KEY POINTS The structures of many solids can be discussed in terms of close-packed arrangements of one atom type in which the tetrahe- dral or octahedral holes are occupied by other atoms or ions. The ratio
FIGURE 4.16 The structure of solid C60 showing the packing of C60 polyhedra in an fcc unit cell.
FIGURE 4.14 The cubic close-packed (ccp) unit cell of the ABC . . . polytype. The colours of the spheres correspond to the layers in Fig 4.12d.
EXAMPLE 4.3 Calculating the occupied space in a close-packed array Calculate the percentage of unoccupied space in a close-packed
arrangement of identical spheres.
Answer Because the space occupied by hard spheres is the same in the ccp and hcp arrays, we can choose the geometrically simpler structure, ccp, for the calculation.
Consider Fig. 4.15. The spheres of radius r are in contact across the face of the cube and so the length of this diagonal is r +2r + r = 4r. The side of such a cell is √8r from Pythagoras’ theorem (the square of the length of the diagonal, (4r)2, equals the sum of the squares of the two sides of length a, so 2 × a2 = (4r)2, giving a = √8r), so the cell volume is (√8r)3 = 83/2r 3. The unit cell contains 81 of a sphere at each corner (for 8× =18 1 in all) and half a sphere on each face (for 8× =18 3 in all), for a total of 4.
Because the volume of each sphere is 43πr3, the total volume occupied by the spheres themselves is 4× π = π43 r3 163 r3. The occupied fraction is therefore (163πr3)/(83/2 3r )= π163 /83/2, which evaluates to 0.740.
Self-test 4.3 Calculate the fraction of space occupied by identical spheres in (a) a primitive cubic cell and (b) a body-centred cubic unit cell. Comment on your answers in comparison with the value obtained for close-packed structures.
of spheres to octahedral holes to tetrahedral holes in a close-packed structure is 1:1:2.
The feature of a close-packed structure that enables us to extend the concept to describe structures more complicated
than elemental metals is the existence of two types of hole, or unoccupied space between the spheres. An octahedral hole lies between two triangles of spheres on adjoining layers (Fig. 4.17a); note that two triangles rotated by 60°
relative to each other define two opposite faces of an octa- hedron. The octahedral holes lie halfway between two par- allel close-packed layers at all the points where the two, staggered triangles are directly adjacent in neighbouring layers.
For a crystal consisting of N spheres in a close-packed structure, there are N octahedral holes. The positions and distribution of these holes in an hcp unit cell are shown in Fig. 4.18a and those in a ccp unit cell in Fig. 4.18b. These illustrations also show that the hole has local octahedral
(a)
(b) FIGURE 4.17 (a) An octahedral hole and (b) a tetrahedral hole
formed in an arrangement of close-packed spheres.
(0,1)
ẵ (0,1)
ẵ
(a) (b)
(ẳ,ắ)
(0,1)
ẵ
ẵ
FIGURE 4.18 (a) The location (represented by a hexagon) of the two octahedral holes in the hcp unit cell and (b) the locations (represented by hexagons) of the octahedral holes in the ccp unit cell. Positions of close-packed spheres in neighbouring unit cells are shown as dotted circles in the hcp case to illustrate the octahedral coordination; dotted lines show the coordination geometry for one octahedral hole in each structure type.
symmetry in the sense that it is surrounded by six nearest- neighbour spheres with their centres at the corners of an octahedron. If each hard sphere has radius r, and if the close- packed spheres are to remain in contact, then each octahedral hole can accommodate a hard sphere representing another type of atom with a radius no larger than 0.414r.
