References for this Topic

✿Clegg, “Crystal Structure Determination”. Lattices, Reciprocal Lattices ✿ Stout & Jensen, “X-Ray Structure Determination”, 2nd Edition. Instrumentation discussion is completely and Diffraction outdated, but still a good text on the subject. ✿ A more authoritative general reference: Giacovazzo, et Chem 634 al., “Fundamentals of Crystallography”, IUCr Texts on T. Hughbanks Crystallography. ✿ MIT has a good site (MIT Open Courseware): http:// ocw.mit.edu/OcwWeb/Chemistry/ 5-841Fall-2006/LectureNotes/index.htm

1 2

Bravais Lattices • Direct Lattice: Reciprocal Lattices – A regular, periodic array of points with a • Reciprocal Lattice: the basis vectors of the reciprocal spacing commensurate with the unit cell lattice are defined as dimensions. The environment around each b × c c × a a × b points in a lattice is identical. a* = ; b* = ; c* = ; V = (a × b)• c • The set of direct lattice points can be written as V V V (vectors): R = ta + ub + vc t,u,v integers (Note: In physics texts, a factor of 2π is usually included b × c • V = volume of unit cell e.g., a* = 2π ; ! ) V V = a ⋅ b × c = • The set of Reciprocal Lattice Vectors (RLVs) are ( ) written as * * * K i = ha + kb + lc b⋅(c × a) = c⋅(a × b) h,k,l = integers

3 4 Side by Side (2-D) b b∗ Reciprocal Lattices a a∗ Reciprocal Lattice • Properties of RLVs: DirectLattice a i a* = b i b* = c i c* = 1 a* i b = a* i c = b* i a = b* i c = c* i b = c* i a = 0 b∗

Alternatively, these can be regarded as definitions. a∗ b In 2 dimensions, we define the RLVs a a* ⊥ b , b* ⊥ a , a i a* = b i b* = 1 • Although they “live in separate universes”, the direct and reciprocal lattices are in rotational “lock-step”.

5 6

Lattice & Reciprocal Lattice Lattice & Reciprocal Lattice

Example: BCC treated as Primitive Example: BCC treated as Primitive ^ ^ (a/2)(–x^ + y^ + ^z) (1/a)(y + z)

a ^ 1 a = (xˆ + yˆ − zˆ) az a* = (xˆ + yˆ ) 2 a (1/a)z^ a 1 b = −xˆ + yˆ + zˆ b* = yˆ + zˆ 2 ( ) b a a ( ) c ^ ^ b* a 1 (1/a)(x + z) c* 1/a c = (xˆ − yˆ + zˆ) c* = (xˆ + zˆ) ^ 2 (a/2)(x^ – y^ + z^) a (1/a)y ay^ (1/a)x^ a* Use to find Recip. Lattice Basis: ax^ Exercise: Find the reciprocal lattice of the reciprocal lattice! (1/a)(x^ + y^) (a/2)(x^ + y^ – z^) b × c c × a a × b a* = ; b* = ; c* = a If we use the conventional cell (which leaves out half the lattice V V V points, the reciprocal lattice includes the ‘open’ points too: V = a × b • c ( ) xˆ yˆ zˆ a = axˆ ; b = ayˆ ; c = azˆ ⇒ a* = ; b* = ; c* = a a a

7 8 Interrelationships Interrelationships • The direct lattice can be partitioned such that all of the lattice points lie on sets of planes. Crystallographers classify these sets of planes using the intercepts on the unit cell axes cut by the plane adjacent to the plane through the origin.

