say, the top right anion, along the cube edge. (The distances from the hole to the bottom right anion and to the face-centered anion are the same.) We can express the right edge length as a sum of anion and hole radii as in equation 7-4-1:

(7-4-1): a = 2r- + 2r+

Here r- is the anion radius, r+ the hole radius. (The + designation is used in anticipation of putting a cation in the hole.) Similarly, we express the length of the face diagonal (fd) in terms of anion radii. This is easy, since the anions touch along this line:

(7-4-2): fd = 4r-

Finally, we relate the fd and edge length via the Pythagorean Theorem:

(7-4-3): (fd)2 = 2a2

Take the square root of both sides of 7-4-3:

(7-4-4): fd = a(2)1/2

Substitute 7-4-4 in 7-4-2 and rearrange for a:

(7-4-5): a = 2(2)1/2 r-

Substitute for a in 7-4-1 and solve for r+:

(7-4-6): r+ = ((2)1/2 - 1) r- = 0.414 r-

The octahedral hole is somewhat less than half as large as the anions that create the hole. By similar geometric methods, equations 7-4-7 and 7-4-8, for tetrahedral and cubic holes respectively, can be developed:

(7-4-7): r+ = 0.224 r-
(7-4-8): r+ = 0.732 r-

You should try to obtain these expressions, making use of right triangles and Pythagorean relationships. (In addition to equation 7-4-4, which relates the face diagonal and edge length of a cube, equation 7-4-9 relates the cube body diagonal (bd) to the edge length:

(7-4-9): bd = (3)1/2 a

According to Equations 7-4-6 to 7-4-8, a tetrahedral hole is about 1/2 the size of an octahedral hole and about 1/3 the size of a cubic hole. Only cations that are relatively small compared with the anion can be accommodated in tetrahedral holes. As cation size increases relative to anion size, the cation will prefer a larger hole, with a larger coordination number of anions. Equations 7-4-6 to 7-4-8 are the basis for the following radius-ratio rules.

  1. If the ratio of the cation and anion radii (r+/r-) lies in the range between 0.224 (equation 7-4-7) and 0.414 (equation 7-4-6), the cation prefers to occupy a tetrahedral hole.
  2. If r+/r- lies in the range between 0.414 and 0.732, the cation prefers an octahedral hole.
  3. If r+/r- exceeds 0.732, the cation prefers a cubic hole.

These rules make good sense. If r+/r- < 0.224, the cation is too small for the tetrahedral hole. The anions will be touching, but will not be touching cations. Since it is the cation-anion attractions that stabilize the lattice, whereas anion-anion or cation-cation repulsions tend to destabilize it, this is not a favorable situation. When r+/r- = 0.224, the cation just fits the tetrahedral hole. Although anions are still touching, they now touch cations as well. This is a stable arrangement. As r+/r- becomes > 0.224, the cation is too big for the tetrahedral hole and starts to push the anions apart. The cation-anion attractions are maintained, because cations and anions still touch, but now anion-anion repulsions are decreased because the anions no longer touch. This is an even better arrangement than the perfect fit. This situation continues to improve until r+/r- becomes equal to 0.414. At this value, the cation prefers to occupy an octahedral hole because it can increase its coordination number without increasing anion-anion repulsion to any significant extent. The number of cation-anion attractive interactions increases from 4 to 6, giving a more stable lattice. As r+/r- increases between 0.414 and 0.732, cations in octahedral holes maintain their contact with 6 anions while pushing the anions further and further apart, decreasing repulsions between them. Throughout this range of r+/r-, occupation of an octahedral hole becomes increasingly favorable until the radius ratio becomes 0.732. The cation can now increase its coordination number to 8 anions, again increasing the number of attractive cation-anion interactions. The radius ratio rules are summarized in Table 7-3. Radii for common cations and anions are given in Table 7-4.

Table 7-3: Radius Ratio Rules
Range of r+/r- Lattice Type Example
0.224-0.414 ZnS ZnS
0.414-0.732 NaCl KCl
> 0.732 CsCl CsCl

Ion radii and the radius ratio rules can be used to predict the probable choice of lattice for an ionic compound.


Example 7-1. The radii of the Na+ ion and the Br- ion are 116 and 182 pm, respectively. Which of the three 1:1 lattices will be adopted by NaBr?
Solution. Calculate r+/r-:

r+/r- = 116/182 = 0.64

This falls between 0.414 and 0.732. The cation will prefer the octahedral hole. NaBr should therefore have the rock salt structure. It does.


The radius ratio rules provide useful guidelines to the lattice structures of ionic compounds. It is important to realize, however, that they are not hard and fast; the predictions made using the rules are often wrong. For example, in the zincblende form of ZnS, Zn2+ ions occupy tetrahedral holes in a fcc lattice of S2- ions. In fact, this compound is the prototype for this type of structure. However, the radius ratio rules predict that the Zn2+ ions should occupy octahedral holes! The radius ratio rules fail frequently because the relative size of the ions is not the only factor that is important in determining structure. Of particular importance is the tendency toward at least some covalent bonding between cation and anion. Covalency involves directional requirements, as we have seen in Chapter 3, and may influence the type of hole preferred by the cation.