A tetrahedral hole (Fig. 4.17b) is formed by a planar tri- angle of touching spheres in a close-packed layer capped by a single sphere of an adjacent close-packed layer which lies in the dip between them. This capping atom may be in the close-packed layer above or below the three atoms in the original close-packed plane. As a result the tetrahedral holes in any close-packed solid can be divided into two sets:
in one the apex of the tetrahedron is directed up (T) and in the other the apex points down (T′). In an arrangement of N close-packed spheres there are N tetrahedral holes of each set and 2N tetrahedral holes in all. The position of the tetrahedral holes between the close-packed layers lies closer to the layer from which three atoms are chosen to define the limits of the tetrahedral hole. So the arrangement of close- packed layers and holes can be summarized as the repeating sequence PTOT′PTOT′P . . ., where P is a close-packed layer and O and T are the octahedral and tetrahedral holes.
In a close-packed structure of spheres of radius r, a tetra- hedral hole can accommodate another hard sphere of radius no greater than 0.225r (see Self-test 4.4). The location of tetrahedral holes, and the four nearest-neighbour spheres for one hole, is shown in Fig. 4.20a for an hcp arrangement and in Fig. 4.20b for a ccp arrangement. Individual tetrahe- dral holes in ccp and hcp structures are identical but in the hcp arrangement neighbouring T and T′ holes share a com- mon tetrahedral face and are so close together that they are never occupied simultaneously.
Where two types of sphere of different radius pack together (for instance, when cations and anions stack together), the larger spheres (normally the anions) can form a close-packed array and the smaller spheres occupy the octahedral or tetrahedral holes. Thus simple ionic struc- tures can be described in terms of the occupation of holes in close-packed arrays (Section 4.9).
EXAMPLE 4.4 Calculating the size of an octahedral hole Calculate the maximum radius of a sphere that may be accommodated in an octahedral hole in a close-packed solid composed of spheres of radius r.
Answer The structure of a hole, with the top spheres removed, is shown in Fig. 4.19a. If the radius of a sphere is r and that of the hole is rh, it follows from Pythagoras’s theorem that (r + rh)2 + (r + rh)2= (2r)2 and therefore that (r + rh)2 = 2r2, which implies that r + rh= √2r. That is, rh = (√2 − 1)r, which evaluates to 0.414r. Note that this is the permitted maximum size subject to keeping the
close-packed spheres in contact; if the spheres are allowed to separate slightly while maintaining their relative positions, then the hole can accommodate a larger sphere.
Self-test 4.4 Show that the maximum radius of a sphere that can fit into a tetrahedral hole is rh= 0.225r: base your calculation on Fig. 4.19b. Note that a tetrahedron may be inscribed inside a cube using four non-adjacent vertices and the centre of the hole has the same central point as that of the tetrahedron.
FIGURE 4.19 The distances used to calculate the size of (a) an octahedral hole and (b) a tetrahedral hole.
(a)
(b)
2r
2r
r + rh
r + rh
FIGURE 4.20 (a) The locations (represented by triangles) of the tetrahedral holes in the hcp unit cell and (b) the locations of the tetrahedral holes in the ccp unit cell. Dotted lines show the coordination geometry for one tetrahedral hole in each structure type.
(1/8, 7/8) (3/8, 5/8)
(ẳ,ắ) (0,1)
ẵ
(0,1) ẵ
T
T′
T′ T (a)
(b)
EXAMPLE 4.5 Demonstrating that the ratio of close-packed spheres to octahedral holes in ccp is 1:1 Determine the number of close-packed spheres and octahedral
holes in the ccp arrangement and hence show the ratio is 1 sphere:1 hole.
Answer The ccp unit cell, with the positions of the octahedral holes marked, is shown in Fig. 4.18b. For the close-packed spheres the calculation of the number in the unit cell follows that given in Section 4.1 for lattice points in the F-centred lattice, as there is one sphere associated with each lattice point. The number of close-packed spheres in the unit cell is therefore (8× + + =18) (6 21) 4. The octahedral holes are sited along each edge of the cube
(12 edges in total), shared between four unit cells, with a further hole at the centre of the cube which is not shared between unit cells. So the total of octahedral holes in the unit cell is (6× + =12) 1 4. So the ratio of close-packed spheres to holes in the unit cell is 4:4, equivalent to 1:1. As the unit cell is the repeating unit for the whole structure, this result applies to the complete close-packed array and it is often quoted that ‘for N close-packed spheres there are N octahedral holes’.
