X-ray diffraction Meet in the LGRT lab Again, will hand in worksheet, not a formal lab report Revision exercise – hand in by April 17 th class.

Diffraction summarized The 6 lattice parameters (a,b,c,  ) of a crystal determine the position of x-ray diffraction peaks. The contents of the cell (atom types and positions) determine the relative intensity of the diffraction peaks. If a diffraction peak can be identified with a Miller Index, the unit cell on the phase can usually be determined.

Miller indices Choose origin Pick 1 st plane away from origin in +a, +b direction Find intercepts in fractional coordinates (“none = ∞”). h = 1/(a-intercept) k = 1/(b-intercept) l = 1/(c-intercept) If no intercept, index = 0 Plane equation: h a + k b + l c = 1 Miller index, hkl (2-10) (110) (100) (150) Same set of planes will be described if all 3 Miller indices are inverted (2-10) ≡ (-210) Minus sign → bar

Crystals facets correspond to Miller indices Haüy, 1784 Crystals (like calcite) are made of miniscule identical subunits b a (-120) Woolfson Facets can be described by low-order Miller indices (-120) (-110) (-100) (-1-10) (-2-10) (3-40) (340)

Miller indices For a lattice of known dimensions, the Miller indices can be used to calculate the d-spacing between hkl planes (a,b,c = lattice parameters). This d-spacing will determine where powder diffraction peaks are observed Miller index, hkl (2-10) (-110) (100) (150) Origin at orange dot

Distance between planes (Miller indices)

Bragg’s law derivation (angle of incidence = angle of diffraction) x = d sin  extra distance = 2d sin  = n n = 2d sin    d x  d xx ab = 2d sin  (for x-ray diffraction)

Powder diffractometer

KI (CsCl-type) x-ray pattern

Crystals, diffraction, and Miller indices (100) (010) (001) (0-10)

Coherent scattering from a row of atoms Will only happen when emission from all atoms is simultaneously stimulated. (Solid lines represent spatial regions where phase = 0 deg).

a Laue condition – vector description Extra distance= a cos  - a cos Extra distance = -(a·S 0 ) + (a·S) Extra distance= (a·s)  a S S0S0 a Will have diffraction when: a cos  - a cos = h (h = integer) Will have diffraction when: a · s = h (h = integer) s = (S – S 0 )/ v1v1 v2v2 cos  = v 1 ·v 2

a Laue condition – 2D Must have coherent scattering from ALL ATOMS in the lattice, no just from one row. If color indicates phase of radiation scattered from each lattice point when observed at a distant site P, we see that scattering from rows is in phase while columns is out of phase, making net scattering from all 35 points incoherent and therefore NOT observable S S0S0 b Will have diffraction when: a · s = h (h = integer) Will have diffraction when: b · s = k (k = integer) Will have diffraction when: c · s = l (l = integer) Every lattice point related by translational symmetry will scattering in phase when conditions are met

Single crystal diffraction Used to solve molecular structure –Co(MIMT) 2 (NO 3 ) 2 example Data in simple format – hkl labels + intensity + error

h=2 h=1 h=0 h=-1

l =6 l =7 l =8 h =2 l =9 l =10

3D view of reciprocal lattice Massa (Lines connecting dots are unecessary)

Relation between direct and reciprocal cells Stout and Jensen

Direct vs. reciprocal lattice direct or real reciprocal Lattice vectors a, b, c Vector to a lattice point: d = ua + vb + wc Lattice planes (hkl) Lattice vectors a*, b*, c* Vector to a reciprocal lattice point: d* = ha* + kb* + lc* Each such vector is normal to the real space plane (hkl) Length of each vector d* = 1/d-spacing (distance between hkl planes)

Hexagonal lattice  changes from 120 to 60 deg a b b* a* direct or realreciprocal

Distances between planes (d*) direct or real reciprocal Vector to a reciprocal lattice point: d* = ha* + kb* + lc* Length of each vector d* = 1/d-spacing (distance between hkl planes) |d*| = (d*·d*) 1/2 =(ha* + kb* + lc*) (ha* + kb* + lc*) =(ha*) 2 + (kb*) 2 + (lc*) 2 + 2(ha*) (kb*) + 2(ha*) (lc*) + 2(kb*) (lc*) = h 2 a 2 * + k 2 b* 2 + l 2 c* 2 + 2klb*c*cos  * + 2lhc*a*cos  * + 2hka*cos  *

Density = mass / volume Mass of unit cell = (# Ca)(m Ca ) + (# Ru)(m Ru ) + (# O)(m O ) Volume of unit cell = abc = ( Å) 3 m amu N A = m grams Density of CaRuO 3 M = molar mass (g/mol) n = # f.u. per cell N A = Avagadro’s # (1/mol) V = cell volume