Chapter 3. The structures of simple solids Structures of Solids Crystalline solids Amorphous solids

Transcription

1 Chapter 3. The structures of simple solids Structures of Solids Crystalline solids The atoms, molecules or ions pack together in an ordered arrangement Such solids typically have flat surfaces, with unique angles between faces and unique 3-dimensional shape Examples of crystalline solids include diamonds, and quartz crystals Amorphous solids No ordered structure to the particles of the solid No well defined faces, angles or shapes Often are mixtures of molecules which do not stack together well, or large flexible molecules Examples would include glass and rubber The packing of spheres 2.1 Unit cells and the description of crystal structure Metals and ionic compounds can be treated as 3-d arrays of hard spheres and in many cases satisfactory results for energy calculations can be obtained by completely neglecting the covalent character which is always present but often small. In simple cases it is possible to understand and even predict crystal structures based on the notion that the cations and anions pack together as would hard spheres with certain constraints, e.g. cations (often smaller) tend to be surrounded by anions.when atoms, molecules or ions pack in a regular arrangment which can be repeated "infinitely" in three dimensions a crystal is formed. Acrystalline solid, therefore, possesses a long-range order; its atoms, molecules or ions occupy regular positions which repeat in three dimensions. The network of atoms, molecules or ions is known as a crystal latticeor simply as a lattice. Any location in a crystal lattice is known as a lattice point. Since the crystal lattice repeats in three dimensions, there will be an entire set of lattice points which are identical. That means that if you were able to make yourself small enough and stand at any such lattice point in the crystal lattice, you would not be able to tell which lattice point of the set you were at - they would all appear identical. Of course, you could move to a different site which would look different. This would constitute a different lattice point. For example, when we examine the sodium chloride lattice later, you will notice that the environment of each sodium ion is identical. If you were to stand at any sodium ion and look around, you would see the same thing. If you stood at a chloride ion, you would see a different environment but that environment would be the same at every chloride ion. Thus, the sodium ion locations form one set of lattice points and the chloride ion locations form another set. Because the atoms, molecules or ions pack in a regular array, it is possible, to use x-ray diffraction to determine the location of the atoms in crystal lattice. When such an experiment is carried out we say that we have determined the crystal structureof the substance. The study of crystal structures is known as crystallography and it is one of the most powerful techniques used today to characterize new compounds. You will discuss the principles behind x-ray diffraction in the lecture part of this course. The forces which stabilize the crystal may be ionic (electrostatic) forces, covalent bonds, metallic bonds, van der Waals forces, hydrogen bonds, or a combination of these. The properties

2 of the crystal will change depending upon what types of bonding is involved in holding the atoms, molecules or ions in the lattice. The fundamental types of crystals based upon the types of forces that hold them together are: metallic ionic network molecular metal cations held together by a sea of electrons cations and anions held together by predominantly electrostatic attractions atoms bonded together covalently throughout the solid (also known as covalent crystal or covalent network). collections of individual molecules; each lattice point in the crystal is a molecule Often crystals held together by more than one type of force and thus may have intermediate properties. Since the crystal lattice is made up of a regular arrangement which repeats in three dimensions, we can save ourselves a great deal of work by considering the simple repeat unit rather than the entire crystal lattice. The basic repeat unit is known as the The Unit Cell Crystalline solids often have flat, well-defined faces that make definite angles with their neighbors and break cleanly when struck. These faces lie along well-defined directions in the unit cell. The unit cell is the smallest, most symmetrical repeat unit that, when translated in three dimensions, will generate the entire crystal lattice.

3 Since the crystal is made up of an arrangement of identical unit cells, then an identical point on each unit cell represents an identical environment within the crystal The array of these identical points is termed the crystal lattice The unit cells shown are cubic

4 o All sides are equal length o All angles are 90 The unit cell need not be cubic o The unit cell lengths along the x,y, and z coordinate axes are termed the a, b and c unit cell dimensions o The unit cell angles are defined as:, the angle formed by the b and c cell edges, the angle formed by the a and c cell edges, the angle formed by the a and b cell edges Cubic Cells There are many types of fundamental unit cells, one of which is the cubic cell. In turn, there are three subclasses of cubic: Simple cubic (sc) cell Body-centered cubic (bcc) cell Face-centered cubic (fcc) cell. Several aspects of each cell type: the number of atoms per unit cell the efficiency of the packing of atoms in the volume of each unit cell the number of nearest neighbors (coordination number) for each type of atom The unit cell is normally selected to be the simplest of the possible repeating units. The basic unit cell in three dimensions is a parallelipedwith side langths and angles as defined below. The angles and lengths used to define the size of the unit cell are known as the unit cell parameters.

