474 CHAPTER 12 Intermolecular Forces and the Physical Properties of Liquids and Solids (a) (b) Figure 12.15 Arrangement of identical spheres in a simple cubic cell. (a) Top view of one layer of spheres. (b) Definition of a simple cubic cell. Primitive cubic Body-centered cubic Face-centered cubic Figure 12.16 Three types of cubic cells. The top view makes it easier to see the locations of the lattice points, but the bottom view is more realistic, with the spheres touching one another. Figure 12.17 In the body-centered cubic arrangement, the spheres in each layer rest in the depressions between spheres in the previous layer. The other types of cubic cells, shown in Figure 12.16, are the body-centered cubic cell (bcc) and the face-centered cubic cell (fcc). Unlike the simple cube, the second layer of atoms in the body-centered cubic arrangement fits into the depressions of the first layer and the third layer fits into the depressions of the second layer (Figure 12.17). The coordination number of each atom in the bcc structure is 8 (each sphere is in contact with four others in the layer above and four others in the layer below). In the face-centered cubic cell, there are atoms at the center of each of the six faces of the cube, in addition to the eight corner atoms. The coordination number in the face-centered cubic cell is 12 (each sphere is in contact with four others in its own layer, four others in the layer above, and four others in the layer below). Because every unit cell in a crystalline solid is adjacent to other unit cells, most of a cell's atoms are shared by neighboring cells. (The atom at the center of the body-centered SECTION 12.3 Crystal Structure 475 (a) (b) (c) Figure 12.18 (a) A comer atom in any cell is shared by eight unit cells. (b) An edge atom is shared by four unit cells. (c) A face-centered atom in a cubic cell is shared by two unit cells. cubic cell is an exception.) In all types of cubic cells, for example, each corner atom belongs to eight unit cells whose corners all touch [Figure 12.18(a)]. An atom that lies on an edge, on the other hand, is shared by four unit cells [Figure 12.18(b)], and a face-centered atom is shared by two unit cells [Figure 12.18(c)]. Because a simple cubic cell has lattice points only at each of the eight corners, and because each corner atom is shared by eight unit cells, there will be the equivalent of only one complete atom contained within a simple cubic unit cell (Figure 12.19]. A body-centered cubic cell contains the equivalent of two complete atoms, one in the center and eight shared corner atoms. A face-centered cubic cell contains the equivalent of four complete atoms-three from the six face-centered atoms and one from the eight shared corner atoms. Closest Packing There is more empty space in the simple cubic and body-centered cubic cells than in the face- centered cubic cell. Closest packing, the most efficient arrangement of atoms, starts with the structure shown in Figure 12.20(a), which we call layer A. Focusing on the only atom that is surrounded completely by other atoms, we see that it has six immediate neighbors in its own layer. In the second layer, which we call layer B, atoms are packed into the depressions between the atoms in the first layer so that all the atoms are as close together as possible [Fig- ure 12.20(b)]. There are two ways that a third layer of atoms can be arranged. They may sit in the depres- sions between second-layer atoms such that the third-layer atoms lie directly over atoms in the first layer [Figure 12.20(c)]. In this case, the third layer is also labeled A. Alternatively, atoms in the third layer may sit in a different set of depressions such that they do not lie directly over atoms in the first layer [Figure 12.20(d)]. In this case, we label the third layer C. A (a) B A (b) A B A (c) C B A v " , , (d) Figure 12.20 (a) In a close-packed layer, each sphere is in contact with six others. (b) Spheres in the second layer fit into the depressions between the first-layer spheres. (c) In the hexagonal close-packed structure, each third-layer sphere is directly over a first-layer sphere. (d) In the cubic close-packed structure, each third-layer sphere fits into a depression that is directly over a depression in the first layer. Figure 12.19 Because each sphere is shared by eight unit cells and there are eight comers in a cube, there is the equivalent of one complete sphere inside a simple cubic unit cell. • How Do We Know the Structures of Crystals? Figure 12.21 An arrangement for obtaining the X-ray diffraction pattern of a crystal. The shield prevents the intense beam of undiffracted X-rays from damaging the photographic plate. 