systematic absences and space group...

Systematic Absences and SpaceGroup Determination

Leopoldo SuescunLaboratorio de Cristalografía Química del Estado Sólido y Materiales,

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

[email protected]

November 29th, 2018

The Structure Factor

• SrTiO3 and metallic Pd have very similar unit cell dimentions:SrTiO3: Cubic a=3.89 Å, Space Group Pm3mPd: Cubic a=3.88 Å, Space Group Fm3m

• But their diffraction patterns showsignificant differences:

why?3

The atomic scattering factor

4

Fourier Transform

Atom 𝜌( 𝑟)atomicscatteringfactor 𝑓𝑎( 𝑠)

The Structure Factor*

0 a

b

Atom 1(x1,y1,z1)

Atom 2(x2,y2,z2)

Atom j(xj,yj,zj)

Fourier Transform

j

j

j

jjjj

N

j

j

z

y

x

cbaczbyaxr

rrr

,,

)(1

f1f2

fjlkhF ,,

~

*

*

*

,,***

2exp~

1

,,

c

b

a

lkhclbkahs

rsifFN

j

jjlkh

Re

Im

12exp rsi

* For an ideal crystal (no static or dinamic distortions)

The temperature factor

6

j

Blzkyhxi

ajhkljjjj eefF

2)sin()(2

Thermal motion (and in some cases disorder) produce the atom to look “blured” and instead of occupying it’s true volume they show up occupying a larger one (with reduced electron density).

The temperature factor (isotropic or anisotropic) attempts to account for this effect.

fa=atomic scattering factorB=temperaturefactor

0 a

b

Atom 1(x1,y1,z1)

Atom j(xj,yj,zj)

2)sin(aB

aef

B=2B=5

Atoms after averagingisotropic thermal motion

The Structure Factor*

N

j

jjlkh rsifF1

,, 2exp~

f1f2

fjlkhF ,,

~

Re

Im

12exp rsi

* For an ideal crystal (no static or dinamic distortions)

jjj

j

j

j

j

j

j

j

j

j

j

lzkyhx

z

y

x

lkh

z

y

x

lkh

z

y

x

cba

c

b

a

lkhclbkahrs

,,

100

010

001

,,

,,

*

*

*

,,***

N

j

jjjjlkh lzkyhxifF1

,, 2exp~

The Structure Factor and theDiffracted Intensity

N

j

jjlkh rsifF1

,, 2exp~

f1f2

fjlkhF ,,

~

Re

Im

12exp rsi

2

,,

~lkhhkl FkALPI

The structure factor is the amplitude of the scattered radiationby one unit cell of the crystal. The total scattered amplitude isobtained adding contributions from all cells and the intensity of the scattered intensity is:

The Structure Factor and theDiffracted Intensity

f1f2

fjlkhF ,,

~

Re

Im

12exp rsi

2

,,

~lkhhkl FkALPI

k=scale factor (function of the geometry, crystal size, incident intensity, etc)A=absorption factor (function of m y and geometry of experiment)LP=Lorentz and polarization factors (function of and nature of incident radiation)M=multiplicity factor (for powder diffraction, function of the symmetry and hkl)

Symmetry of the structure factor

• Friedel’s Law:

IN THE ABSENCE OF RESONANT SCATTERING the diffractionpattern is centrosymmetric

N

j

jj

N

j

jjlkh

N

j

jjlkh

rsifrsifF

rsifF

11

,,

1

,,

2exp2exp~

2exp~

f1f2

fjlkhF ,,

~

Re

Im

12exp rsi

lkhF ,,

~

hklhklhkl

lkh

N

j

jjlkh

IFFI

FrsifF

hkl

2*2

*

,,

1

,,

~2exp

~

hklhkl II

IGNORING RESONANT SCATTERING

Symmetry of the structure factor

• Centrosymmetric Structure:

in a centred structure for eachatom at 𝑟 there is another oneat − 𝑟 so the S.F can be written:

quantity real a is 2cos2

2exp~

2

22

1

,,

1

22

1

)(22

1

,,

N

N

jj

N

jj

j

jjlkh

j

rsirsi

j

j

rsirsi

j

N

j

jjlkh

rsifF

eefeef

rsifF

f1

fN-1

lkhF ,,Re

Im

Symmetry of the diffraction pattern

• In the presence of any symmetry operation that transforms (x,y,z) into(x’,y’,z’) it is possible to deduce the effect on the S.F. and on the diffractedintensities.

