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8 SOLIDS
Solids are classified into two groups; crystalline solids in which the constituent particles are orderly arranged, and amorphous solids in which constituent particles lack any complete regularity. The study of crystals has a long history probably because crystals were easier to study than amorphous solids The most important development in the continuing study of crystals is X-ray crystallographic analysis. Initially this technique was barely able to treat materials of even simple structure such as common salt (NaCl). During the last eighty years, however, the progress of X-ray crystallographic analysis has been so rapid that proteins with considerably large molecular weight can now be studied with this technique.
8.1 Crystalline and amorphous solids There are several ways to classify solids, which include a wide range of varieties. However, the simplest classification is to divide solids into two groups: crystalline solids in which the constituent particles are very regularly arrayed and amorphous solids in which little or no regularity is present.
(a) Crystalline solids In some crystalline solids, the constituent particles are arrayed so regularly that their regularity is evident to the naked eye. Familiar crystals such as sodium chloride, hydrated copper sulfate and quartz are typical examples. The locations of the constituent particles of crystalline solids (ions, atoms or molecules) are usually represented by a lattice , and the location of each particle is a lattice point. The smallest repeating unit of a lattice is called a unit cell.
Figure 8.1 The definition of a unit cell. Unit cell is constituted by lattice points and is identified by bold lines. The distances between two points along each axis are defined as a , b and c. The angles made by two axes are defined as α, β and γ.
The simplest unit cell is a cube. The three axes of a cube and of some other unit cells are perpendicular to each other but axes of other unit cells are not perpendicular. The factors that determine the unit cell are the distances between points, and the angles between lattice axes. These are called the lattice constants (Fig. 8.1). In 1848, the French crystallographer Auguste Bravais (1811-1863) classified the crystal lattices based on their symmetries, and found that there are 14 kinds of crystal lattices as indicated in Fig. 8.2. These are called the Bravais lattices. These 14 Bravais lattices are classified into seven crystal systems．In this book, only three well known cubic systems -- simple cubic lattice , body-centered cubic lattice and face-centered cubic lattice --will be treated.
Figure 8.2 Bravais lattices. Crystals are classified into 14 Bravais lattices and 7 crystalline systems.
The magnitude of a unit cell can be determined by Bragg’s condition, which was proposed by the English physicist William Lawrence Bragg (1890-1971) in 1912. In order to obtain the detailed information on the accurate arrangement of particles in crystals, measurements of the intensity of spots in the diffraction pattern is necessary.
(b) Amorphous solids The array of particles in amorphous solids is partly regular and thus resembles crystalline solids to some extent. This regularity is, however, limited and is not present in the whole solid. Many amorphous solids around you--common glass, rubber and polyethylene are some examples—have partial regularity (Fig. 8.3). This feature of the structure of amorphous solids may be regarded as intermediate between that of liquids and that of solids. Recently much attention has been focused on such artificial materials as optical fibers and amorphous silicon (Table 8.1).
coordination number. For the closest packed structure, the coordination number is twelve, which is the maximum. In these cases, four particles are included in one unit cell.
Figure 8.4 Closest packed structures (a) A typical layer. Each sphere is surrounded by twelve spheres. (b) The 2 nd^ layer is similar to the 1 st^ layer. Each sphere occupies the pit formed by three spheres in the 1 st^ layer. (c) Each of the spheres in the 3 rd^ layer lies directly above the spheres of the 1 st^ layer (the aba arrangement). (d) Each sphere of the 3 rd^ layer lies above the pit of the 1 st^ layer which is not used by the spheres of the 2 nd^ layer (the abc arrangement).
Silver crystallizes in a cubic closest packed structure. If the crystal is cut as shown in Fig. 8.5, a sphere is located at the center of each face of the cube. Since a sphere (an atom) is located at the center of each face of the cube, this type of lattice is called the face-centered lattice.
Figure 8.5 face-centered lattice In such a case, the relation between r , the radius of the sphere, and d , the length of the edge of the unit cell, can be determined by the Pythagorean theorem.
