1

Lecture 2Principles of EXAFS and

XANES data analysisHow to avoid common errors in EXAFS and XANES analysis

Tutorials and other Training MaterialBruce Ravel's notes on using FEFFIT for data analysisDaresbury Laboratory lectures on data analysis (EXCURV98)Grant Bunker's XAFS tutorialsFrenkel et al on comparing PCA with other methodsChantler (Uni. Melbourne) on the absolute determination of x-ray

absorption

ProgramsESRF Software catalogXFITEXAFSPAKAthena and Artemis

FEFFIT and IFEFFITDL EXCURVWinXAS

2

where

and

-10

-5

0

5

10

0 2 4 6 8 10 12

EX

AF

S•k

3

k (Å -1 )

Amplitude

coordinationnumber

Frequency

bondlength

Phase

scatterer

Shape

scatterer

k N sAs k S02

kRas2 exp 2k 2as

2 exp 2Ras sin 2kRas as k

k 2me E Eo 2

k E o E o E

E s E

b E

Z ± 10N ± 1R ± 0.02 Å

k N sAs k S02

kRas2 exp 2k 2as

2 exp 2Ras sin 2kRas as k +E0, b ,

Radial structure only

Fourier transformation can be used to visualize frequencies contributing to EXAFS

k N sAs k S02

kRas2 exp 2k 2as

2 exp 2Ras sin 2kRas as k

R2sinA kkk

22

10 RR

Å R

3

Common errors in EXAFS analysis

• Least-squares minimization

• Fourier Filtering

• Resolution

• The job of a least-squares fitting program is to give you the best (smallest deviation) solution, not to give you the right solution.

• If you see something, it tells you something; if you see nothing, it tells you nothing.

4

R

Carbonscattering

Iterative refinements are especially susceptible to multiple minima

Pd

P P

CCH2C

Me

ca b Me

N

Phosphorusscattering

Clot, et al., J. Am. Chem. Soc 2004, 126, 9079-9084

Minima give very different structures but nearly identical EXAFS

M-P

M-C

sum

Nfree vs. Nobs

Nobs ~ 300

(k = 0.05 Å; kmax= 15 Å)

Nfree ~ (2 k R ) /

5

Anatoly Frenkel

EXAFS Data Collection and Analysis Workshop, NSLS

Further explorations of the difficulty of multiple minima

N and 2

R and E0

# Pt foil, T=200 K

guess S02 = 0.9 guess ss1 = 0guess dr1 = 0guess th1 = 0guess e0 = 0

data = ptfoil-200avk.chiout = ptfoil-200avk

rmin = 2.1 rmax = 3.3kmin = 2 kmax= 20 w = 2 dk=2

!% 1st path: e0shift 1 e0 amp 1 S02path 1 p1.datid 1 SS Pt-Pt(1), r=2.7719delr 1 dr1sigma2 1 abs(ss1)third 1 th1

0 2 4 6 8 10 12 14 16 18 20 22-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5 200 K

k2 (k

), Å

-2

k, Å-1

0 2 4 6 8 10 12 14 16 18 20 22-1.0

-0.5

0.0

0.5

1.0

673 K

k2 (k

), Å

-2

k, Å-1

How to model metal (Pt) foil data:

)(2sin

e)()(

333

4

2eff2

20

22

kkCkr

kfkr

NSk k

6

0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

FT

Mag

nitu

de,

Å-3

r, Å0 2 4 6 8 10

0.0

0.2

0.4

0.6

0.8

1.0

FT

Mag

nitu

de,

Å-3

r, Å

S02 = 0.747364 0.044827ss1 = 0.009313 0.000385 dr1 = 0.005673 0.007571 th1 = 0.000259 0.000076e0 = 8.069036 0.594447

0 2 4 6 8 100

1

2

3

4

5

FT

Mag

nitu

de,

Å-3

r, Å0 2 4 6 8 10

0

1

2

3

4

5

FT

Mag

nitu

de,

Å-3

r, Å

S02 = 0.874046 0.033009ss1 = 0.003609 0.000106dr1 = -0.009912 0.003630th1 = -0.000030 0.000019e0 = 8.286025 0.590203

This is not physically reasonable….

