An Introduction to X-Ray

Diffraction by Single Crystals

and Powders

Patrick McArdle NUI, Galway, Ireland

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The Nature of Crystalline Materials

• Crystalline materials differ from amorphous materials in that they have

long range order. They also exhibit X-ray powder diffraction patterns.

• Amorphous materials may have very short range order (e.g. molecular

dimers) but do not have long range order and do not exhibit X-ray powder

diffraction patterns.

• The packing of atoms, molecules or ions within a crystal occurs in a

symmetrical manner and furthermore this symmetrical arrangement is

repetitive throughout a piece of crystalline material.

• This repetitive arrangement forms a crystal lattice. A crystal lattice can be

constructed as follows:

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A 2-dimensional Lattice

Pick any position within the 2 dimensional lattice in Fig. 1(a) and note

the arrangement about this point. Place a dot at this position and then

place dots at all other identical positions as in Fig. 1(b). Join these

lattice points using lines to give a lattice grid. The basic building block

of this lattice (unit cell) is indicated in Fig. 1(c).

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Unit Cells may centred on one or more faces or at the centre of the cell. When the

unit cells listed above are combined with centring there are 14 different Bravais

lattices.

In general, six parameters are required to define the shape and size of a unit cell,

these being three cell edge lengths (conventionally, defined as a, b, and c),

and three angles (conventionally, defined as , , and ). In the strict mathematical

sense, a, b, and c are vectors since they specify both length and direction.

is the angle between b and c, is the angle between a and c, is the angle

between a and b. The unit cell should be right handed. Check the cell above with

your right hand

Unit Cell Types and The Seven Crystal Systems

Cubic a = b = c. = = = 90º.

Tetragonal a = b c. = = = 90º.

Orthorhombic a b c. = = = 90 º.

Monoclinic a b c. = = 90º, 90º.

Triclinic a b c.. 90º.

Rhombohedral a = b = c. = = 90 º.

(or Trigonal)

Hexagonal a = b c. = = 90º, = 120º.

Orthorhombic

a

cb

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Cubic

Tetragonal

Orthorhombic

Monoclinic

Triclinic

Primitive Cell

Body

Centred

Cell

Face

Centred

Cell

End Face Centred

Cell

P

I F

C

Trigonal

Hexagonal

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The 14 Bravais Lattices

5

Primitive and Centered cells

On the previous slide you can see that that a Monoclinic unit cell can be primitive

or centred (by convention on the c face). These are referred to as Monoclinic P or

Monoclinic C.

Choice of Unit CellA unit cell can be any unit of a lattice array which when repeated in all directions,

and always maintaining the same orientation in space, generates the

lattice array.There is no unique way of choosing

a unit cell. For example, each of the

cells (A to D) in Fig. 2 are OK.

However, the cell favoured by

crystallographers is smallest most

orthogonal cell that displays all of

the symmetry of the lattice.

Thus, cells C and A are the

preferred unit cells for the lattices

of Figs. 2 and 3 respectively.

A B

C D

A

B

Fig. 1 Fig. 2Fig. 2 Fig. 3pma 2019

6

1. Every crystal system has a primitive Bravais lattice or P type.

2. The distribution of lattice points in a cell must be such as to maintain the

total symmetry of the crystal system.

Thus, the cubic system cannot have a C-type cell.

3. The fact that a unit cell meets the symmetry requirements of a crystal

system does not guarantee its inclusion within the crystal system.

This could result if the lattice it generated could be equally well represented

by a unit cell type which is already included within the crystal system.

The C-type cell for the tetragonal system (see Fig. 4) provides a good

example.

Fig. 4

C - cell

P - Cell

Four important points on crystal lattices:

4. If you apply point 3 to the orthorhombic

system you will find that the primitive cell you

generate will not have 90º angles. This would not

be orthorhombic and thus orthorhombic C is

included in the orthorhombic system. This is

shown on the next slide.

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Fig. 4

C - cell

P - Cell

A simplified view down c-axis can be

used to illustrate points 3 and 4 on the

previous slide.