476 Virtually all we know about crystal structure has been learned from X-ray diffraction studies. X-ray diffraction is the scattering of X rays by the units of a crystalline solid. The scattering, or diffraction patterns, produced are used to deduce the arrangement of particles in the solid lattice. In Section 6.1 we discussed the interference phenomenon associated with waves (see Figure 6.4). Because X rays are a form of electromagnetic radiation (i.e., they are waves), they exhibit interference phenomena under suitable conditions. In 1912 Max von Laue 4 correctly suggested that, because the wavelength of X rays is comparable in magnitude to the distances between lattice points in a crystal, the lattice should be able to diffract X rays. An X-ray dif- fraction pattern is the result of interference in the waves associated with X rays. Figure 12.21 shows a typical X-ray diffraction setup. A beam of X rays is directed at a mounted crystal. When X-ray photons encounter the electrons in the atoms of a crystalline solid, some of the incoming radiation is reflected, much as visible light is reflected by a mirror; the process is called the scattering of X rays. To understand how a diffraction pattern arises, consider the scattering of X rays by the atoms in two parallel planes (Figure 12.22). Initially, the two incident rays are in phase with each other (their maxima and minima occur at the same positions). The upper wave is scat- tered, or reflected, by an atom in the first layer, while the lower wave is scattered by an atom in the second layer. For these two scattered waves to be in phase again, the extra distance traveled by the lower wave (the sum of the distance between points Band C and the distance between points C and D) must be an integral multiple of the wavelength ("-) of the X ray; that is, Equation 12.5 BC + CD = 2d sin 8 = n"- n = 1, 2, 3, where 8 is the angle between the X rays and the plane of the crystal and d is the distance between adjacent planes. Equation 12.5 is known as the Bragg equation, after William H. Bragg and Sir William L. Bragg. s The reinforced waves produce a dark spot on a photographic film for each value of that satisfies the Bragg equation. x tube X-ray beam Shield Crystal Photographic plate 4. Max Theodor Felix von Laue (1879- 1960). German physicist. Von Laue received the Nobel Prize in Physics in 1914 for his discovery of X-ray diffraction. 5. William Henry Bragg (1862-1942) and Sir William Lawrence Bragg (1890-1972). English physicists, father and son. Both worked on X-ray crystallography. The younger Bragg formulated the fundamental equation for X-ray dif- fraction. The two shared the Nobel Prize in Physics in 1915. Incident rays Reflected rays • • • • • c d sin e • • • • • • Figure 12.22 Reflection of X rays from two layers of atom s. The lower wave travels a distance 2d sin 0 longer than the upper wave does. For the two waves to be in phase again after reflection, it must be true that 2d sin 0 = 11"-, where "- is the wavelength of the X ray and 11 = 1, 2, 3, The sharply defined spots in Figure 12.21 are observed only if the crystal is large enough to consist of hundreds of parallel layers. X rays of wavelength 0.154 nm strike an aluminum crystal; the rays are reflected at an angle of 19.3°. Assuming that 11 = 1, calculate the spacing between the planes of aluminum atoms (in pm) that is responsible for this angle of reflection. . . . . . . . . Strategy Use Equation 12.5 to solve for d. Setup 0 = 19.3 °,11 = 1, and "- = 154 pm. Solution d = 11"- 2 sin 0 154pm , '-:: ::-:- = ? 