Solids are substances that tend to have a definite shape indicating an extremely well-organized and regular structure (crystalline structure). The crystalline structure is studied through various means, including the use of x-rays. These studies are used to make predictions relating to the stability of the structure when under stress, the changes that occur with the addition of other substances, and the prediction of the characteristics of other crystalline substances.
The arrangement of the simpler particles in a crystalline array is called a lattice. Every lattice is a three-dimensional stacking of identical building blocks called unit cells. The properties of a crystal, including its overall symmetry, can be understood in terms of the unit cell. Each different unit cell (14 of them) can fill in space by being arranged adjacent to an identical unit with others placed above, below, and in the other adjacent positions (think of stacking boxes in a warehouse). Figure 10-1 contains sketches of three unit cells for your inspection; these are the only unit cells we discuss in detail in this book.
The unit cells in Fig. 10-1 are the only ones having cubic symmetry. The lattice points (corners of the cubes and centers of the edges or faces) represent the centers of the atoms or ions occupying the lattice. The atoms or ions by themselves are not points (displayed as circles), but are three-dimensional objects which usually are in contact with each other. The representations of these objects in Fig. 10-1 are purposely out of scale (shrunken) for simplicity. The length of the cube edge is designated by a. A crystal having any of these three lattices can be thought of as a three-dimensional stack of unit cell cubes, packed face-to-face so that the space occupied by the crystal is completely filled.
Fig. 10-1 Unit cells of symmetry.
The crystal classes that are less symmetrical than the cubic have unit cells that may be thought of as more or less distorted cubes, the opposite faces of which are parallel to each other. Such shapes are generally known as parallelipiped. Crystals with hexagonal symmetry, like snow or ice, have unit cells that are prisms with a vertical axis perpendicular to a rhombus-shaped base, the equal edges of which are at 60° and 120° with respect to each other. A typical hexagonal unit cell is shown in Fig. 10-2. The letters designate the length of each of the edges. Although this doesn’t look like a hexagonal prism, three unit cells joined adjacent to each other will produce the structure.
Fig. 10-2 A hexagonal unit cell.
Keeping in mind that a crystal is composed of many of its unit cells and assuming there is no contamination, the density of a crystal can be computed from the properties of the unit cell. It is necessary to apportion the mass of the crystal among the various unit cells, and then to divide the mass apportioned to one unit cell by the volume of the unit cell. In computing the mass of a unit cell, it is important to assign to the cell only that fraction of an atom which lies completely in that cell. If an atom is shared by cells, the portion of the atom in each unit is credited to that unit cell.
If a unit cell is cubic, we see from Fig. 10-3, since a corner atom is shared by eight unit cells, that
Fig. 10-3 Stack of eight cubic unit cells.
Note that the formula holds whether or not all the atoms belong to the same species and also holds for noncubic cells.
The coordination number of an atom in a crystal is the number of nearest-neighbor atoms. The coordination number is constant for a given lattice [see Problems 10.1(b) and 10.14(d)].
Two simple lattice structures allow a high degree of packing of the atoms in a crystal. The first is called a close-packed structure. It is a structure in which identical spheres (the atoms) occupy the greatest fraction of the total space. It can be achieved by packing on top of each other two-dimensional close-packed layers. In each of the layers, every sphere is surrounded by a regular hexagonal arrangement of six other spheres, as in Fig. 10-4. In the figure, the large circles represent the spheres in one such layer. The squares surround the centers of spheres in the adjoining superimposed layer. If the third and every odd-numbered layer is made up of spheres directly over the spheres in the first layer and if every even-numbered layer is made up of spheres directly over the spheres in the second layer, the structure is a hexagonal close-packed structure. The corresponding unit cell for the centers of the spheres is given in Problem 10-15.
Fig. 10-4
In the case of the second structure allowing a high degree of packing, if the structure is a regular alternation of three kinds of layers, the third being made up of spheres whose centers are enclosed by small circles in Fig. 10-4, the structure is a cubic close-packed structure. The unit cell of this structure is the face-centered cubic. The close-packed layers are perpendicular to a body diagonal of the unit cell [see the dashed triangles in Fig. 10-1(b)].
