Crystal Structure

Solids are divided into 2 categories:

crystalline – possesses rigid and long-range order; its atoms , molecules or ions occupy specific positions, e.g., ice

amorphous – lack a well-defined arrangement and long-range molecular order, e.g., glass.

Unit cell – basic repeating structural unit of a crystalline solid.

Each sphere represents an atom, ion or molecule, called a lattice point.

Unit cell is described by 6 parameters.

Axes are a, b and c and the angles between the axes are , and .

( is the angle between b and c, etc.)

The position of an atom in the unit cell is given by fractions of the unit cell axes, (x/a, y/b, z/c), called fractional coordinates.

c

a b

An origin is defined in the unit cell and has position (0,0,0).

Other coordinates are given relative to this origin, e.g., an atom in the centre of the unit cell has fractional coordinates of (½,½,½).

Unit cells are chosen by a variety of ways, but the most preferable one is the smallest cell that shows greatest symmetry.

Fig 1

N.B. combining these 14 Bravais lattices with all possible symmetry elements  230 different space groups

Consider the following figures:

Fig 2.1 Fig 2.2

In the 2D pattern shown, there are a variety of unit cells, each of which repeats the exact contents of the box.

Fig 2.2 is preferred to Fig 2.1 because it is half the area.

simple cubea = b = c

α = β = γ = 900

tetragonal

a = b ≠ c α = β = γ = 900

orthorhombic a ≠ b ≠ c

α = β = γ = 900

rhombohedrala = b = c

α = β = γ ≠ 900

monoclinic triclinic

hexagonala ≠ b ≠ c

γ ≠ α = β = 900

a ≠ b ≠ c

α ≠ β ≠ γ = 900 a = b ≠ c

α = β = 900 , γ = 120o

Every crystalline solid can be described in terms of one of seven types of unit cells.

We will only focus on cubic systems (watch the videos)

Note: it is difficult to draw 3-D figures; use 2-D projection along a specific direction.

For a simple cube:

1,0

1,0

1,0

1,0Fig 4.1.2

Fig 4.1 Fig 4.1.1

Unit cell contains 1 lattice point (8 on corners counting ⅛ each).

More complex lattice types are body-centred and face-centred with two and four lattice points in each unit cell respectively (Fig 4.2 and 4.3)

Fig 4.2 Fig 4.3

For body centered cube (bcc):

1,0 1,0

1,0 1,0

Fig 4.2.1 Fig 4.2.2

Lattice point at each corner and one at the centre of the unit cell.

i.e., 8 lattice points on corners, each shared by 8 unit cells, ⅛ per unit cell.

Total: 8 x ⅛ + 1 = 2 lattice points in bcc.

Exercise

Show that the face-centered lattice has 4 lattice points.

Atomic Radius

Fig 5.1 sc Fig 5.2 bcc Fig 5.3 fcc

Forsc a = 2r

bcc a =4r

(3)½

fcc a = [(8)½]r

Example 1

Gold (Au) crystallizes in a cubic close-packed structure (fcc) and has a density of 19.3 g cm-3. Calculate the atomic radius of gold in picometres.

Solution

In a fcc unit cell, the number of atoms is given by

(8 x ⅛) + (6 x ½) = 4

1 mol of Au contains 6.022 x 1023atomsx mol of Au contains 4 atoms

x = 6.64 x 10-24 mol

Mass of unit cell = 6.64 x 10-24 mol x 197.0 g Au

= 1.31 x 10-21 g/unit cell

Now from ρ = m/v

v = m/ρ =1.31 x 10-21 g

19.3 g cm-3= 6.79 x 10-23 cm3

Because volume is length cubed, we take the cube root of the volume of the unit cell to obtain the edge length (a) of the cell.

a = (v)⅓ = (6.79 x 10-23 cm3)½ = 4.08 x 10-8 cm

Now, a = (8)½r r = a (8)-½ = 4.08 x 10-8 cm (8)-½

= 1.44 x 10-8 cm

= 144 pm

Homework Exercise

1. Metallic iron crystallizes in a cubic lattice. The unit cell edge length is 287 pm. The density of iron is 7.87 g cm-3. How many iron atoms are within a unit cell?