• 2[001](0,0,z):

(x,y,z) (-x,-y,z) 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙and using Friedel 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘 𝑙

• m[001](x,y,0):

(x,y,z) (x,y,-z) 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘 𝑙

and using Friedel 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘 𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙


• 2/m (eje único b):

(x,y,z) (-x,y,-z) 𝐼ℎ𝑘𝑙 = 𝐼 ℎ𝑘 𝑙

(x,y,z) (x,-y,z) 𝐼ℎ𝑘𝑙 = 𝐼ℎ 𝑘𝑙and using Friedel 𝐼ℎ𝑘𝑙 = 𝐼ℎ 𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼 ℎ𝑘 𝑙

The symmetry of the diffraction pattern is equal for a compound with a 2-fold axis or mirror symmetry or thecombination of both symmetry elements.


• The symmetry of thediffraction pattern takinginto account the diffractedintensities, is the same as the point group of thecrystal.

• In some cases whereresonant scattering can be ingored the symmetry of the diffraction patter willcoincide with the Laue classof the crystal.


• In the ideal case, ignoring resonant scattering, the point group symmetry of the crystal iscombined with a center of symmetry to givethe diffraction pattern symmetry.

• We can always determine the lattice type and cystal system just by considering thesymmetry of the diffraction pattern.


- Cell parameters were extracted from a diffraction pattern of a crystal withabc, bc90° and the following symmetry of the intensities:

𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘 𝑙 = 𝐼ℎ 𝑘𝑙 = 𝐼 ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼 ℎ𝑘 𝑙 = 𝐼ℎ𝑘𝑙

Find the point group of the diffraction pattern.

- Find the point group of another diffraction pattern from a different crystalwith the same cell parameters relation but the intensities have thefollowing symmetry:

𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘𝑙 = 𝐼ℎ𝑘 𝑙

- When temperatura changes, the first crystal suffers a transformation thatmakes a b but the symmetry of the intensities remains unchanged. Whatis the LAUE symmetry of the crystal.

EXERCISE

The S.F. and Bragg’s Law

17

2

d


18

sen2 hkld

2

d

A

B

C D

2’

11’

d

sin2

sinsin

d

ABAB

BDCBr

The S.F. and Bragg´s Law

• Consider the structure of SrTiO3

with a Cubic Primitive unit cell.

• Atomic Positions:Sr: (0,0,0), Ti (1

2,1

2,1

2)

O: (1

2,1

2, 0), (

1

2, 0,

1

2), (0,

1

2,1

2)

c

bSr O

OO

Ti

SrO plane

TiO2 plane

SrO plane

TiO2 plane

SrO plane


• Consider the reflection of X-rays by planes of the (001) family and the Bragg´s Lawapplied.

SrO plane

TiO2 plane

TiO2 plane

SrO plane

d001

2


SrO plane

TiO2 plane

SrO plane

d001

2

• Considering that this is an ideal structure and that the scattering power of Ba, Ti and O atoms are fBa, fTi and fO respectively.

• Scattered X-rays from (001) planes contain contributions to the S.F. by one Ba and one O atom per unit cell, all Ba and O atoms of BaO plane scatter in phase.

𝐹001 = 𝑓𝑆𝑟 + 𝑓𝑂

but wait!!!


• WHAT ABOUT X-RAYS SCATTERED BY ATOMS BETWEEN 001 PLANES!!!

