Diffraction Experiments, Datacloud.crm2.univ-lorraine.fr/pdf/Bogota2018/Suescun_Diffraction.pdf · Representación esquemática de un cristal y su diagrama de difracción. Powder

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Diffraction Experiments, Data processing and reduction

Leopoldo SuescunLaboratorio de Cristalografía Química del Estado Sólido y Materiales,

Facultad de Química, Universidad de la República, Montevideo, Uruguay.

[email protected]

November 29th, 2018

Bragg’s Law

send2n hkl

2

d

2

2

2

2

2

2

1

c

l

b

k

a

hdhkl

For Orthogonal unit cells only

Bragg’s Law

send2n hkl

2

d

A

B

C D

2’

11’

d

2

2

2

2

2

2

1

c

l

b

k

a

hdhkl

For Orthogonal unit cells only

Phase shift:CB+BD

d

A

B

CD

CB=d*sin()

CB+BD=2d*sin()

For constructive interferenceCB+BD = n

Bragg’s Law & Laue Equations

• Another equation for Braggs law:

A

11’

2 2

0S

S

0ˆˆ SS

s



2

0S

S

0ˆˆ SS

s

)(2)90cos(

ˆ

2

0 senss

Ss

s

2

)(2

)(2

dsen

sens

ds

sens

1

)(1

2

𝑠 𝑑

s



Reciprocal lattice

a*

c*

b*

a

b

c

hkld

''' lkhd

Antitransform

FourierTransform

Crystal Structure Diffraction pattern

hkl

hklhkld

sr1*

'''

'''

*

'''

1

lkh

lkhlkhd

sr

0

V

bac

V

acb

V

cba

***

Reciprocal lattice

Nomenclature:

a*

c*

b*

a

b

c

V

bac

V

acb

V

cba

***

𝑟ℎ𝑘𝑙∗ = ℎ 𝑎

∗+ 𝑘𝑏

∗+ 𝑙 𝑐

∗= 𝑠ℎ𝑘𝑙

𝑠ℎ𝑘𝑙 = ℎ 𝑎∗+ 𝑘𝑏

∗+ 𝑙 𝑐

∗

𝑠ℎ𝑘𝑙 = 𝑠ℎ𝑘𝑙

𝑠ℎ𝑘𝑙 = 𝑠ℎ𝑘𝑙 =1

𝑑ℎ𝑘𝑙

Reciprocal lattice and Ewald sphere

hkls

1

Rayos Xincidentes

hkl

hkld

s1

0Origen R.R.

Rayos Xdifractados

2

cristal

hklshkl

hkld

s1

0

2

hkl

hkld

s1

1

1

hkl

hkl

d

d

2

1

2)sin(

1*

)sin(2 hkld


hklds

1)sin(

2

hkl

hkld

s1


The Ewald sphere is the geometric site where the reciprocal lattice vectors fulfill the Laue equations or the family of planes fulfill Bragg’s Law therefore X-rays exit the crystal throught the reflection sphere at the point where the vector touches the sphere.

hkls

1

Rayos Xincidentes

hkl

hkld

s1

0Origen R.R.

Rayos Xdifractados

2

cristal

1

1

1

Rayos Xincidentes 0

RetículoRecíproco

cristal

Rayos Xdifractados


***1

)sin(2

clbkahd

shkl

hkl

Shkl Shkl

Esfera de ewald

hkls

1

Rayos X

incidentes

hkl

hkld

s1

Rayos X

difractados

2

Cristal

0Origen R.R.

1

Esfera de ewald

Diffraction condition: when a RL point intersects the Reflection sphere X-ray will emerge from the cristal through that point.

1

Rayos X

incidentes 0

Retículo

Recíproco

cristal

Rayos X

difractados

Ewald Sphere or Reflection Sphere

Diffraction condition: when a RL point intersects theReflection sphere X-ray will emerge from the cristal throughthat point

1

Rayos X

incidentes 0

Retículo

Recíproco

cristal

Rayos X

difractados

Rotate the crystal and the reciprocal lattice

Limiting Sphere

Could all points of the Reciprocal Lattice be made toto intersect the reflection sphere for a given ?.