2. When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 408.7 pm. Calculate the density of silver.

Packing Spheres

The way spheres are arranged in layers determines the type of unit cell.

Simplest case, layer of spheres can be arranged as shown.

xA 3-dimensional structure can be obtained by placing layers directly above and below this layer.

Focusing on sphere ‘x’, see that it is in contact with 6 spheres, 4 around , 1 above and 1 below.

Each sphere in this arrangement is said to have a coordination number of 6 because it has 6 immediate neighbours.

Coordination number is defined as the number of atoms (or ions) surrounding an atom (or ion) in a crystal lattice.

Its value gives us a measure of how tightly the spheres are packed together, the larger the coordination number, the closer the spheres are to each other.

The basic, repeating unit in the array of spheres described is called a simple cubic cell (scc).

The other types of cubic cells are the body-centred cubic cell (bcc) and the face-centred cubic cell (fcc).

Simple cubic

bcc

A body-centred cubic arrangement differs from a simple cube in that the second layer of spheres fits into the depressions of the first layer and the third layer into the depressions of the second layer.

The coordination number of each sphere in this structure is 8 (each sphere is in contact with 4 spheres in the layer above and 4 spheres in the layer below.

fcc

In the face-centred cubic cell, there are spheres at the centre of each of the 6 faces of the cube, in addition to the 8 corner spheres.

In both the structures, each sphere has a coordination number of 12 (each sphere is in contact with 6 spheres in its own layer, 3 spheres in the layer above and 3 spheres in the layer below)

The unoccupied space in a close-packed structure amounts to 26% of the total volume.

Calculating the unoccupied space in a close-packed array

Since the space occupied by hard spheres is the same for ccp and hcp arrays, choose the geometrically simpler structure, ccp, for the calculation.

The side of such a cell = (8)½ r, where r is the radius of each sphere.

Volume = {(8)½ r}3 = {(8)½ }3 r3

Ccp (fcc) has 4 lattice points.

Volume of each sphere = 4/3πr3 , total volume of spheres = 4(4/3πr3)

The occupied fraction is therefore

4(4/3πr3)

{(8)½ }3 r3

= 0.740

The unoccupied fraction is therefore 0.26 corresponding to 26%.

Homework Exercise

Calculate the fraction of space occupied by identical spheres in a primitive cubic unit cell.

fccClosest Packing

Most efficient arrangement of spheres.

Start with the structure which we call layer A.

Layer AFocusing on x, can see that it has 6 immediate neighbours in the layer.

x

In the second layer (which we call layer B), spheres are packed into the depressions between the spheres in the first layer so that all the spheres are as close as possible to each other.

There are 2 ways that a third layer sphere may cover the second layer to achieve closest packing.

The spheres may fit into depressions so that each third layer sphere is directly over the first layer sphere.

Since there is no difference between the arrangement of the first and third layers, we call the third layer, layer A.

The ABAB….. arrangement that is obtained is called an hexagonal close-packing (hcp) structure.

Alternatively, the third-layer spheres may fit into the depressions of the second layer.

In this case, we call the third layer, layer C.

This ABCABC…. arrangement is called cubic close-packing (ccp).

Summary

Calculations based on a knowledge of the crystal arrangement

Non-metallic species also crystallize out into these simple arrangements

Holes in close-packed structures

2 types of holes or unoccupied space between the spheres:

(i) An octahedral hole lies between two planar triangle of spheres in adjoining layers.(Fig A)

Fig A

Fig B

For a crystal consisting of N spheres, there are N octahedral holes.

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 the hole with the top spheres removed is shown in the opposite figure.

If the radius of the sphere is r and that of the hole is rh, it follows from Pythagoras’ Theorem that:

(r + rh)2 + (r + rh)2 = (2r)2

(r + rh)2 = 2r2

r + rh = (2)½ r

rh = (2)½ r - r or rh = {(2)½ - 1} r = 0.414 r

(ii) A tetrahedral hole, T, shown in the figure below, is formed by a planar triangle of touching spheres capped by a single sphere lying in the dip between them.