Both the hexagonal close-packed and cubic close-packed lattices have the same 74 percent filling of space by spheres in contact [see Problem 10.1(d)], and the two lattices also have the same coordination number, which is 12.
There are commonly void spaces (holes) in a crystal that can sometimes admit foreign particles of a smaller size than the hole. An understanding of the geometry of these holes becomes an important consideration as characteristics of the crystal will be affected when a foreign substance is introduced. In the cubic close-packed structure, the two major types of holes are the tetrahedral and the octahedral holes. In Fig. 10-1(b), tetrahedral holes are in the centers of the indicated minicubes of side a/2. Each tetrahedral hole has four nearest-neighbor occupied sites. The octahedral holes are in the body center and on the centers of the edges of the indicated unit cell. Each octahedral hole has six nearest-neighbor occupied sites.
The strength of the forces holding crystalline substances together varies considerably. In molecular crystals, such as CO2 and benzene (both nonpolar, both in the solid state), each molecule is almost independent of all the others and retains practically the same internal geometry (bond lengths and angles) that it has in the gaseous or liquid state. These crystals are held together by van der Waals forces, a very weak intermolecular attractive force. The melting and boiling points for these substances are never very high as compared to the substances described next.
For substances that can undergo hydrogen bonding, the intermolecular forces in the crystal can be great enough to impose a noticeable change in the molecular geometry. Hydrogen bonding is the attraction between the positive charge-carrying hydrogen of a polar bond in one molecule (or part of a molecule) and the negative charge-carrying atom of a polar bond of another molecule (or part of the same molecule—a common condition in proteins). The most important polar bonds that allow hydrogen bonding are those between hydrogen and the most electronegative element: fluorine, oxygen, nitrogen, and chlorine. Water is an example of extensive hydrogen bonding; the angle between the two O—H bonds in the vapor is about 105°, but equals the tetrahedral 109°28′ in the crystal conforming with the crystalline spatial requirements, rather than the molecular arrangement.
In metals, a special type of crystalline force comes into play which is characterized by a largely nondirectional nature. Fixed bond angles do not play a major role in metals and the more stable crystal structures for most elementary metals are those with the densest packing. Important exceptions are the Group IA metals and iron, which have the body-centered cubic structure. In the covalent crystals, like diamond or silicon carbide, the crystal is held together by a three-dimensional network of covalent bonds, the mutual angles of which are determined largely by the covalent bonding requirements of the individual atoms.
The attractive forces operating in ionic crystals are mostly electrostatic in nature, the classical attraction between oppositely charged particles. To avoid the repulsion between similarly charged particles, ionic substances crystallize in structures in which a positive and a negative ion can come within touching distance, while ions of like charge are kept away from each other. In fact, the dimensions of most simple purely ionic crystals can be understood by assuming an ionic radius (Chapter 9) for each ion that is valid for all compounds of that ion and taking the smallest cation-anion separation to be the sum of the ionic radii of the cation and anion. Radii for some elementary ions are listed in Table 10-1.
The forces that bind atoms, ions, and molecules in liquids are the same as in those that result in solids. The difference is that the strengths of these forces are not sufficient to hold the particles in as rigid a structure as is seen in the solid state. Although, just above the melting point, these forces are insufficient to restrain the atoms, ions, or molecules to their lattice positions, in most cases they are strong enough to prevent vaporization.
Liquid metals and salts, especially those that are ionic, are not common except in industrial technology and occur at much higher temperatures than normally encountered. Since the internal forces are so strong, the temperatures required for melting are high and the temperature range for the liquid state is very great as compared with the melting points for covalent substances.
