ΣΣΣ - York University · Each reflection has a unique set of indices hkl. Angles give the unit...

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Basics of Crystallography 1. Diffraction & Bragg’s Law 2. Symmetry Bravais Lattices Space Groups 3. A Fourier Series describes the e-density in crystals f(xyz). rho = 1/V ΣΣΣ F HKL cos(2π(hx + ky + lz)) sum over all h,k, and L 4. Structure Factors F HKL (coefficients in series above) /Fobs/ from reflection intensities /Fobs/ = kI HKL phases + or - in centrosymmetric case from Direct Methods Fcalc = Σ f (I,Φ) cos(2π(hx i + ky i + lz i )) summed over all atom locations 5. Solving a structure. (Expt # 5) learn by doing.

Transcript of ΣΣΣ - York University · Each reflection has a unique set of indices hkl. Angles give the unit...

Basics of Crystallography1. Diffraction & Bragg’s Law2. SymmetryBravais LatticesSpace Groups

3. A Fourier Series describes the e-density in crystals f(xyz).

rho = 1/V Σ Σ Σ FHKLcos(2π(hx + ky + lz)) sum over all h,k, and L

4. Structure Factors FHKL (coefficients in series above)/Fobs/ from reflection intensities /Fobs/ = k√IHKL

phases + or - in centrosymmetric case from Direct Methods

Fcalc = Σ f(I,Φ) cos(2π(hxi + kyi + lzi )) summed over all atom locations

5. Solving a structure. (Expt # 5) learn by doing.

Each reflection has a unique set of indices hkl.

Angles give the unit cell parameters a,b,c,α, β, γ

Intensities Ihkl contain info about unit cell content/atom locations (if phasing known).

HKL planes intersect a unit cell at 1/h, 1/k, and 1/L in a,b, and c directions

respectively.

Bragg’s Law

n λ = 2 dhkl sin θhkl (n =1)

(Mo radiation λ = 0.71073 Å , Cu radiation λ = 1.542 Å)

dhkl = intraplanar d spacing

cubic crystal orthorhombic

d100 =d 010= d001= a d100 =a , d010 = b ,d001 = c

d200 = a/2

d2hkl = a2/(h2+k2+ l2) 1/d2

hkl = (h/a)2 + (k/b)2 + (l/c)2

Bragg angle θ = sin-1 (λ/2d)

sin(θ) = opp/hypot = x/d

path diff = 2x = 2 dsin(θ) must be an integral

number of λ’s 2d sin(θ)= nλ

x x

The 14 Bravais Lattices – consist of the 7 basic unit cell shapes (crystal systems) plus various types of centering.

1. Cubic a=b=c all angles 90 P, F , I 2. Tetragonal a=b ≠ c all angles 90 P, I 3. Orthorhombic a ≠ b ≠ c all angles 90 P, I, F, (A,B,C)4. Monoclinic a ≠ b ≠ c only β ≠ 90 P, C5. Triclinic a ≠ b ≠ c no angles 90 P6. Rhombic a = b = c no angles 90 P7. Hexagonal a =b ≠ c α =β = 90 and γ = 120 P

P = primitive - object only at (x,y,z) I = body centered -objects at (x,y,z) and ( ½ + x , ½ +y, ½ + z ) C = C centered -objects at (x,y,z) and ( ½ + x , ½ +y, z) For A centered extra object is on A face (x, ½ + y, ½ + z), and for B

centering, on the B face. F = face centered - objects at (x,y,z), ( ½ + x , ½ + y , z),

(x, ½ + y, ½ + z), ( ½ + x, y, ½ + z)

Z =1 Z =2 Z = 4

Counting : Z = 1/8 (corner) + ½ (face + ¼ (edge)

NaCl Fm3m z=4

CsCl Pm3m Z =1

α-Fe Im3m Z =2

Space Groups

In addition to: E = 1, C2 = 2 , S4 = -4 or 4bar, i = -1, σ = m

1. Translation : For every object Q at (x,y,z) another Q is found by a full unit cell translation in any direction (x+1,y,z), ( x,y+1,z), (x+1, y+1, z) etc. The unit cell repeats itself in all 3 dimensions ad infinitum.