• The intercepts are of the form (1/h, 0, 0), (0,1/k, b b* 0), and (0,0,1/l). a 2a* + 3b* • These planes are normal to the RLVs a*

intercept: (1/2,0) * * * -3a+ 2b intercept: (0,1/3) K hkl = ha + kb + lc note: (-3a + 2b)•(2a* + 3b*) = 0

9 10

Interrelationships Interrelationships Reciprocal Lattice lattice planes DirectLattice b

K hkl

ˆ K hkl K hkl = K hkl θb - aunitvector (0,1/k,0) θa a c b (1/h,0,0) ˆ ˆ dhkl =(1/h)a•K hkl = (1/h)a K cosθa 2a* + b* ˆ ˆ a (hkl) plane dhkl =(1/k)b•K hkl = (1/k)b K cosθb (0,0,1/l)

intercept: (1/2,0) (0,1/k, 0) intercept: (0,1) -(1/h)a + (1/l)c b

(1/h,0,0) a -(1/h)a + (1/k)b

11 12 hkl Planes are normal to Khkl Kj•Ri products involve no cross terms and always yield integers c

(hkl) plane • For any DLV Ri and any RLV Kj (0,0,1/l)

(0,1/k, 0) R K ta ub vc ha* kb* lc* -(1/h)a + (1/l)c i i j = ( + + )⋅( + + ) b (1/h,0,0) = th + uk + vl = m, an integer a -(1/h)a + (1/k)b • Proof: Consider the plane that cuts the cell axes at (1/h,0,0), e2πiRiiK j 1, for all R and all K (0,1/k,0) and (0,0,1/l). ⇒ = i j 1 1 1 1 • Two vectors lie in the plane: − a + b and − a + c . h k h l ⎛ 1 1 ⎞ h k K a b a a* b b* 1 1 0 hkl ⋅⎜ − + ⎟ = − ⋅ + ⋅ = − + = ⎝ h k ⎠ h k ∴K hkl ⊥ to both vectors ⎛ 1 1 ⎞ h k ∴K ⊥ to the family of planes K ⋅ − a + c = − a ⋅a* + c⋅c* = −1+1= 0 hkl hkl ⎝⎜ h l ⎠⎟ h k

13 14

1 1 To Prove: d = = hkl K K Diffraction hkl hkl • Assume K is the shortest RLV that points in the exact direction hkl • In handling scattering of x-rays (or electrons) by matter, we need it points (i.e., h,k, and l have no common factor.) to characterize the x-ray beam by specifying its wavelength and • Khkl is normal to the “hkl family of planes”. We can relate the direction of propagation: E E xˆ distance between planes to the magnitude of Khkl: = 0 H H yˆ lattice planes = 0 k = kzˆ ˆ Eˆ Hˆ kˆ 1 a K a b 0 × 0 = d a Kˆ hkl cos cos hkl = ⋅ hkl = θa = θa K H H exp 2 i(k r t) h h h hkl = 0 { π ⋅ −ν } ReH H cos 2 (k r t) Where Kˆ is a unit vector parallel to K . = 0 { π ⋅ −ν } hkl hkl ˆ K hkl 1 K hkl = k = * * * E = E exp 2πi(k ⋅r −νt) K ha + kb + lc K hkl 0 { } λ ˆ hkl θb now, K hkl = = a - aunitvector ReE E cos 2 (k r t) (0,1/k,0) θa = 0 { π ⋅ −ν } Khkl Khkl • Note: The usual physics definition is 1 * * * (1/h,0,0) ˆ ˆ so d = a ⋅ ha + kb + lc d =(1/h)a•K = (1/h)a K cosθa 2π hkl hK ( ) hkl hkl k hkl ˆ ˆ = dhkl =(1/k)b•K hkl = (1/k)b K cosθb λ 1 ∴dhkl = Khkl

15 16 Additional Comments Traveling wave packet • By “adding” up a narrow range of frequencies, • In order to speak of the localization of an electron (or photon), we must ‘blur’ the specification of the electron’s (photon’s) a “wave packet” can be constructed and a wavelength, and allow that there is some uncertainty in the photon visualized. wavelength (and frequency and energy):

