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An Introduction to X-Ray Diffraction by Single Crystals and Powders Patrick McArdle NUI, Galway, Ireland LATTICE ARRAYS AND BRAVAIS LATTICES Crystalline materials differ from amorphous materials in that in the former there is
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Patrick McArdle NUI, Galway, Ireland
Crystalline materials differ from amorphous materials in that in the former there is
order in the arrangement of the molecular contents whereas in the latter there is
no order or at best a tendency for a short-range order.
The packing of atoms, molecules or ions within a crystal occurs in a symmetrical
manner and furthermore this symmetrical arrangement is repetitive.
A most important common characteristic that crystals may share is the manner
in which repetition occurs. This will be expressed in a common lattice array.
A lattice array is constructed from the arrangement of
atomic material within the crystal as follows:
Pick any position within the 2 dimensional lattice in Fig. 1(a) and note the
arrangement about this point. The chosen position can be indicated by
a point (a lattice point). In view of the repetitive arrangement, there will
be a 2 dimensional array of identical positions and if these are also
marked by a point a 2-dimensional lattice will result if the points are
In a real 3-dimensional crystal lattice the same ideas apply.
When crystal structures are represented by lattices, it transpires that all crystals
break down into one of fourteen three dimensional lattice arrangements.
Bravais demonstrated mathematically that there are only fourteen ways
in which repetitive symmetry can occur and the fourteen lattices representing
the ways in which repetition can occur are referred to as the Bravais lattices.
A 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
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 the one of
smallest volume 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.
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.
Hexagonal a = b c. = = 90º, = 120º.
between a and b. The unit cell should be right handed. Check the cell above with your right hand
When these unit cells are combined with possible “centering” there are
14 different Bravais lattices.
End Face Centred
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.
Four simple points on crystal lattices:
1. Every crystal system has a primitive Bravais lattice.
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.
4. If you repeat 3. within the orthorhombic
system you will find that the primitive cell you
generate will not have 90º angles. This is not
orthorhombic and thus orthorhombic C is
included in the crystal system.
a ≠ b ≠ c, a = b = = 90º
a = b ≠ c, a = b = = 90º
Smaller cell is Tetragonal P
Angle not 90° smaller cell
A simplified view down c-axis can be used to illustrate points 3 and 4
Point group theory is not dealt with here. What follows is just a summary. Point group symmetry defines the symmetry of an isolated object or
group of objects, whereas space group symmetry further defines the systematic
fashion in which an object, or group of objects is repeated in space to
generate an infinite periodic array in 3D.
Point group symmetry is quantified in terms of symmetry elements
(existing within the object or group of objects) and their associated operations.
Four symmetry elements are used to quantify point group symmetry
Symmetry Element Symmetry Operation
Rotation axis (n-fold) Rotation
Mirror plane Reflection
Centre of Symmetry Inversion
Rotor-reflection axis (n-fold) Rotation and reflection
Rotor-inversion axis (n-fold) Rotation and inversion
To generate a 3D lattice from an object it is necessary to add translational
symmetry to point group symmetry. The two important space group
symmetry operations which move objects are glide planes and screw
axes. These operations combine translation and reflection and translation
and rotation respectively.
The pentagons on the
left are related by
In 5(b) the pentagon
on the top left of the
cell is related to the
one in the centre by
translation a/2 followed
by either reflection or
rotation. Centres of
inversion in 5(b) are
marked with tiny
BRAVAIS LATTICES (14)
SPACE GROUPS (230)
C and A
P and R
The 230 Space Groups
There is an infinite number of combinations of the four symmetry elements.
However, if there is a restriction on the order of the rotation axes to
2, 3, 4, and 6, as is the case for repetitive symmetry (crystallographic
symmetry) this leaves only 32 unique combinations. These are the 32
crystallographic point groups. Adding screw axes and glide planes gives
the 230 space groups. The overall breakdown of symmetry for crystals
then is as shown in Fig. 9.
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.
Space Group determination is an important step in crystal structure determination.
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
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 (d) 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”.
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.
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.
In (a) above 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.
SINGLE CRYSTAL DIFFRACTION TECHNIQUES
N.B. The crystal must be a single crystal.
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 will have
given rise to reflection.
In order to solve a crystal structure it is necessary
to record a large number of reflections.