7-5 X-Ray Crystallography. In 1912, von Laue predicted that crystals could serve as diffraction gratings if the wavelength of radiation used were approximately the same as the distance between planes of atoms in the crystals. This spacing was known to be on the order of 1 * 10-10 m, corresponding to wavelengths in the x-ray region of the electromagnetic spectrum. Subsequently, WH Bragg discovered that when X-rays of appropriate wavelength are directed onto the smooth face of a crystal, the X-rays are diffracted and a pattern characteristic of the lattice structure of the crystal is obtained. This observation is the foundation for the modern technique of X-ray Crystallography, the most powerful method that we have for determining molecular structure. Like the types of spectroscopy discussed in Chapter 4, x-ray crystallography is based on the interaction of light with matter. Unlike spectroscopy, however, the light is not absorbed by the matter in this case. Instead, it is scattered from the atoms of the crystal in a very definite way. Analysis of the scattering enables the arrangements of atoms within the crystal to be deduced. The analysis of x-ray diffraction patterns is complex, even for materials of simple structure. However, thanks to the efforts of many crystallographers over many years, it is today done quite quickly and routinely. Once a suitable crystal of a substance is grown, the modern crystallographer is able to mount it in a fully automated x-ray diffractometer, the instrument that is used for the x-ray diffraction study. The diffractometer automatically obtains all of the necessary data for the determination of the lattice structure. The data are then fed to one of a number of computer programs that have been developed and refined for the analysis of such data, and the structure is "solved." The output of the program reveals not only the lattice structure, but also the locations of all of the atoms of a molecule of the substance. Structures of substances ranging from simple salts or molecular substances to huge complex proteins and enzymes can be determined using this approach. The structures of two molecules, obtained from x-ray diffraction data, are shown in Figure 7-12. Figure 7-12a shows the arrangement of atoms in the relatively simple molecule, phthalocyanine. Figure 7-12b shows part of the structure of the protein, hemoglobin, which has the function of carrying oxygen from the lungs to the tissues in living organisms.

We now look at a couple of examples of simple applications of crystallographic data.


Example 7-2. Aluminum crystallizes in a cubic lattice. X-ray diffraction reveals that the edge length of the unit cell is 405.0 pm. The density of Al is 2.7 g/cm3. Which cubic lattice does Al adopt?
Solution. We adopt the following strategy:

unit cell mass ---> Number Al atoms per uc ---> lattice type

We can obtain the unit cell mass from the density by calculating unit cell volume from the cube edge length.

Volume of uc = a3 = (4.05 * 10-8 cm)3
= 6.643 * 10-23 cm3/uc

Mass of unit cell = r x Volume uc = 2.7 g/cm3 * 6.643 * 10-23 cm3/uc
= 1.794 * 10-22 g

Moles Al per uc = mass/molar mass = 6.647 * 10-24 moles Al

Atoms Al per uc = 6.647 * 10-24 moles * 6.022 * 1023 atoms/mol
= 4.0

The unit cell is face-centered cubic. This example shows the connection between macroscopic and microscopic properties. Note that in a reversal of the above procedure, knowledge of the type of cubic lattice could be used to experimentally measure Avogadro's number, No.


Example 7-3. In early nuclear reactors, fuel rods were made of uranium metal, with a cubic close packed lattice. These suffered from the problem that the fuel rods frequently burst out of their protective sheaths, which could obviously cause major safety problems. Why did this occur, and what could be done to prevent it?
Solution. Nuclear fission (the spontaneous breakup of a radioactive nucleus) produces 2 nuclei for each one that decays. One possible decay route is shown below:

235U ---> 90Sr + 143Xe + 2 1n

In the ccp U lattice, the holes are too small to accommodate these extra nuclei. The resulting expansion of the lattice caused bursting of the protective sheaths. The solution of the problem was to make the fuel rods from UO2 (yellow cake), which consists of ccp U atoms with O atoms in the tetrahedral sites. The larger octahedral holes in this lattice were able to accommodate the fission fragments without expansion.

The Bragg Equation. Figure 7-13 shows two planes of atoms in a crystal. A source of X-rays at the left emits radiation, two beams of which are shown impinging on the planes of atoms and scattering off to the right, where they are "seen" at a detector. The angles made by the incident beam and the deflected beam with the planes of atoms are the same. A diffraction pattern is a series of alternating light and dark lines (sometimes spots), as shown in Figure 7-14. The white spots result from the arrival at the detector of reflected beams that are all in-phase. That is, their waves are matched up peak for peak and trough for trough. Waves that are in phase add together to give a wave with larger peaks and larger troughs -- i.e., more intensity. This is called constructive interference. The white spots in the diffraction pattern correspond to deflection angles that give in-phase waves. Similarly, the dark areas correspond to angles that give out-of-phase waves. Here the peak of one wave arrives at the detector at the same time as a trough of a second wave arrives. Such waves cancel each other, and very little or no signal is seen at the detector. This is called destructive interference.

The criterion that must be met in order that two waves arrive at the detector in phase is that the difference in the distances travelled by them must correspond to an integer number of wavelengths. This will give the maximum intensity at the detector. If the difference in distance travelled corresponds to a half-integral number of wavelengths, cancellation will occur, and there will be no intensity registered at the detector. The difference in the distances travelled by the two deflected beams in Figure 7-13 is 2x. Since x is related to the angle of incidence (deflection) and to the separation of the planes of scattering atoms as x = dsinq, and since 2x must be an integer number, n, of wavelengths in order to maximize intensity, equation 7-5-1 results.