5 This unit cell has no symmetry in that the cell parameters and and angles may take any values. An increasing level of symmetry produces relationships between the various cell parameters and leads to the seven crystal classes. Crystal Class Triclinic Monoclinic Orthorhombic Restrictions on Unit Cell Parameters a is not equal tob is not equal toc is not equal to is not equal to. a is not equal to b is not equal to c = is not equal to. Highest Type of Symmetry Element Required no symmetry is required, an inversion center may be present highest symmetry element allowed is a C 2 axis or a mirror plane a is not equal to b is not equal to c has three mutually perpendicular == mirror planes and/or C 2 axes Tetragonal a =b is not equal to c == has one C 4 axis Cubic a =b =c == has C 3 and C 4 axes Hexagonal, Trigonal a =b is not equal to c == 120 C 6 axis (hexagonal); C 3 axis (trigonal) Rhombohedral* a =b =c ==is not equal to C 3 axis (trigonal) It is possible to have a number of different choices for the unit cell. By convention, the unit cell that reflects the highest symmetry of the lattice is the one that is chosen. A unit cell may be thought of as being like a brick which is used to build a building. Many bricks are stacked together to create the entire structure. Because the unit cell must translate in three dimensions, there are certain geometrical constraints placed upon its shape. The main criterion is that the opposite faces of the unit cell must be parallel. Because of this restriction there are only six parameters that we need to define in order to define the shape of the unit cell. These include three edge lengths a, b and c and three angles, and. Once these are defined all other distances and angles in the unit cell are set. As a result of symmetry, some of these angles and edge lengths may be the same. There are only seven different shapes for unit cells possible. These are given in the chart above. *There is some discussion about whether the rhombohedral unit cell is a different group or is really a subset of the trigonal/hexagonal types of unit cell. In contrast amorphous solids do not possess any long-range order, so they do not typically form

6 well-defined faces nor do they fracture evenly. Glass is an example of an amorphous solid. Amorphous solids have very interesting properties in their own right that differ from those of crystalline materials. We will not consider their structures in this laboratory exercise. 2.2 The close-packing of spheres Closest Packing (Close-packing) of Spheres (Section 2.2) A crystal is a repeating array. In describing this structure we must distinguish between the pattern of repetition (the lattice type) and what is repeated (the unit cell) described below. Packing and Geometry Imagine a stack of balls in a box. This was actually how structures were first visualised before computers (BC). The balls will pack together to fill up all the space. This is called close packing - you can see how it works if you look at a pile of oranges in the supermarket. Notice how the oranges form a pattern. Each orange labelled A will be surrounded by six other oranges within one layer. Notice the holes labelled B and C. We can place a second layer of close-packed oranges on either the B-sites or the C-sites (but not both). In this way we can build up a 3D structure. The closest packing of identical spheres forms the basis for the structures of a number of metals, and the arrangement of the largest ions (often the anions) in a number of simple ionic structures. Close-packing of spheres is one example of an arrangement of objects that forms such an extended structure. Extended close-packing of spheres results in 74% of the available space being occupied by spheres (or atoms), with the remainder attributed to the empty space between the spheres. This is the highest space-filling efficiency of any spherepacking arrangement. The nature of extended structures as well as close-packing, which occurs in two forms called hexagonal closest packing (hcp) and cubic closest packing (ccp), will be explored in this lab activity. Sixty-eight of the ninety naturally occurring elements are metallic elements. Forty of these metals have three-dimensional submicroscopic structures that can be described in terms of closepacking of spheres. Another sixteen of the sixty-eight naturally occurring metallic elements can be described in terms of a different type of extended structure that is not as efficient at spacefilling. This structure occupies only 68% of the available space. This second largest subgroup exhibits a sphere packing arrangement called body-centered cubic (bcc). There are two packing methods: Hexagonal close-packed (hcp) Type ABABAB... (Figs. 2.2a and 2.3a)

7 Actually there is another common form of close-packing, corresponding to layers with stacking AB.AB... or AC.AC... (these are equivalent). This is called hexagonal close-packing HCP, and the competition between CCP and HCP is determined by longer range forces between the atoms. This is the structure of sodium at low temperatures. No, we can't transform sodium to gold by stacking the atoms differently! For such simple materials, the different properties are mainly due to the differences between the sodium and gold atoms themselves. Cubic close-packed (ccp) or face-centred cubic (fcc) Type ABCABCABC... (Figs. 2.2b and 2.3b)

8 Does this structure correspond to anything in nature (apart from oranges in supermarkets)? Of course! A stack of layers of types ABC.ABC... represents the cubic close-packed CCP atomic structure of gold as determined by X-rays. Atoms lie on the corners of a cube, with additional atoms at the centers of each cube face: for that reason it is often called face centered cubic or FCC. There is one host atom at each corner, one host atom in each face, and the host atoms touch along the face diagonal (a = r, Z = 4).Many simple metals have this FCC structure, whose symmetry is described as Fm-3m where F means Face-centered,m signifies a mirror-plane (there are two) and -3 tells us that there is a 3-fold symmetry axis (along the body diagonal) as well as inversion symmetry. where A, B and C are hexagonally packed layers of spheres. Both fill 74% of the space in an extended structure. Simple Structures which are not Close-packed (Section 2.5) There are two other simple packing types which are found: Body-centered cubic - 68% filled (Fig 2.6)

