an introduction to x-ray diffraction by single crystals ... · long range order. they also exhibit...
TRANSCRIPT
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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