3 ~ pm :2 sin 19.30 - J Practice Problem A X rays of wavelength 0.154 nm are diffracted from a crystal at an angle of 14.17°. Assuming that 11 = 1, calculate the di stance (in pm) between layers in the crysta l. Practice Problem B At what angle will X rays of wavelength 0.154 nm be diffracted from a crystal if the distance (in pm) between layers in the crystal is 188 pm? (Assume 11 = 1. ) The X-ray diffraction technique offers the most accurate method for determining bond lengths and bond angles in molecules in solids. Because X rays are scattered by electrons, chemists can construct an electron-density contour map from the diffraction patterns by using a complex mathematical procedure. Basically, an electron-density contour map tells us the relative electron densities at various locations in a molecule. The densities reach a maximum near the center of each atom. In this manner, we can determine the positions of the nuclei and hence the geometric parameters of the molecule. • 1 nm = 1000 pm . Think About It The distance between layers of atoms in a crystal should be similar in magnitude to the wavelength of the X rays diffracted by the crystal (compare 0.154 nm with 0.233 nm). 477 478 CHAPTER 12 Intermolecular Forces and the Physical Properties of Liquids and Solids The noble gases, which are monatomic, crystallize in the ccp structure, with the exception of he l ium, which crystallizes in the hcp structure. (a) ( • (b) Figure 12.23 Exploded views of (a) a hexagonal close-packed structure and (b) a cubic close-packed structure. This view is tilted to show the face-centered cubic unit cell more clearly. Note that this arrangement is the s ame as the face-centered unit cell. Figure 12.23 shows the exploded views and the structures resulting from these two arrange- ments. The ABA arrangement [Figure 12.23(a)] is known as the hexagonal close-packed (hcp) structure, and the ABC arrangement [Figure 12.23(b)] is the cubic close-packed (ccp) structure, which corresponds to the face-centered cube already described. In the hcp structure, the spheres in every other layer occupy the same vertical position (ABABAB ), while in the ccp structure, the spheres in every fourth layer occupy the same vertical position (ABCABCA ). In both struc- tures, each sphere has a coordination number of 12 (each sphere is in contact with six spheres in its own layer, three spheres in the layer above, and three spheres in the layer below). Both the hcp and ccp structures represent the most efficient way of packing identical spheres in a unit cell, and the coordination number cannot exceed 12. Many metals form crystals with hcp or ccp structures. For example, magnesium, titanium, and zinc crystallize with their atoms in an hcp array, while aluminum, nickel, and silver crystal- lize in the ccp arrangement. A substance will crystallize with the arrangement that maximizes the " , stability of the solid. Figure 12.24 summarizes the relationship between the atomic radius r and the edge length a of a simple cubic cell, a body-centered cubic cell, and a face-centered cubic cell. This relationship can be used to determine the atomic radius of a sphere in which the density of the crystal is known. see a = 21' (a) bee b 2 = a 2 + a 2 c 2 =a 2 +b 2 = 3a 2 c = f3a = 4,. 41' a =-::- f3 (b) fee b = 41' b 2 = a 2 + a 2 16,.2 = 2a 2 a = hr (c) Figure 12.24 The relationship between the edge length (a) and radius (r) of atoms in the (a) simple cubic cell, (b) body-centered cubic cell, and (c) face-centered cubic cell. SECTION 12.3 Crystal Structure 479 Sample Problem 12.4 illustrates the relationships between the unit cell type, cell dimen- sions, and density of a metal. Go ld crystallizes in a cubic close-packed structure (face-centered cubic unit cell) and has a density of 19.3 g/cm 3 . Calculate the atomic radius of an Au atom in angstroms (A). Strategy Using the given density and the mass of gold contained within a face-centered cubic unit cell, determine the vo lu me of the unit ce l l. Then, use the volume to determine the value of a, and use the eq uation supplied in Figure 12.24(c) to find r. Be sure to use consistent units for mass, length, and volume. Setup The face-centered cubic unit cell contains a total of four atom s of gold [six faces, each shared by two unit cells, and eight corners, each shared by eight unit cells- Figure 12.24(c) ]. D = m/ Vand V = a 3 . Solution First, we determine the mass of gold (in grams) contained within a unit cell: 4 ~ 1 J1Wi 197.0 g Au 21 m = X X - = 1.31 X 10 - g/unit cell unit cell 6.022 X 10 23 ~ 1 .J].1GJ-krf Then we calculate the volume of the unit cell in cm 3 : m 1.31 X 10 - 21 is V = - = = 6.78 X 10 - 23 cm 3 d 19.3 g/ cm 3 Using the calculated volume