(7-5-1): n = 2dsinq

This is known as the Bragg equation. It is the basis of modern x-ray crystallography. It shows that if we measure the angles at which constructive interference occurs, we can calculate d, the spacing between atom planes in the crystal.


Example 7-4. A set of planes in a crystal is separated by 250 pm (2.50 * 10-10 m). What is the maximum x-ray wavelength that could be used to measure this distance?
Solution. The largest scattering angle possible is 90o (Figure 7-13). Since sin 90o = 1, n = 2d = 5.00 * 10-10 m. q is largest for n = 1, giving = 5.00 * 10-10 m. A wavelength longer than this cannot fit even once into the distance, 2d, so cannot be used to measure the distance. X-rays with less than 5.00 * 10-10 m must be used.

Lattice Imperfections. Perfect crystals, in which every atom or molecule is in place, are an idealization, like the ideal gas. Real crystals contain fairly large numbers of imperfections, called defects. Three major types of defect are called point defects, line defects, and plane defects. Common point defects in simple lattices (in which all lattice points are identical) are vacancies (in which an atom is missing); atoms out of place, occupying what is normally a hole in the lattice; substitutional impurities, in which foreign atoms occupy occasional sites; and interstitial impurities, in which foreign atoms occupy holes in the lattice. Ionic (compound) lattices exhibit Schottky defects, in which a cation or anion is missing; and Frenkel defects, in which an ion leaves its site and enters a lattice hole. Line defects result from missing rows of atoms. Particularly common are edge defects, in which an extra partial plane of atoms is inserted near an edge of the crystal. This is illustrated in Figure 7-15.


Example 7-5. Bending a paper clip. Take an ordinary paper clip and try to straighten it out. Are you able to get it perfectly straight in the regions where it was bent? Why not? Now try to rebend the paper clip to its original shape. Are you able to do it? Why not?
Solution. When the clip was manufactured, it was annealed following bending. Annealing is the process of heating the clip to soften it enough that atoms can diffuse into the gaps left in the bending process. In other words, annealing produces edge defects. When you attempt to straighten the clip, the edge defects prevent the alignment of atomic planes, so complete straightening is impossible. However, straightening the clip introduces new edge defects on what was originally the inside curve of the bend, so that when you attempt to rebend the clip, the new defects interfere.

Plane defects in crystals occur at the interface between two differently-oriented crystalline regions. As might be expected, such defects greatly diminish the resistance of the crystal to fracture.

Crystal defects are important for a number of reasons. A problem with defects, especially of the line and plane type, is that they diminish the structural strength of the solid. Annealing is used to offset this. Rusting of iron often occurs at defect sites on the surface, which may result when the metal is worked or penetrated by a rivet or bolt. However, defects can be advantageous as well. Defects are sites of high energy (i.e., relative instability). Reactions that take place at the surfaces of crystals often take place at these sites. This is involved in heterogeneous catalysis by solid materials, which is used on a huge industrial scale in our economy. Electronic components for modern computers require the reliable production of silicon that has a purity of 99.999999%. Silicon crystals of such purity contain very few defects, which would diminish their functional effectiveness. Silicon of this purity is produced by a process called zone refining. In this method, a bar of silicon of high purity is passed very slowly past a very narrow heating element, which melts a narrow region of the silicon bar. Impurities in the silicon dissolve in the melted part, and follow along with the melted region as the bar moves past the heating element. Eventually the fused region, containing all of the impurities, reaches the end of the bar, where it is solidified and cut off. The remaining silicon has the purity required in the manufacture of electronic components.

7-6 Properties of Liquids. We take up the liquid phase last because it is the most complex and least understood of the phases of matter. Liquids have properties intermediate between those of the solid and gas phases:

These properties are readily observed visually. Several other properties of liquids are more subtle:

We begin with a molecular model for the liquid phase, based on its observable properties.

7-7 Model of the Liquid Phase. The small molar volume and incompressibility of liquids suggest that the molecules are close together. However, flow and the lack of a fixed shape indicate molecular mobility, despite the closeness. We propose the model in Figure 7-16 to explain this.

A feature of the model that is not evident from the figure is that the molecules are in constant motion, as in the gas and solid, colliding frequently with neighbors. As for the gas and the solid phases, the kinetic energies of the particles of the liquid are described by the Maxwell-Boltzmann distribution, and the average KE per particle is 3/2 kT. Each molecule is fairly closely surrounded by other molecules, but there are definite gaps (holes) in the structure. Molecules use these gaps to slip and slide easily past one another, which manifests macroscopically in the ability to flow. There are regions in the liquid that are quite ordered, similar to the solid, but the regions are constantly shifting position as molecules move to close gaps and open new ones. The gaps prevent the order from extending over long distances. Thus liquids have short-range order, in contrast to the long-range order of crystalline solids. The gaps allow diffusion of a molecule through the liquid, but frequent collisions with neighboring molecules makes diffusion slow. Contrast this with the rapid diffusion in the gas phase, where a molecule travels a long distance between collisions; and the extremely slow diffusion of the solid phase, where a molecule is locked into its lattice location. Diffusion rates in solid, liquid, and gas are best understood in terms of the mean free path, the average distance travelled by a molecule between collisions. The mean free path in the solid is virtually zero. In the liquid, it is a fraction of the molecular diameter. But in the gas, the mean free path may be 10-100 molecular diameters, depending on gas pressure.