Most familiar liquids consist of molecular substances (water, alcohols, benzene, bromine, etc.) which freeze into molecular solids. The molecules are bound to their neighbors by weak forces, the strongest of which, by far, are hydrogen bonds. Others are known collectively as van der Waals forces. All atoms and molecules are attracted to one another. If a molecule has a permanent dipole moment (Chapter 9), dipole-dipole attraction makes a significant contribution to the weak force. But even in the absence of permanent dipoles, weak forces known as London forces exist. London forces are due to the presence of transient dipoles of very short duration, as explained by Fritz London. Generally, the greater the atomic number of the atoms in contact, the greater the area of contact and the greater the London forces between molecules.
The strength of the weak forces is revealed by the volatility of a substance—the greater the intermolecular attractive forces, the higher the boiling point. An example is Group VIIIA, the noble gases (inert gases). These elements display a steady increase in boiling point as the atomic number increases due to the increase in London forces. Additionally, comparing SiCl4 (B.P. 57.6°C) to PCl3 (B.P. 75.5°C, we see the contribution of the dipole-dipole attractions in the latter case. The contributions of dipole-dipole attractions and hydrogen bonding are very important in comparing organic compounds, as illustrated by C2H6 (ethane’s B.P. –89°C, London forces only), CH3F (methyl fluoride’s B.P. –78°C, dipole), and CH3OH (methyl alcohol’s B.P. 65°C, H bonding). Among the compounds of formula CnH2n+2 the straight-chain compound always has a boiling point higher than any of its branched chair isomers because of the increased area of contact of such a molecule with its surroundings.
10.1. Metallic gold crystallizes in the face-centered cubic lattice. The length of the cubic unit cell [Fig. 10-1(b)] is a = 407.0 pm. (a) What is the closest distance between gold atoms? (b) How many “nearest-neighbors” does each gold atom have at the distance calculated in (a)? (c) What is the density of gold? (d) Prove that the packing factor for gold, the fraction of the total volume occupied by the atoms themselves, is 0.74.
(a) Consider the corner gold atom in Fig. 10-1(b). The closest distance to another corner atom is a. The distance to an atom at the center of a face is one-half the diagonal of that face, as follows:
and, by substitution, the closest distance between atoms is
(b) The problem is to find how many face-centers are equidistant from a corner atom. Point A in Fig. 10-3 may be taken as the reference corner atoms. In that same figure, B is one of the face-center points at the nearest distance to A. In the plane ABD in the figure there are three other points equally close to A: the centers of the squares in the upper right, lower left, and lower right quadrants of the plane, measured around A. Plane ACE, parallel to the plane of the paper, also has points in the centers of each of the squares in the four quadrants around A. Additionally, plane ACF, perpendicular to the plane of the paper, has points in the centers of each of the squares in the four quadrants around A. Adding them up, there are 12 nearest-neighbors in all, the number expected for a close-packed structure.
The same result would have been obtained by counting the nearest-neighbors around B, a face-centered point.
(c) Let us use m for the mass of a single gold atom and the atomic mass of 197.0 g/mol Au. For the face-centered cubic structure, with 8 corners and 6 face-centers,
and
then
The reverse of this type of calculation can be used for a precise determination of Avogadro’s number, provided the lattice dimensions, the density, and the atomic mass are known precisely.
(d) Since the atoms at closest distance are in contact in a close-packed structure, the closest distance between centers calculated in (a), a/
, must equal the sum of the radii of the two spherical atoms, 2r. Going one step further, r = a//2 = a/23/2. From (c), there are four gold atoms per unit cell. Then,
and
Note that the parameter a for the gold unit cell canceled and that the result holds for any cubic close-packed structure. The metallic radius calculated above, r = a/23/2 = 143.9 pm, is different from both an ionic radius and a covalent radius.
10.2. Show that the tetrahedral and octahedral holes in gold are appropriately named. Find the closest distance between an impurity atom and a gold atom if the impurity atom occupies (a) a tetrahedral hole, (b) an octahedral hole. How many holes of each type are there per gold atom?
