Systematic Absences and SpaceGroup Determination
Leopoldo SuescunLaboratorio de Cristalografía Química del Estado Sólido y Materiales,
Facultad de Química, Universidad de la República, Montevideo, Uruguay.
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
Symmetry of the diffraction pattern
• 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.
Symmetry of the diffraction pattern
• 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.
Symmetry of the diffraction pattern
• 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.
Symmetry 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
The S.F. and Bragg’s Law
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
The S.F. and Bragg’s Law
• 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
The S.F. and Bragg’s Law
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!!!
The S.F. and Bragg’s Law
• 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
The S.F. and Bragg’s Law
• 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~
The S.F. and Bragg’s Law
• 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
The S.F. and Bragg’s Law
• 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
The S.F. and Bragg’s Law
• F(001)=0
• F(011)=0
• F(021)=0
(001)
(011)(021)
The S.F. and Bragg’s Law
• F(002)=4fPd
• F(022)=4fPd
• F(111)=4fPd
(002)
(022)
(111)
The S.F. and Bragg’s Law
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
𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)
R-centered (obverse) -h+k+l=3n𝐹ℎ𝑘𝑙 = 3
𝑗=1
𝑁 3
𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)
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
𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)
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
𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)
R-centered (obverse) -h+k+l=3n𝐹ℎ𝑘𝑙 = 3
𝑗=1
𝑁 3
𝑓𝑗𝑒−2𝜋𝑖( 𝑠∙ 𝑟𝑗)
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
Glide planes
Some zonal reflection conditions:
* For a complete list of conditions of different glide planes in differen orientations see ITA Table 2.2.13.2
e glide planes are not a symmetry operation, they are just a name choice when two glide planes with different translation vectors share the same geometric element (the mirror).
Glide plane type* Normal to plane Existence condition
a [010] | [001] h0l, h=2n | hk0, h=2n
b [100] | [001] 0kl, k=2n | hk0, k=2n
c [100] | [010] 0kl, l=2n | h0l, l=2n
n [100] | [010] | [001] 0kl, k+l=2n | h0l, h+k=2n | hk0, k=2n
d [100] | [010] | [001] 0kl, k+l=4n | h0l, h+k=4n | hk0, k=4n
General and Special reflection conditions
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
General and Special reflection conditions
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
General and Special reflection conditions
• 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.
General and Special reflection conditions
Determine if Wyckoff positions 1a and 2c of Mois’s favorite space groupproduce special refection conditions and in the case they do which.
EXERCISE
General and Special reflection conditions
Determine if Wyckoff positions 1a and 2c of Mois’s favorite space groupproduce special refection conditions and in the case they do which.
EXERCISE
General and Special reflection conditions
• 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.
General and Special reflection conditions
• 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.
General and Special reflection conditions
General and Special reflection conditions
General and Special reflection conditions
• 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]
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