Orthorhombic

a ≠ b ≠ c, a = b = = 90º

a

b

Angle not 90° thus the smaller cell is

not orthorhombic and must be rejected

Smaller cell is Tetragonal P and is preferred

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Tetragonal C and Orthorhombic C centred cells

Tetragonal

a = b ≠ c, a = b = = 90º

b

a

8

Symmetry - Point Groups and Space Groups

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C2

C

O

HH

z

y

x

sv

sv’

• Point Groups describe symmetry about

a point.

• Formaldehyde has the symmetry of the

C2v point group.

• The symmetry operations of C2v are C2

(rotation about z of 360/2), two planes

of symmetry sv and sv’ (vertical planes)

and the identity operation.

• C2v is the Schoenflies symbol for the

point group

• mm2 is the Hermann-Mauguin symbol

for this point group.

• Stereographic projections can be used

to represent point groups

• There are 32 crystallographic point

groups (also called classes)

• Point groups cannot describe a crystal

lattice – Space Groups are required.

+

+

+

+

Stereographic projection

of C2v or mm2 (pick any

+ and apply C2 and s to

get the others)9

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Crystal System 32 Crystallographic Point Groups

Triclinic 1 1

Monoclinic 2 m 2/m

Orthorhombic 222 mm2 mmm

Tetragonal 4 4 4/m 422 4mm 4 2𝑚 4/mmm

Trigonal 3 3 32 3m 3 m

Hexagonal 6 6 6/m 622 6mm 6 2𝑚 6/mmm

Cubic 23 𝑚3 432 4 3𝑚 𝑚3 𝑚

The reduction of a space group to a point

group is described on slide 16

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Space Groups from Point Groups.

Point Group symmetry operations (sym.ops.)

• Identity x,y,z → x,y,z

• Inversion x,y,z → -x,-y,-z

• Mirror e.g. xy plane mirror x,y,z → x,y,-z

• Rotation axis rotation by 360/n n = 1,2,3,4 or 6.

Space Group sym.ops. also have translational symmetry – screw axes

and glide planes

A screw axis is represented by nm where n is the rotation (360/n) and m/n

is the fraction of the unit cell length of the translation e.g. a 21 along b

• 21 along b x,y,z → -x,1/2+y, -z

A glide plane has translation (often ½) and a reflection

• b glide with a yz mirror x,y,z → -x,½+y,z

• Combining these symmetry operations with the 32 point groups leads to

the 230 possible 3d Space Groups.

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The distribution of Space Groups among the Bravais lattice type is shown in

Fig. 9.

The 230 Space Groups

CRYSTAL SYSTEMS (7)

Cubic

Tetragonal

Orthorhombic

Monoclinic

Triclinic

Rhombohedral

Hexagonal

BRAVAIS LATTICES (14)

P

F

I

P

I

P

F

I

P

C

P

P

SPACE GROUPS (230)

49

19

30

9

C and A 15

5

8

5

2

27

P and R 25

68

59

13

2

25

27

36

15

11

10

Fig. 9

Space Group determination is an

important step in crystal

structure determination.

The International Tables for

Crystallography list the symmetry

properties for all 230 Space Groups.

The 2nd edition was in one volume and

edited by Kathleen Lonsdale. The

current edition runs to 7 volumes.

The CSD or Cambridge Data Base is a

repository for the structures of organic

and organometallic compounds which

in 2019 exceeded 1000000 entries.

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The ABSEN program within Oscail can provide Bar Charts

of the contents of the Cambridge Data Base (CSD)

The number of entries by crystal system Entries in the first 25 space groups

No 14 most entries

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Space Group No. 14 P21/c

• This monoclinic space group has the most entries on the CSD

• Read its name as “p21 upon c”

• The full name is P 1 21/c 1 (there is no symmetry on a or c)

• There are 4 general positions 1 x,y,z 2 -x,1/2+y,1/2-z 3 -x,-y,-z 4 x,1/2-y,1/2+z

4

3

2

1

Stereographic view down b View down a

21 screw axis glide plane at 1/4b inversion centre

glide normal to screen at 1/4b

(21 at 0,y,1/4) Inversion at (0,0,0) Glide at (x,1/4,z)