Self-test 4.5 Show that the ratio of close-packed spheres to tetrahedral holes in ccp is 1:2.
The structures of metals and alloys
TABLE 4.2 The crystal structures adopted by metals under normal conditions
Crystal structure Element
Hexagonal close-packed (hcp) Be, Ca, Co, Mg, Ti, Zn Cubic close-packed (ccp) Ag, Al, Au, Cd, Cu, Ni, Pb, Pt Body-centred cubic (bcc) Ba, Cr, Fe, W, alkali metals Primitive cubic (cubic-P) Po
X-ray diffraction studies (Section 8.1) reveal that many metallic elements have close-packed structures, indicating that the bonds between the atoms have little directional covalent character (Table 4.2, Fig. 4.21). One consequence of this close packing is that metals often have high densities because the most mass is packed into the smallest volume.
Indeed, the elements deep in the d block, near iridium and osmium, include the densest solids known under normal
3 4 5 6 7 8 9 10 11 12
13 14
1 2
2
3
4
5
6
7
bcc hcp ccp
FIGURE 4.21 The structures of the metallic elements at room temperature. Elements with more complex structures are left blank.
conditions of temperature and pressure. Osmium has the highest density of all the elements at 22.61 g cm−3, and the density of tungsten, 19.25 g cm−3, which is almost twice that of lead (11.3 g cm−3), results in its use as weighting mate- rial in fishing equipment and as ballast in high-performance cars.
EXAMPLE 4.6 Calculating the density of a substance from a structure
Calculate the density of gold, with a cubic close-packed array of atoms of molar mass M = 196.97 g mol−1 and a cubic lattice parameter a = 409 pm.
Answer Density is an intensive property; therefore the density of the unit cell is the same as the density of any macroscopic sample. We represent the ccp arrangement as a face-centred lattice with a sphere at each lattice point; there are four spheres associated with the unit cell. The mass of each atom is M/NA, where NA is Avogadro’s constant, and the total mass of the unit cell containing four gold atoms is 4M/NA. The volume of the cubic unit cell is a3. The mass density of the cell is ρ = 4M/NAa3. At this point we insert the data:
ρ = × ×
× − ×− × −− = × −
4 (196.97 10 kgmol )
(6.022 10 mol ) (409 10 m)23 1 3 112 3 1.91 10 kgm4 3 That is, the density of the unit cell, and therefore of the bulk metal, is 19.1 g cm−3. The experimental value is 19.2 g cm−3, in good agreement with this calculated value.
and B, can move easily relative to each other, but considera- tion of the ccp structure, when drawn as in Fig. 4.14, shows that the planes ABC can be chosen in different orthogonal directions, allowing the closed-packed layers of atoms to move easily in multiple directions.
A close-packed structure need not be either of the com- mon ABAB . . . or ABCABC . . . polytypes. An infinite range of close-packed polytypes can in fact occur, as the layers may stack in a more complex repetition of A, B, and C layers or even in some permissible random sequence. The stacking cannot be a completely random choice of A, B, and C sequences, however, because adjacent layers can- not have exactly the same sphere positions; for instance, AA, BB, and CC cannot occur because spheres in one layer must occupy dips in the adjacent layer. Cobalt is an exam- ple of a metal that displays this more complex polytypism.
Above 500°C, cobalt is ccp but it undergoes a transition when cooled. The structure that results is a nearly ran- domly stacked set (for instance, ABACBABABC . . .) of close-packed layers of Co atoms. In some samples of cobalt the polytypism is not random, as the sequence of planes of atoms repeats after several hundred layers. The long-range repeat may be a consequence of a spiral growth of the crys- tal that requires several hundred turns before a stacking pattern is repeated.