1

Rayos X

incidentes 0

Retículo

Recíproco

cristal

Limiting sphere

1

Rayos X

incidentes 0cristal

Retículo

Recíproco

Limiting sphere

1

Rayos X

incidentes 0cristal

Retículo

Recíproco

Limiting Sphere

1

Rayos X

incidentes 0cristal

Retículo

Recíproco

2

Those reciprocal lattice points that lay within 2/ of the origin (that have |s|< 2/ or d>/2)can be accessed in a diffraction experiment for a given .

• Using Bragg’s Law we can also calculate the maximum resolution (minimum d spacing) that we can measure in a diffraction experiment.

𝜆 = 2𝑑𝑠𝑖𝑛 𝜃 ⇒ 𝑑𝑚𝑖𝑛 =𝜆

2𝑠𝑖𝑛 𝜃𝑚𝑎𝑥=

𝜆

2𝑠𝑖𝑛 90=

𝜆

2

the minimum interplanar spacing accessible in the experiment depends on the wavelength. At 2=180 the minimum resolution is obtained, we cannot measure diffraction data beyond that angle.

Limiting Sphere

0

1cristal

2

Limiting Sphere

0

maxmin

2sd

Limiting Sphere

maxmin '''

2sd

’ > ,

d’min>dmin, s’max<smax

The shorter the better resolution for the same 2 angle.

0

1cristal

2

Limiting SphereIf we assign each reciprocal lattice node the volume of the reciprocal unit cell, we can determine how many nodes are within the Limiting sphere, or the total number of reflections Nmax

accessible by the experiment.

𝑁𝑚𝑎𝑥 = 𝑉𝐿𝑆𝑉∗ =

43𝜋

2𝜆

3

𝑉∗=32𝜋𝑉

3𝜆3

Limiting Sphere

BaTiO3 at room temperature is tetragonalwith space group P4mm, a=3.999 Å, c=4.017 Å and V=64.269 Å3

Calculate the number of accessible reflections and the minimum dhkl available (maximum resolution) if:- (MoKa)=0.71073 Å- (CuKa)=1.5418 Å

Idem if you consider the structure of hen-egg lysozyme with space group P43212 with a=60.01 Å, c=261.8 Å and V=942938 Å3


43𝜋

2𝜆

3

𝑉∗=32𝜋𝑉

3𝜆3

Experimental resolution

When we perform an experiment, reaching the maximum available resolution requires to move the detector up to 2=180.

Normally we would limit the resolution of the experiment to a diffractometer-limited angle (140 to 160 depending on the maker and limitations of the instrument).

If we limit the scan to 2max, smax=/[2sin(max)] and dmin=2sin(max)/

the number of measurable reflections reduces to:


43𝜋

1𝑑𝑚𝑖𝑛

3

𝑉∗=

4𝜋𝑉

3𝑑𝑚𝑖𝑛3

1crystal

The diffraction experiment

beam-stop

0

Incident

X-rays,

1crystal

The diffraction experiment

beam-stop

0

Incident

X-rays,

Single crystal data collection

• Modern CCD/CMOS area detectors allow for simultaneous collection of hundreds/thousands of diffraction intensities (reflections).

• Data is collected in frames produced by rotation of the crystal by a fraction of degree.

• Programs automatically integrate the images into individual indexed intensities.

X rays ()

Single crystal intensities

• For each indexed reflection (dots inside the circles) an intensity value will be determined by a numerical integration procedure to provide the output in a h,k,l, Ihkl, sIhkllist that is used for structure determination and refinement.


• what should I do now?

Reflections, all, unique, observed

• Part of the reflection list for the compound CH4N2S.

• During data collection, some reflections are measured more than once (red).

• Some are equivalent by symmetry (blue).

• Some are not observed (I<2sI) (grey).

• Etc…


For the compound CH4N2S:a=7.6527 Å, b=8.5439 Å c=5.4858 Å, a=b=g=90°:=0.71073 Å, 2max=51.7°, smax=1.227 Å, dmin=0.815 Å

– 4119 intensities were integrated.