Sample exercise 8.1 Density of a metal The atomic radius of a silver atom is 0.144 nm. Knowing that silver crystallizes in a cubic closest packed structure, calculate the density (g cm-3) of silver. Answer The packing of silver atoms is shown in Fig. 8.5. You need to determine the volume and the number of silver atoms in the unit cell. Since the length of the diagonal is 4 r , d can be determined by the Pythagorean theorem. d^2 + d^2 = (4 r ) 2 ∴ d = r √ 8 = 0.144√8 = 0.407 nm The number of silver atoms in the unit cell can be obtained from Fig. 8.5. Thus, there are six half-sphere and eight of eighth-sphere. Altogether there are four atoms in the unit cell. The mass m of a silver atom is: m = 107.9 (g mol -1) / 6.022 x 10^23 (atom mol -1) = 1.792 x 10 -22^ (g atom -1) Since the density is given by (mass/volume), the density of silver d Ag is; d Ag = [4(atom) x 1.792 x 10 -22^ (g atom -1)]/(0.407 x 10-7) 3 (cm 3 ) = 10.63 (g cm -3) The experimental value is 10.5 (g cm -3) (20°C)
(b) Body-centered cubic lattice Some metals, such as alkaline metals, crystallize in a body-centered cubic lattice in which the center of a sphere is located at the center and at each apex (there are eight) of the unit cell as shown in Fig. 8.6. This packing is called the body-centered cubic lattice.
Figure 8.6 Body-centered cubic lattice
Sample exercise 8.2 Packing of the metallic crystals With the aid of Fig. 8.6, answer the following questions.
optical paths of X-rays reflected by the atoms in the 1 st^ layer and by the atoms in the 2 nd^ layer is 2 d sinθ, the waves reinforce each other to give a diffraction pattern. Thus, the intensity of the diffraction pattern will be maximum when
n λ = 2 d sinθ (8.1)
This equation is called Bragg’s condition.
Bragg’s condition can be applied for two purposes. If the interatomic distance d is known, the wavelength of X-rays can be determined by measuring the angle of diffraction. Moseley used this method when he determined the wavelength of the characteristic X-rays of various elements. On the contrary, if the wavelength of X-rays λ is known, the interatomic distances can be determined by measuring the angle of diffraction. This is the basic principle of X-ray crystallographic analysis.
Sample exercise 8.3 Bragg’s condition X-rays of wavelengh 0.154 nm were used to analyze the aluminum crystal. A diffraction pattern was obtained at θ = 19.3°. Determine the distance d of two atomic planes, assuming n = 1. Answer d = n λ/2sinθ = (1 x 0.154)/(2 x 0.3305) = 0.233 (nm)
8.3 Various crystals
So far, crystals have been classified by the mode of packing of particles. Crystals can also be classified by the type of particles that constitute the crystal or by the interaction that combines the particles together (Table 8.2).
Table 8.2 Types of metals
Metallic ionic molecular covalent Li 38 LiF 246.7 Ar 1.56 C(diamond) 170 Ca 42 NaCl 186.2 Xe 3.02 Si 105 Al 77 AgCl 216 Cl 2 4.88 SiO2 433 Fe 99 ZnO 964 CO 2 6. W 200 CH 4 1. The values are the energy required to break the crystals apart into their constituent atoms (or ions, molecules, etc ) (kcal mol-1)
(a) Metallic crystals The lattice of metallic crystals is constituted by the metallic bond. The valence electrons in metallic atoms are readily removed (because of small ionization energy) to yield cations. When two metallic atoms approach, their outer atomic orbitals overlap to form a molecular orbital. When a 3rd atom approaches the two atoms, interaction between the orbitals takes place and a new molecular orbital is formed. Thus, a great number of molecular orbitals will be formed by many metallic atoms, and the resultant molecular orbitals will spread in three-dimensions. This was already mentioned in Ch. 3.4 (Fig. 3.8). Since atomic orbitals overlap repeatedly, the electrons in the outer shell of each atom will be affected by many other atoms. Such electrons do not necessarily belong to one particular atom, but move freely in the lattice formed by all these atoms. Thus, these electrons are called free electrons. Useful properties of metals such as malleability, ductility, electric conductivity, and the metallic luster of freshly cut metals are related to the properties of the metallic bonds. For instance, metals can maintain their structure even if some deformation takes place. This is because there exists a strong interaction in various directions between atoms and the free electrons around them (Fig. 8.8).
Figure 8.8 Deformation of the structure of metals Metals will be deformed if a strong force is applied, but they are not readily broken. This property is due to the strong (electric) interaction between metallic ions and free electrons.