(a) Examine Fig. 10-1(b) and imagine a hole in the center of the upper left front minicube. This hole is equidistant from the four occupied corners of the minicube, the common distance being half a body diagonal of the minicube, or
These four occupied corners define a regular tetrahedron (see Problem 9.18) with the center of the tetrahedron being the point equidistant from the corners, which we showed as the location of the hole. This justifies the name “tetrahedral hole.” As the unit cell contains 8 tetrahedral holes, one in each minicube, and 4 gold atoms (Problem 10.1), there are 8 ÷ 4 = 2 tetrahedral holes per gold atom.
(b) Now, consider the hole at the center of the unit cell of Fig. 10-1(b). This hole is equidistant from the centers of all six faces of the unit cell, all of which are the nearest occupied sites to the hole. These six points are the vertices of an eight-faced figure; the faces are congruent equilateral triangles (whose edges are face diagonals of the minicube). Such a figure is a regular octahedron and the hole is at its center; “octahedral hole” is correct.
The distance between the hole and the nearest-neighbor atom is a/2. A similar proof can be made for an octahedral hole on the center of an edge of the unit cell in Fig. 10-1(b) if we note that the actual crystal lattice consists of a three-dimensional stack of unit cells, as in Fig. 10-3. Each such edge-center hole is shared by four unit cells and there are 12 edges in a cube, so that the number of octahedral holes per unit cell is
The ratio of octahedral holes to gold atoms is 4:4, or 1:1, which can be simply stated as 1.
There are competing advantages of tetrahedral and octahedral holes for housing impurities or second components of an alloy. If the crystal forces, whatever their nature, depend mostly on interactions between nearest-neighbors, the octahedral hole has the advantage of having more nearest-neighbors with which it interacts (6 instead of 4). However, the tetrahedral hole has a shorter near-neighbor distance 
as opposed to 0.500a), giving it the advantage of a greater potential interaction with any one host atom. An octahedral hole, with its larger nearest-neighbor distance, can accommodate a larger impurity or alloying atom than can a tetrahedral hole, without straining the host lattice.
10.3. CsCl crystallizes in a cubic structure that has a Cl– at each corner and a Cs+ at the center of the unit cell. Use the ionic radii listed in Table 10.1 to predict the lattice constant, a, and compare with the value of a calculated from the observed density of CsCl, 3.97 g/cm3.
Figure 10-5(a) shows a schematic view of the unit cell where the shaded circles are the Cs+ cation and the unshaded circles represent Cl–. The circles are made small with respect to the unit cell length, a, in order to show more clearly the locations of the various ions. Figure 10-5(b) is a more realistic representation of the right triangle ABC showing anion-cation-anion contact along the diagonal AC.
Fig. 10-5
Let us assume that the closest Cs+-to-Cl– distance is the sum of the ionic radius of Cs+ and Cl–, which is 169 + 181 = 350 pm. This distance is one-half the cube diagonal, or a 
/2. Then
The density can be used to calculate a if we count the number of ions of each type per unit cell. The number of assigned Cl– ions per unit cell is one-eighth of the number of corner Cl– ions, or 
(8) = 1.
The only Cs+ in the unit cell is the center Cs+, so that the assigned number of cesium ions is also 1. (This type of assignment of ions or atoms in a compound must always agree with the empirical formula of the compound, as the 1:1 ratio does in this case.) The assigned mass per unit cell is that of 1 formula unit of CsCl by
then,
This value, based on the experimental density, is to be considered the more reliable, since it is based on a measured property of CsCl, while the ionic radii are based on averages over many different compounds. Unit cell dimensions can be measured accurately by x-ray diffraction and from them the theoretical density can be calculated. The measured density is usually lower because most samples that are large enough to measure are not perfect single crystals and contain empty spaces in the form of grain boundaries and various crystalline imperfections.
The CsCl structure is not described as body-centered, since the particle occupying the center is different from the particles occupying the corners of the unit cell. There are two ways of describing the structure. One way is to say that Cs+ occupies the central holes of the Cl– simple cubic lattice. Another way is to say that the structure is made up of two interpenetrating simple cubic lattices, one made up of Cl– and one of Cs+. The Cs+ lattice is displaced from the Cl– lattice along the direction of the unit cell diagonal by one-half the length of the unit cell diagonal.