7-8 Interesting Physical Properties. Diffuse X-ray Diffraction Patterns. Figure 7-16 indicates regions of short-range order in the liquid. Within these regions, molecules are almost close-packed. However, the gaps allow fluctuation in the distance of separation of the molecules. The nearly close-packed regions of the liquid diffract x-rays according to the Bragg equation, n = 2dsintheta . Fluctuations in d, a consequence of the gaps, cause fluctuations in theta, giving x-ray patterns that are diffuse, or smeared out.

Surface Tension. This is a consequence of intermolecular forces. Consider a molecule at the surface of a liquid (Figure 7-17). In contrast to a molecule in the bulk, which is attracted equally from all directions, the surface molecule suffers an imbalance of forces because there are no molecules above it. The net force on it is directed toward the interior of the liquid. This pulling in of the liquid surface is called surface tension, symbolized z. The surface molecule has higher potential energy than a bulk molecule because of the lesser force on it. To minimize energy, the liquid adopts a shape that minimizes the number of surface molecules. The shape with the smallest surface area-to-volume ratio is the sphere. Thus surface tension explains the spherical shape of raindrops and water droplets on a freshly waxed car: the spherical shape minimizes potential energy. It is a safe prediction that manufacture of ball bearings will someday be carried out in space, where gravitational forces will not distort the sphericity that surface tension imposes on a drop of molten metal.

As temperature increases, molecular kinetic energy can overcome the attraction of intermolecular forces, and surface tension decreases according to equation 7-8-1.

(7-8-1): z = a - bT

The constants a and b are characteristic of a particular liquid. For ethanol, a = 24.05 * 10-3 J/m2 and b = 0.0832 * 10-3 J/m2-oC. The units of surface tension are energy per area, consistent with the relation between potential energy and surface area discussed above.

A drop of water on a clean glass surface spreads and flattens, forming a film on the surface. This phenomenon of spreading and flattening is called wetting. Wetting occurs because forces between the surface molecules and molecules of the liquid exceed the intermolecular forces within the liquid. The liquid spreads to maximize the area of contact with the surface. Forces between unlike molecules are called adhesive forces; those between like molecules are called cohesive. Wetting occurs because adhesive forces exceed cohesive forces. If the reverse is true, a liquid will not wet a surface, but instead will sit on the surface in spherical droplets to minimize the area of contact with the surface. One visible consequence of wetting is the formation of a meniscus at the top of a column of liquid in a tube, such as a buret or pipet. Figure 7-18 shows the menisci formed by water and mercury. The adhesive forces between the glass surface and water molecules are stronger than the cohesive forces between water molecules. Consequently, water achieves a maximum surface area of contact with the glass by bending up at the edges and sagging in the center. The reverse is true for mercury, which minimizes the surface area of contact with the glass and forms a spherical surface shape to maximize cohesive forces.

The phenomenon of capillary action results from competition between surface tension and wetting. When one end of a capillary tube (a glass tube with inner diameter of approximately 1mm) is submerged in a liquid, a difference in height develops between the levels of liquid in the capillary and in the bulk liquid. A liquid that wets glass rises in the tube; a liquid that does not wet glass falls. Figure 7-19 shows this behavior, called capillary action, for water and mercury. As water wets the inside wall of the capillary, the surface area of the water is increased. However, surface tension tends to minimize the number of molecules at the surface, and causes the water to move up the tube, against the force of gravity, to offset the surface area increase resulting from wetting. The result is a capillary rise, as in Figure 7-19a. For mercury, adhesive forces are weak, so the level in the capillary falls to minimize contact with the glass surface, and the surface of mercury in the capillary is pulled into a sphere by the strong cohesive forces. The result is capillary fall, as in Figure 7-19b. Capillary action is important in the operation of blotters and sponges and is the mechanism by which water penetrates soils.

Viscosity. The viscosity of a liquid is a measure of its resistance to flow. A mobile liquid is one that flows readily (e.g., water). A liquid that flows with difficulty (e.g., molasses) is said to be viscous. Liquids composed of small, roughly spherical molecules are mobile, because the molecules can roll over one another smoothly, like ball bearings. Viscous liquids are composed of long, string-like molecules that easily become tangled during flow. For example, the molecules of high viscosity motor oils consist of long chains of carbon atoms that intertwine and impede flow of the liquid. An extreme example of molecular tangling occurs in silly putty, a polymer that appears to be a solid but is in reality a liquid in which molecular tangling is so severe that hours are required for the putty to spontaneously change shape to that of the container. Like surface tension, viscosity decreases (flow becomes easier) as temperature increases due to more rapid thermal motion of the molecules.

Vapor Pressure. Vapor pressure is the most fascinating physical property of liquids. It provides remarkable insight into the laws governing the behavior of physical systems. We begin with a purely experimental approach to the phenomenon. Consider the 0.500-L box shown in the upper left corner of Figure 7-20. The box is attached to a closed-end manometer and is connected via a stopcock to a vacuum pump. A thermometer is inserted in the box and there is a septum-capped port (opening) that can be penetrated with a syringe needle. The stopcock to the vacuum pump is opened and the air is pumped out until the two arms of the manometer are at the same level. The stopcock is then closed. This is the initial situation, state 1, shown at the top left of the figure. Several experiments will now be performed in sequence.

