Basic CrystallographyPart 1

Theory and Practice of X-ray Crystal Structure Determination

We have a crystal…

… we want a structure!

How do we get there ?

The Unit Cell Concept

Ralph Krätzner

a

b

c

ab

g

Unit Cell Description in terms of Lattice Parameters

n a ,b, and c define the edge lengths and are referred to as the crystallographic axes.

n a, b, and g give the angles between these axes.

n Lattice parameters àdimensions of the unit cell.

A

A crystal is… a homogenous solid formed by a repeating, three-dimensional pattern of atoms.

AAAAAAAAAAAAAAAA

The unit cell is the repeating unit with dimensions of a, b, c and angles a, b and g. A crystal can be described completely by translations of the unit cell along the unit cell axes.

A

a

b

A+ a

A+ a

A+ b

A+ a

Choice of the Unit Cell (sometimes confusing)

Choice of the Unit Cell

No symmetry - many possible unit cells. A primitive cell with angles close to 90º (C or D) is preferable.

C D

A

B

C

A

The conventional C-centered cell (C) has 90º angles, but one of the primitive cells (B) has two equal sides.

B

The shortest possible introduction to crystallographyThere are seven types of unit cells (crystal systems).

Crystal systems:triclinicmonoclinicorhorhombictetragonaltrigonal/hexagonal cubic

Bravais lattices:aPmP, mCoP, oA, oI, oFtP, tIhP, hRcP, cI, cF

P : primitive, A,B,C : face centeredI : body centered F : (all-)face centeredR : rhombohedral centered

a = g = 90°

a = b = g = 90°

a = b = g = 90°a = b

a = b = g = 90°a = b = c

a = b = 90°g = 120°

a = b

Combined with centering, we obtain the 14 Bravais lattices.

7 Crystal Systems - Metric Constraints

n Triclinic - none

n Monoclinic - a = g = 90°, b ³ 90°

n Orthorhombic - a = b = g = 90°

n Tetragonal - a = b = g = 90°, a = b

n Cubic - a = b = g = 90°, a = b = c

n Trigonal - a = b = 90°, g = 120°, a = b (hexagonal setting) ora = b = g , a = b = c (rhombohedral setting)

n Hexagonal - a = b = 90°, g = 120°, a = b

Bravais Lattices

n Within each crystal system, different types of centering produce a total of 14 different lattices.

n P – Simplen I – Body-centeredn F – Face-centeredn B – Base-centered (A, B, or C-centered)

n All crystalline materials can have their crystal structure described by one of these Bravais lattices.

Bravais Lattices

Space groups (monoclinic):

n The monoclinic space groups:

n P2 P21 C2n Pm Pc Cm Ccn P2/m P21/m C2/m C2/c

2: two fold axis21: screw axis or improper axism: mirror planec: sliding mirror or improper mirror

Crystal Families, Crystal Systems, and Lattice Systems

Generation of and Properties of X-rays

Ka

La

Kb

K

L

M

EmissionPhotoelectron

Electron

Generation of Characteristic Radiation

n Incoming electron knocks out an electron from the inner shell of an atom.

n Designation K,L,M correspond to shells with a different principal quantum number.

Bohr`s model

Generation of Characteristic Radiation

n Not every electron in each of these shells has the same energy. The shells must be further divided.

n K-shell vacancy can be filled by electrons from 2 orbitals in the L shell, for example.

n The electron transmission and the characteristic radiation emitted is given a further numerical subscript.

Emission Spectrum of an X-Ray Tube

Interaction of X-rays with matter- Thomson-Diffusion -

e-

• The interaction with an electromagnetic field induces the oscillation of an electron

• Being an accelerated charged particle, the electron emitsanother electromagnetic wave.

hn hn

Constructive and Deconstructive Interference

0 100 200 300 400 500 600 700 800 900 1000

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Y Axis

Title

X Axis Title

0 100 200 300 400 500 600 700 800 900 1000

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Y Ax

is Tit

le

X Axis Title

0 100 200 300 400 500 600 700 800 900 1000

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Y Axis Ti

tle

X Axis Title

0 100 200 300 400 500 600 700 800 900 1000

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Y Axis

Title

X Axis Title

0 100 200 300 400 500 600 700 800 900 1000

-1.4

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Y Axis

Title

X Axis Title

=

=

q2cos2422

4

crmeII iTh =

e-

For polarisation in y- direction:

The intensity of the diffracted X-ray beam depends on the diffusion angle.

yx

The picture can't be displayed.

The picture can't be displayed.