This implies accurately measuring their intensities and
recording their directions with respect to crystal orientation
and initial X-ray beam direction.
Many experimental techniques have been devised
to achieve this. The steps involved in a crystal structure
determination are summarised in the flow chart.
The first crystallographic data collection systems used
photographic methods. These
were replaced by automateddiffractometers which measuredreflections one at a time. A typical
data collection took several days.
modern systems use area detectors
which measure 100s at a time.
The crystal is oscillated over < 2°
while an image is collected then
rotated by the same amount
and oscillated again. The process is
repeated over a total range of about 180°.
Each image is exposed for < 100s.
Thus if readout time is ignored total
data collection time is often < 3 hr.
A typical image shown to the left.
A computer program is used to predict
the unit cell from several images.
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 much thought.
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
CSD Total for all SGs at end of 2006 385035
In this case P21/c is the only choice offered by Oscail and this is likely to be correct.
Notice the symmetry operations move to the right when present in the data.
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
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.
After many years of study statistical trends were observed in reflection phases.
It was also observed that some reflections are more important than others.
The Direct Methods approach tries to guess the reflection phases and then
awards a Figure of Merit to the guess. The most popular is the Combined Figure
of Merit or CFOM). It is normal practice to accept solutions with CFOM values
in the range 0.03-0.15 and to try to use these phases to generate an
electron density map.
The Patterson or heavy atom method.
It is the electrons that scatter X-rays. Atoms that are heavier than all others
in the crystal will give rise to the strongest reflections. Heavy is a relative term
however, all atoms after Rb are normally considered to be heavy. Iodine for example will scatter a lot of X-rays compared to carbon. Using the
Patterson method it should be possible to calculate the position of an iodine
atom and this could be used to phase most of the reflections and calculate
an electron density map.
The charge flipping method works well if the data has good resolution.
There is an example in the Oscail tutorials. All non-H atoms are often identified correctly in the solution.
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
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 n = 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.
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
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”.
R Factor – The R factors used are Rw and wR2. Rw should be < 8% and
wR2 should be <15%. The lower the better. If the R or error is greater then
these values the structure is not much use. Rs are of the form Sum[(I0-Ic)/Io]
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.
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.
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.
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 counter
moves. This is the /2 or Bragg
Notice the plot contains 2 on
the X-axis and X-ray intensity
on the y-axis.
In 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.
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
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
For an orthogonal system (a = b = = 90°) the relationship between
interplanar spacing (d) and the unit cell parameters is given by the
This is the expression for an orthorhombic crystal.
For the tetragonal system it reduces to
and, for the cubic system, it further reduces to
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
Important Cubic Lattice Types
Two of the most important cubic lattice types are the NaCl type and the
NaCl crystallizes in the Space Group Fm-3m
Stoichiometry (formula) from the Unit Cell
Site Na+ Cl-
Central 0 1
Face 6/2 0
Edge 0 12/4
Corner 8/8 0
Total 4 4
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
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
Since in any specific case a and l are constant and if l2/4a2 = A
Diffraction data have been collected on a powder diffractometer for a series of
compounds that crystallise in the cubic system.
In the practical course each person will calculate the unit cell parameter and
density from an X-ray powder diffraction pattern.
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 .
Index the diffraction data and calculate the density of aluminium.
can be used to obtain sin,
The Bragg equation,
The ccp lattice is an F type lattice and the only reflections observed are those
with all even or all odd indices.
that are allowed
Thus the only values of sin2 in
are 3A, 4A , 8A, 11A, 12A,16A and 19A for the first eight reflections.
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.
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 aluminium
a3 = 66.40356 Å3 = 66.40356 x 10-24 cm3.
If the density of aluminium is r (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.
r x 66.40356 x 10-24 = 179.209 x 10-24
p = 2.6988 g.cm-3.
wavelength 1.54Å is shown below. The peaks are labelled with 2θ values
Since values are available sin2 values can be calculated and inserted
in a table.
From Sin2 = A(h2 + k2 + l2) the possible values are:
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
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.
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
It is possible to calculate the theoretical diffraction pattern if the
crystal structure is known.
There are no preferred orientation effects here as all reflections have their
There is clear preferred orientation here.
The 002 is the flat face exposed when the
needles lie down on a flat plate.