Expt 1. The box is thermostatted at 25oC, and 5.000 g (approximately 5 mL) of water is injected through the septum by syringe. We observe that a pressure develops within the box. Over a brief time (perhaps 1 minute) the pressure rises to 23.8 torr and remains there. The change of pressure with time is as shown in Figure 7-21a. Note that the rate of change of the pressure (the slope of the curve) is largest just after injection of the liquid, and gradually decreases as the pressure smoothly approaches the final value. There is still liquid at the bottom of the box. This is the situation at state 2 in Figure 7-20. We conclude that some water has evaporated to form water vapor, which is responsible for the observed pressure. Since the pressure is caused by the vapor formed by evaporation of liquid, it is called the vapor pressure, Pvap. Assuming that water vapor behaves as an ideal gas, we can use the ideal gas law to calculate the moles of water in the vapor phase. The result is 6.40 x 10-4 moles, or 0.0115 g, of water in the vapor phase.

Expt 2. With T = 25oC and the vapor pressure steady at 23.8 torr, we inject 5.000 g more liquid water to the box. This time we observe no change in the pressure registered at the manometer. This is the situation at state 3 of the diagram. We conclude that the value of Pvap does not depend on how much liquid water is in the box.

Expt 3. This experiment will be done using the box in Figure 7-22. It is identical in all respects to the first box except that its volume is 1.00 L. In state 1, Figure 7-22, the 1.00-L box has been evacuated by pumping until the Hg level in both arms of the manometer is the same. With the box at 25oC, we inject 5.000 g H2O. As for Expt 1, the pressure rises over several minutes to a final value of 23.8 torr, according to a curve like that in Figure 7-21a. The moles of water in the vapor phase must be twice as great as for the original box, because the volume is twice as great. But the pressure is the same. We conclude that Pvap is independent of the container size.

Expt 4. Return to state 3 in Figure 7-20. The box contains 6.40 * 10-4 moles of water vapor, exerting a vapor pressure of 23.8 torr, and a little less than 10.00 g liquid water, all at 25oC. We now quickly raise the temperature of the box to 35oC. Over a period of a few minutes, the pressure rises to 42.18 torr. The change of pressure with time is as shown in Figure 7-21b. This is the situation in state 4 of Figure 7-20. By drawing off the liquid and weighing it, we find that 9.980 g of liquid remains. We conclude that raising the temperature has caused more liquid to evaporate, giving a larger Pvap.

Expt 5. Beginning with state 4, we raise T to 45oC. The pressure rises again, levelling off at 71.88 torr. Change of pressure with time follows a curve like that in Figure 7-21b. This is the situation at state 5.

Expt 6. Beginning with state 5, we lower T to 25oC. Over several minutes, pressure decreases to 23.8 torr, following a time curve like that in Figure 7-21c. Simultaneously, some water vapor condenses to liquid, as we ascertain by weighing the liquid. We are now back to state 3.

Experiments 4-6 tell us that Pvap increases with temperature. Moreover, the increase is non-linear: the increase between 35 and 45o (29.7 torr) is much greater than that between 25 and 35o (18.38 torr). We will do one last experiment using water.

Expt 7. Remove all liquid water from the 0.500-L box, and pump out all water vapor. The box is empty with equal levels of Hg in the manometer arms. This is state 6. We inject 0.00700 g H2O into the box. This time, the pressure increases to a final value of 14.4 torr and all of the liquid disappears. We conclude that after injection, evaporation occurs as the water attempts to achieve a pressure of 23.8 torr, the characteristic Pvap at 25oC. However, before this pressure can be reached, all of the water has evaporated. The amount of liquid injected was insufficient to achieve the characteristic Pvap. The minimum quantity of water needed to achieve P = 23.8 torr was found in Expt 1 to be 0.0115 g. Any amount less than this will evaporate entirely, giving a final pressure equal to the value calculated using the injected number of moles in the ideal gas law.

Expt 8. Empty the 0.500-L box and inject 5.000 g of a different liquid, say ethanol (ethyl alcohol, C2H6O) at 25oC. We observe the same pattern of behavior as with water, except that the final vapor pressure is 54.24 torr. We conclude that all liquids have a tendency to evaporate to some extent in a closed container, and establish a vapor pressure that depends only on T and the identity of the liquid. The vapor pressure phenomenon is a universal property of liquids.

Before summarizing the results of our thought experiments, we deal with a related matter. Suppose we put 10 mL of water in a beaker and leave it open to the atmosphere. Over a few days, the water evaporates completely. (Acetone and ethyl alcohol would completely evaporate in a few hours; diethyl ether in a few minutes.) Why? This situation differs from that involving the water in the box because the beaker system is not closed -- that is, there is no fixed boundary on top of the beaker to contain the water vapor. Water in the beaker evaporates in an attempt to establish the characteristic Pvap (23.8 torr at 25oC). Water vapor develops over the surface of liquid in the beaker, but begins to diffuse away because there is no confining boundary on top. This diffusion, aided by air currents in the room, moves vapor molecules away from the space over the beaker. This reduces the local water vapor pressure, and more water evaporates to make up for the loss. This process continues until all liquid is gone. We will see in Chapter 10 that this is a situation in which equilibrium cannot be attained between liquid and vapor.