9 The third common metallic structure is called body-centered cubic BCC, and consists of a unit cube with atoms at its corners and center. The BCC structure is slightly less closely packed than FCC or HCP and is often the high temperature form of metals that are closepacked at lower temperatures. For example sodium changes from HCP to BCC above degrees C! The structure of iron (Fe) can be either CCP or BCC depending on its heat treatment, while metals such as chromium are always BCC. Metals which are BCC are, like chromium, usually harder and less malleable than close-packed metals such as gold. When the metal is deformed, the planes of atoms must slip over each other, and this is more difficult in the BCC structure. Note that there are other important mechanisms for hardening metals, and these involve introducing impurities or defects which also block slipping. (The body-centred cubic type is relatively common, but only -Po is know to adopt the primitive cubic structure.) There is one host atom at each corner of the cubic unit cell and one atom in the cell center. Each atom touches eight other host atoms along the body diagonal of the cube (a = r, Z = 2). Primitive (or simple) cubic - 52% filled(fig 2.7) Simple Cubic (SC) - There is one host atom ("lattice point") at each corner of a cubic unit cell. The unit cell is described by three edge lengths a = b = c = 2r (r is the host atom radius), and the angles between the edges, alpha = beta = gamma = 90 degrees. There is one atom wholly inside the cube (Z = 1). Unit cells in which there are host atoms (or lattice points) only at the eight corners are called primitive. You should be able to calculate the % of void space using simple geometry.

10 When spherical objects of equal size are packed in some type of arrangement, the number of nearest neighbors to any given sphere is dependent upon the efficiency of space filling. The number of nearest neighbors is called the coordination number and abbreviated as CN. The sphere packing schemes with the highest space-filling efficiency will have the highest CN. Coordination number will be explored in this lab activity. A useful way to describe extended structures, which, in principle, can be infinitely large, is to conceive of a three-dimensional parallelepiped, which is a six-sided solid having parallel faces. This represents a unit cell which can be moved in three directions to duplicate the entire structure of the crystal; for a cubic unit cell, the cell is shifted in the three perpendicular X, Y and Z directions. The unit cell is the repeating three-dimensional pattern for extended structures. A unit cell has a pattern for the objects as well as for the void spaces. Unit cells will be explored in this lab activity. The remaining unoccupied space in any sphere packing scheme is found as void space. This void space occurs between the spheres and gives rise to so-called interstitial sites. 2.3 Holes in close-packed structures Holes ("Interstices") in Closest Packed Arrays Tetrahedral Hole - Consider any two successive planes in a closest packed lattice. One atom in the A layer nestles in the triangular groove formed by three adjacent atoms in the B layer, and the four atoms touch along the edges (of length 2r) of a regular tetrahedron; the center of the tetrahedron is a cavity called the Tetrahedral (or Td) hole; a guest sphere will just fill this cavity (and touch the four host spheres) if its radius is r.

11 Octahedral Hole - Adjacent to the Td hole, three atoms in the B layer touch three atoms in the A layer such that a trigonal antiprismatic polyhedron (a regular octahedron) is formed; the center of the octahedron is a cavity called the Octahedral (or Oh) hole. A guest sphere will just fill this cavity (and touch the six host spheres) if its radius is r. It can be shown that there are twice as many Td as Oh holes in any closest packed bilayer. coordination number The coordination number is the number of particles surrounding a particle in the crystal structure. In each packing arrangement (simple cube), a particle in the crystal has a coordination number of 6 In each packing arrangement (body centerd cube), a particle in the crystal has a coordination number of 8 In each packing arrangement (hexagonal close pack, cubic close pack), a particle in the crystal has a coordination number of 12 Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell, as shown in the figure below. An atom on a face is shared by two unit cells, so only half of the atom belongs to each of these cells. An atom on an edge is shared by four unit cells, and an atom on a corner is shared by eight unit cells. Thus, only one-quarter of an atom on an edge and one-eighth of an atom on a corner can be assigned to each of the unit cells that share these atoms.

12 If nickel crystallized in a simple cubic unit cell, there would be a nickel atom on each of the eight corners of the cell. Because only one-eighth of these atoms can be assigned to a given unit cell, each unit cell in a simple cubic structure would have one net nickel atom. Simple cubic structure: 8 corners x 1/8 = 1 atom If nickel formed a body-centered cubic structure, there would be two atoms per unit cell, because the nickel atom in the center of the body wouldn't be shared with any other unit cells. Body-centered cubic structure: (8 corners x 1/8) + 1 body = 2 atoms If nickel crystallized in a face-centered cubic structure, the six atoms on the faces of the unit cell would contribute three net nickel atoms, for a total of four atoms per unit cell. Face-centered cubic structure: (8 corners x 1/8) + (6 faces x 1/2) = 4 atoms Because they have different numbers of atoms in a unit cell, each of these structures would have a different density. Let's therefore calculate the density for nickel based on each of these structures and the unit cell edge length for nickel given in the previous section: nm. In order to do this, we need to know the volume of the unit cell in cubic centimeters and the mass of a single nickel atom. The volume (V) of the unit cell is equal to the cell-edge length (a) cubed. V = a 3 = ( nm) 3 = nm 3

13 Since there are 10 9 nm in a meter and 100 cm in a meter, there must be 10 7 nm in a cm. We can therefore convert the volume of the unit cell to cm 3 as follows. The mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadro's number. The density of nickel, if it crystallized in a simple cubic structure, would therefore be 2.23 g/cm 3, to three significant figures. Simple cubic structure: Because there would be twice as many atoms per unit cell if nickel crystallized in a bodycentered cubic structure, the density of nickel in this structure would be twice as large. Body-centered cubic structure: There would be four atoms per unit cell in a face-centered cubic structure and the density of nickel in this structure would be four times as large. Face-centered cubic structure:

14 The experimental value for the density of nickel is 8.90 g/cm 3. The obvious conclusion is that nickel crystallizes in a face-centered cubic unit cell and therefore has a cubic closest-packed structure. The structures of metals Important properties of the unit cells are The type of atoms and their radii R. cell dimensions (side a in cubic cells, side of base a and height c in HCP) in terms of R.

