Transcript Document 7433097

Structural Analysis
Apurva Mehta
Physics of Diffraction
X-ray Lens not
very good
Mathematically
Intersection of
Ewald sphere with Reciprocal Lattice
outline
Information in a Diffraction pattern
Structure Solution
Refinement Methods
Pointers for Refinement quality data
What does a diffraction
pattern tell us?
 Peak Shape & Width:
crystallite size
Strain gradient
 Peak Positions:
Phase identification
Lattice symmetry
Lattice expansion
 Peak Intensity:
Structure solution
Crystallite orientation
niasin(a)
A(a) = S e
Sample Diffraction
a
A=
= FT(a)
A = S cos(nf)
+ i sin(nf)
S
iwt
Fe
e
A = S F(a)
i nf
iwt + inf
e
iwt + inf
F(a) e
a
e
iwt
asin (a) = f
Laue’s Eq.
F(a) eiwt + i3f
F(a) eiwt + i2f
iwt + if
F(a) e
F(a)e
iwt
Sample  Diffraction
Diffraction Pattern ~ {FT(sample) } {FT(sample) }
=
x
Sample size
(S)
*
Infinite Periodic
Lattice (P)
M
o
t
i
f
(M)
Sample  Diffraction
FT(Sample) = FT((S x P)*M)
Convolution theorem
FT(Sample) = FT(S x P) x FT(M)
FT(Sample) = (FT(S) * FT(P)) x FT(M)
FT(S)
X

Y
FT(P)

FT (S x P) = FT(S) * FT(P)
x
*
=
y
FT(M)

FT(sample) = FT(S x P) x FT(M)
Along X direction
x
X

What does a diffraction
pattern tell us?
 Peak Shape & Width:
crystallite size
Strain gradient
 Peak Positions:
Phase identification
Lattice symmetry
Lattice expansion
 Peak Intensity:
Structure solution
Crystallite orientation
Structure Solution
 Single Crystal
 Powder
 Protein Structure
 Due to small crystallite size
kinematic equations valid
 Sample with heavy Z
problems Due to
 Many small molecule
structures obtained via
synchrotron diffraction
 Absorption/extinction
effects
 Mostly used in Resonance
mode
 Site specific valence
 Orbital ordering.
 Peak overlap a problem – high
resolution setup helps
 Much lower intensity – loss on
super lattice peaks from small
symmetry breaks. (Fourier
difference helps)
Diffraction from Crystalline Solid
 Long range order ----> diffraction pattern periodic
 crystal rotates ----> diffraction pattern rotates
Pink beam laue pattern
Or intersection of a large
Ewald Sphere with RL
From 4 crystallites
From Powder
Powder Pattern
 Loss of angular information
 Not a problem as peak
position = fn(a, b & a )
 Peak Overlap :: A problem
 But can be useful for precise
lattice parameter
measurements
Peak Broadening
 ~ (invers.) “size” of the sample
Crystallite size
Domain size
Strain & strain gradient
Diffractometer resolution should be better than
Peak broadening But not much better.
Diffractometer Resolution
Wd2 = M2 x fb2 + fs2
M= (2 tan q/tan qm -tan qa/ tan qm -1)
Where
f = divergence of the incident beam,
b
f = cumulative divergences due to slits and apertures
s
q, q and q = Bragg angle for the sample, analyzer and the monochromator
a
m
Powder Average
Single crystal – no intensity
Even if Bragg angle right,
But the incident angle wrong
Fixed 2q
Q +/- d(Q) = q +/- d(q)
d(q) = Mosaic width ~ 0.001 – 0.01 deg
d(Q) = beam dvg ~ >0.1 deg for sealed tubes
~ 0.01- 0.001 deg for
d(q)
synchrotron
q
2q
Q
For Powder Avg
Need <3600 rnd crystallites – sealed tube
Need ~ 30000 rnd crystallites - synchrotron
Powder samples must be prepared carefully
And data must be collected while rocking the sample
Physics of Diffraction
No X-ray Lens
Mathematically
Phase Problem
rxyz = Shkl Fhkl exp(-2pi{hx + ky + lz})
Fhkl is a Complex quantity
Fhkl(fi, ri): (Fhkl)2 = Ihkl/(K*Lp*Abs)
 rxyz
= Shkl CIhkl exp(-(f + Df))
Df = phase unknown
Hence Inverse Modeling
Solution to Phase Problem
 Must be guessed
And then refined.
How to guess?
Heavy atom substitution, SAD or MAD
Similarity to homologous compounds
Patterson function or pair distribution analysis.
Procedure for
Refinement/Inverse Modeling
Measure peak positions:
Obtain lattice symmetry and point group
Guess the space group.
 Use all and compare via F-factor analysis
Guess the motif and its placement
Phases for each hkl
Measure the peak widths
Use an appropriate profile shape function
Construct a full diff. pattern and compare with
measurements
Inverse Modeling Method 1
7000
6000
Data
Intensity
Reitveld Method
5000
4000
3000
2000
1000
0
40
60
80
100
120
140
160
q
Model
Profile
shape
Background
Refined Structure
Inverse Modeling Method 2
7000
6000
Fourier Method
Data
Intensity
5000
4000
3000
2000
1000
0
40
60
80
100
120
140
160
q
subtract
Background
Profile
shape
Integrated
Intensities
Model
Refined Structure
phases
Inverse Modeling Methods
 Rietveld Method
 More precise
 Yields Statistically reliable
uncertainties
 Fourier Method
 Picture of the real space
 Shows “missing” atoms,
broken symmetry,
positional disorder
Should iterate between Rietveld and Fourier.
Be skeptical about the Fourier picture if Rietveld
refinement does not significantly improve the fit with
the “new” model.
Need for High Q
Many more reflections at higher Q.
Therefore, most of the structural information is at higher Q
Profile Shape function
Empirical
Voigt function modified for axial divergence
(Finger, Jephcoat, Cox)
Refinable parameters – for crystallite size, strain
gradient, etc…
From Fundamental Principles
Collect data on Calibrant
under the same conditions
Obtain accurate wavelength and
diffractometer misalignment parameters
Obtain the initial values for the profile
function (instrumental only parameters)
Refine polarization factor
Tells of other misalignment and problems
Selected list of Programs
CCP14 for a more complete list
http://www.ccp14.ac.uk/mirror/want_to_do.html
GSAS
Fullprof
DBW
MAUD
Topaz – not free - Bruker – fundamental
approach
Structure of MnO
MnO @ 6530eV
7000
Scattering
density
6000
structural fit
In te n sity
5000
Intensity
4000
3000
2000
fMn(x,y,z,T,E)
fO(x,y,z,T,E)
0
2000
72
74
76
2q
1000
0
-1000
40
60
80
100
2q
120
140
160
Resonance Scattering
Fhkl = Sxyz fxyz exp(2pi{hx + ky + lz})
7000
6000
Intensity
5000
4000
3000
2000
1000
0
40
60
80
100
q
120
140
160
fxyz = scattering density
Away from absorption edge
a electron density
Anomalous Scattering Factors
fxyz = fe{fiexyzT} f = Thomson scattering for an electron
fi = fi0(q) + fi’(E) + i fi”(E)
m(E) = E * fi”(E)
e
Kramers -Kronig :: fi’(E) <-> fi”(E)
f ' ( E0 ) =
2