and the relationship V = a 3 (rearranged to solve for a), we determine the l ength of a side of a unit cell: a = 3 .yv = 3~6.78 X 10- 23 cm 3 = 4.08 X 10 - 8 cm Using the relationship provided in Figure 12.22(c) (rearranged to solve for r), we determine the rad ius of a gold atom in centimeters: r = a = 4.08 X 10 - 8 em = 1.44 X 10 - 8 cm -V8 -V8 Finally, we convert centimeters to angstroms: 1.44 X 10 - 8 cm X 1 X 10 - 2 m X 1 A = 1.44 A 1 em 1 X 10 - 10 m Practice Problem A When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 4.087 A. Calculate the density of silver. Practice Problem B The density of sodium metal is 0.971 g/ cm 3, and the unit cell edge length is o 4.285 A. Determine the unit cell (simple, body-centered, or face-centered cubic ) of sodium metal. Checkpoint 12.3 Crystal Structure 12.3.1 Nickel has a face-centered cubic unit cell with an edge length of 352.4 pm. Calculate the density of nickel. a) 2.227 g/ cm 3 b) 4.455 g/ cm 3 c) 38.99 g/ cm 3 d) 8.908 g/ cm 3 e) 11.14 g/ cm 3 12.3.2 At what angle would you expect X rays of wa velength 0.154 nm to be reflected from a crystal in which the distance between layers is 312 pm? (Assume n =1. ) a) 1.6° b) 29.6° c) 0.25° d) 6.8° e) 14.3° Think About It Atomic radii t end o • to be on the order of 1 A, so thIS answer is reasonable. 480 CHAPTER 12 Intermolecular Forces and the Ph ysical Properties of Liquids and Solids It is important to realize that the lattice points u sed to define a unit cell must all be identicaL We can d efine the unit cell of Na( 1 based on the pos iti ons of the Na + ions or the positions of t he (1 - i ons. It is a common mistak e to identify the (5(1 struc ture as b od y-c entered cubic. Remember th at the lattice po i nts used to define a unit ce ll m ust a ll be id enticaL In this case, they are all (l - i ons . ( 5( 1 has a simp le cubic unit cell . Figure 12.25 The unit cell of an ionic compound can be defined eit her by (a) the positions of anions or (b) the positi on s of cations. Figure 12.26 Cr ystal s tructur es of (a) CsCl , (b) ZnS, and (c) CaF 2 . In ea ch ca se, the s maller sphere repre se nt s the cation. .' .'. 12.4 - ~ . _ . . - - Types of Crystals The s tructure s and properties of cr ys talline s olids , s uch as melting point , density, and hardness, are determined by the kind s of force s that hold the particles together. We can cla ssify any crystal as one of four ty pe s: ionic , co v alent , molecular , or metallic. Ionic Crystals Ionic cry stals are compo s ed of charged spheres (cations and anions) that are held together by Coulom- bic attraction. Anion s typically are con siderably bigger than cations [I •• Section 7.6] , and the relative sizes and relative number s of the ions in a c ompound determine how the ions are arranged in the solid lattice. NaCl adopts a face-centered cubic arrangement as s hown in Figure 12.25. Note the positions of ions within the unit cell, and within the lattice overall. Both the Na + ions and the Cl- ions adopt face- centered cubic arrangements, and the unit cell defined by the arrangeme nt of cations overlaps with the . . . . . . . . . , . . . . . . . . unit cell defined by the arrangement of anions. Look closely at the unit cell s hown in Figure 12.25(a). It is defined as fcc by the po sitions of the Cl- ions. Recall that there is the equivalent of four spheres con- tained in the fcc unit cell ( half a sphere at each of six faces and one-eighth of a sphere at each of eight comer s). In this ca se the spheres are Cl- ion s, so the unit ceU of N aC1 contains four Cl- ions. Now look at the positions of the Na + ions. Ther e are Na + ions centered on each edge of the cube, in addition to one Na + at the cente r. Each sphere on the cube's edge is shared by four unit cells, and there are 12 such ed ges. Thu s, the unit cell in Figure 12.25(a) also contains four Na + ions (one-quarter sphere at each of 12 edges, giving three spheres, and one sphere at the center). The unit cell of an ionic compound always contains the same ratio of cations to anions as the empirical formula of the compound. Figure 12.26 s ho ws the cry s tal s tructure s of three ionic compound s: CsCl, ZnS, and CaF 2 . ." . . . . . ." . . . . - ." .' . . . Ce s ium chloride [Figur e 12.26 (a)] ha s the s imple cubic lattice. Despite the apparent