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Space Group No. 14 P21/c, Example Benzoic acid CSD BENZAC

2

34

1

• The molecules in the unit cell of

benzoic acid illustrate the

positions in the stereographic

projection of the space group

• 1 and 2 are related by a 21 screw

• 1 and 3 by an inversion centre

• 1 and 4 by a c glide

1

2

3

4

0 a

c

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Converting a Space Group to a Point Group

• When the translational parts of the symmetry

operations are removed the Space Group is

reduced to a point group

• The P21/c symm ops are;

1, x,y,z 2, -x,½+y,½-z 3 -x,-y,-z 4 x,½-y,½+z

• Removing the ½ s will remove the translations

leaving

• 1, x,y,z 2 -x,y,-z, 3 -x,-y,-z 4 x,-y,z

• With respect to 1 symm ops 2,3 and 4 now are

a 2-fold axis along b, inversion and a mirror

normal to b. Thus the HM symbol for this is 2/m

and the Schoenflies symbol is C2h

• Thus P21/c is reduced to the point group 2/m.

2/m is the only centrosymmetric monoclinic

point group

Stereographic

projection of 2/m

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Crystal Planes and Miller Indices

The use of crystal planes to describe the

structure of crystals goes back to the

start of crystallography and crystal

planes were used by Bragg to explain

diffraction as will be seen later.

Crystal planes are defined by the

intercepts they make on the

crystal axes of the unit cell.

The inverse of these fractions are

the Miller Indices of the planes.

In (a) the intercepts are ½, ½, 1 and

the Miller Indices are (2 2 1).

In (c) the intercepts on b and c are at

infinity the inverse of which is 0 and

the plane is the (2 0 0).

In (f) the plane cuts the negative c

axis at -1 and thus is (1 1 -1). In

crystallography -1 is often written ī

and pronounced “Bar 1”. 17

incidentbeam

reflectedbeam

x

y

m n

d

o o

o o

o oA B

C D

a

incidentbeam

reflectedbeam

x

y

m n

d

o o

o o

o oA B

C D

d

aUNITCELL

Z

O P

(a)

(b)

Fig. 11

(2,0,0)

(1,0,0)

E F

UNIT

CELL

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Diffraction and the Bragg Equation

Max von Laue was the first to

suggest that crystals might

diffract X-rays and he also

provided the first explanation

for the diffraction observed.

However, it is the explanation

provided by Bragg that is simpler

and more popular.

In the Bragg view crystal

planes act a mirrors.

Constructive interference

is observed when the path

difference between the two

reflected beams in (a) = nl.

The path difference in (a) is

2my. Since my/d = sin

2my = 2dsin = nl

where d is the interplanar spacing.

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l =sin2 )0,0,1(d

l 2sin2 )0,0,1( =d

On the previous slide in (a) it is clear that the planes are the (1,0,0)

set of planes. If the path difference is simply one wavelength the

Bragg condition can be stated as

This is a first order reflection. If the path difference is

two wave lengths the Bragg condition becomes

and the reflection is a second order reflection.

Bragg Reflection Order

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CRYSTAL SYSTEMand

UNIT CELL DIMENSIONS

FULL DATA SET

COLLECTION

BRAVAIS LATTICE

SPACE GROUP

CONSTRUCT AN

ELECTRON DENSITYMAP

LOCATE ATOMPOSITIONS

STRUCTUREREFINEMENT

SELECT A SUITABLE

CRYSTAL

A

B

C

D

E

F

G

Step by Step Single Crystal Structure

Solution using X-ray Diffraction

• Bragg's equation specifies that, if a crystal is rotated

within a monochromatic X-ray beam, such that every

conceivable orientation of the crystal relative to the

beam is achieved, each set of planes will have had the

opportunity to satisfy the Bragg equation and to give

rise to a reflection.

• In order to solve a crystal structure it is necessary to

record a large number of reflections.

• Many experimental techniques have been devised to

achieve this. The steps involved in a crystal structure

determination are summarised in the flow chart on the

right.