• Scattered X-rays from atoms in-between (100) planes contain contributions to theS.F. that are out of phase respect to contributions from BaO planes. In particular TiO2 planes are half-way between BaO planes so its contribution is exactly ½ wave out of phase, so TiO2 contributions subtract from BaO contributions:

𝐹001 = 𝑓𝑆𝑟 + 𝑓𝑂 − 𝑓𝑇𝑖 − 2𝑓𝑂

SrO plane

TiO2 plane

SrO plane

d001

2


• What do we get for 𝐹001 if we apply the definition of the Structure Factor?

𝐹001 = 𝑓𝑆𝑟 − 𝑓𝑇𝑖 − 𝑓𝑂

d001

2SrO plane

TiO2 plane

SrO plane

N

j

jjjjlkh lzkyhxifF1

,, 2exp~


• Now lets consider the case of Pd:

F-centered cell with Pd at (0,0,0), (1

2,1

2,0), (

1

2,0,

1

2) and (0,

1

2,1

2)

d001

2


• As for barium titanate we have in-phase contributionsfor 2 Pd atoms per cell at (001) planes (at z=0, 1, etc.) and

contributions out of phase for Cd atoms between theplanes (at positions with z=1/2, 3/2, etc.)

𝐹001 = 2𝑓𝑃𝑑 𝐹001 = 2𝑓𝑃𝑑 − 2𝑓𝑃𝑑 𝐹001 = 2𝑓𝑃𝑑 − 2𝑓𝑃𝑑=0

d001

2


• F(001)=0

• F(011)=0

• F(021)=0

(001)

(011)(021)


• F(002)=4fPd

• F(022)=4fPd

• F(111)=4fPd

(002)

(022)

(111)


Using the definition of the structure factor for a general

In general, if we have an F-centered cell to represent the structure, for eachatom j at (xj,yj,zj) there will always be three equivalent atoms (obtained bylattice centering translation operations) with positions at: (xj+1/2, yj+1/2,zj), (xj+1/2,yj,zj+1/2) and (xj,yj+1/2,zj+1/2).

The S.F. of an structure represented in an F unit cell will be:

𝐹ℎ𝑘𝑙

𝐹ℎ𝑘𝑙 = 𝑓𝑃𝑑 1 + 𝑒𝑥𝑝 𝜋𝑖(ℎ + 𝑘) + 𝑒𝑥𝑝 𝜋𝑖(ℎ + 𝑙) 𝑒𝑥𝑝 𝜋𝑖(𝑘 + 𝑙) == 𝑓𝑃𝑑 1 + −1 ℎ+𝑘 + −1 ℎ+𝑙 + −1 𝑘+𝑙

𝐹ℎ𝑘𝑙 = 4𝑓𝑃𝑑 𝑖𝑓 ℎ + 𝑘, ℎ + 𝑙 𝑎𝑛𝑑 𝑘 + 𝑙 𝑒𝑣𝑒𝑛0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

N

j

jjjjlkh lzkyhxifF1

,, 2exp~

Systematic absences

The S.F. of a structure represented in an F unit cell will be:

otherwise0

parity same ,,2exp4~4

1,,

lkhlzkyhxifF

N

j

jjjjlkh

Fourier Transform

020010

100 200

110220

220

220

420

000

111001

002022

222202

Direct lattice representedwith an F-centered cell

Reciprocal space lattice representedwith an I-centered cell

Systematic absences

- Find the reflection existence conditions for a C-centered structure.

- (a): Find the reflection existence conditions for an I-centered cell.

(b): Indicate what kind of unit cell can be associated with the R.L.

EXERCISE

Systematic Absences

• The existence of additional lattice translations in the structure

(apart from 𝑎, 𝑏 and 𝑐) imply that the periodicity of the lattice, in certain directions, is smaller than indicated by the basisvectors.

• This causes the absence of certain Reciprocal Lattice points(Fhkl or Ihkl is null) that correspond to non-integral reflectionconditions in the direction perpendicular to the reflectionplanes considered.