The tetrahedral hole in any close-packed solid can be divided into two sets:

• in one, the apex of the tetrahedron is directed up (T).• in the other, the apex points down (T’).

In an arrangement of N close-packed spheres, there are N tetrahedral holes in each set, and 2N tetrahedral holes in all.

In a close-packed structure of spheres of radius r, a tetrahedral hole can accommodate another hard sphere of radius no greater than 0.225 r.

(prove!)

Ionic Solids

In general, anions tend to be larger than cations.

- -

- -

+

+

+++

The exact type of structure adopted depends on the relative sizes of the anions and the cations and the charges on each.

Each type of ionic solid has one thing in common.

The ions are in fixed positions and therefore ionic solids are poor

conductors of electricity.

Simple Ionic Crystals

The formation of Ions

The basic process involved in the formation of an ionic compound is the transfer of one or more electrons from one type of atom to another.

The resulting ions are held together by electrostatic attraction.

The arrangement of the ions in the solid is the one which gives the highest electrostatic energy.

To see what factors determine such an arrangement, let us consider the process of bringing up successive anions around a given cation.

If there are already n anions surrounding the cation, the addition of a further anion produces a number of repulsions between its charge and the charges on the n anions already present.

Thus, there are two opposing tendencies:

One is to increase the attractive forces by making the coordination number of the cation as large as possible.

Balance this by addition of more and more anions to increase the repulsive forces.

When the two tendencies balance, the final structure results.

The coordination numbers in a solid of a given formula, such as AB, depends on the number of the larger ions which may be packed around the smaller one.

The stoichiometry in this example , 1:1, determines the coordination number of the larger ion.

Generally, cations are smaller in size than anions and it is the number of anions which can pack around the cation that determines the coordination numbers and structures.

For example, sodium chloride crystallizes in a structure where the sodiun ion is surrounded by six chloride ions and the chloride ion has six sodium ions around it.

The larger cesium cation allows a coordination number of eight in the structure of cesium chloride.

The number of anions which can pack around a given cation may be determined from the ratio of the radii of the cation and anion.

The radius ratio, r+ / r- , may be used to give an indication of the likely coordination number for a salt of a given formula type.

NaCl CsCl ZnS

For example, in six coordination, a cross section through a site in the lattice is:

When the anions just touch BAB’ = 90o

Then,

AB = AB’ = (r+ + r-)

Can we relate this relative size to any useful predictions:

Radius ratio rule

BB’ = 2r-

AB

BB’

r+ + r-

2r-

= =1

(2)½cos45o =

Therefore

r+ = (2)½ r- - r-

orr+ / r- = (2)½ - 1 = 0.41

Similar calculations may be carried out for all the coordination numbers.

The results of which indicate the range of values for the radius ratio within which different coordination numbers should be stable.

These are shown below:

r+ / r- : 0.155 to 0.23 to 0.41 to 0.73 to higher values

C.N. : 3 4 6 8

The validity of this simple method of predicting the coordination number may be assessed by examining the structures of some AB and AB2 compounds.

Ionic solids of the formula AB generally adopt either the sodium chloride or cesium chloride structure.

The common AB2 structures are those of rutile (titanium dioxide) in which the coordination is 6:3 or of fluorite (calcium fluoride) where the coordination is 8:4.

Rutile fluorite

A number of other structures adopted by crystals of more complex stoichiometry are:

Wurtzite structure (ZnS)

Rhenium oxide

oxygen

rhenium

Spinel structure

perovskite

oxygen

calcium

titanium

oxide ionmetal ion in Oh

sitemetal ion in Th site

Lattice enthalpy and the Born-Haber cycle

The lattice enthalpy, ΔHLo, is the standard molar enthalpy change

accompanying the formation of a gas of ions from the solid:

MX(s) M+(g) + X-(g) ΔHLo

Because lattice disruptions is always endothermic, lattice enthalpies are always positive.