The results of our experiments with vapor pressure can be summarized as follows:

The thought experiments, which can be verified readily in the laboratory, provide much descriptive information about the vapor pressure phenomenon. But the questions of why liquids tend to evaporate, and why they establish characteristic vapor pressures, are unanswered by the experiments. Answers to these questions are forthcoming in Chapter 10, where we attempt a molecular (microscopic) interpretation of vapor pressure and introduce the concept of phase equilibrium.

7-9 The Dissolving Power of Liquids. Liquids are invaluable to chemists as solvents. They are able to accommodate molecules of solute readily within the gaps in their structures to produce solutions that are themselves liquids. Such solutions are readily poured or otherwise transferred from one container to another; are conveniently measured out quantitatively using volumetric apparatus; and provide a uniform medium in which a chemical reaction can occur. The unique ability of liquids to act as solvents is due to their dissolving power -- the molecular motion of the liquid dislodges ions or molecules from the surface of crystals of a solid, then accommodates the separated ions or molecules in gaps. In this section we briefly discuss the dynamics of the dissolution process at the molecular level.

Figure 7-23 shows a microscopic view of two adjoining faces of a crystal of NaCl. A Na+ ion on a face of the crystal differs from those in the interior because it experiences forces of attraction from only 5 nearest-neighbor Cl- ions, rather than 6. Facial Cl- ions are in a similar situation. Such ions are therefore in shallower potential wells than ions in the interior of the crystal and should be more easily dislodged. Edge ions, such as the Na+ ion labelled "a" in the figure, are attracted by only 4 nearest neighbors, and should be even more readily dislodged than face ions. Water molecules from the solvent continually bump against the faces and edges of the solid crystal. As they approach the crystal, their dipoles become oriented due to interactions with the face ion that they approach. Water molecules approaching cations have the O atom oriented toward the crystal face; those approaching anions put a H atom forward. When a rapidly moving water molecule strikes a face or edge ion, sufficient kinetic energy can be transferred to dislodge the ion, which moves away from the crystal into a gap in the solvent structure (e.g., Na+ ion "b" in the figure). It becomes immediately surrounded by a cage of water molecules, with oxygen atoms nearest the Na+ ion to maximize ion-dipole forces. The solvent cage insulates it from the crystal and makes return to the crystal face unlikely. Since edge ions are more readily dislodged than facial ions, and because they can be approached by solvent from several directions, dissolution occurs more rapidly at crystal edges than faces. The initially sharp edges of a dissolving crystal become rounded as dissolution proceeds.

7-10 Phase Transformations. A phase transformation is a conversion of a pure substance from one phase to another. There are 6 types, each involving two of the three phases:

The phase changes are illustrated in Figure 7-24. Since the solid and liquid phases have small molar volumes due to close packing of molecules, they are referred to as condensed phases.

The two variables that determine the phase in which a substance exists are the temperature and the pressure to which the substance is subjected. Phase transformations may therefore be accomplished in two limiting ways: by changing the temperature of the substance while keeping it under constant pressure; or by changing the pressure on the substance while keeping it at constant temperature. The first approach is the more familiar of the two. We discuss that now and postpone a discussion of pressure-induced phase changes to Chapter 10.

We make three assertions about temperature-induced phase changes:

  1. At a constant pressure, a phase transformation occurs at a constant temperature that is characteristic of the substance.
  2. To convert a given amount of substance from solid to liquid to gas requires addition of energy. This is usually accomplished by adding heat. The energies of the phases are therefore in the following order:

    Es < El < Eg

  3. Conversion from solid to liquid to gas decreases the order (degree of organization) of the substance.

To clarify these assertions, we take a thought experimental approach to the phase changes of a substance, much as we did for vapor pressure. The results of these thought experiments can be readily verified in the laboratory. We start with a quantity of solid water (ice) in a syringe, as shown in Figure 7-25, state 1. The initial temperature of the ice and the inside of the syringe is -10oC. We plan to slowly add energy to the ice by controlled heating, which can be done by submerging the syringe in a large antifreeze bath with a hot plate under it. As we slowly add heat, we monitor the temperature of the water in the syringe with a thermometer inserted through an air-tight seal in the syringe. We begin at state 1.

Process 1. Turn on the hot plate and the bath stirrer. As the ice warms slowly from -10 to 0oC, nothing seems to happen. Temperature changes slowly and steadily. When the ice reaches 0oC, it begins to melt, and we see some liquid water form in the syringe. This is state 2 of Figure 7-25.

Process 2. We continue to gently heat the bath. The ice gradually melts, but the temperature of the ice-water mixture remains constant at 0oC, despite the addition of heat. The temperature of the bath rises above 0oC, but as long as both ice and liquid water are in the syringe, T in the syringe remains at 0o. State 3 is reached when the last ice crystal melts. All water in the syringe is liquid, at T = 0oC.

Process 3. Once the ice has melted, the temperature of the water in the syringe begins to rise. As we slowly warm the bath, the water temperature slowly rises. Nothing happens in the syringe other than a slight increase in the volume of the liquid until T = 100oC. Then water begins to vaporize. This is the situation at state 4.

Process 4. We continue to slowly add heat to the bath and observe that temperature in the syringe remains constant at 100oC as long as both liquid and vapor are present, even though the bath temperature rises above 100oC. We also observe that as vapor forms, it pushes the plunger out. After all liquid has vaporized, at state 5, the temperature of the water in the syringe once more starts to rise.