Systematic Absences

• Integral reflection conditions:

Cell Type Existence condition Structure Factor value

Primitive No condition -

C-centered h+k=2n𝐹ℎ𝑘𝑙 = 2

𝑗=1

𝑁 2

𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)

A-centered k+l=2n Idem C-centered unit cell

B-centered h+l=2n Idem C-centered unit cell

I-centered h+k+l=2n Idem C-centered unit cell

F-centered h+k=2n, h+l=2n and k+l=2nh,k,l all even or all odd

𝐹ℎ𝑘𝑙 = 4

𝑗=1

𝑁 4


R-centered (obverse) -h+k+l=3n𝐹ℎ𝑘𝑙 = 3

𝑗=1

𝑁 3


H-centered h-k=3n Idem C-centered unit cell

Systematic Absences

• Integral reflection conditions:– They are a consequence of our decision to choose the unit cell parameters

NOT TO BE the shortest periodicity vectors, but along the main symmetrydirections of the lattice.

– When we change theunit cell choice (P to Sor S to I or I to F) thenames of reflections(hkl) change to fulfillthe existence condition.

– Every structure can berepresented in anyof the defined centeredunit cells, it is just a matterof convenience.

Cell Type Existence condition Structure Factor value

Primitive No condition -

C-centered h+k=2n𝐹ℎ𝑘𝑙 = 2

𝑗=1

𝑁 2


A-centered k+k=2n Idem C-centered unit cell

B-centered h+l=2n Idem C-centered unit cell

I-centered h+k+l=2n Idem C-centered unit cell

F-centered h+k=2n, h+l=2n and k+l=2nh,k,l all even or all odd 𝐹ℎ𝑘𝑙 = 4

𝑗=1

𝑁 4


R-centered (obverse) -h+k+l=3n𝐹ℎ𝑘𝑙 = 3

𝑗=1

𝑁 3


H-centered h-k=3n Idem C-centered unit cell

Systematic Absences

• The C-centered monoclinic unit cell with lattice parameters (a b c)(b unique) can be converted in an I centered monoclinic unit cellwith (a’ b’ c’)=(a+c b -a) with matrix P:

P =1 0 −10 1 01 0 0

Knowing that:

- (hkl) convert to (h’k’l’) using P

- In a C-centered cell only reflections with h+k=even are observed.

FIND THE REFLECTION CONDITIONS FOR THE I CENTERED UNIT CELL OBTAINED BY THE TRANSFORMATION

EXERCISE

Symmetry operations with translation components.

• Additionally to centering translations, cristal structures may show symmetry operations that include translation components:

• Glide planes (g) named a, b, c, n, e, d

• Screw axes (np) 21, 31, 32, 41, 42, 43, 61, 62, 63, 64, 65.

• In particular conditions these translations may give rise to additionalsystematic absences.

t=T*p/n

j=360/nt=T/2

,

Screw axes

• Consider the following crystalline structure:

C8H10O2N4.H2OMonoclinic, P21/aa=14.8(1) Åb=16.7(1) Åc=3.97(3) Åb=97.0(5)Z=4 Caffeine hydrate

b

a

Screw axes

• Let´s apply Bragg´s Law to determine F010:

2For each molecule at b≈1/4

There is another one at b≈3/4

Screw axes

• The structure contains: 21[010](1/4,y,0).

• A molecule centered at (x,y,z) has an equivalentmolecule centered at (1/2-x,y+1/2,-z) relatedby the screw axis.

• This is very similar to the case of Pd whereequivalent atoms were placed at (0,0,0) and(0,1/2,1/2) therefore their contributions tothe scattered intensity would cancel each-other.

b

a

• Contributions from the molecule at (x,y,z) and at (1/2-x,y+1/2,-z) are exactly out of phase when X-rays reflecting from (010) plane are

considered, therefore F010=0

b

a

Screw axes

2

• From the point of view of diffraction from (010) family of planes the total structure looks like the projection of the crystal along [010] direction.

b

a

Screw axes

2

• The periodicity along b axis is half of the lattice translation b.

• The true periodicity is b/2 so the first allowed reflection is 020.