p
0
f " ( E ) E 2 E- E 2 dE
0
Resonance Scattering vs Xanes
7000
6000
5000
Intensity
4000
3000
-2
MnO2
2000
5000
1000
4000
-3
40
60
80
100
120
-4
140
2q
3000
Intensity
from resonance scattering
0
-5
2000
-6
1000
9000
0
8000
-7
80
100
120
7000
140
2q
6000
-8
5000
Intensity
60
f'
40
-9
4000
3000
2000
1000
-10
1.0
0.8
40
from KK transform of XANES
60
80
100
2q
-11
0.6
m = f"/E
0
-12
0.4
-13
6440
0.2
6460
6480
6500
6520
6540
Energy(eV)
0.0
6500
6520
6540
6560
Energy (eV)
6580
6600
6560
6580
6600
120
140
XANE Spectra of Mn Oxides
0.8
Absorption
0.8
Mn Valence
MnO2
Mn
MnO
0.4
Mn3O4
0.0
6540
6570
Mn2O3
6600
Energy(eV)
Mn2O3
Absorption
0.4
Mn3O4
MnO2
0.0
6540
6560
Energy (eV)
Mn(II)?
Mn(II)?
Mn(I)?
MnO
Mn(I)?
Mn
Avg.
Actual
F’ for Mn Oxides
f' (electrons)
-3
f ' ( e le c t r o n s )
-6
-6
M n 2O 3:1
M n 2O 3:2
M nO
2
M nO
-8
-9
M n 3O 4:2
M n 3O 4:1
-1 0
C r o m e r -L ib e r m a n M n
6530
6540
6450
6550
E n e rg y (e V )
6560
6500
Energy (eV)
6550
Why Resonance Scattering?
Sensitive to a specific crystallographic
phase. (e.g., can investigate FeO layer growing on
metallic Fe.)
Sensitive to a specific crystallographic site
in a phase. (e.g., can investigate the tetrahedral
and the octahedral site of Mn3O4)
Mn valences in Mn Oxides
•Mn valence of the two sites in
Mn2O3 very similar
d(f')/dE
•Valence of the two Mn sites in
Mn3O4 different but not as
different as expected.
-3
CL
0
Mn3O4
-4
-5
f'
-6
-7
-9
6540
Mn1
Mn2
-8
6480
6548
X Axis Title
A. Mehta, A. Lawson, and J. Arthur
6460
6544
6500
Energy (eV)
6520
6540
6552
6556