2q

Interaction of X-rays with matter- Thomson-Diffusion -

Amorphous sample

Primary X-ray beam

X-ray experiment

Why do we need a single crystal ?

→ diffuse reflection

Crystalline sample Amorphous sample → diffuse reflection

X-ray experiment

Why do we need a single crystal ?

→ localised reflections

Coherent Scattering by an Atom

n Coherent scattering by an atom is the sum of this scattering by all of the electrons.

n Electrons are at different positions in space, so coherent scattering from each generally has different phase relationships.

n At higher scattering angles, the sum of the coherent scattering is less.

2q

Localised reflections - Laue constructionInteractions of an X-ray beam with several diffracting centers

a·cos µ

a·cos n

The intensity is only different from zero, when:

D = a·cos n + a·cos µ = n·l

au

µ

Bragg construction

Glancing reflections at the lattice planes hkl of the crystal, which obey the Laue condition.

Bragg law:

2dhkl·sinq = n·l (n = 1, 2, 3 …)

qq

q dhkl

d·sinq

Thomson diffusion is coherent: jTh = ji + 180°

Bragg’s Law

We can think of diffraction as reflection at sets of planes running through the crystal. Only at certain angles 2θ are the waves diffracted from different planes a whole number of wavelengths apart (i.e., in phase). At other angles, the waves reflected from different planes are out of phase and cancel one another out.

nl = 2d sin(q)

dqq

Reflection Indices

n These planes must intersect the cell edges rationally, otherwise the diffraction from the different unit cells would interfere destructively.

n We can index them by the number of times h, k and l that they cut each edge.

n The same h, k and l values are used to index the X-ray reflections from the planes.

zy

x

Planes 3 -1 2 (or -3 1 -2)

Examples of Diffracting Planes and their Miller Indices

a

b

c

n Method for identifying diffracting planes in a crystal system.

n A plane is identified by indices (hkl) called Miller indices, that are the reciprocals of the fractional intercepts that the plane makes with the crystallographic axes (abc).

Miller indices of lattice planes

a

b

a

b

110

• The plane nearest to the origin (not the plane through the origin), intersects the axes a, b and c at 1/h, 1/k and 1/l.

• An index of 0 indicates a plane parallel to an axis. • hkl are the “Miller indices” of the lattice planes• The higher the indices, the smaller the lattice spacing dhkl.

b = 1/k = 1 ð k = 1

a = 1/l = 1 ð l = 1

Miller indices of lattice planes

a

b

a

b

b = 1/k = 1 ð k = 1

a

b

a=1/2

b=1/1

210

a

ba=1/0 = ∞

b=1/3

a

b

a=1/-1

b=1/1

324a

b

c

a = 1/l = 1 ð l = 1

a=1/3b=1/2

c=1/4

x

x

From the position of primary and diffracted beam: → Orientation of the lattice planes in the crystal (perpendicular to the bisecting of the two beams)

Reflection angle q : → Distance between lattice planes

Knowing the distances between lattice planes (dhkl) and their orientations, we obtain the unit cell.

180°-2q

qqq dhkl

X-ray experiment

Very fast: the reciprocal lattice

2

2

2

2

2

2

2

1cl

bk

ah

d++=

The distance d (dhkl) between lattice planes can be calculated from the unit cell parameters:

(orthorhombic system)

With the reciprocal values d* = 1/d, a* = 1/a, b* = 1/b, c* = 1/c we obtain:

2*22*22*22* clbkahd ++=Each reflection hkl can thus be described as a vector d* = (h k l) in the reciprocal space formed by the basis vectors a*, b* and c*.

From the orientation of the primary and reflected beams, we obtain the direction of d* for each reflection, from the reflection angle theta the lattice spacing d and thus d* = 1/d. “Indexing” is the art to find a set of basis vectors a*, b*, c* which allow the description of each reflection with integer values of h, k and l.

• Finding the longest vectors which can describe all reflections

• A certain error must be allowed• If necessary, move to shorter basis vectors• From a*, b* and c*, the unit cell parameters and

the Miller indices are known.

020

220

110

100200

Bragg Law:

2dhkl·sinq = l Û sinq = ½ l / dhkl

Tetragonal unit cell:

a=b=5 Å, c=10 Å a=b=5 Å, c=20 Åh k l dhkl sinq q h k l dhkl sinq q

0 0 1 20 Å 0.039 2°0 0 1 10 Å 0.077 4° 0 0 2 10 Å 0.077 4°

0 0 3 6.7 Å 0.116 7°0 0 2 5 Å 0.154 9° 0 0 4 5 Å 0.154 9°

0 0 5 4 Å 0.193 11°0 0 3 3.3 Å 0.231 13° 0 0 6 3.3 Å 0.231 13°

0 0 7 2.9 Å 0.270 16°0 0 4 2.5 Å 0.308 18° 0 0 8 2.5 Å 0.308 18°

0 0 9 2.2 Å 0.347 20°0 0 5 2 Å 0.385 23° 0 0 10 2 Å 0.385 23°

0 0 11 1.8 Å 0.424 25°

What happens if we increase the size of our unit cell?