What caused S02 to be different at 200 K and 673 K?

- correlation with other fit variables:

)(2sin

e)()(

333

4

2eff2

20

22

kkCkr

kfkr

NSk k

Fit Results

)(2sine)()( 333

42eff2

20

22

kkCkrkfkr

NSk k

0 2 4 6 8 10 12 14 16 18 20 22-2.5

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5 200 K 300 K 473 K 673 K

k2(k) of Pt foil: temperature dependence

k2 (k

), Å

-2

k, Å-1

One possible solution:a multiple-data-set (mds) fit.

What variables are not expected to change at different temperatures?

How to break the correlation?

E0, N

)exp(1

)exp(1

2 E

E2

222

T

Td

ds

2s E,

7

0 2 4 6 8 100

1

2

3

4

5

200 KF

T M

agn

itude

, Å

-3

r, Å

0 2 4 6 8 100.0

0.5

1.0

1.5

2.0

2.5

3.0

FT

Ma

gnitu

de,

Å-3

r, Å

300 K

0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

FT

Ma

gnitu

de,

Å-3

r, Å

673 K

MDS fit results

ss011 = 0.000533 0.000093 theins1 = 189.743073 2.311668

s02 = 0.836704 0.017830

dr11 = -0.011222 0.002248 dr12 = -0.009361 0.003034 dr13 = -0.000354 0.003642 dr14 = 0.006588 0.004801

th11 = -0.000035 0.000013 th12 = -0.000017 0.000022 th13 = 0.000113 0.000033 th14 = 0.000267 0.000060

e0 = 8.064717 0.271896

Physical (chemical, engineering, mat.science, life science etc.)

reality checks:

1) Debye temperature: 240 K for PtAs obtained (through ): 253(3) K

2) Static disorder : ~03) Corrections to model distances: ~0

4) Thermal expansion: evident

5) S02: reasonable (between 0.7 and 1.0)

2s

E

0 2 4 6 8 100.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

1.6

FT

Mag

nitu

de,

Å-3

r, Å

473 K

How to model XAFS data in nanoparticles?

A priori knowledge or a working hypothesis must exist(the “zero” approximation)

otherwise: the transferability of amplitude/phase will not work!)

1) Hemispherical2) Crystal order3) Size: about 20 Å

What information can be obtained from 1st shell EXAFS analysis?

1) Size of the particle (via N)2) Distances, thermal vibration,

expansion3) Static disorder (icosahedral?

surface tension?)

Rel

ativ

e A

bund

ance

Nanoparticle Diameter (A)

0 5 10 15 20 25 30 35 40 45 50

Ave

rage

Fir

st-S

hell

Coo

rdin

atio

n

4

5

6

7

8

9

10

11

12

(Å)

Ave

rage

CN

8

0 2 4 6 8 100.0

0.5

1.0

1.5

2.0

2.5

200 K

FT

Ma

gnitu

de

, Å

-3

r, Å0 2 4 6 8 10

0.0

0.2

0.4

0.6

0.8

1.0

FT

Ma

gn

itud

e,

Å-3

r, Å

473 K

0 2 4 6 8 100.0

0.5

1.0

1.5

FT

Mag

nitu

de,

Å-3

r, Å

300 K

0 2 4 6 8 100.0

0.2

0.4

0.6

FT

Ma

gni

tud

e, Å

-3

r, Å

673 K

MDS fit (1shell) to the nanoparticles EXAFS

- Coordination number is now guessed (a variable)

- is fixed to be equal to that in Pt foil EXAFS

- E0 is fixed to be equal to that in Pt foil EXAFS

20S

ss011 = 0.001676 0.000177 theins1 = 191.842209 3.893480

n1 = 7.879327 0.197850

dr11 = -0.015809 0.003938 dr12 = -0.011870 0.002064 dr13 = -0.008558 0.003883 dr14 = -0.000845 0.004875 th11 = -0.000017 0.000030 th12 = 0.000055 0.000019th13 = 0.000159 0.000047th14 = 0.000421 0.000079