• When you have had a look at the introduction to single

crystal X-ray diffraction given here you can look the

worked examples in Oscail tutorials on Crystallogtaphy.

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Single Crystal X-Ray Data Collection

X-ray Beam

Beam Stop

Diffracted Beam & Spot

Image Plate / CCD

Crystal

X-ray Beam

Beam Stop

Diffracted Beam & Spot

Image Plate / CCD

Crystal

In a typical setup a crystal is oscillated

over < 2° while an image is collected

the crystal is then rotated by the same

amount and oscillated again. The

process is repeated over a total range

of up to 180 or 360°.

Data collection time depends on the

crystal size, quality and other factors

and may be from 40 mins to several

hours on a lab diffractometer and just

seconds on a synchrotron.

The first crystallographic

data collection systems used

photographic methods. Modern

diffractometers use electronic

area detectors which measure

hundreds of reflections at a

time.

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Determination of the Lattice type and Space Group

High symmetry can lead to reflections being systematically absent from the

data set. Absent reflections have no measurable intensity. There are two types

of absences, General Absences and Special Absences.

The general absences determine the lattice type;

Primitive (P) has no general absences and no restrictions on h, k or l.

End Cantered (C) h+k=2n+1 are all absent.

Face Cantered (F) only h, k, l, all even or all odd are observed.

Body Cantered (I) h+k+l=2n+1 are all absent.

The special absences refer to specific sets of reflections and are used to

detect the presence of glide planes and screw axes. Some Space Groups

are uniquely determined by special absences but in many cases several

Space Groups will have to be considered.

Computer programs are able to lay out the data in tables with absences

indicated and possible Space Groups can be suggested however the choice of

Space Group will often need to be carefully considered.

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Group Cond. Op. All Odd Cut1 Cut2 Cut3 Op. No.

h00 h=2n+1 21.. 18 10 8 8 8 1

0k0 k=2n+1 3 1 1 0 0 .21. 2

00l l=2n+1 11 6 0 0 0 ..21 3

0kl k=2n+1 b.. 95 53 49 44 37 4

0kl l=2n+1 c.. 49 43 40 33 5

0kl k+l=2n+1 n.. 40 34 30 28 6

h0l h=2n+1 .a. 412 211 96 89 81 7

h0l l=2n+1 211 1 1 0 .c. 8

h0l h+l=2n+1 .n. 212 95 88 81 9

hk0 h=2n+1 ..a 168 84 67 60 49 10

hk0 k=2n+1 ..b 84 76 69 48 11

hk0 h+k=2n+1 ..n 86 71 69 55 12

hkl k+l=2n+1 A.. 1591 1196 1084 902 13

hkl h+l=2n+1 .B. 1638 1271 1151 915 14

hkl h+k=2n+1 ..C 1651 1285 1145 921 15

hkl h+k+l=2n+1 I 1637 1288 1148 943 16

hkl not all odd/even F 2440 1876 1690 1369 17

Reflection Analysis I/sI Cut1 Cut2 Cut3 = 3.0 6.0 12.0

P21/c (14)

CSD Total for all SGs in 2019 has exceeded 1000000

In this case P21/c is the only choice offered and this is likely to be correct.

Notice the symmetry operations move to the right when present in the data.23

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h0l zero level reflections for a P21/c example

300 reflection

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Solving the Structure

The unit cell, the Space Group and the reflection intensities cannot be used

to generate the structure as there is no reflection phase information in the

data set. This is the phase problem.

If the reflection phases were known then an electron density map could be

calculated using a Fourier series. If (x,y,z) is the electron density at x,y,z then

= .)cos(coscoscos1

),,( )0,1,1()1,0,0()0,1,0)0,0,1( etcb

y

a

xF

c

zF

b

yF

a

xF

Vzyx

),,(

2

),,( lkhlkh IF =The F here is the square root of the measured intensity

When intensity is measured it is measured without sign and the phase is lost.

There are three ways to solve the phase problem:

1. The Patterson or heavy atom method

2. Direct Methods (Hauptman and Karle 1985 Nobel prize)

3. The Charge flipping method is a recent development

The ShelxT program combines these methods automatically, solves the

structure and suggests space groups.