15 n, number of atoms per unit cell. For an atom that is shared with m adjacent unit cells, we only count a fraction of the atom, 1/m. CN, the coordination number, which is the number of closest neighbors to which an atom is bonded. APF, the atomic packing factor, which is the fraction of the volume of the cell actually occupied by the hard spheres. APF = Sum of atomic volumes/volume of cell. Unit Cell n CN a/r APF SC BCC FCC HCP The closest packed direction in a BCC cell is along the diagonal of the cube; in a FCC cell is along the diagonal of a face of the cube. Unit Cells: Calculating Metallic Estimates of the radii of most metal atoms can be found. Where do these data come from? How do we know, for example, that the radius of a nickel atom is nm? Nickel crystallizes in a face-centered cubic unit cell with a cell-edge length of nm to calculate the radius of a nickel atom. One of the faces of a face-centered cubic unit cell is shown in the figure below. According to this figure, the diagonal across the face of this unit cell is equal to four times the radius of a nickel atom.

16 The Pythagorean theorem states that the diagonal across a right triangle is equal to the sum of the squares of the other sides. The diagonal across the face of the unit cell is therefore related to the unit-cell edge length by the following equation. Taking the square root of both sides gives the following result. We now substitute into this equation the relationship between the diagonal across the face of this unit cell and the radius of a nickel atom: Solving for the radius of a nickel atom gives a value of nm: Length face diagonal = a(2) 1/2 = 4r Use this information to calculate the density of an fcc metal. Density Calculations Example calculation.

17 Al has a ccp arrangement of atoms. The radius of Al = 1.423Å ( = 143.2pm). Calculate the lattice parameter of the unit cell and the density of solid Al (atomic weight = 26.98). Solution: Because Al is ccp we have an fcc unit cell. Cell contents: 4 atoms/cell [8 at corners (each 1/8), 6 in faces (each 1/2)] Lattice parameter: atoms in contact along face diagonal, therefore 4r Al = a(2) 1/2 a = 4(1.432Å)/(2) 1/2 = 4.050Å. Density (= Al ) = Mass/Volume = Mass per unit cell/volume per unit cell g/cm 3 Mass of unit cell = mass 4 Al atoms = (26.98)(g/mol)(1mol/6.022x10 23 atoms)(4 atoms/unit cell) = x g/unit cell Volume unit cell = a 3 = (4.05x10-8 cm) 3 = 66.43x10-24 cm 3 /unit cell Therefore Al = {1.792x10-22g/unit cell}/{66.43x10-24 cm 3 /unit cell} = g/cm 3 Because they have different numbers of atoms in a unit cell, each of these structures would have a different density. Let's therefore calculate the density for nickel based on each of these structures and the unit cell edge length for nickel given in the previous section: nm. In order to do this, we need to know the volume of the unit cell in cubic centimeters and the mass of a single nickel atom. The volume (V) of the unit cell is equal to the cell-edge length (a) cubed. V = a 3 = ( nm) 3 = nm 3 Since there are 10 9 nm in a meter and 100 cm in a meter, there must be 10 7 nm in a cm. We can therefore convert the volume of the unit cell to cm 3 as follows. The mass of a nickel atom can be calculated from the atomic weight of this metal and Avogadro's number.

18 The density of nickel, if it crystallized in a simple cubic structure, would therefore be 2.23 g/cm 3, to three significant figures. Simple cubic structure: Because there would be twice as many atoms per unit cell if nickel crystallized in a bodycentered cubic structure, the density of nickel in this structure would be twice as large. Body-centered cubic structure: There would be four atoms per unit cell in a face-centered cubic structure and the density of nickel in this structure would be four times as large. Face-centered cubic structure: The experimental value for the density of nickel is 8.90 g/cm 3. The obvious conclusion is that nickel crystallizes in a face-centered cubic unit cell and therefore has a cubic closest-packed structure. How is the ideal density of a substance calculated from unit cell data? The ideal density of a general crystalline unit cell is calculated by the following method: D is the density expressed in grams per cubic centimeter (g / cm 3 ), Z is the number of atoms or molecules per unit cell, FWT is the formula weight of the atoms or molecules expressed in grams (g), Na is Avogadro's Number = x atoms or molecules V is the volume of the unit cell expressed in cubic centimeters (cm 3 ).