Summary

• The more variables you control, the more likely you are to obtain a unique solution

• Multiple data sets (elements, temperature, concentration, time, etc.) almost always help

• Conclusions are only as good as your model

9

Fourier filtered data can be distorted

M-O at 2.0 and 2.2 Å give two apparently well resolved peaks in FT

0

5

10

15

20

25

30

35

40

0 1 2 3 4

Fou

rier

tran

sfor

m m

agni

tude

R + (Å)

10

Fitting each filtered peak gives the appearance of M-O and M-S EXAFS

-4

-2

0

2

4

6

8

10

4 6 8 10 12

EX

AF

S•k

3

k (Å-1)

"M-O" peak "M-S" peak

Effect of limited resolution on EXAFS

11

EXAFS resolution is ~ /2k =

0.13 Å

-2

0

2

4

6

8

10

12

0 2 4 6 8 10 12

EX

AF

S•k

3

k (Å-1)

R1=1.75 Å, R

2=2.00 Å

R1=1.85 Å, R

2=2.00 Å

R1=1.85 Å, R

2=2.00 Å

R=1.875 Å

R=1.925 Å

R=1.925 Å,damped

Clark-Baldwin, et al. "The limitations of X-ray absorption spectroscopy for determining the structure of zinc sites in proteins. When is a tetrathiolate not a tetrathiolate?" J. Am. Chem. Soc. 1998, 120, 8401-8409.

A case study in data under-determination

12

Zn EXAFS is remarkably insensitive to changes in ligation.

-10

-5

0

5

10

15

20

25

30

2 4 6 8 10 12

EX

AF

S•k

3

k (Å-1)

ZnS2N

2

ZnS3N

ZnS4

0

10

20

30

40

50

0 1.5 3 4.5 6 7.5

FT

Mag

nitu

de

R + (Å)

ZnS2N

2

ZnS3N

ZnS4

ZnSR

SR

L1

L2

Z ± 10 ???

XANES spectra are sensitive to ligation

0

100

200

300

400

9650 9660 9670 9680 9690 9700

Nor

mal

ized

Abs

orpt

ion

(cm

2 /g)

Energy (eV)

Peptide

ZnS2N

2

ZnS3N

ZnS4

0

100

200

300

400

9650 9660 9670 9680 9690 9700

Nor

mal

ized

Abs

orpt

ion

(cm

2 /g)

Energy (eV)

Inorganic

ZnS2N

2

ZnS3N

ZnS4

but show greater variation between different compounds than with changes in ligation.

13

Zn-S and Zn-N EXAFS signals are approximately out of phase

-3

-2

-1

0

1

2

3

2 4 6 8 10 12 14

k3 E

XA

FS

k (A-1)

Zn-S Zn-N

The observed EXAFS for mixed S/N sites is dominated by Zn-S scattering

-15

-10

-5

0

5

2 4 6 8 10 12 14

k3 E

XA

FS

k (A-1)

Zn-S Zn-N ZnS2N

2

14

One solution is to measure data over wide k range(ZnS2N2 inorganic)

0

10

20

30

40

50

60

70

0 1.5 3 4.5 6 7.5

FT

Mag

nitu

de

R + (Å)

k=2-11 Å-1

k=2-12 Å-1

k=2-13 Å-1

k=2-14 Å-1

k=2-15 Å-1

k=2-16 Å-1

Note – R ~ 0.25 /2k = 0.25 Å kmin ~ 6.3

High resolution EXAFS is required to reliably distinguish Zn-S from Zn-N …

-15

-10

-5

0

5

2 4 6 8 10 12 14

k3 E

XA

FS

k (A-1)

Zn-S Zn-N ZnS2N

2

15

…and even with high resolution data, extremely high signal/noise ratios are required

to detect Zn-N in the presence of Zn-S

-15

-10

-5

0

5

2 4 6 8 10 12 14

k3 E

XA

FS

k (A-1)

Zn-S Zn-N ZnS2N

2

-100

-50

0

50

20 30 40 50 60 70 80 90 100% sulfur

P i(%)

Peptide

ZnS2N

2

ZnS3N

ZnS4

-100

-50

0

50

20 30 40 50 60 70 80 90 100

Pi(%

)

% sulfur

Inorganic

ZnS2N

2ZnS

3N

ZnS4

It is possible to reliably distinguish between ZnS4, ZnS3N, and ZnS2N2

if variable para-meters are carefully controlled.