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Data ResolutionIt is the amount of 4s data at 1Å that determines the success of direct methods

this example has sufficient 4s data for direct methods.

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Refining a Structure

It should be possible to “see” atoms in an electron density map if it has good

resolution i.e. at least 1Å resolution. The steps in refining a structure are.

1. Use whatever atoms you have that look OK to generate an electron density

map.

2. The known atoms are subtracted from this to generate a difference map.

3. Any atoms that have been missed should be in the difference map.

4. The refinement process minimises the difference between observed and

calculated reflection intensities.

5. In the final difference map there should be no peaks larger than a H atom

i.e. > 1e/Å3. (A H atom has a volume of about 1Å3 and has 1 e.)

Resolution The resolution of a crystal structure is usually quoted in

Angstroms, Å. Standard small molecule structures should always be at least

of 1 Å resolution to give accurate bond lengths. Resolution can be related to

Bragg angle at any wavelength through the Bragg equation nl = 2d sin.

Using the value of the reflection with the largest Bragg angle in a data set

then d = l/2sin gives the resolution. The pattern shown on slide 15 has a

resolution of 0.98Å at the edge.

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Final stages of refinement.

There are many was in which a structure can be “improved”. The two most

important considerations are addition of hydrogen atoms and anisotropic

refinement of the non-hydrogen atoms.

Addition of hydrogen atoms – Hydrogen atoms have only 1 electron and are

often not seen in difference maps. It is best to include them at calculated

positions. This is easy to do and it will improve the “R factor”.

Anisotropic refinement of the non-hydrogen atoms – In the early stages

atoms are refined as if they were spheres. Since atoms vibrate in a way that

is controlled by chemical bonds and interactions with their neighbours,

it is better to refine then as ellipsoids. One parameter (the radius) is enough

to define a sphere this with x,y,z means that isotropic refinement requires

4 parameters per atom. An ellipsoid needs 6 parameters thus an anisotropic

atom requires 9 parameters.This is an example of an anisotropic atom

R Factor – The R factors used are Rw and wR2. Rw should be < 8% and

wR2 should be <15%. The lower the better. Rs are of the form Sum[(I0-Ic)/Io]

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The Need for Single Crystals - In order to carry out a detailed X-ray structure

determination, it is essential to have a crystal of the material in question.

Many compounds cannot be crystallised and thus are not amenable to

diffraction studies. There are also commercially important materials such

as glasses and many ceramics which owe their unique properties to their

amorphous nature. Being amorphous (no long range order), the structures

of these materials cannot be investigated in detail by diffraction techniques.

Locating Hydrogen atoms - Hydrogen atoms make extremely small contributions

and for this reason X-ray crystallography is not a good technique for accurately

locating hydrogen atom positions. If the location of hydrogen atoms is of specific

interest (e.g. in the study of hydride structures and hydrogen bonding interactions)

use has got to be made of the much more expensive and less available

technique of neutron diffraction. The theory of neutron diffraction is very similar

to that for X-ray diffraction but an essential difference is that hydrogen atoms

scatter neutrons as effectively as many other atoms and for this reason they

can be located with good accuracy in the structure determination.

Problems with single crystal X-ray Crystallography

Low Temperature Structure Determination – When X-ray data are collected at

low temperature (<-150 ºC) thermal ellipsoids are smaller and better defined.

N.B. bond lengths show very little variation with temperature.

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Using CheckCif to examine the quality of a structure

• The IUCr CheckCif service should be used to

check the quality of crystal structures

• It is available online at https://checkcif.iucr.org/

• Problems/Alerts are ranked A, B, C etc.

• A and B alerts should be examined and fixed if

possible.

• C alerts often indicate ways in which structures

can be improved.

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A Really Good Structure.In this structure the diff map peaks are drawn as lime green spheres and you

can see that they are all "bond blobs" caused by the effects of chemical bonding

which slightly distorts the spherical atoms. This is the limit of what can be

achieved using the spherical atom model.