19 2.4 Polyptism The structure of metals often changes as a function of temperature - phase transitions occur - this is called polymorphism or polytypism 2.5 Structures that are not close-packed Simple Structures which are not Close-packed (Section 2.5) There are two other simple packing types which are found: Body-centered cubic - 68% filled (Fig 2.6) The third common metallic structure is called body-centered cubic BCC, and consists of a unit cube with atoms at its corners and center. The BCC structure is slightly less closely packed than FCC or HCP and is often the high temperature form of metals that are closepacked at lower temperatures. For example sodium changes from HCP to BCC above degrees C! The structure of iron (Fe) can be either CCP or BCC depending on its heat treatment, while metals such as chromium are always BCC. Metals which are BCC are, like chromium, usually harder and less malleable than close-packed metals such as gold. When the metal is deformed, the planes of atoms must slip over each other, and this is more difficult in the BCC structure. Note that there are other important mechanisms for hardening metals, and these involve introducing impurities or defects which also block slipping. (The body-centred cubic type is relatively common, but only -Po is know to adopt the primitive cubic structure.) There is one host atom at each corner of the cubic unit cell and one atom in the cell center. Each atom touches eight other host atoms along the body diagonal of the cube (a = r, Z = 2). Primitive (or simple) cubic - 52% filled(fig 2.7) Simple Cubic (SC) - There is one host atom ("lattice point") at each corner of a cubic unit cell. The unit cell is described by three edge lengths a = b = c = 2r (r is the host atom radius), and the angles between the edges, alpha = beta = gamma = 90 degrees. There is one atom wholly inside the cube (Z = 1). Unit cells in which there are host atoms (or

20 lattice points) only at the eight corners are called primitive. 2.6 Polymorphism of metals Polymorphism of metals (Section 2.6) Many metals adopt one of the common packing types described above and many adopt more than one: Crystal Packing of Metals (ccp) (bcc) Cubic Close-Packed Body-Centred Cubic Hexagonal Close- Packed Disordered ccp/hcp Where two or more symbols are used, the largest indicates the room temperature form.

21 The body-centred cubic structure is typically a higher temperature structure, though for some elements, e.g. sodium the phase transition occurs below room temperature. The closer packed structures also seem to be favoured by metals with larger numbers of valence electrons. 2.7 Atomic radii of metals Atomic radii which are mostly metallic radii follows the periodic trends. Metallic radii (12-coordinate) (pm) Elem. Rad. Elem. Rad. Elem. Rad. Elem. Rad. Ag Fe Nb Sn Al Ga Nd Sr Au Gd Ni Ta Ba Hf Os Tb Be Hg Pb Tc Bi 170 Ho Pd Th Ca In Pm Ti Cd Ir Pr Tl Ce K Pt Tm Co La Rb U 156 Cr Li Re V Cs Lu Rh W Cu Mg Ru Y Dy Mn Sb 159 Yb Er Mo Sc Zn Eu Na Sm Zr Source:Teatum,E., Gschneidner,K., & Waber,J. (1960) Compilation of calculated data useful in predicting metallurgical behaviour of the elements in binary alloy systems, LA-2345, Los Alamos Scientific Laboratory.

22 2.8 Alloys Alloys are form by mixing different types of metals together: Substitutional alloys: one metal replaces another metal at some lattice sites. This requires that both types of metals be roughly the same size. Interstitial alloys: the new atoms are introduced into the hole sites (called interstices) of the host lattice. The new atoms can occupy either T d or O h holes. Copper and Copper Alloys Copper Alloys The most common way to catalog copper and its alloys is to divide them into six families: coppers, dilute copper alloys, brasses, bronzes, copper nickels and nickel silvers. The first family, the coppers, is essentially commercially pure copper, which ordinarily is soft and ductile and contains less than about 0.7% total impurities. The dilute copper alloys contain small amounts of various alloying elements that modify one or more of the basic properties of copper. Solid Solution Alloys. The most compatible alloying elements with copper are those that form solidsolution fields. These include all elements forming useful alloy families (Zn, Sn, Al,Si ). Hardening in these systems is great enough to make useful objects without encountering brittleness associated with second phases or compounds. Tin and Tin Alloys Tin in Alloys. Solders account for the second largest use of tin (after tinplate). Tin is an important constituent in solders because it wets and adheres to many common base metals at temperatures considerably below their melting points.tin is alloyed with lead to produce solders with melting points lower than those of either tin or lead. Small amounts of various metals, notably antimony and silver, are added to tin-lead solders to increase their strength. These solders can be used for joints subjected to high or even subzero service temperatures. Check this lick for many other alloys Shape-Memory Alloys Ionic solids 2.9 Characteristic structures of ionic solids Geometries of Crystal Lattices (Section 2.9) There are a few very common crystal structures adopted by large numbers of compounds. A few of them are shown below. For simple binary compounds which are predominantly ionic, the structure which is observed can often be predicted based on the stoichiometry (1:1, 1:2, etc.) and on the relative sizes of the ions. Anions tend to be larger than cations so, in an ideal structure, there is contact between adjacent anions, while the cations are in contact with the neighbouring anions, but there is no cation - cation contact.