similarity of (a) (b) (a) (b) (c) SECTION 12.4 Types of Crystals 481 the fonnulas of CsCl and NaCl, CsCl adopts a different arrangement because the Cs + ion is much larger than the Na+ ion. Zinc sulfide [Figure l2.26(b)] has the zincblende structure, which is based on the face-centered cubic lattice. If the So- ions occupy the lattice points, the smaller Zn 2 + ions are arranged tetrahedrally about each So- ion. Other ionic compounds that have the zincblende structure include CuCl, BeS, CdS, and HgS. Calcium fluoride [Figure 12.26(c)] has the fluorite structure. The unit cell in Figure 12.26(c) is defined based on the positions of the cations, rather than the positions of the anions. The Ca2+ ions occupy the lattice point s, and each F- ion is surrounded tetrahedrally by four Ca2+ ion s. The compounds SrF 2 , BaF z, BaClz, and PbF 2 also have the fluorite structure. Sample Problems 12.5 and 12.6 show how to determine the number of ions in a unit cell and the density of an ionic crystal, respectively. How many of each ion are contained within a unit cell of ZnS? Strategy Determine the contribution of each ion in the unit cell based on its position. Setup Referring to Figure 12.26, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2 - ions at each of the eight corners and at each of the six faces. Interior ions (those completely contained within the unit cell) contribute one, tho se at the corners each contribute one- eighth, and those on the faces each contribute one-half. Solution The ZnS unit cell contains four Zn 2+ ions (interior) and four S2- ions [8 X J (corners) and 6 X ~ (faces)]. Practice Problem A Referring to Figure 12.26, determine how many of each ion are contained within a unit cell of CaF 2 . Practice Problem B Referring to Figure 12.26, determine how many of each ion are contained within a unit cell of CsC!. Sample Problem 12.6 The edge length of the NaCl unjt cell is 564 pm. Determine the density of NaCl in g/cm 3 . Strategy Use the number of Na + and CC ions in a unit cell (four of each) to determine the mass of a unit cel!. Calculate volume using the edge length given in the problem statement. Density is mass divided by volume (d = m/V). Be careful to u se units consistently. + . . Setup The ma sses of Na and CI - ions are 22.99 amu and 35.45 amu, respectively. The conversion factor from amu to grams is 1 g 6.022 X 10 23 amu so the ma sses of the Na + and CI - ions are 3.818 X 10- 23 g and 5.887 X 10- 23 g, respectively. The unit cell length is 564ym X 1 X 1O - 12 fi{ x ' 1 cm = 5.64 X 10- 8 cm lym 1 X 1O- 2 m Solution The mass of a unit cell is 3.882 X 10- 22 g (4 X 3.818 X 10- 23 g + 4 X 5.887 X 10- 23 g). The volume of a unit cell is 1. 794 X 10- 22 cm 3 [(5.64 X 10- 8 cm )3 ]. Therefore, the density is given by 3.882 X 10- 22 g d = = 2.16 a/cm 3 -?O 3 <> 1.794 X 10 cm Practice Problem A LiF has the same unit cell as NaCl (fcc). The edge length of the LiF unit cell is 402 pm. Determine the density of LiF in g/cm3 Practice Problem B NiO also adopts the face-centered-cubic arrangement. Given that the density of NiO is 6.67 g/ cm 3, calculate the length of the edge of its unit cell (in pm ). ~I __ ~ _________________________________________________ " · • • • • • Think About It Make sure that the ratio of cations to anions that you determine for a unit cell matches the ratio expressed in the compound's empirical formula. Note tha t the mass of an atomic ion is treated the same as the mass of th e parent atom. In the se cases, the mass of an electron is not significant [ ~ Section 2.2, Table 2.1]. · Think About It If you were to hold a cubic centimeter (1 cm 3 ) of salt in your hand, how heavy would you expect it to be? Common errors in this type of problem include errors of unit conversion- especially with regard to length and volume. Such errors can lead to results that are off by many orders of magnitude. Often you can use common sense to gauge whether or not a calculated answer is reasonable. For instance, simply getting the cen timeter-meter conversion upside down would result in a calculated density of 2.16 X 10 12 g/ cm 3 ! You know that a cubic centimeter of salt doesn't have a mass that large. (T hat 's billions of kilograms!) If the magnitude of a result is not rea sonable, go back and check your work. 482 CHAPTER 12 Intermolecular