– 2227 were different– 2391 were observed (I>2sI)– 340 were unique

(unrelated by point group symmetry)

– 265 of the unique were observed.


• Identification of useful intensities is required before structure determination since we need to know the degree of over-determination of the problem to know if the structure determination procedure will succeed.

• For the compound CH4N2S a=7.6527 Å, b=8.5439 Å c=5.4858 Å, a=b=g=90°: – 264 of the unique were observed.

– With this number of unique observed intensities only a small number of atomic parameters can be determined accurately. It is a requisite of the IUCr Journals to have 10 reflections per structural parameters so only around 25 structural parameters can be determined.


• Even though we don’t need 3855 of the 4119 collected intensities for the structure determination and refinement, the redundancy allows for symmetry determination and data consistency checks, for the determination of other effects such as absorption or radiation damage of the crystal and to reduce the total uncertainty of the refined parameters due to a reduction in the uncertainty of the observed intensities.

• Rint (internal consistency of symmetry-related reflections) and Rsigma (average I/sI for all reflections) are good measures of the consistency between equivalent reflections (symmetry-related reflections that should have equal intensities) and the overall uncertainty of the recorded intensities respectively.


• Since the reciprocal space is the Fourier Transform of the crystal and the crystal is (assumed) periodic, all the information in the reciprocal space is condensed in the reciprocal lattice points .

• Therefore the intensities of the RL points contain all the structural information and should allow to determine the structure.

• We will start by determining the type of space group symmetry of the crystal.

• Intensities in the reciprocal lattice are related to the structure factors (Fhkl) through a series of corrections that should be accounted for correctly, so one can determine the structure from the intensities after proper data Processing and Reduction.

Powder Intensities

• If one set of reciprocal lattice points can be associated to each crystal in the sample, consider a polycrystalline (powder) sample:– A very large number (ideally > 107 crystallites)– Of (ideal) homogeneous size between 1 and 10 mm– Ideally randomly oriented in space and packed homogeneously

• The Reciprocal Space becomes populated with a large number of copies of the same RL but randomly rotated around the origin of the reciprocal space.

• RL vectors shkl with length |shkl|=1/dhkl are distributed randomly around the origin defining the radii of a sphere.

• Intersection of the spheres with the Ewald sphere produce circles through where diffracted rays come out of the sample forming diffraction cones.

1

Rayos X

incidentes 0

Retículo

Recíproco

cristal

Rayos X

difractados

Powder Intensities

• For one crystal we found:

One nanocrystal illuminated by an

electron beam will give a diffraction

pattern corresponding to a single

crystal (the difference between ED

and XRD are a matter of other

school).

Powder Intensities

Representación esquemática de un cristal y su diagrama

de difracción.

Powder Intensities

One small crystals will define a Reciprocal lattice with an orientatioconsisten with the relative orientation of the crystal with the X-ray beam.

Four crystals in different orientations will diffract together but theirReciprocal lattices points are rotated by the same angle of the crystals.

Powder Intensities

40 crystallites diffracting together4 crystallites

Powder Intensities

40 crystallites 200 crystallites

Powder Intensities

• The Fourier Transform of a polycrystallines sample is the sum of the FTsfor each of the crystallites in the sample (ideally >107 crystals of equalsize in the 1-10 mm range with random orientations)

X rays ()

Difracción de rayos X de polvo

45

Fourier Transform

dhkl

shkl =1/d

• A policrystalline sample (powder) is composed of an “infinite” (1010) number of micrometric (1-10 mm) crystal with a perfectly randomorder and its Fourier Transform

Rayos X ()

46

Fourier Transform

dhkl

d*hkl =1/d=shkl

Powder diffraction data collection

• The powder pattern is obtained from the intersection of thereciprocal lattice of the powder (spheres) and the Ewaldsphere.

Rayos X ()

47

1/


Intersections are circles and X-rays come from the sample exiting the Ewaldsphere in cones of 4 angle.

2

4

• Depending on the detector system geometry (and the sample shapeand orientation respect to the beam different kinds of diffractionpatterns are obtained.