Screw axes

2

b

b/2

• As well as for Pd the scattering from (020) family of planes contains in-

phase contributions from every molecule of the structure so F0200

b

a

Screw axes

b

a

2

Screw axes

Compute F010 for caffeine hydrate considering that the centroidof the molecules (caffeine+H2O) is at (xc,yc,zc) and the total scattering power of the pair of molecules is f .

EXERCISE

Screw axes

• In general, if there is a 21 screw axis along b going through the origin of the cell: (x,y,z) (-x,y+1/2,-z) and the S.F. is:

N

j

jjjjlkh lzkyhxifF1

,, 2exp~

2/

1

421

N

j

lzhxiiklzkyhxi

jjjjjj eeef

2/

1

2222

2/

1

)()()(22

2

21

1N

j

lzhxilzkyhxi

j

N

j

zlykxhilzkyhxi

j

jk

jjjj

jjjjjj

eef

eef

Screw axes

• For a general hkl this equation gives no special number

• But for hkl with h=l=0:

2/

1

421

~N

j

lzhxiiklzkyhxi

jhkljjjjj eeefF

2/

1

22/

1

2

00 )1(11~

N

j

kkyi

j

N

j

ikkyi

jkjj efeefF

evenk for 2

oddk for 0~ 2/

1

200

N

j

kyi

jk jefF

Screw axes

• In the presence of a 2-fold screw axis paralell to certain lattice direction, all reflections with odd index along that direction will have null S.F. because the true periodicity of the projection of the structure in thedirection of the 2-fold screw axis is half of the lattice translation in thatdirection.

• 21[100],

• 21[010],

• 21[001],

• It is not important if the screw axis is at or away from the origin of the unitcell since this point is arbitrary, therefore we can always place the 21 on it.

Fh00=0 for h odd will be absent

F0k0=0 for k odd will be absent

F00l=0 for l odd will be absent

Screw axes

Identify the extinction condition produced by a 31 screw axis paralell to [001] (hexagonal axes) in the trigonal space group P31. (#144).

EXERCISE

Systematic Absences

Serial reflection conditions:

Screw axis type Direction of axis Existence condition

21 [100] | [010] | [001] | [110] h00, h=2n | 0k0, k=2n | 00l, l=2n | hh0 h=2n

31/32 [001] (hexagonal basis) 00l, l=3n

41/43 [100] | [010] | [001] h00, h=4n | 0k0, k=4n | 00l, l=4n

42 [100] | [010] | [001] h00, h=2n | 0k0, k=2n | 00l, l=2n

61/65 [001] 00l, l=6n

62/64 [001] 00l, l=3n

63 [001] 00l, l=2n

Glide planes

• Consider the same cristal structure of caffeine hydrate:

• There is a glide plane a[010](x,1/4,z) that will reduce the periodicity along[100]

2

Glide planes

• And for all the vectors in the plane of the glide plane [h0l] the projectionof the structure on it has a shorter period than the lattice translation as shown for [101]

Glide planes

• For a glide plane a[010](x,0,z) a pair of equivalent coordinates will be (x,y,z) and (x+½, -y,z) so we can calculate the structure factor for h0l reflections:

N

j

jjjjlkh lzkyhxifF1

,, 2exp~

2/

1

421

N

j

kyiihlzkyhxi

jjjjj eeef

2/

1

222

2/

1

22

2/

1

)()(22

2

2

21

1N

j

kyilzkyhxi

j

N

j

lzkyhxilzkyhxi

j

N

j

lzykxhilzkyhxi

j

jh

jjj

jjh

jjjj

jjjjjj

eef

eef

eef

Glide planes

• This structure factor is non-zero for any combination of hkl:

• But for h0l reflections

2/

1

421

~N

j

kyiihlzkyhxi

jhkljjjj eeefF

2/

1

2

0 1~

N

j

ihlzhxi

jlh eefF jj

even ish if2

odd ish if0~ 2/

1

20

N

j

lzhxi

jlh jjefF

Glide planes

Determine the extinction condition produced by the followingglide planes:

b[100](0,y,z)

n[001](x,y,0)

e[010](x,0,z)

EXERCISE

General and Special reflection conditions


• An atom in the general position will have n-1 equivalent ones (n is thegroup multiplicity) also in general positions as indicated by the spacegroup symmetry.