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X-Ray Powder Diffraction (XRPD)

A crystalline powder sample will

diffract X-rays but since the

orientations of the individual

crystals are random the data set

produced is a plot of intensity v.s.

diffraction angle or Bragg angle .

Here the sample is sitting on a

flat plate and the plate is turned

about the centre of the

diffractometer at half the rate

through which the detector

moves. This is the /2 or Bragg

scan method.

Notice the plot contains 2 on

the X-axis and X-ray intensity

on the y-axis.

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Uses of X-ray Powder DiffractionIn general, powder diffraction data are unsuitable for solving crystal structures.

Some advances have recently been made using the Rietveld method. However

this is far from trivial and it works best in relatively simple cases. It is very difficult

to be sure that the unit cell is correct as the reflections overlap and are difficult to

resolve from one another. There may also be problems with preferred orientation

of crystallites.

Important advantages and uses of powder diffraction:1. The need to grow crystals is eliminated.

2. A powder diffraction pattern can be recorded very rapidly and the technique

is non-destructive.

3. With special equipment very small samples may be used (1-2mg.)

4. A powder diffraction pattern may be used as a fingerprint. It is often superior to

an infrared spectrum in this respect.

5. It can be used for the qualitative, and often the quantitative, determination of

the crystalline components of a powder mixture.

6. Powder diffractometry provides an easy and fast method for the detection of

crystal polymorphs. Powder patterns are provided when a drug is being

registered with the FDA. (Polymorphs are different crystal forms of the same

substance.)

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Preferred Orientation Effects in XRPD

It is possible to calculate the theoretical diffraction pattern if the

crystal structure is known.

Calculated pattern

Observed pattern

There are no preferred orientation effects here as all reflections have their

expected intensity.

Nifedipine

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XRPD of Benzoic acid

Calculated pattern

Observed pattern

There is clear preferred orientation here.

The 002 is the flat face exposed when the

needles lie down on a flat plate.

002

004

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Some points relating to preferred orientation effects.

• Preferred orientation effects are often observed for needles and

plates.

• Preferred orientation effects can be reduced by sample rotation

and sample grinding.

• When an indexed calculated pattern is compared to that of a

sample showing preferred orientation it may be possible to to

index the faces of plate like crystals.

• Deviations from calculated patterns can be used to monitor the

crystal morphology of production batches.

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2

2

2

2

2

2

2),,(

1

c

l

b

k

a

h

d lkh

=

2

2

2

22

2),,(

)(1

c

l

a

kh

d lkh

=

2

222

2),,(

)(1

a

lkh

d lkh

=

For an orthogonal system ( = = = 90°) the relationship between

interplanar spacing (d) and the unit cell parameters is given by the

expression:

This is the expression for an orthorhombic crystal.

For the tetragonal system it reduces to

and, for the cubic system, it further reduces to

Calculations using X-ray powder diffraction patterns.

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Site Na+ Cl-

Central 0 1

Face 6/2 0

Edge 0 12/4

Corner 8/8 0

Total 4 4

Important Cubic Lattice Types

Two of the most important cubic lattice types are the NaCl type and the

CsCl type.

Stoichiometry (formula) from the Unit Cell

In the CsCl structure both ions have coordination numbers of 8 and the structure

is a simple primitive one with no centring.

Formula Cs at centre = 1

8 x 1/8Cl = 1 = CsCl

NaCl crystallizes in the Space Group Fm-3m

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The unit cell of a cubic close packed

Metal has a face cantered or F type lattice

The formula of the unit cell is:

6 x ½ + 8 x 1/8 = 4

Cubic close packed spheres

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l sin2dn =

The Bragg equation may be rearranged (if n=1)

from to l 2

2

2

sin4

=d

If the value of 1/(dh,k,l)2 in the cubic system equation above is inserted into

this form of the Bragg equation you have

)(4

sin 222

2

22 lkh

a=

l

Since in any specific case a and l are constant and if l2/4a2 = A

)(sin 2222 lkhA =

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Analysis Cubic XRPD patterns

Example 1

Aluminium powder gives a diffraction pattern that yields the following eight

largest d-spacings: 2.338, 2.024, 1.431, 1.221, 1.169, 1.0124, 0.9289

and 0.9055 Å. Aluminium has a cubic close packed structure and its

atomic weight is 26.98 and l = 1.5405 A .