A unit cell with two times the volume has twice the number of reflections in the same q region. Can we thus increase the number of reflections by increasing the size of our unit cell?

0 0 1 0 0 2

On doubling the axis length a reflection {0 0 1} becomes { 0 0 2}.

What about the “new” reflection { 0 0 1} of our increased unit cell?

What happens if we increase the size of our unit cell?

0 0 ½ 0 0 1

We find systematically another atom at dhkl/2. (In other words, we introduced by doubling of the unit cell a new translation operation x, y, z+0.5).

What about the “new” reflection { 0 0 1} of our increased unit cell?

Due to the additional translational symmetry, only reflections with {h k l} with l = 2n are present (path length difference = 2nl). For reflections {h k l} with odd h, the reflections are systematically absent, since the atom z+0.5 causes destructive interference.

The path length difference to this atom is ½ l.

What happens if we increase the size of our unit cell?

Tetragonal unit cell: a=b=5 Å, c=10 Å a=b=5 Å, c=20 Åh k l dhkl sinq q h k l dhkl sinq q

0 0 1 20 Å 0.039 2°0 0 1 10 Å 0.077 4° 0 0 2 10 Å 0.077 4°

0 0 3 6.7 Å 0.116 7°0 0 2 5 Å 0.154 9° 0 0 4 5 Å 0.154 9°

0 0 5 4 Å 0.193 11°0 0 3 3.3 Å 0.231 13° 0 0 6 3.3 Å 0.231 13°

0 0 7 2.9 Å 0.270 16°0 0 4 2.5 Å 0.308 18° 0 0 8 2.5 Å 0.308 18°

0 0 9 2.2 Å 0.347 20°0 0 5 2 Å 0.385 23° 0 0 10 2 Å 0.385 23°

0 0 11 1.8 Å 0.424 25°

An artificial increase of an axis length adds new reflections, which are, however, systematically absent (I = 0).

What happens if we increase the size of our unit cell?

Systematic absences and centeringThe introduction of a translational symmetry (x+0.5, y, z) causes the

systematic absence of all reflections {h k l} with h ≠ 2n.

All translational symmetries introduces systematic absences. Since all

centerings introduce additional translations, centerings are associated with

the presence of systematic absences, which we can use to DETERMINE

the centering from the reflection list.:

translation Z restrictionsP - 1 -

A x, y+0.5, z+0,5 2 k+l = 2n

B x+0.5, y, z+0.5 2 h+l = 2n

C x+0.5, y+0.5, z 2 h+k = 2n

F A + B + C 4 h+k = 2n, h+l = 2n, k+l = 2n

I x+0.5, y+0.5, z+0.5 2 h+k+l = 2n

Thus from investigating systematic absences in the reflection list, we can determine the Bravais lattice.

Systematic absences

In the same way, symmetry elements which include translations also cause systematic absences:

Screw axes (axes hélicoïdal):

translation restrictions21, 42, 63 || a x+0.5 h00 with h = 2n

|| b y+0.5 0k0 with k = 2n|| c z+0.5 00l with l = 2n

31, 32, 62, 64 || c z+1/3 00l with l = 3n

41, 43 || c z+0.25 00l with l = 4n

61, 65 || c z+1/6 00l with l = 6n

Systematic absences

Glide planes:

translation zonal restrictionsb ┴ a y+0.5 0kl avec k = 2nc ┴ a z+0.5 0kl avec l = 2nn ┴ a y+0.5, z+0.5 0kl avec k+l = 2nd ┴ a y+0.25, z+0.25 0kl avec k+l = 4n (F)

c ┴ b z+0.5 h0l avec l = 2na ┴ c x+0.5 hk0 avec h = 2n

c ┴ [110] z+0.5 hhl avec l = 2n (t, c)c ┴ [120] z+0.5 hhl avec l = 2n (trigonal)d ┴ [110] x+0.5, z+0.25 hhl avec 2h+l = 4n (t, cI)

Small Molecule Example – YLIDSpace Group Determination

hk0 layer 0kl layer

The structure factor F

F H F H F H

F H F H F H

F H F H F H

F

H

F

H

F

H

F

H

F

H

F

H

d100d010

F100 = fF + fH F010 = fF - fH

The intensity of an X-ray beam diffracted at an hkl-plane depends on the structure factor Fhkl for this reflection. The structure factor is the sum of all the formfactors (atomic scattering factors) in the unit cell.