Process 5. Continue heating the bath to T = 110oC. The temperature of the vapor in the syringe rises from 100o at state 5 to 110o at state 6, while its volume increases according to the ideal gas law. The pressure of the water vapor is maintained constant at 1 atm by movement of the syringe plunger.

These experiments verify the assertions at the beginning of the section. First, phase changes occur at constant T. This is shown by processes 2 and 4. The fixed, characteristic temperature at which a substance melts is called the melting point (s to l) or freezing point (l to s), and is symbolized Tf. The normal melting point is the temperature of melting under a pressure of 1.00 atm. For water, this is 0oC. The characteristic temperature at which a substance boils under 1.0 atm pressure is called the normal boiling point. For water, this is 100oC. Second, process 2 shows that an input of energy is required to convert a solid to liquid. The same is true for conversion of liquid to gas (process 4). For each of these processes, the change in energy, DE (Efinal - Einitial) is positive. Processes 2 and 4 may be represented as follows:

(7-10-1): H2O(s), 0oC ---> H2O(l), 0oC [DE > 0]
(7-10-2): H2O(l), 100oC ---> H2O(g), 100oC [DE > 0]

The process in equation 7-10-1 occurs at constant T. Therefore the change in kinetic energy of the molecules is 0 (recall that an average molecule has KE of 3kT/2 at temperature T, regardless of phase). DE is therefore a change in potential energy, DPE. Thus the PE of liquid is higher than that of solid water at 0oC. Similarly, the PE of vapor is higher than that of liquid water at 100oC. Combining these conclusions gives equation 7-10-3:

(7-10-3): PEg > PEl > PEs at any T

DPE > 0 for sublimation, fusion, and vaporization; DPE < 0 for the reverse processes. Third, the amount of disorder at the molecular level increases in our sequence of processes (for reasons that will become clear later, we phrase this in terms of disorder rather than order). With addition of thermal energy (heat), the molecules convert from the long-range organized arrangement of the solid to the completely chaotic and random molecular distribution of the gas. Even warming the gas from 100 to 110oC decreases the order, because the volume available for the molecules to bounce around in increases. This conclusion is summarized in equation 7-10-4.

(7-10-4): disorder s < disorder l < disorder g

Phases and the Potential Well, Revisited. In Chapter 6 we discussed the origin of the potential well in terms of electromagnetic forces within or between the fundamental units of substances. We concluded that forces are strongest and potential energy is minimized when fundamental particles form an orderly close-packed arrangement, and we recognized this as the situation in the solid phase (see the first half of this chapter). Heating the solid phase increases the kinetic energies of the particles of the solid phase until they are able to partially overcome the close-packed forces and separate somewhat. At this point, solid melts to liquid. Separation has increased the potential energy of the particles. Further heating eventually gives the particles sufficient kinetic energy that they may escape the forces in the liquid into the space above the liquid; i.e., vaporization occurs. In the vapor phase, molecules are so widely spaced that forces between them are negligible and potential energy has increased to zero (Figure 7-26). We thus see that phase conversion from solid to gas can be viewed in terms of particles climbing out of the potential well. Similarly, in the reverse conversion, particles fall into the potential well. The decrease in potential energy of the molecules is released as heat.

Applications

7-1. In the 2-dimensional lattice, outline a primitive unit cell and a non-primitive unit cell. Assuming that there is an atom at each lattice point, calculate the number of atoms contained within each of your unit cells.

7-2. Draw a simple cubic unit cell; a body-centered cubic unit cell; a face-centered cubic unit cell. Show how to determine the number of atoms per unit cell in each lattice.

7-3. Give the coordinates of the tetrahedral holes in a fcc unit cell.

7-4. Give the coordinates of the octahedral holes in a fcc unit cell.

7-5. Determine the number of atoms per hexagonal unit cell.

7-6. Show that in a face-centered cubic lattice, 74% of the available space is occupied by atoms.

7-7. What fraction of the available volume is occupied by atoms in a simple cubic unit cell? In a body-centered cubic unit cell?

7-8. Calcium crystallizes in a face-centered cubic lattice. The length of the edge of the unit cell is 0.556 nm (1 nm = 10-9 m). What is the radius of a calcium atom?

7-9. Sodium oxide, Na2O, has a lattice consisting of fcc O2- ions with Na+ ions in the tetrahedral holes. Show that there are 4 formula units of the compound per unit cell. Show that the coordination number of Na+ is 4, whereas that of O2- is 8.

7-10. How many carbon atoms are there per unit cell of the diamond structure? What is the coordination number of each C atom?

7-11. A set of planes in a crystal is separated by 645 pm. What is the longest x-ray wavelength that can be used to measure this distance?

7-12. Iron(II) oxide is an example of a non-stoichiometric compound. This means that the ratio of iron to oxygen atoms is non-integral. This situation can occur whenever the cation can exhibit more than one stable oxidation state. In this case, iron exhibits oxidations states +2 and +3. A sample of iron(II) oxide is prepared with the composition Fe0.85O. What fraction of the iron atoms are Fe(III)?

7-13. A crystal of very pure silicon suitable for semiconductor devices contains only 0.00001% defects. Silicon has the same crystal structure as diamond. How many unit cells contain defects in a 1-mole crystal of silicon?

7-14. When a perfectly cubic crystal of sodium chloride, NaCl, is placed in water, it begins to dissolve. As it does so, the corners and edges of the cube become rounded, but the faces remain flat. Explain these observations in atomic/molecular terms.