∞ E (z,t) = E (k′)exp 2πi(k′z −νt) dk′ k ∫−∞ 0 { } where we have assumed, for simplicity, that the wave is propagating along the z-axis and E0(k) is a function strongly peaked near a particular ‘approximate’ wavenumber k. • Electrons and photons are different in that wave packets for electrons spread out over time, while photon wave packets retain their width as they propagate. For more details, see:

http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html

http://farside.ph.utexas.edu/teaching/qmech/lectures/node1.html

17 18

Interference - “Time Lapse” pictures Elastic Scattering

• When an x-ray is scattered by an atom it may emerge in any direction, but unless absorption occurs, it has the same wavelength: 1 k = k′ = λ

• X-ray diffraction for structure determination is concerned with interference effects that result from scattering by periodic arrays of atoms.

This is my own heuristic exercise, not “reality”.

19 20 Atomic Form Atomic X-ray Atomic Form Factors: f (2θ) scattering Factors: • Forward scattering is more intense in “real f(2θ) atoms” with diffuse electron clouds. If all the electron density were concentrated very close to the nucleus, 2θ dependence would disappear.

• f (2θ = 0) = N (number of electrons)

water ripples

• The “water ripples” concept is not actually accurate; forward Atomic X-ray scattering is actually more scattering intense. Clegg, p. 24.

21 22

“Thermal” effects: f ′(2θ) Interference between scatterers

• Vibrational motion “spreads out” average electron Consider the interference between two scatterers, separated by a vector d: density and causes effective scattering at higher 1 2 angles to fall off more rapidly; U is the isotropic k displacement parameter (units of Å2 ).

20 2 2 ⎛ ⎞ d d −8π U sin θ ˆ k – f ′(2 ) f (2 ) •exp k = Cl j θ = j θ ⎜ 2 ⎟ λ k 15 ⎝ ⎠ ˆ k′ k′ = 3 4 k′ k f j (2θ) Again, scattering at higher angles 10 1 is attenuated by spreading out the k = k′ = or kˆ = λk λ f ′(2θ) electron density k ′ 5 j d d

0.2 0.4 0.6 0.8

sinθ/λ (Å–1) Clegg, p. 24.

23 24 Interference between Interference between scatterers scatterers Consider the interference between two scatterers, separated by a • Amplitude of scattered wave vector d:

∝ f1(2θ) + f2 (2θ)⋅( phase shift factor) k kˆ = k • The phase shift factor depends on the path length difference = k′ kˆ ′ = k′ kˆ ⋅d + (−kˆ ′)⋅d = (kˆ − kˆ ′)⋅d 1 k = k′ = or kˆ = λk λ • For completely constructive interference, where n is some integer:

(kˆ − kˆ ′)⋅d = nλ or (k − k′) i d = n

• For completely destructive interference 1 (k − k′) i d = n + Clegg, Sec. 1-6. 2

25 26

• In a crystal, all pairs of translationally equivalent • For the general case the amplitude of the atoms are separated by some direct Lattice scattered wave: Vector (Ri). A f (2 ) f (2 )cos 2 (k k ) •d ∝ 1 θ + 2 θ ⎣⎡ π − ′ ⎦⎤ If all atoms in the crystal that are related by cos⎣⎡2π (k − k′) •d⎦⎤ = 1 when (k − k′) •d = n ∴ translational symmetry are to scatter X-rays to 1 cos⎡2π (k − k′) •d⎤ = −1 when (k − k′) •d = n + give constructive interference simultaneously, ⎣ ⎦ 2 • In general we use a complex form: we must have the following: A f (2 ) f (2 )exp 2 i(k k ) •d ∝ 1 θ + 2 θ { π − ′ } 2πi(k – k′)•R Ri •(k – k′) = n or e = 1

where Ri is any DLV.

27 28 The beam is diffracted by a RLV! Bragg View

This places conditions on the possible vectors k – k′ that are exactly the same as the conditions used to define RLVs. We therefore obtain the Laue condition for diffraction: k – k′ = K where K is some RLV j j k

k´

k

K hkl Clegg, Secs. 1.4-1.5.