10.4. The CsCl structure (Fig. 10-5) is observed in alkali halides only when the radius of the cation is sufficiently large to keep its eight nearest-neighbor anions from touching. What minimum value for the ratio of the cation-to-anion radii, r+/r–, is needed to prevent this contact?
In the CsCl structure, the nearest cation-anion distance occurs along the diagonal of the unit cell cube, while the nearest anion-anion distance occurs along a unit cell edge. This relationship is shown in Fig. 10-5(b). In the figure,
If we assume anion-cation contact along AC, then 
= 2(r+ + r–) = a. In the limiting case, where the anions touch along the edge of the unit cell, 2r– = a. Dividing the former equation by the latter,
If the ratio were less than this critical value, anions would touch (increasing repulsive forces). Also, the cation and anion would be separated (decreasing attractive forces). Both effects would tend to make the structure unstable.
10.5. Ice crystallizes in a hexagonal lattice. At the low temperature at which the structure was determined, the lattice constants were a = 453 pm and c = 741 pm (Fig. 10-2). How many water molecules are contained in a unit cell?
The volume, V, of the unit cell in Fig. 10-2 is
Although the density of ice at the experimental temperature is not stated, it could not be very different from the value at 0°C, which is 0.92 g/cm3.
This is close to 4 times the molecular mass of water; we conclude that there are four molecules of water per unit cell. The discrepancy between 73 u and the actual mass of four molecules, 72 u, is undoubtedly due to the uncertainty in the density at the experimental temperature.
10.6. BaTiO3 crystallizes in the perovskite structure. This structure may be described as a barium-oxygen face-centered cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying the face-centers, and titanium ions occupying the centers of the unit cells. (a) If titanium is described as occupying holes in the Ba-O lattice, what type of hole does it occupy? (b) What fraction of the holes of this type does it occupy? (c) Suggest a reason why it occupies those holes of this type but not the other holes of the same type?
(a) These are octahedral holes.
(b) The octahedral holes at the centers of the unit cells constitute just one-fourth of all the octahedral holes in a face-centered cubic lattice (see Problem 10.2).
(c) An octahedral hole at the center of a unit cell has six nearest-neighbor oxide ions and is occupied by a titanium ion. The other octahedral holes are located at the centers of the edges of the unit cell and have six nearest-neighbors each, as is the case with any octahedral hole. However, two of the six neighbors are barium ions (at the unit-cell corners terminating in a given edge) and four are oxide ions. The proximity of two cations, Ba2+ and Ti4+, would be electrostatically unfavorable.
10.7. The melting point of quartz, a crystalline form of SiO2, is 1610°C, and the sublimation point of CO2 is –79°C. How similar do you expect the crystal structures of these two substances to be?
The big difference in melting points suggests a difference in type of crystal binding. The intermolecular forces in solid CO2 must be very low to be overcome by a low-temperature sublimation. CO2 is actually a molecular lattice held together only by the weak van der Waals forces between discrete CO2 molecules. SiO2 is a covalent lattice with a three-dimensional network of bonds; each silicon atom is bonded tetrahedrally to four oxygen atoms and each oxygen is bonded to two silicon atoms.
10.8. In the hexagonal ice structure (Fig. 10-6 shows the position of the oxygen atoms), each oxygen is coordinated tetrahedrally with four other oxygens. There is an intervening hydrogen between adjoining oxygen atoms. The ΔH of sublimation of ice at 0°C is 51.0 kJ/mol H2O. It has been estimated by comparison with non-hydrogen-bonded solids having intermolecular van der Waals forces similar to those in ice that the ΔH of sublimation would be only 15.5 kJ/mol if ice were not hydrogen-bonded. Estimate the strength of the hydrogen bond in ice from these data.
Fig. 10-6
The excess of ΔH of sublimation above that of a non-hydrogen-bonded solid can be attributed to hydrogen bonds.