7-15. A particular set of planes in a crystalline solid has a spacing distance of 4.38 * 10-10 m (0.438 nm). X-rays of wavelength 0.1886 nm are diffracted from this set of planes. At what angles will reflections from these planes be observed?

7-16. Explain each fact:
  1. Motor oil is viscous at 25oC, but flows readily in an operating engine.
  2. Water bugs are denser than water, but can scoot around on its surface.
  3. A glass can be filled "fuller than full" if water is added carefully.
  4. Tap water beads up on a dirty glass but forms a film on a clean glass.
  5. Wet grains of sand clump together.
7-17. Spontaneous mixing of two liquids is more rapid at high than at low T. Explain this observation.

7-18. Relative humidity indicates the water content of the atmosphere. 50% relative humidity at a certain T means that the partial pressure of H2O(v) is 50% of the vapor pressure of water at that T. In which case is the partial pressure of water vapor larger: 50% relative humidity at 25oC, or 40% relative humidity at 30oC? Pvap(H2O) = 23.8 torr at 25oC, 31.8 torr at 30oC.

7-19. 2.0 g liquid H2O is injected into an evacuated box with V = 0.521 L. The pressure in the box rises to a final value of 23.8 torr at 25oC. Calculate
a. Moles of water in the vapor phase.
b. Moles of water in the liquid phase.

7-20. A 0.750-L glass bulb contains a mixture of He(g) and H2O(g) at 22oC and a pressure of 52 torr. A side arm on the bulb is cooled to -195oC and the water vapor is frozen out completely as H2O(s). H2O(s) has a negligible vapor pressure at -195oC. He remains as a gas. The pressure in the bulb is then found to be 18.5 torr. How many moles of He are in the bulb and what was the partial pressure of water at 22oC?

7-21. A gas mixture in a 0.500-L bulb contains N2(g), H2O(g), and CO2(g) at a total pressure of 632 torr at 25oC. CO2(s) has a high vapor pressure (1.013 bar) at dry ice temperature (-78oC) but the vapor pressure of water is negligible at -78oC. Neither CO2(s) nor H2O(s) has a significant vapor pressure at the temperature of liquid nitrogen (-195oC), but N2(l) has a vapor pressure of 1.013 bar (atmospheric pressure) at -195oC. When the bulb sidearm is cooled to -78oC, the pressure falls to 610 torr. When the sidearm is cooled to -195oC, the pressure falls to 18.0 torr. How many moles of each gas are in the bulb?

7-22. Pvap of H2O(l) at 0oC is 4.54 torr. 5.00 x 10-3 g of H2O(l) is frozen at a very low temperature in the bottom of the sidearm tube of a gas sample bulb of volume 1.5 L. All of the air is pumped out and the stopcock is closed. The water at the bottom of the tube is melted by heating it to 0oC. The vessel is kept at 25oC. What phases of H2O will be present at the end of this process, and how much of each will there be? MM(H2O) = 18.015 g/mole.

7-23. 0.0082 g of ethyl alcohol, C2H5OH, is injected into an evacuated 0.493-L bulb at 31oC. Pvap of ethyl alcohol is 74.86 torr at 31oC. What will be the final pressure in the bulb? What mass of liquid ethanol will be present?

7-24. A liquid is in equilibrium with its vapor in a cylinder/piston apparatus. Show graphically how the rates of evaporation and condensation vary with time.

  1. From time t0 to time t1, the system is at equilibrium.
  2. At time t1, T is instantly increased to some higher value. The system returns to equilibrium.
  3. At time t2, the piston is instantly moved down to point B. The system returns to equilibrium in the smaller volume.
  4. At time t3, T is instantly decreased. The system returns to equilibrium.
7-25. 5.00 g of H2O(l) is injected into a 1.50 L box at 35oC. Pvap(H2O) at 35oC = 42.18 torr. Calculate the mass of H2O in the g and l phases after Pvap has been established.

7-26. A liquid is in equilibrium with its vapor in a cylinder/piston apparatus. On the axes below, show how the rates of evaporation and condensation vary with time in response to each of the indicated processes.

  1. From time t0 to time t1, the system is at equilibrium.
  2. At time t1, T is instantly decreased. The system returns to equilibrium.
  3. At time t2, the piston is instantly moved up to point A. The system returns to equilibrium in the larger volume.
  4. At time t3, additional vapor is instantly injected into the cylinder. The system returns to equilibrium.
  5. At time t4, some vapor is instantly withdrawn from the cylinder. The system returns to equilibrium.
7-27. The compound, Na3AlF6, is called cryolite. It is important in the production of aluminum. Which description below is consistent with the stoichiometry of cryolite.

  1. Cubic close packed F- ions with Na+ ions in the octahedral holes and Al3+ in the tetrahedral holes.
  2. Cubic close packed AlF63- ions with Na+ ions in all of the octahedral and tetrahedral holes.
  3. Cubic close packed Na+ and Al3+ ions with F- ions in the octahedral holes.
  4. Cubic close packed Na+ ions with AlF63- ions in the tetrahedral holes.
  5. Simple cubic AlF63- ions with Na+ and Al3+ ions at the cube centers.
7-28. Shown below are the front faces of the cubic unit cells of two materials. Based on the diagrams and the additional information given, determine which of the following statements is true.

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