Lattice enthalpies are determined from enthalpy data by using a Born-Haber cycle, a closed path of steps that includes lattice formation as one stage, such as that shown:

K+(g) + e-(g) + Cl(g)

K+(g) + e-(g) + ½Cl2(g)

K+(g) + Cl-(g)

-355

122

K(g) + ½Cl2(g)

425

89

K(s) + ½Cl2(g)

438

x

KCl(s)

The Born-Haber cycle for KCl, lattice enthalpy (kJ mol-1) is equal to -x

The standard enthalpy of decomposition of a compound into its elements is the negative of its standard enthalpy of formation, ΔfHo:

M(s) + X(s) MX(s) ΔfHo

MX(s) M(s) + X(s) -ΔfHo

Likewise, the standard enthalpy of lattice formation from the gaseous ions is the negative of the lattice enthalpy:

MX(s) M+(g) + X-(g) ΔHLo

M+(g) + X-(g) MX(s) -ΔHLo

For a solid element, the standard enthalpy of atomization, ΔatomHo, is the standard enthalpy of sublimation, as in the process

K(s) K(g) ΔsubHo = +89 kJ mol-1

For a gaseous element, the standard enthalpy of atomization is the standard enthalpy of dissociation, as in

Cl2(g) 2Cl(g) ΔdisHo = +244 kJ mol-1

The standard enthalpy for the formation of ions from their neutral atoms is the enthalpy of ionization (for the formation of cations) and the electron-gain enthalpy(for anions).

Two examples are

K(g) K+(g) + e- ΔionHo = +425 kJ mol-1

Cl(g) + e- Cl-(g) ΔegHo = -355 kJ mol-1

The value of the lattice enthalpy, the only unknown in a well chosen cycle, is found from the requirement that the sum of the enthalpy changes round a complete cycle is zero.

Example

Calculate the lattice enthalpy of KCl(s) using a Born-Haber cycle and the following information:

ΔH/(kJ mol-1)

Sublimation of K(s) +89

Ionization of K(g) +425

Dissociation of Cl2(g) +244

Electron gain by Cl(g) -355

Formation of KCl(s) -438

Solution

The required cycle was shown previously.

The sum of the enthalpy changes around the cycle is zero,so

ΔHLo = (+438 + 89 + 425 + 122 – 355) kJ mol-1 = 719 kJ mol-1

Homework Exercise

Calculate the lattice enthalpy of magnesium bromide from the data shown below:

ΔH/(kJ mol-1)

Sublimation of Mg(s) +148

Ionization of Mg(g) to Mg2+ +2187

Vaporization of Br2(l) +31

Dissociation of Br2(g) +193

Electron gain by Br(g) -331

Formation of MgBr2(s) -524

Calculation of lattice enthalpies

To calculate the lattice enthalpy of an ionic solid, we need to take into account several contributions to its enthalpy.

These include the attractions and repulsions between the ions.

This calculation yields the Born-Mayer equation for the lattice enthalpy at T = 0:

where

do = r+ + r- is the distance between centres of neighbouring cations and anions

NA is the Avogadro’s constant

Ad

d

d

ezzNH

ooo

BAA

L

1

4

2

zA and zB the charge numbers of the cation and anion

e is the fundamental charge

o the vacuum permittivity

d is a constant (normally 34.5 pm) used to represent the repulsion between ions at short range

A is called the Madelung constant

Madelung constants

Structural type A

Caesium chloride 1.763

Fluorite 2.519

Rock salt 1.748

Rutile 2.408

Sphalerite 1.638

Wurtzite 1.642

Example

Using the Born-Mayer equation, estimate the lattice enthalpy of sodium chloride.

Solution

For sodium chloride, zNa+ = +1 and zCl

- = -1 and from the table, A = 1.745.

do = rNa+ + rCl

- = 282 pm and using fundamental constants:

748.1282

5.341

)1082.2()10854.8()4(

)10602.1()1()1()10023.6(1012112

219123

xpm

pmx

mxxmCJxx

CxxxmolxHL

= 7.56 x 105 J mol-1

Note: this value compares reasonably well with the experimental value from the Born-Haber cycle, 788 kJ mol-1

Exercise for the Idle Mind

Estimate the lattice enthalpy of CsCl using the experimental do = 356 pm