Each H2O is hydrogen-bonded to four other water molecules through O—H—O linkages (indicated in Fig. 10-6 only for the two interior molecules). Each such hydrogen-bonded linkage is shared by two water molecules (to which the two oxygen atoms belong). Our conclusion is that on the average, each water can be assigned 4 halves, or 2 hydrogen bonds.
Then, from Fig. 10-6,
Four water molecules are assigned to the unit cell, agreeing with the result from Problem 10.5.
10.9. Which would have the higher melting point, (a) V or Ca, (b) MgO or KCl? Explain.
(a) Vanadium is expected to have a much greater charge density in the electron sea since it has three 3d electrons in addition to the two 4s electrons to contribute; calcium only has two electrons in the outside orbit. Further, the ionic cores, being smaller, will be closer together. As a result of these factors, the crystal forces will be greater in vanadium than calcium. The actual melting points are 1890°C for vanadium and 845°C for calcium. The relative closeness of the cores is indicated by the densities, which are 6.11 g/cm3 for V and 1.55 g/cm3 for Ca.
(b) MgO will have a higher melting point. The first consideration is that the smaller cation and anion in MgO allows for closer approach. Further, both the cation and the anion carry twice as much charge as K+ and Cl–. This becomes clear when we consider that the electrostatic force is proportional to the product of the charges and inversely proportional to the square of the distance between them. The actual melting points are 2800°C for MgO and 776°C for KCl.
10.10. In each case, indicate which liquid will have the higher boiling point and explain your choice.
(a)CO2 or SO2
(b) (CH3)2CHCH(CH3)2 or CH3CH2CH2CH2CH2CH3
(c) Cl2 or Br2
(d) C2H5SH or C2H5OH
(a) SO2. The molecule is angular and has a permanent dipole moment; CO2 is linear and not polar. Note that CO2 is never liquid at 1 atm pressure; further, the solid sublimes (goes directly to the gaseous state without passing through the liquid state).
(b) CH3CH2CH2CH2CH2CH3. Assuming the straight-chain isomer, there is a greater area of contact between neighbors.
(c) Br2. The higher the atomic number, the greater the London force attraction between molecules.
(d) C2H5OH. Hydrogen bonding is present due to the oxygen; the attractions in the sulfur compound are weak enough to not be a consideration.
10.11. The crystalline structure of lead (207.2 g/mol) is face-centered cubic; lead’s density is 11.34 g/cm3. Calculate the length of a unit cell.
Ans. 4.95 Å
10.12. One of the crystalline forms of plutonium (Pu, 244 g/mol) is assumed to be body-centered cubic. The density of this form is 16.51 g/cm3; calculate
(a) the mass of a unit cell;
(b) the length of one side of the unit cell in Å; and
(c) the radius of one plutonium atom in Å.
Ans. (a) 8.10 × 10–22 g; (b) 3.66 Å; (c) 2.59 Å
10.13. Potassium crystallizes in a body-centered cubic lattice (unit cell length a = 520 pm).
(a) What is the distance between nearest-neighbors?
(b) What is the distance between next-nearest-neighbors?
(c) How many nearest-neighbors does each potassium atom have?
(d) How many next-nearest-neighbors does each potassium atom have?
(e) What is the calculated density of crystalline potassium?
Ans. (a) 450 pm; (b) 520 pm; (c)8; (d)6; (e) 0.924 g/cm3
10.14. The hexagonal close-packed lattice can be represented by Fig. 10-2 if 
There is an atom at each corner of the unit cell and another atom which can be located by moving one-third the distance along the diagonal of the rhombus base, starting at the lower left-hand corner and moving perpendicularly upward by c/2. Magnesium crystallizes in this lattice and has the density of 1.74 g/cm3.
(a) What is the volume of the unit cell?
(b) What is a?
(c) What is the distance between nearest-neighbors?
(d) How many nearest-neighbors does each atom have?
Ans. (a) 46.4 × 106 pm3; (b) 320 pm; (c) 320 pm; (d)12
10.15. The NaCl lattice has the cubic unit cell shown in Fig. 10-7. KBr also crystallizes this lattice.
(a) How many K+ ions and how many Br– ions are in each unit cell?