The high thermal conductivity of metals can also be explained by free electrons. When one end of a metal piece is heated, the kinetic energy of electrons around that part is increased. The increased kinetic energy is quickly transferred to other free electrons. Electric conductivity is explained in the same manner. If voltage is applied to the both ends of a metal piece, the cloud of free electrons flow toward the positive charge. Metallic luster is due to the numerous molecular orbitals of metallic crystals. Since there are so many molecular orbitals, the gap between energy levels is very small. When the surface of the metal is irradiated by light, electrons will absorb its energy and be excited. As a result, the range of wavelengths for such absorbed light is very wide. When the excited electrons discharge the acquired energy and return to their original energy levels, light with a wide range of wavelengths is emitted, which is observed as metallic luster.
(b) Ionic crystals Ionic crystals such as sodium chloride (NaCl) are formed by the electrostatic attraction between positively charged and negatively charged ions. Ionic crystals generally have high melting points and low electrical conductivity. However, in solution or in a molten state, ionic crystals are dissociated into ions and hence possesses electrical conductivity. It is assumed that bonds are formed between cations and anions. In crystals of sodium chloride, sodium ions and chloride ions are bonded by ionic bonds. Contrary to covalent bonds, ionic bonds have no directionality, and as a result, a sodium ion interacts with all chloride ions in the crystal although the intensity of interaction varies. Similarly, a chloride ion interacts with all sodium ions in the crystal. The most stable array of an ionic crystal is the one in which the amount of contact between oppositely charged particles is largest, or in other words, the coordination number is largest. However, the size of cations is different from that of anions, and as a result, there is a tendency for the larger anions to pack in a closest packed structure, and for the smaller cations to locate in the gaps between anions. In the case of sodium chloride, the chloride anion (ionic radius is 0.181 nm) form a closest packed face-centered cubic lattice with somewhat elongated interatomic distances so that smaller sodium cations (0.098 nm) can be readily accommodated in the space (Fig. 8.9(a)). Each sodium ion is surrounded by six chloride anions (coordination number = 6). Similarly, each chloride ion is surrounded by six sodium ions (coordination number = 6) (Fig. 8.9(b)). Thus, a 6:6 coordination is achieved.
Sample exercise 8.4 Packing in ionic crystals Using the values of ionic radii (nm) given below, propose the packing pattern of lithium fluoride LiF and rubidium bromide RbBr. Li+^ = 0.074, Rb +^ = 0.149, F -^ = 0.131, Br -^ = 0. Answer For LiF, r C/ r A = 0.074/0.131 = 0.565. This is a six-coordinated crystal as are other halides of alkaline metal elements. For RbBr, r C/ r A = 0.149/0.196 = 0.760. It is an eight-coordinated crystal and has a structure of the cesium salt type.
(c) Molecular crystals Crystals in which molecules are bound by an intermolecular force such as the van der Waals force are called molecular crystals. Crystals thus far discussed are made of some kind of chemical bond between atoms or ions. However, crystals can be formed, without the aid of bonding, by weak interactions between molecules. Even rare gases crystallize at an extremely low temperature. Argon crystallizes by virtue of extremely weak van der Waals interactions, and its melting point is –189.2°C. Since argon is a monatomic molecule, it can be regarded as a sphere like a metal atom. In fact the structure of solid argon is the closest packed cube. A diatomic molecule such as iodine cannot be approximated as a sphere. Though regularly arrayed in crystals, the direction of the molecules alternates (Fig. 8.11). However, because of their simple structure, the surfaces of the crystals are regular. This is the reason why crystals of iodine have a luster.
Fig. 8.11 Crystal structure of iodine The structure is an orthorhombic face-centered lattice. The molecules at the center of each face are black-colored.
The packing pattern of crystals of organic compounds with more complicated structures are investigated exclusively by means of X-ray crystallographic analysis. Their shapes are in most cases very similar or essentially identical with those of isolated molecules in gaseous phases or in solution.
(d) Covalent crystals Many crystals have polymer-like or giant molecule-like structures. In such crystals all constituent atoms (not necessarily one type) are repeatedly bonded by covalent bonds to such an extent that the resultant particles are observable to the naked eye. Diamond is a typical example of this type, and its hardness originates from a strong network formed by covalent bonds between sp^3 hybridized carbon atoms (Fig. 8.12). Diamond is stable up to 3500°C, and at this temperature or above it sublimes. Crystals such as silicon carbide (SiC) (^) n or boron nitride (BN) (^) n have much the same structure as diamond. A well-known example is silicon dioxide (quartz; SiO 2 ) (Fig. 8.13). Silicon is tetravalent as is carbon, and binds to four oxygen atoms to form a tetrahedron. Each oxygen atom binds to another silicon atom. The melting point of quartz is 1700°C.