Forces and the Physical Properties of Liquids and Solids Most ionic crystals have high melting points, an indication of the strong cohesive forces holding the ions together. A measure of the stability of ionic crystals is the lattice energy [ ~~ Sectio n 8.2]; the higher the lattice energy, the more stable the compound. Ionic solids do not conduct electricity because the ions are fixed in position. In the molten (melted) state or when dis- solved in water, however, the compound's ions are free to move and the resulting liquid conducts electricity. Covalent Crystals In covalent crystals, atoms are held together in an extensive three-dimensional network entirely by covalent bonds. Well-known examples are two of carbon's allotropes: diamond and graphite. In diamond, each carbon atom is Sp3 -hybridized and bonded to four other carbon atoms [Figure 12.27(a)]. The strong covalent bonds in three dimensions contribute to diamond's unusual hard- ness (it is the hardest material known) and very high melting point (3550°C). In graphite, carbon atoms are arranged in six-membered rings [Figure 12.27(b)]. The atoms are all sp2-hybridized, and each atom is bonded to three other atoms. The remaining unhybridized 2p orbital on each carbon atom is used in pi bonding. In fact, each layer of graphite has the kind of delocalized molecular orbital that is present in benzene [ ~~ Sect ion 9.7] . Because electrons are free to move around in this extensively delocalized molecular orbital, graphite is a good conductor of electricity in direc- tions along the planes of the carbon atoms. The layers are held together by weak van der Waals forces. The covalent bonds in graphite account for its hardness; however, because the layers can slide past one another, graphite is slippery to the touch and is effective as a lubricant. It is also used as the "lead" in pencils. Another covalent crystal is quartz (Si0 2 ). The arrangement of silicon atoms in quartz is similar to that of carbon in diamond, but in quartz there is an oxygen atom between each pair of Si atoms. Because Si and 0 have different electronegativities, the Si -0 bond is polar. Nevertheless, Si0 2 is similar to diamond in many respects, such as being very hard and having a high melting point (161O°C). 335 pm (a) (b) Figure 12.27 Structures of (a) diamond and (b) graphite. Note that in diamond, each carbon atom is bonded in a tetrahedral arrangement to four other carbon atoms. In graphite, each carbon atom is bonded in a trigonal planar arrangement to three other carbon atoms. The distance between layers in graphite is 335 pm. SECTION 12.4 Types of Crystals 483 .' Molecular Crystals o· .' .' •• . ' .' . , /. In a molecular crystal, the lattice points are occupied by molecules, so the attractive forces between them are van der Waals forces and/or hydrogen bonding. An example of a molecular crystal is solid sulfur dioxide (S02), in which the predominant attractive force is a dipole-dipole interaction. Intermolecular hyqrogen bonding is mainly responsible for maintaining the three-dimensionallat- tice of ice (Figure 12.28). Other examples of molecular crystals are 1 2 , P4, and Ss. Except in ice, molecules in molecular crystals are generally packed together as closely as their size and shape allow. Because van der Waals forces and hydrogen bonding are usually quite weak compared with covalent and ionic bonds, molecular crystals are more easily broken apart than ionic and covalent crystals. Indeed, mo st molecular crystals melt at temperatures below 100°C. Metallic Crystals Every lattice point in a metallic crystal is occupied by an atom of the same metal. Metallic crystals are generally body-centered cubic, face-centered cubic, or hexagonal clo se -packed. Consequently, metallic elements are usually very den se . The bonding in metals is quite different from that in other types of crystals. In a metal, the bonding electrons are delocalized over the entire crystal. In fact, metal atom s in a crystal can be imagined as an array of po sitive ions immersed in a sea of delocalized valence electrons (F igure 12.29). The great cohesive force re sulting from delocalization is re sponsible for a metal's strength, whereas the mobility of the delocalized electrons makes metals good conductors of heat and elec- tricity. Table 12.4 summarizes the