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# Objectives By the end of this section you should:

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Objectives By the end of this section you should:
know how atom positions are denoted by fractional coordinates be able to calculate bond lengths for octahedral and tetrahedral sites in a cube be able to calculate the size of interstitial sites in a cube know what the packing fraction represents be able to define and derive packing fractions for 2 different packing regimes

Fractional coordinates
Used to locate atoms within unit cell 0, 0, 0 ½, ½, 0 ½, 0, ½ 0, ½, ½ 1. 2. 3. 4. Note 1: atoms are in contact along face diagonals (close packed) Note 2: all other positions described by positions above (next unit cell along)

Octahedral Sites Coordinate ½, ½, ½ Distance = a/2
In a face centred cubic anion array, cation octahedral sites at: ½ ½ ½, ½ 0 0, ½ 0, ½

Tetrahedral sites Relation of a tetrahedron to a cube:
i.e. a cube with alternate corners missing and the tetrahedral site at the body centre

Can divide the f.c.c. unit cell into 8 ‘minicubes’ by bisecting each edge; in the centre of each minicube is a tetrahedral site

So 8 tetrahedral sites in a fcc

important dimensions in a cube
Bond lengths important dimensions in a cube Face diagonal, fd (fd) = (a2 + a2) = a 2 Body diagonal, bd (bd) = (2a2 + a2) = a 3

Bond lengths: Octahedral: half cell edge, a/2 Tetrahedral: quarter of body diagonal, 1/4 of a3 Anion-anion: half face diagonal, 1/2 of a2

Sizes of interstitials
fcc / ccp Spheres are in contact along face diagonals octahedral site, bond distance = a/2 radius of octahedral site = (a/2) - r tetrahedral site, bond distance = a3/4 radius of tetrahedral site = (a3/4) - r

Summary f.c.c./c.c.p anions
4 anions per unit cell at: 000 ½½0 0½½ ½0½ 4 octahedral sites at: ½½½ 00½ ½00 0½0 4 tetrahedral T+ sites at: ¼¼¼ ¾¾¼ ¾¼¾ ¼¾¾ 4 tetrahedral T- sites at: ¾¼¼ ¼¼¾ ¼¾¼ ¾¾¾ A variety of different structures form by occupying T+ T- and O sites to differing amounts: they can be empty, part full or full. We will look at some of these later. Can also vary the anion stacking sequence - ccp or hcp

Packing Fraction We (briefly) mentioned energy considerations in relation to close packing (low energy configuration) Rough estimate - C, N, O occupy 20Å3 Can use this value to estimate unit cell contents Useful to examine the efficiency of packing - take c.c.p. (f.c.c.) as example

So the face of the unit cell looks like:
Calculate unit cell side in terms of r: 2a2 = (4r)2 a = 2r 2 Volume = (162) r3 Face centred cubic - so number of atoms per unit cell =corners + face centres = (8  1/8) + (6  1/2) = 4

Packing fraction The fraction of space which is occupied by atoms is called the “packing fraction”, , for the structure For cubic close packing: The spheres have been packed together as closely as possible, resulting in a packing fraction of 0.74

Group exercise: Calculate the packing fraction for a primitive unit cell

Primitive

Close packing Cubic close packing = f.c.c. has =0.74
Calculation (not done here) shows h.c.p. also has = equally efficient close packing Primitive is much lower: Lots of space left over! A calculation (try for next time) shows that body centred cubic is in between the two values. THINK ABOUT THIS! Look at the pictures - the above values should make some physical sense!

Summary By understanding the basic geometry of a cube and use of Pythagoras’ theorem, we can calculate the bond lengths in a fcc structure As a consequence, we can calculate the radius of the interstitial sites we can calculate the packing efficiency for different packed structures h.c.p and c.c.p are equally efficient packing schemes

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