Note that fit quality always improves for mixed ligation fits.

Clark-Baldwin, K et al, J. Am. Chem. Soc. 1998, 120, 8401-8409.

16

In addition to Pi, 2 depends on ligation

-2

0

2

4

6

8

10

20 30 40 50 60 70 80 90 100

2 (Å

2 x10-3

)

% sulfur

Threshold energy changes apparent ligation

50

60

70

80

90

100

4 6 8 10 12 14 16

Sm

ax

ZnS2N

2

ZnS3N

ZnS4

E0,S

(eV)

Calibrated E0

5 eV is enough to change a sulfur into a nitrogen!

17

Distinction between S and N rests largely on phase, which depends on E0

-8

-6

-4

-2

0

2

4

4 6 8 10 12

k3 E

XA

FS

k (A-1)

Zn-S Zn-N

Geometry determination

Multiple scattering in EXAFS

18

0

2

4

6

8

10

12

14

0 1 2 3 4 5 6 7

Fou

rier

Tra

nsfo

rm M

agni

tude

Radius + (Å)

N

NH

Zn

Outer shell scattering can provide ligand identification and geometric information

Multiple scattering makes EXAFS sensitive to angular arrangement of ligands

N

NH

Zn

19

M CC

C

C

C

N

N

N

N

N

C

N

M N

O

O

N

O

OC

O R1

R2

R3

R4

1

2

Pa

ram

ete

rsR

Info

rma

tion

However, ability to reliably determine geometry is limited

XANES spectra contain useful information regarding structure

• Quantitative comparisons (e.g., titration) requires accurate normalization.

• Correction for various artifacts (self-absorption) requires accurate normalization.

• Common normalization procedures were developed for extracting EXAFS and do not necessarily work well for XANES.

20

Normalization Schemes

Conventional

MBACK

Conventional normalization is sensitive to background shape

MBACK shows much weaker sensitivity

21

Conventional normalization is sensitive to range of data

0

0.5

1

1.5

9300 9400 9500 9600 9700 9800 9900 10000 10100

980098209840986098809900992099409960998010000100201004010060

Energy (eV)

MBACK shows only slight sensitivity for Emax150 eV above edge

-50

0

50

100

150

200

250

300

350

9300 9400 9500 9600 9700 9800 9900 10000 10100

980098209840986098809900992099409960998010000100201004010060

Ab

sorp

tion

(cm

2 /g)

Energy (eV)

22

Conventional normalization misses changes in XANES

MBACK reveals subtle changes when thiolate is added

4 possible difference specta –should all be the same

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

9700 9750 9800 9850 9900 9950 10000 10050 10100

Conventional

Energy (eV)

23

With new normalization, difference signal is detectable

-10

-5

0

5

10

9700 9750 9800 9850 9900 9950 10000 10050 10100

Energy (eV)

Dependence of XANES on Oxidation State

0

400

800

6535 6555 6575 6595Energy (eV)

Nor

mal

ized

Abs

orpt

ion

II/II

II/III

III/III

III/IV

24

Edge “energy” is poorly defined

-80

-60

-40

-20

0

20

40

0

200

400

600

800

1000

1200

d(A

bs)/

dEA

bsorption

Simon Bare, NSLS workshop

25

Factor Analysis in Chemistry, 2nd Ed. John Wiley & Sons, NY, 1991

Coulston et al., Science 275 (1997) 191-193

T

1

T

k

l

kkk

l sX

VSUX

vu

is the closest l-rank matrix to X(see http://public.lanl.gov/mewall/kluwer2002.html)

EnergyTime