23 Structures where the cations is large enough to push apart the anions can occur, but structures where the cation would be "loose" in their sites are undesirable. Therefore it is possible to calculate ranges over which a particular structure is likely based on the ratio of cation to anion radius. This illustrated below for three common 1:1 structures. Rock Salt (Sodium Chloride) This structure is based on cubic close-packed anions with cations in all the octahedral holes, that is, the cations a 6-coordinate.

24 The ideal radius ratio (either r + /r_ or r_/r + ) is calculated by eliminating d from the following equations: 2r_ = 2d r + + r_ = d which gives: r + /r_ = 0.41 or r_/r + = 2.44 Cesium Chloride This structure is based on a primitive cubic packing of the ions. Each type (anion or cation) is 8-coordinate.

25 The ideal radius ratio (either r + /r_ or r_/r + ) is calculated by eliminating d from the following equations: 2r_ = d r + + r_ = 3d/2 which gives: r + /r_ = 0.72 or r_/r + = 1.37 Zinc Blende (Cubic Zinc Sulphide) This strucure is based on cubic close-packed anions with cations in half the tetrahedral holes, that is, the cations are 4-coordinate. The ideal radius ratio (either r + /r_ or r_/r + ) is calculated by eliminating d from the following equations: 2r_ = 2d r + + r_ = 3d/2 which gives: r + /r_ = 0.23 or r_/r + = 4.44 Other structures The other simple structures described in class are: Wurtzite - another form of zinc sulphide based on hexagonally close packed sulphide ions with zinc ions in one half of the tetrahedral holes, just like the zinc blende structure. The same ideal radius ratio applies.

26 Flourite and Antifluorite - The fluorite typified by CaF 2 structure is based on a facecentred cubic flouride arrangement with calcium ions in all the tetrahedral holes, and the anifluorite structure typified by K 2 O has the reverse arrangement. Rutile - The compound from which the name comes is TiO 2. The ion Ti 4+ cannot exist in a truly ionic compound: there will always be significant covalency because the ion would be very small and highly charged, strongly polarizing the counter anions. The structure is not based on cubic or hexagonal close packing. The cations are 6-coordinate and the anions are 3-coordinate. Perovskite - This is an example of a more complicated structure for compounds of the form ABX 3. The named mineral is CaTiO 3. It was used as an example of how to deduce the formula of an ionic compound from a picture of the unit cell The rationalization of structures SolidsRationalization of Structures (Section 2.10) Ionic Radii and the Ideal Radius Ratios table Bonding Configurations in Ionic solids. In reality an ideal fit of a cation into the close packed anion arrangement almost never occurs. Now consider what would be the consequence of placing a cation that is (a) larger than the ideal, (b) smaller than the ideal, into the cation sites. Example: The following compounds have similar empirical formulas. Use the radius ratio rules and the table of ionic radii in the appendix to explain why they have different structures. (a) NaCl (b) ZnS (c) CsCl Radius Ratio rules. The discussion of tetrahedral, octahedral, and cubic holes in the previous section suggests that the structure of an ionic solid depends on the relative size of the ions that form the solid. The

27 relative size of these ions is given by the radius ratio, which is the radius of the positive ion divided by the radius of the negative ion. The relationship between the coordination number of the positive ions in ionic solids and the radius ratio of the ions is given in the table below. As the radius ratio increases, the number of negative ions that can pack around each positive ion increases. When the radius ratio is between and 0.414, positive ions tend to pack in tetrahedral holes between planes of negative ions in a cubic or hexagonal closest-packed structure. When the radius ratio is between and 0.732, the positive ions tend to pack in octahedral holes between planes of negative ions in a closest-packed structure. Radius Ratio Radius Ratio Rules Coordination Number Holes in Which Positive Ions Pack tetrahedral holes octahedral holes cubic holes 1 12 closest-packed structure The table above suggests that tetrahedral holes aren't used until the positive ion is large enough to touch all four of the negative ions that form this hole. As the radius ratio increases from to 0.414, the positive ion distorts the structure of the negative ions toward a structure that purists might describe as closely-packed. As soon as the positive ion is large enough to touch all six negative ions in an octahedral hole, the positive ions start to pack in octahedral holes. These holes are used until the positive ion is so large that it can't fit into even a distorted octahedral hole. Eventually a point is reached at which the positive ion can no longer fit into either the tetrahedral or octahedral holes in a closest-packed crystal. When the radius ratio is between about and 1, ionic solids tend to crystallize in a simple cubic array of negative ions with positive ions occupying some or all of the cubic holes between these planes. When the radius ratio is about 1, the positive ions can be incorporated directly into the positions of the closest-packed structure. Coordination numbers for cations and anions are the same when there are equal numbers of ions e.g. NaCl. MX : Rock-salt (NaCl) r+ / r- = 0.52 Octahedral, 6 Coordination The relative sizes of the anions and cations required for a perfect fit of the cation into the octahedral sites in a close packed anion array can be determined by simple geometry:the basic premise for making predictions is that it is acceptable if the smaller ion (usually the cation) is too