(b) Assuming that ionic radii are additive, what is a?
(c) Calculate the density of a perfect KBr crystal.
(d) What minimum value of r+/r– is needed to prevent anion-anion contact in this structure?
Ans. (a) 4 each; (b) 656 pm; (c) 2.80 g/cm3; (d) 0.414
10.16. MgS and CaS both crystallize in the NaCl-type lattice (Fig. 10-7). From the ionic radii listed in Table 10.1, what conclusion can you draw about anion-cation contact in these crystals?
Fig. 10-7 NaCl unit cell.
Ans. Ca2+ and S2– can be in contact, but Mg2+ and S2– cannot. In MgS, if Mg2+ and S2– were in contact, there would not be enough room for the sulfide ions along the diagonal of a square constituting one-quarter of a unit cell face. In other words, for MgS, r+/r– is less than 0.414 [see Problem 10.15(d)].
10.17. Each rubidium halide (Group VIIA element) crystallizing in the NaCl-type lattice has a unit cell length 30 pm greater than that for the corresponding potassium salt of the same halogen. What is the ionic radius of Rb+ computed from these data?
Ans. 148 pm
10.18. Iron crystallizes in several modifications. At about 910°C, the body-centered cubic α-form undergoes a transition to the face-centered cubic γ-form. Assuming that the distance between nearest-neighbors is the same in the two forms at the transition temperature, calculate the ratio of density of γ-iron to that of α-iron at the transition temperature.
Ans. 1.09
10.19. The ZnS zinc blende structure is cubic. The unit cell may be described as a face-centered sulfide ion sublattice with zinc ions in the centers of alternating minicubes made by partitioning the main cube into eight equal parts.
(a) How many nearest-neighbors does each Zn2+ have?
(b) How many nearest-neighbors does each S2– have?
(c) What angle is made by the lines connecting any Zn2+ to any two of its nearest-neighbors?
(d) What minimum r+/r– ratio is needed to avoid anion-anion contact, if closest cation-anion pairs are assumed to touch?
Ans. (a)4; (b)4; (c) 109°28′; (d) 0.225
10.20. Why does ZnS not crystallize in the NaCl structure? (Hint: Refer to Problem 10.15.)
Ans. The r+/r– ratio is 0.402, too low to avoid anion-anion contact in the NaCl structure.
10.21. Calculate the packing factor of spheres occupying (a) a body-centered cubic and (b) a simple cubic structure, where closest neighbors in both cases are in contact.
Ans. (a) 0.680; (b) 0.524
10.22. Many oxide minerals can be visualized as a face-centered oxide ion lattice with cations distributed within the tetrahedral and octahedral holes. Calculate the lattice constant, a, for a face-centered O2– lattice. If cations occupy all the octahedral holes in MgO and CaO, calculate a for these minerals. Use data in Table 10-1.
Ans. For an oxide lattice, a = 396 pm. With magnesium and calcium ions in the octahedral holes, anion-anion contact is broken and a expands to 410 pm and 478 pm, respectively.
10.23. Lithium iodide crystallizes in the NaCl lattice in spite of the fact that r+/r– is less than 0.414. Its density is 3.49 g/cm3. Calculate from these data the ionic radius of the iodide ion.
Ans. The calculated ionic radius is 224 pm. Note that the value in Table 10-1 is 216 pm.
10.24. Thallium(I) bromide crystallizes in the CsCl lattice. Its density is 7557 kg/m3 and its unit cell edge-length, a, is 397 pm. From these data, estimate Avogadro’s number.
Ans. 6.01 × 1023 molecules/mol
10.25. Aluminum can be used in wiring and often is for high-tension lines because it is a reasonably good conductor and is light-weight. The atomic radius for Al is 1.431 Å; a crystal of Al is face-centered cubic. (a) Calculate the cube edge length of a crystal. (b) What is the number of nearest-neighbors for each atom in the crystal?