The structure factor Fhkl thus contains information on the spatial distribution of atoms.

å=

×+×+××=N

j

zlykxhijhkl

jjjefF1

)(2p

Structure factor

The structure factor Fhkl depends on the spatial distribution of the atoms and, more specifically, on their

distance to the reflection plane.

D

F

D

21 22

21

21 2cosD×D× ×+×=

×D+=ii efef

ffFpp

p

F = f + f F = f - f

F = f

0 1

The structure factor F

D for general lattice plane

The phase problemThe value of each structure factor Fhkl depends on the distribution of the atoms in the unit cell (i. e. the electronic density). We can thus obtain this electronic density, from the combination of all structural factors using a Fourier transformation

å ++-×=hkl

lzkyhxihkl eFVzyx )(21),,( pr

hkljjj ihkl

N

j

zlykxhijhkl eFefF ap ×=×=å

=

×+×+×

1

)(2

The factor Fhkl takes the form of a cosinus function with an amplitude |Fhkl| and a phase ahkl. These two, |Fhkl| and ahkl, vary for each reflection hkl and depend on the atomic positions relative to the hkl plane.

)(21),,( lzkyhxi

hkl

ihkl eeFVzyx hkl ++-××= å par

The only formula you have to know by heart!

|Fhkl|

ahkl

The phase problem

)(21),,( lzkyhxi

hkl

ihkl eeFVzyx hkl ++-××= å par

One small problem: Since we can determine |Fhkl| = √Ihkl, we do not know the phases ahkl !

),,( zyxr hklihklhkl eFF a×=

FT

FT

known unknown

A (very) short introduction to phases

h,k,l = 2,3,0; Centers on planes, a230 = 0º, strong reflection

h,k,l = 2,1,0; centers between planes, a210 = 180º, strong reflection

A (very) short introduction to phases

The phase f of a reflection where the atoms are situated on thehkl planes has a phase of approximately 0º; if the atoms arefound between the planes the phase is approximately 180º. Forrandomly distributed atoms, we cannot predict the phase andthe reflection is weak due to strong destructive interference.

h,k,l = 0,3,0; weak reflection, a030 = ?

Do we really need the phases?

First estimation of phases (Patterson, direct methodes):

FTr1(x,y,z) = Atomic coordinates

Manual confirmation

rc(x,y,z)Refinement

We want to optimise: r(xyz) Fc = |Fc| ahkl

Optimisation criterium:M = ∑w(|Fo|2-|Fc|2)2

ahkl

|Fo| = √I

ahkl

r2(x,y,z) Dr=1/V∑(Fo-Fc)e-2pi(hx+ky+lz)

Manual confirmation

Difference Fourier map

Fhkl = |Fo| · ahkl

Fhkl = |Fo| · ahkl rc(x,y,z)

The phases improve with each

cycle

TF

TF

Experiment

Structure solution

FT

For some centers of electron density an element is already proposed. You have to verify if they are correctly assigned.

Initial solution

S

BF5C6 C6F5

C6F5

Initial solution

Several centers of electron densities are not related to your structure.

Do not hesitate to delete them! A deleted electron density will show up again in later cycles, if it is correct.

If a structure is difficult to identify, look for regular structures, such as aromatic rings etc.

Q-peaks are sorted after their electronic density with Q1 being the highest electron density.

The peaks with the smallest numbers have the highest electron density and are thus fluorine atoms.

S

BF5C6 C6F5

C6F5

Initial solution

Now you have to assign correct elements to the remaining electron densities:

• Carbon

Initial solution

Now you have to assign correct elements to the remaining electron densities:

• Carbon• Fluor

Initial solution

At the ends no Q-peaks must be left!

Now you have to assign correct elements to the remaining electron densities:

• Carbon• Fluor• Boron

Initial solution

Summary of Part 1

n Introduction to crystals and crystallographyn Definitions of unit cells, crystal systems, Bravais lattices n Introduction to concepts of reciprocal space, Fourier

transforms, d-spacing, Miller indicesn Bragg’s Law – geometric conditions for relating the

concerted coherent scattering of monochromatic X-rays by diffracting planes of a crystal to its d-spacing.

n Part 2 – We will use these concepts and apply them in a practical way to demonstrate how an X-ray crystal structure is carried out.