2. Screw axis 21 rotation by 360/2 and translation ½ unit cell along it in a, b or c direction.

For 21 along a (xyz) ���� ( ½ + x, -y, -z) 41 along c would rotate 90 and translate ¼ along z (xyz) ���� (-y,x,z + ¼)

3. Glide Plane a,b,c,n, or d reflect through a plane and then translate in a direction parallel to it.a glide perpendicular to b (xyz) ���� (½ + x ,-y, z)

21/c means a twofold screw perpendicular to a c glide.

Symmetry Operations & Coordinates

inversion -1 xyz � -x,-y,-z

mirror m ╧ a xyz � -x,y,z ╧ b x,-y, z

C2 2 along x xyz � x,-y,-z etc

screw 21 along x xyz � x + ½ ,-y,-z

along z xyz � -x, -y, z+ ½

glide c glide ╧ a xyz � -x, y, z + ½

n glide ╧ a xyz � -x, y + ½ , z + ½

A mirror or glide flips one coordinate

A C2 or 21 flips two , inversion flips all three.

ab

Symmetry elements hereObjects

here

Special positions here, all require -1 symm

inverted object screw ¼ up z

general positions here

c glide

from 21

c glide

All space group diagrams at http://img.chem.ucl.ac.uk/sgp/large/sgp.htm

No special positions here

Note 21Screw symbols

they lie ¼ way up each axis

Note 21 and 2 symbols.

You can’t sit on a screw but you can on a -1 or 2.

commas denote inverted objects

To save space , you are left to add this to below. C centering.

• Coordinates are fractions of the unit cell dimensions Thus (½ ,0, 0) lies a/2 from origin in x direction

(½, ½,0) lies on the middle of the c face• translational equivalents (-½,0,0) is the same as (½, 0,0) in the next unit cell (-0.15,0,0) is the same as (0.85,0,0) in the next cell

• the unit cell and its content repeats itself in a,b, and c directions

• Any object placed in a general position generates n objects where n = number of general positions

• An object placed in a special position generates fewer equivalents - but may lie there ONLY if it contains the proper symmetry & that element is aligned correctly.

It is a simple matter to generate the space group tables for lowsymmetry cases. Consider triclinic space groups.

P1 - no symmetry general position is Z =1 (x,y,z)

P-1 inversion (-1) at origin general Z =2 (x,y,z) (-x,-y,-z) special Z =1 -1 imposed at (0,0,0) and various ½’s

Then try the monoclinic C centred space group #8Cm mirror along x axis and at y = ½general Z =4 (x,y,z) (x,-y,z) and (1/2,1/2,0)+ each special Z =2 mirror (x,0,z) and (½+x, 1/2, z) note that glides are present at y=1/4 and ¾

Pmma On the following slide Connect pairs of objects with the symmetry element that relates them

on the next slide. Identify which general position corresponds to which operation on (x,y,z).

Deduce all special positions, #, sym coordinates.

Symmetry in FlatlandYou could easily go batty trying to follow 3 dimensional operations on a two dimensional piece of paper so let us travel to flatland.

There are 17 2D space groups. Crystal systems are:

oblique - angle not 90o

rectangular 90o and a ≠ b

square and hexagonal

Exercises

1. Generate space group diagrams for Pm (mirror along y)

Pg glide along y

Cm mirror along y

Pmm mirrors along x and y

2. Draw stickman or mickey mouse in each of these with Z = 2 or Z =4

3. Show how Cu(NH3)4(NO3)2 -5H2O could crystallize in Pmm Z =1.

Space Group from Systematic Absences

Crystal 1 present 020 120 130 040 013 023absent 050

Crystal 2 present 020 130 040 023absent 120 050 013

Crystal 3 present 020 120 130 040 013absent 050 023

P212121 h00, 0k0, or 00l h,k,and l =2n

Pbcn as above and 0kl, k =2n; hk0, h+k =2n; h0l l=2n ;

Pnnm as above except hk0, no conditions; h0l, h+l = 2n; and 0kl, k+l =2n

P212121

Pbcn

Pnnm

Body centered

α-Fe Im3m z =2

100 - absent

200 present