S. G. P121/a1 (P21/c unique axis b, cell choice 3)

1/4

+

, -

, -

1/2+

½-,

, -

1/2+Cell choice 3


1/4

+

½-,

, -

1/2+

(000)

(½,½,0)

General and Special reflection conditions1/4

+

½-,

, -

1/2+

(000)

(½,½,0)

If we only keep the atoms at Wyckoff position 2a the atomicarrangement looks like a C-centered one


• In the majority of cases where special positions exist, special conditionsfor existence of certain intensities will be observed, in general “simulating” translation operations not present in the space group.


Determine if Wyckoff positions 1a and 2c of Mois’s favorite space groupproduce special refection conditions and in the case they do which.

EXERCISE


• Atoms in special positions may be related by additionaltranslation operations not applicable to the general position that will generate special reflection conditions.

• This may imply that the diffraction pattern of a structurethat should de described in a conventional primitivelattice (e.g. P21/a with atoms siting only at 2a Wyckoffsite) looks like the one we would expect for a centeredlattice.

• But most importantly, an atom sitting at a specialposition with special reflection conditions will notcontribute to any of the reflections that do not fulfill thatcondition.


• An atom sitting at a special position with special reflection conditions willnot contribute to any of the reflections that do not fulfill that condition.

• Cubic MOF Gd2Ca3(oda)6.xH2O• Space Group: Fd-3c, a = 26.5954(7) Å

• Atomic positions (origin choice 2):

Gd: 32b ¼, ¼, ¼

Ca1:32c 0,0,0

Ca2: 16ª 1/8, 1/8, 1/8

10 C, H, O atoms at general positions

The structure was refined from single cristaland syncrhrotron powder x-ray diffraction data

Powder Diffraction Journal (2012), 4, 232-242.Suescun et al.


• Gd and Ca (heavy) atoms in the structure, only contributeto reflections with h+k+l=4n, and Gd only to the smallgroup for which h,k and l is even.

• From all allowed reflections:F-cell, h,k,l all even or all oddthe all odd reflections containno information on the heaviestatoms, while only some containsome information about oneof the three Ca atoms per A.U.

• The special reflection conditionsallowed to refine this structurewith significant precission evenwith powder diffraction data.

Gd

Ca

Ca

S.G. determination (example 1)

• Using systematic absences and intensity statistics the space group could be univocally determined if the data quality is high enough.

• Example: Orthorhombic cellReflection list shows the following Intensity Statistics <E2-1>

systematic absences:

hkl: no absences

h0l: no absenceshk0: h odd

0kl: k+l oddh00: h odd0k0: k odd00l: l odd

hkl: 0.867

h0l: 0.981hk0: 0.774

0kl: 0.743

ambiguousacentriccentricacentric

P cella glidecmbn glideaa glide

Possible Space groups Pn21a and Pnma

or 2//bor 21//b

n gliden glide

Pn21a (Pna21)

S.G. determination (Example 2)

• Using systematic absences and intensity statistics the space group could be univocally determined if the data quality is high enough.

• Example: Tetragonal cellReflection list shows the following Intensity Statistics <E2-1>

systematic absences: hkl: h+k+l odd hkl: 0.756hk0: h+k odd hk0: 0.967h0l: h+l odd h0l: 0.9820kl: k+l odd

hhl: l odd

h00: h odd

0k0: k odd

00l: l non-multiple of 4

hh0: no extinction

acentriccentriccentric

I celln glidec or I cell

n glideb or I cell

I cellPossible Space groups: I41 or I4122

21/a or I cell21/b or I cell

I4122

n glidea or I cell

41[001]

systematic absences and space group...

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