Ans. (a) 21.02 Å (b)12
10.26. Calculate the theoretical density of aluminum and compare it to the published density, 2.702 g/cm3. Refer to the previous problem.
Ans. 2.7023, calculated. The theoretical density is the same as the published density, when considered at the same number of significant digits.
10.27. In solid ammonia, each NH3 molecule has six other NH3 molecules as nearest-neighbors. The ΔH of sublimation of NH3 at the melting point is 30.8 kJ/mol, and the estimated ΔH of sublimation in the absence of hydrogen bonding is 14.4 kJ/mol. What is the strength of a hydrogen bond in solid ammonia?
Ans. 5.5 kJ/mol
10.28. Which of the two crystals in each of the following cases has the higher melting point. Why? (a) Cs or Ba; (b)Si or P4; (c) Xe or Kr; (d) MgF2 or CaCl2.
Ans. (a) Ba—denser electron sea; (b) Si—covalently bonded network versus molecular crystal; (c) Xe—higher atomic number means stronger London forces; (d) MgF2—cations and anions both smaller.
10.29. Account for the differences in melting point between (a) and (b), between (c) and (d), and between these two differences as shown in Fig. 10-8.
Fig. 10-8
Ans. The crystal forces in (b) and (d) are largely van der Waals. Compounds (a) and (c), containing the polar hydroxide group, are capable of hydrogen bonding. In the case of (c), the hydrogen bonding is from the hydroxide of one molecule to the doubly bonded oxygen of the neighboring molecule; and the resulting intermolecular (between molecules) attraction leads to a very large increase in the melting point as compared with (d), the non-hydrogen-bonded control. In the case of (a), the molecular structure allows intramolecular (within a molecule) hydrogen bonding from the hydroxide group of each molecule to the doubly bonded oxygen of the same molecule; in the absence of strong intermolecular hydrogen bonding the difference in melting point as compared with the reference substance (b) should be small, related perhaps to differences in crystal structure or to the van der Waals forces, which should be slightly larger for (b) than for (a) because of the extra CH3 group.
10.30. Which one of each of the following pairs of liquids has the higher boiling point? Why?
(a) CH3CH2CH2OH or HOCH2CH2OH; (b) CH3CH2CH3 or CH3CH2F; (c) Xe or Kr; (d) H2O or H2S; (e) CH3CH2CH2CH2CH2CH3 or CH3CH2CH2CH3.
Ans. (a) HOCH2CH2OH, twice as many H-bonds per molecule; (b)CH3CH2F, strong dipole moment; (c) Xe, higher atomic number, greater London forces; (d) H2O, strong H-bonds; (e) CH3CH2CH2CH2CH2CH3, longer molecule, greater area of intermolecular contact
10.31. Which of the following pairs of liquids is/are miscible (can be mixed)? Why, or why not? (a) butane (C4H10) and pentane (C5H12); (b) butane and water; (c) 1-butanol (C4H9OH) and water.
Ans. (a) Miscible; forces of attraction between like and unlike molecules are about the same. (b) Not miscible; mixing would disrupt the strong H-bonds in water; there is no especially strong attraction between unlike molecules to compensate. (c) Miscible; both components have hydrogen bonding. Breaking of H-bonds in water is compensated for by the formation of H-bonds between unlike molecules.
10.32. Explain why UF6 (molecular mass 352) is more volatile than SbCl5 (molecular mass 299).
Ans. The intermolecular forces are greater for SbCl5 because the outer atoms have higher Z values providing for greater London forces than in UF6.
10.33. Explain why water dissolves in acetone (CH3COCH3), but not in hexane (C6H14).
Ans. When water dissolves, quite a lot of energy is required to break its hydrogen bonds. This is compensated for in acetone by hydrogen bonding between water and the oxygen atom of CH3COCH3, but there is no compensating strong interaction with hexane.
10.34. Consider the structure of the methylamine, H2N—CH3, and predict if it is water-soluble.
Ans. The electron pair associated with the nitrogen can attract the hydrogen from water (hydrogen bonding), providing a strong attraction. The conclusion is that this compound is soluble.