Rayos X ()

48

1/


Transmission Geometry (frequently found in Synchrotron labs)

Sample

1/

Reflection Geometry (frequently found in conventional labs)

49

• Depending on the detector system geometry (and the sample shapeand orientation respect to the beam different kinds of diffractionpatterns are obtained.


Rayos X ()


OFFON

50

Rayos X ()



Position-sensitive detector with/without energy discrimination

• Debye-Scherrer Camera with cylindrical film

51

Rayos X ()


Transmission Geometry (Debye Scherrer Camera)

2

• En función de la geometría del detector de rayos X se obtienen los distintos tipos de diagramas de polvo. Con una pantalla plana (Image Plate)

52

Rayos X ()



• We can integrate the detector counts into a diffraction pattern

53


• We can numerically integrate the detector counts into a diffraction pattern

54


• Depending on the geometry of the diffractometer different procedures for data collectionare performed.

55


2Sample

Source

Modern

laboratory powder diffractometer

Powder intensities

Powder Intensities

• Due to either systematic properties or experimental conditions in a powder pattern a large number of reflections overlap. – Symmetry overlap of equivalent reflections (all showing

identical |shkl| or dhkl)

– Metric overlap due to equal d-spacing of reflections with different hkl but similar dhkl

– Convolution overlap for independent reflections with D2<instrumental resolution

Sometimes this makes it impossible to determine individual intensities

Powder Intensity Extraction

• In modern times two methods are systematically used for intensity extraction from Powder Data:– Pawley fit

– Le Bail fit

• In both cases the procedure provides a set intensities corresponding to hkl indices of the corresponding contributions to each peak.

• In the case of severe overlap (no true multiplet is visible) the intensity extraction gives equal intensity to all the components of the fitted peak.


• Both Pawley and LeBail method are based on the Rietveld equation:

j hkl

hklihkljbiobsi 22LPAIsSyy

Pawley:Ihkl is refined as a variable

intensity Ihkl to fit the corresponding peak/s.

Le Bail:Ihkl is calculated from refined

|Fobs|2 . |Fhkl|=1 is the starting

value and then replaced by |Fobs| in successive LS cycles to achieve the best fit and best Ihkl.


• After a Powder Intensity Extraction an h,k,l,Ihkl list is obtained but all available hkl values are those non-overlaping so they are all “unique” reflections.

• Accidental peak overlap reduces the number of observations further by a percentage up to 25% for large-cell/low-symmetry materials.

Powder vs Single Crystal intensities

• Conventional X-ray powder diffraction data for Thiourea(CH4N2S) down to a resolution of 0.84 Å (2=133° for CuKa) provides ~342 unique reflections of which ~249 do not overlap significantly (D2<FWHM/2) of which ~224 are observed (I>2sI criterion).

• Conventional X-ray single crystal diffraction data for Thiourea (CH4N2S) down to a resolution of 0.84 Å (2=50° for MoKa) provides 340 unique reflections (none overlaping) of which 265 are observed (I>2sI

criterion).

Peak overlap produces a reduction of at least 40% of the candidate observed intensities, these calculations are made for a small unit cell wilthrelatively high symmetry so things get worse for low-symmetry/large-cell problems.

Powder vs Single Crystal intensities

• In either case for Thiourea (CH4N2S) we ended up with files containing h, k, l, Ihkl and sIhkl lists that have to be correctly processed before submitting the problem for structural determination.

Intensity-Structure Factor relation

• The intensities can be related to the structure factors of the crystal by the equation:

Ihkl=kALp|Fhkl|2

k= scale factorA= absorption correctionL= Lorentz correctionp= Polarization correction


Ihkl=kALp|Fhkl|2

k= scale factorDepends on:

– Incident beam intensity– Data collection time per point/frame– Absorption of x-rays in the diffraction path– Detector efficiency– Diffractometer geometry (slits, filters,

monochromators, etc)– …


Ihkl=kALp|Fhkl|2

A= absorption correction m

eII 0

Cylindric powder sample:

m es el coeficiente de absorción lineal de la muestra y R el radio del cilindro (capilar).