properties of the four different types of crystals discussed. Note that the data in Table 12.4 refer to the solid pha se of each s ub sta nce listed. Type of Crystal Ionic Covalent Molecular t Metallic Cohesive Forces Coulombic attraction and dispersion forces Covalent bonds Dispersion and dipole-dipole forces, hydrogen bond s Metallic bonds *Diamond is a good conductor of heat. tlncluded in this category are crystals made up of individual atoms. General Properties Hard, brittle, high melting point , poor conductor of heat and electricity Hard, brittle, high melting point, poor conductor of heat and electricity Soft, low melting point, poor conductor of heat and electricity Variable hardness and melting point, good conductor of heat and electricity Figure 12.28 The three- dimensional structure of ice. The covalent bonds are shown by short solid lines and the weaker hydrogen bonds by long dotted lines between 0 and H. The empty space in the structure accounts for the low density of ice, relati ve to liquid water. Figure 12.29 A cross section of a metallic crystal. Each circled po sitive charge represents the nucleus and inner electrons of a metal atom. The grey area surrounding the positive metal ions indicates the mobile "sea" of electrons. Examples NaCl, LiF, MgO, CaC0 3 C ( diamond ),* Si0 2 (quartz) All metallic elements, such as Na, Mg, Fe, Cu [...]... copper(II) oxide (CuO), yellow glass contains uranium(IV) oxide (UO z), blue glass contains cobalt(Il) and copper(II) oxides (CoO and CuO), and red glass contains small particles of gold and copper Phase Changes A phase is a homogeneous part of a system that is separated from the rest of the system by a welldefined boundary When an ice cube floats in a glass of water, for example, the liquid water is... is involved, the amount of energy required to assimilate snow is much greater than the amount necessary to assimilate an equal mass of water even if the water is ice-cold This can contribute to hypothermia, a potentially dangerous drop in body temperature Sample Problem 12.8 -.~ (a) Calculate the amount of heat deposited on the skin of a person burned by 1.00 g of liquid water at 100.0°C and... SUMMARY • Cubic unit cells may be simple cubic, body-centered cubic, or Section 12.1 • • • • • The particles (atoms, molecules, or ions) in the condensed phases (solids and liquids) are held together by intermolecular forces Intermolecular forces are electrostatic attractions between opposite charges or partial charges Intermolecular forces acting between atoms or molecules in a pure substance are called... substance has the opposite effect of heating it If we remove heat from a gas sample at a steady rate, its temperature decreases As the liquid is being formed, heat is given off by the system, because its potential energy is decreasing For this reason, the temperature of the system remains constant over the condensation period (D • C) After all the vapor has condensed, the temperature of the liquid begins... is the same whether the substance changes directly from the solid to the vapor phase or if it changes from the solid to the liquid and then to the vapor phase SECTION 12.6 Phase Changes 489 Bringing Chemistry to Life The Dangers of Phase Changes If you have ever suffered a steam burn, you know that it can be far more serious than a burn caused simply by boiling water even though steam and boiling... contain H - N, H -0, or H - F bonds Section 12.5 Ion-dipole interactions are those that occur (in solutions) between ions and polar molecules Section 12.4 crystalline solid has characteristics determined in part by the types of interactions holding it together • Amorphous solids such as glass lack regular three-dimensional structure Section 12.6 Section 12.2 • • • The possible phase changes are melting or . copper(II) oxides (CoO and CuO), and red glass contains small particles of gold and copper. Phase Changes A phase is a homogeneous part of a system that is separated from the rest of the. equal mas s of water even if the water is ice-cold. This can contribute to hypothermia, a potentially dangerous drop in body temperature. I Sample Problem 12.8 ~ . . where "- is the wavelength of the X ray and 11 = 1, 2, 3, The sharply defined spots in Figure 12.21 are observed only if the crystal is large enough to consist of hundreds