Fig. 8.12 Crystal structure of diamond The ∠C-C-C angle is the tetrahedral angle, and each carbon atom is surrounded by four other carbon atoms.
Figure 8.13 Crystal structure of silicon dioxide If oxygen atoms are ignored, silicon atoms form a diamond-like structure. Oxygen atoms are present between silicon atoms.
Sample question 8.5 Classification of solids The crystals given below are either metallic, ionic, covalent or molecular crystals. Identify the type of crystal for each.
solid m.p.(°C) solubility in water electroconductivity A 150 not soluble no conductivity B 1450 not soluble with conductivity C 2000 not soluble no conductivity D 1050 soluble no conductivity
Answer A = molecular crystal, B = metallic crystal, C = covalent crystal, D = ionic crystal
Crystals are usually classified as in sample exercise 8.5. In another method, crystals are classified depending on their constituent particles, i.e ., atoms, molecules or ions. Crystals made of atoms include metallic crystals, covalent crystals such as diamond, and molecular crystals such as
8.15. Liquid crystals are widely used for practical purposes such as displays on watches or TVs.
Figure 8.15 The order in liquid crystals The order is three-dimensional in crystals. It may be said that the order is two-dimensional in smectic liquid crystals and one-dimensional in nematic liquid crystals. T is the transition temperature.
Exercises
8.1 Bragg’s condition A crystal was examined with X-rays (λ = 0.1541 nm) to obtain a diffraction pattern at θ = 15.55°. Determine the distance between layers when n = 1. 8.1 Answer 0.2874 nm
8.2 Crystals of sodium chloride type In ionic crystals of sodium chloride type (coordination number = 6), a cation is surrounded by six anions. Determine the ratio of the radius of the cation to the radius of anion ( r C/ r A ) when the two ions just contact. 8.2 Answer r C + r A = √ 2 r A. ∴ r C/ r A = 0.414. If the ratio r C/ r A is smaller than this value, there is no contact between cations and anions.
8.3 Density of crystals The radius of a nickel (Ni) atom is 1.24 x 10-10^ m, and the crystals are cubic closest packed (face-centered cube). Calculate the density of crystals of nickel. 8.3 Answer Take the length of one edge of the unit cell of crystals of nickel as d , then (4 r ) 2 = 2 d^2 by the
Pythagorean theorem. ∴ d = √ 2 r. The volume of a unit cell V is: V = 22.63 r^3. In a unit cell, four atoms are included. Hence, the weight of a unit cell w is: w = (4 x 58.70 (g mol -1))/(6.022 x 10^23 (mol-1)) = 3.900 x 10 -22^ (g). ∴ d = w / V = 9.04 g cm -3. The experimental value is 8.90 g cm -3. A good agreement.
8.4 Body-centered cubic lattice The crystals of titanium have a body-centered cubic lattice structure and its density is 4.50 g cm -3. Calculate the length of the edge of the unit cell l , and the atomic radius r of a titanium atom. In a body-centered cubic lattice, the atoms contact each other along the diagonal of the unit cell. 8.4 Answer Since two atoms are contained in one unit cell of the body-centered cubic lattice, the following equation is obtained. 4.50 (g cm -3) = (2 x 47.88 (g mol -1))/( 6.022 x 10^23 (mol-1) x l^3 (cm 3 )) ∴ l = 3.28 x 10 -8^ cm. By the Pythagorean theorem, (4 r ) 2 = (3.28 x 10 -8) 2 + [√2(3.28 x 10 -8)]^2 ∴ r = 1.42 x 10 -8^ cm.
8.5 Classification of crystals Classify the crystals given below based on the classification shown in Table 8.2. (a) dry ice (CO 2 ); (b) graphite; (c) CaF; (d) MnO 2 ; (e) C 10 H 8 (naphthalene); (f) P 4 ; (g) SiO 2 ; (h) Si; (i) CH 4 ; (j) Ru; (k) I 2 ; (l) KBr; (m) H 2 O; (n) NaOH; (o) U; (p) CaCO 3 ; (q) PH 3. 8.5 Answer (a) molecular; (b) covalent; (c) ionic; (d) ionic; (e) molecular; (f) molecular; (g) covalent; (h) metallic; (i) molecular; (j) metallic; (k) molecular; (l) ionic; (m) molecular; (n) ionic; (o) metallic; (p) ionic; (q) molecular.