28 big for the hole in the close packed array of larger ions (usually the anions) so that the anions are forced apart. However, the smaller ion must not be a loose fit in its site. In this way, for 1:1 compounds we predict that: Compounds with r + /r - from 0.22 to 0.41 will adopt the ZnS (blende or wurtzite) structures. Compounds with r + /r - from 0.41 to 0.72 will adopt the NaCl structure. Compounds with r + /r - greater than 0.72 will adopt the CsCl structure. I r + /r - is less than 0.22, the copounds are likely to be quite covalent. When the actual structures are compared to what is predicted, the results are quite bad, especially for the ZnS structure, which persists in cases where r + /r - is much larger than the presumed limit of Only a general trend can be detected. One difficulty is the difficulty in assigning reliable ratii to the ions. Structure Maps These are plots of the electronegativity difference between the elements of which the compound is formed, and the average principal quantum number of the elements. On such scatter diagrams (Figures 2.22 and 2.23) quite well defined zones of one structural type are revealed. This is not unexpected because there is a close link between ion size and the electronegativity and between size and the period in which the ion is found. The Bond Triangle The covalent-ionic continuum described above is certainly an improvement over the old covalent -versus - ionic dichotomy that existed only in the textbook and classroom, but it is still only a one-dimensional view of a multidimensional world, and thus a view that hides more than it reveals. The main thing missing is any allowance for the type of bonding that occurs between more pairs of elements than any other: metallic bonding. Intermetallic compounds are rarely even mentioned in introductory courses, but since most of the elements are metals, there are a lot of them, and many play an important role in metallurgy. In metallic bonding, the valence electrons lose their association with individual atoms; they form what amounts to a mobile "electron fluid" that fills the space between the crystal lattice positions occupied by the atoms, (now essentially positive ions.) The more readily this electron delocalization occurs, the more "metallic" the element. Thus instead of the one-dimension chart shown above, we can construct a triangular diagram whose corners represent the three extremes of "pure" covalent, ionic, and metallic bonding.

29 We can take this a step farther by taking into account collection of weaker binding effects known generally as van der Waals forces. Contrary to what is often implied in introductory textbooks, these are the major binding forces in most of the common salts that are not alkali halides; these include NaOH, CaCl 2, MgSO 4. They are also significant in solids such as CuCl 2 and solid SO 3 in which infinite covalently-bound chains are held together by ion-induced dipole and similar forces. The only way to represent this four-dimensional bonding-type space in two dimensions is to draw a projection of a tetrahedron, each of its four corners representing the "pure" case of one type of bonding. Note that some of the entries on this diagram (ice, CH 4, and the two parts of NH 4 ClO 4 ) are covalently bound units, and their placement refers to the binding between these units. Thus the H 2 O molecules in ice are held together mainly by hydrogen bonding, which is a van der Waals force, with only a small covalent contribution. Note: the triangular and tetrahedral diagrams

30 above were adapted from those in the excellent article by William B. Jensen, "Logic, history and the chemistry textbook", Part II, J. Chemical Education 1998: The energetics of ionic bonding The Born-Haber Cycle: Experimental Approaches to Lattice EnergySection 2.11 Born-Haber cycles represent, on paper, different routes for the formation of a ionic crystal structure from its elements. As an example, the cycle for the formation of Al 2 O 3 from its elements, metallic aluminum and gaseous oxygen is given below. Usually such cycles are used to calculate a missing enthalpy, for example, Electron capture enthalpies which are difficult to measure directly by experiment. Apart from the lattice energy, the other quantities can be obtained reasonably precisely. The lattice energy can be obtained by calculation, as described in the next section. The Lattice Energy of Sodium Chloride Section 4-2 The crystal lattice of the mineral rock salt (sodium chloride) is used as an example of the method of calculation. There are two significant terms which contribute to the energy: 1. Coulombic (electrostatic) attractions and repulsions. 2. Short range repulsions.

31 In addition, there is a little vibrational energy and other minor sources which will not be considered. The diagram below shows a small part of the structure of the sodium chloride lattice. The electrostatic potential energy (E) of any single ion is given by the sum of the series: E coul = 1/(4 o )(- 6e 2 /d + 12e 2 /2d - 8e 2 /3d + 6e 2 /4d - 24e 2 /5d...) The first term in the series corresponds to the 6 counter ions in the immediate coordination sphere of the chosen ion, and is negative because of the opposite charges. The second term goes with the 8 nearest similar ions, and so on. The coefficients in the series are not predictable in any straightforward way, and because of the alternation of signs, the series converges rather slowly. Nevertheless, the job has been done for a number of crystal structures. The general formula for a mole of ions is: E coul = NAz A z B e 2 /4 o d N is avagadro's number, and A is the Madlung constant. z A and z B are the charges on the ions. (For NaCl, z A = 1 and z B = -1 so E will be negative, as expected.) A selection of values of A is given below: Structure A NaCl CsCl ZnS (blende) ZnS (wurtzite)

32 CaF TiO CDI Al 2 O Notice how A is more or less constant for structures with the same stoichiometry and is very approximately proportional to the number of ions in the formula unit. The other term contributing to the energy of the lattice comes from the short range repulsion between the cations and anion as their electron clouds start to overlap. The formula for this repulsion energy is: E rep = NB/d n B is a constant, and n varies with the nature of the ions, specifically their configuration: Ion Configuration n He 5 Ne 7 Ar 9 Kr 10 Xe 12 For mixed cases, an average can be used for n. The total lattice energy (U) is given by:

33 U = E coul + E rep = NAz A z B e 2 /4 o d + NB/d n At the equilibrium distance (where d = d o ) du/dd = 0: du/dd = NAz A z B e 2 /4 o d o 2 - NB/d o n+1 = 0 B = Az A z B e 2 d o n-1 /4 o n U = (NAz A z B e 2 /4 o d o )(1-1/n) When all the constants are given their numerical values: U = 1389(z A z B M/d o )(1-1/n) (kj mol -1 ) Note that the calculated values of U are quite insensitive to the value of n. Generalization of the Lattice Enery Calculation The derivation above produces an equation which requires precise knowledge of the crystal structure. Two other equations which are easier to use deserve mention: The Born-Meyer Equation U = (NAz A z B e 2 /4 o d o )(1-d*/d o ) In this equation the invariant n is replaced. The value d* = 34.5 pm works for most crystals. The Kapustinski Equation U = Knz A z B e 2 /4 o d o )(1-d*/d o ) Here K = 1.21 MJ Å mol -1 and n is the number of ions in the formula empirical formula of the compound, e.g. n = 5 for Al 2 O 3. The value of d to be used is (r + + r_) where these radii are the

34 radii for 6-coordinate ions found in many tables. This equation works because it is found that (A/nd) is approximately constant from one structure to another Consequences of lattice enthalpies Ionic Consequences of Lattice Enthalpies On the Existence or Non-existence of Simple Ionic Compounds Consider the thermodynamic data (in kj) in the table below: Salt Hf = U + H EA + H sub + H diss + H IP NaCl "NaCl 2 " "CaF" CaF Note: The lattice energies for "NaCl 2 " is estimated assuming it would crystallize in the CaCl 2 structure, and the lattice energy for "CaF" is estimated assuming it would crystallize in the NaF structure. The hypothetical compound "NaCl 2 " is strongly endothermic due to the very large second ionization potential of Na + Na 2+ (4562 kj mol -1 ) which is only partly compensated by the much larger lattice energy for a 1:2 stoichiometry. Any attempt to make this compound from Na and Cl 2 would simply result in left-over Cl 2. Both CaF 2 and the hypothetical "CaF" are exothermic compounds, but CaF 2 is far more stable. Once again the dominant terms are the lattice energies and the ionization potentials, but the second ionization potential: Ca + Ca 2+ (1145 kj mol -1 ) is much smaller than for Na +. The compound "CaF", if it could be formed, would be very unstable with respect to CaF 2 and excess Ca. The Thermal Decomposition of Carbonates of the Alkaline Earth Metals (Group II) (Section 2.12 (a) The reaction is: MCO 3 (s) MO(s) + CO 2 (g) and is the temperature at which the equilibrium produces a pressure of 1 atm. Consider the thermodynamic data in the table below: MgCO 3 CaCO 3 SrCO 3 BaCO 3 G o kj mol H o kj mol

35 Note the following points: S o J K -1 mol o C The temperature of decomposition rises from MgCO 3 to BaCO 3. The equilibriun constant will be unity (K = 1) when G = 0, and at this point: T = H o /S o (bcause G = H - T.S) The entropy change, S o, remains roughly constant, since it is mostly due to the evolution of the very disordered CO 2 gas. Therefore the free energy change, G o, is largely determined by changes in the enthalpy, H o. In turn the enthalpy is partly determined by the change in lattice energy going from the carbonate to the oxide. While total enthalpy change is positive, the lattice energy change makes a negative contribution to it, so the more more stable the oxide lattice relative to the carbonate the less positive the enthalpy will be. Since the stoichiometry of the compounds remains 1:1 and the charges remain the same (cation 2+, and anion 2-) the only other factor in the lattice energy calculation is the interionic separation. These are estimated in the Table below for the Mg and Ba compounds. Compound Distance Compound Distance % Change MgCO MgO BaCO BaO The (6-coordinate) radii of Mg 2+ and Ba 2+ are 0.72 and 1.75 Å respectively. The radius of O 2- is 1.40 Å, and the effective ("thermodynamic") radius of CO 3 2- is 1.85 Â. Compound Lattice Energy Compound Lattice Energy Lattice Energy Difference

36 MgCO MgO BaCO BaO The lattice energies are calculated using the Kapustinski equation and the interionic distantances from the table above. Energies are in kj mol -1 The percentage change in the cation - anion distance, and hence the change in lattice energy, is largest for MgCO 3, and decreases to BaCO 3, so the enthalpy change for MgCO 3 will be smallest, and the decomposition temperature lowest. Recommended Questions from Shriver and Atkins: "Exercises" 2.1, 2.2 You should be able to answer this. 2.3 Not covered, Fall You should be able to do these. You should be able to recognize the new structure if you draw 2.13 diagram of 8 Unit cells of CsCl, and then delet the appropriate ions You should be able to do these. 2.18, 2.19 May not be covered, Fall "Problems" You should be able to do these. 2.5 May not be covered, Fall This is just geometry - you should be able to do it. 2.7 A bit too philosophical for this course. 2.8 You shoul dbe able to answer this. 2.9 May not be covered, Fall 2001.