Problem - Index the diffraction data and calculate the density of aluminium.

)(sin 2222 lkhA =

l sin2d=The Bragg equation, can be used to obtain sin,d2

sinl

=

The ccp lattice is an F type lattice and the only reflections observed are those

with all even or all odd indices.

Thus the only values of sin2 in that are allowed

are 3A, 4A , 8A, 11A, 12A,16A and 19A for the first eight reflections.

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d/Å Sin Sin2 Calc. Sin2 (h, k, I)

2.338 0.32945 0.10854 (1,1,1)

2.024 0.38056 0.14482 0.14472 (2,0,0)

1.431 0.53826 0.28972 0.28944 (2,2,0)

1.221 0.63084 0.39795 0.39798 (3,1,1)

1.169 0.65890 0.43414 0.43416 (2,2,2)

1.0124 0.76082 0.57884 0.57888 (4,0,0)

0.9289 0.82921 0.68758 0.68742 (3,3,1)

0.9055 0.85063 0.72358 0.72360 (4,2,0)

Insert the values into a table and compute sin and sin2.

Since the lowest value of sin2 is 3A and the next is 4A the first

Entry in the Calc. sin2 column is (0.10854/3)*4 etc.

The reflections have now been indexed.

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For the first reflection (for which h2 + k2 + l2 = 3)

sin2 = 3A = 3 ( l2 / 4a2 )

a2 = 3l2 / 4sin2

a = 4.04946 Å = 4.04946 x 10-8 cm.

Calculation of the density of aluminiuma3 = 66.40356 Å3 = 66.40356 x 10-24 cm3.

If the density of aluminium is (g. cm.-3), the mass of the unit cell is

x 66.40356 x 10-24 g.

The unit cell of aluminium contains 4 atoms.

The weight of one aluminium atom is 26.98/(6.022 x 1023) = 4.48024 x 10-23

and the weight of four atoms (the content of the unit cell) is 179.209 x 10-24.

x 66.40356 x 10-24 = 179.209 x 10-24

p = 2.6988 g.cm-3.

Calculation of a

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Calculate the following:

On the basis that the structure is cubic and of either the NaCl or CsCl type

1. Index the first six reflections. , 2. Calculate the unit cell parameter, 3.

Calculate the density of AgCl.(Assume the following atomic weights: Ag, 107.868; Cl, 35.453;

and Avogadro’s number is 6.022 x 1023)

Example 2

The X-ray powder diffraction pattern of AgCl obtained using radiation of

wavelength 1.54Å is shown below. The peaks are labelled with 2θ values

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Since values are available sin2 values can be calculated and inserted

in a table.

1. for a face centred lattice 3A, 4A , 8A, 11A, 12A and 16A

2. for a primitive lattice 1A, 2A, 3A, 4A, 5A and 6A

2 Sin2

27.80 13.90 0.0577

32.20 16.10 0.0769

46.20 23.10 0.1539

54.80 27.40 0.2118

57.45 28.73 0.2310

67.45 33.73 0.3083

The second option is not possible as the first 2 are not in the ratio of 1:2.

To test the first option, divide the first by 3 and multiply the result by 4, 8 etc.

Calc. Sin2

0.07693

0.1539

0.2116

0.2308

0.3077

From Sin2 = A(h2 + k2 + l2) the possible values are:

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Since sin2 = l2(h2 + k2 + l2)/4a2

a2 = (1.54)2.(16)/4(0.3083) using the largest (most accurate) 2

a2 = 30.7692

a = 5.547Ǻ (1Ǻ = 10-8 cm)

Formula wt. of unit cell = 4AgCl = 573.284g

This is the weight of 4 moles of AgCl.

The weight of 4 molecules is 573.284 / (6.02 x 1023)

Density = 573.284 / (6.02 x 1023)(5.547 x 10-8)3

A is in Ǻ thus the answer should be multiplied by 1 / 10-24

Density = 5.580 g/cm3

Density of AgCl

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