X-rays ()

43

32

210 RkRkRkRk

eAmmmm

...

0456.26)(sen*109561.0e)(sen*01911.099978.25(k

697653.1k

)(sen*024514.02/1

1

0

2

Absorption profile of a plate-shaped single crystal of a Pb and I-containing sample sorted by run number.

Calculated from the Lambert-Beer Law


Ihkl=kALp|Fhkl|2

L= Lorentz correction

)2(sen

1L

L accounts for the area (volume) difference of the intersection of the Ewald Sphere (a thick line due to wavelength dispersion) and each diffraction spot or sphere with 2 angle.


Ihkl=kALp|Fhkl|2

p= Polarization correction

Sealed tube with filter: p=1/2+cos2(2)/2

Synchrotron radiation: p=1

Sealed tube with IBM: p=1/2+cos2(2m)cos2 (2)/2

2

))2(cos1(P

2


• For a spherical (or low absorbing m<1) crystal:

|Frel(hkl)|=(I(hkl)/Lp)1/2=(ILp(hkl))1/2

k|Frel(hkl)|2=kILp(hkl)=|F(hkl) |2

|Frel(hkl)| is the relative structure factor modulus

(determined assuming k includes all unknown corections to the intensity that are independent of hklor .

Wilson Plot

• With a little help of mathematics we can estimate k to get thestructure factors to its correct magnitude, and as a bonus wecan estimate the average B for the crystal:

j

)lzkyhx(i2

aj

B

ajaj

j

B)lzkyhx(i2

aj

jjj

2)sin(j

2)sin(jjjj

eg)hkl(F

efg

eef)hkl(F

Atomic scattering factor

Atomic displacement (thermal, Debye-Waller)parameter

69

Wilson Plot

• With a little help of mathematics we can estimate k to get thestructure factors to its correct magnitude, and as a bonus wecan estimate the average B for the crystal:

2)sin(aB

aa efg

B=2B=5

af

Wilson Plot

• With a little help of mathematics we can estimatek to get the structure factors toits correct magnitude, and as a bonus we can estimate theaverage B for the crystal:

klnsin

B2f

hklIln

2

ave

j

2

aj

Lp

ln(k)

Intensity scaling and statistics

• Normalized Structure Factor E(hkl):

< |E(hkl)|2>=< |F0(hkl)|2>/Sjf2

aj=Sjf2

aj/Sjf2

aj=1

j

2

aj

2

2

)(f

)hkl(F)hkl(E

72

j

)lzkyhx(i2

ajjjjeg)hkl(F


• The Normalized Structure Factors E(hkl) are a way of expressing the structure factors considering only the effect of amplitude decay by interference among radiation scattered byatoms and thermal motion.

• With the calculation of E(hkl) we eliminate the effect of scattering amplitude decay with the scattering angle.

73

2)sin(aB

aa efg

B=2B=5

a

a

fg

1

0


• E(hkl) are fundamental for structural determination both fromsingle crystal or powder diffraction data by Direct Methods orthe Patterson Method.

• The distribution of E(hkl) values with 2 is also useful since itfollows the symmetry of the structure.

• A number of statistic indicators based on E(hkl) are in general used to distinguish among centric and acentric space groups(task that cannot be performed only by inspecting thesystematic absences in many cases) or to analyze thesymmetry of special projections of the structure that may beuseful to solve disorder problems or other rarities of crystalstructures.

74


• We have shown that <|E(hkl)|2> =1

• It is possible to demonstrate that if a structure is:

– Centrosymmetric: <|E(hkl)2-1|> =0.968

– Non-centrosymmetric: <|E(hkl)2-1|> =0.736

• This is not only valid for every hkl but for special groups of them such as 0kl, h0l and hk0 thatare the projections of crystalstructure along [100], [010] and [001] respectively.

75

In the afternoon we will check how symmetry affects the diffraction pattern and extract the space group from the analysis of diffraction intensities.

Diffraction Experiments, Datacloud.crm2.univ-lorraine.fr/pdf/Bogota2018/Suescun_Diffraction.pdf · Representación esquemática de un cristal y su diagrama de difracción. Powder

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