29 30

Optical Transforms; Fraunhofer Diffraction Bragg View • The diffraction phenomenon can demonstrated with monochromatic visible light by use of optical transforms. • The phenomenon relies on interference effects, like X-ray diffraction, though the scattering mechanism is different. • The interference phenomenon occurs when the spacing between scatterers is on the same order of magnitude as the wavelength of the light used.

31 32 33 34

35 36 Ewald Construction

• Start with reciprocal lattice and the incident wavevector, k, originating at (h,k,l) = (0,0,0). • Draw a sphere with the tip of k at the center that contains k the origin, (0,0,0), on the k′ (0,0,0) sphere’s surface. K hkl • As the direction of k is ′ K hkl =k – k changed, other RLVs (Kj’s) move through the surface – which is the condition that a diffraction peak is observed. Stout & Jensen, Sec. 2.4.

37 38

k k Ewald Construction Limitations on the Data k k • Reflections (Kj’s) lie on the Ewald sphere to meet the the (0,0,l) diffraction condition. • As the relative orientation of c k′ k′ k′ the crystal and the incident X- ray beam are varied, the

k direction (but not the length) a k k (0,0,0) 1 of k changes, and k can in k k principal point any direction. k 2 • Therefore, the “full sphere” of possible reflections has radius 2|k| = 2k = 2/λ. The spacings between the RLVs (RL points) k′ k′ ∝ 1/a (or 1/b or 1/c). (h,0,0) k′

39 40 Getting more Data - k − k′ = K ⇒ − k′ = K − k shorter λ j j Use to get ⇒ k′ i k′ = (K j − k) i (K j − k) Laue equations • Reflections (K ’s) are limited to j 2 2 2 those RLVs no further from the k = K j − 2k i K j + k′ origin than 2|k| = 2k = 2/λ. 2 ∴2k i K j = K j • The spacings between the RLVs 2 2 (∝ 1/a or 1/b or 1/c) are 2k K cos(π 2 −θ) = K ⇒ 2k sinθ = K ⇒ sinθ = K j !##"##$ j j λ j determined by the unit cell sin(θ ) dimensions. 2 2 k k Mo • For example, if the X-ray ⇒ sinθ = λ ⇒ sinθ = λ ⇒ 2d sinθ = nλ Bragg's Law Cu K n K hkl wavelength is shortened, more j hkl (0,0,0) reflections fall “in the In the last line we've used the fact that K j is a multiple of K hkl , which is the sphere” (e.g., λ = 1.54056 Cu shortest RLV for which h, k, and l have no common factors. (Kα1), λMo = 0.71073 (Kα1). i.e., K j = nK hkl , where n = integer.

Stout & Jensen, Sec. 2.4.

41 42

Laue Equations Laue Equations, cont. nλ = 2dhkl sinθ; 2 2 2 * * * Monoclinic 2 2 2 Khkl = (4 λ )sin θ where K hkl = ha + kb + lc (***) K = (4 λ )sin θ sin2 θ = Ah2 + Bk 2 + Cl2 − Dhl hkl 2 2 2 2 2 * * * or [1 d ] = (4 λ )sin θ, since Khkl = [1 d ] where K = ha + kb + lc (***) A = λ 2 (4a2 sin2 β); B = λ 2 4b2 ; hkl The “Laue Equations” for each crystal class can be derived from the form (***). C = λ 2 (4c2 sin2 β); D = (λ2 cosβ) (2acsin2 β) Cubic or [1 d 2] = h2 (a2 sin2 β) + k 2 b2 + l2 (c2 sin2 β) − 2hlcosβ (acsin2 β) sin2 θ = A(h2 + k 2 + l2 ); A = λ2 4a2 or [1 d 2 ] = (h2 + k 2 + l2 ) a2

Triclinic Tetragonal 2 2 2 2 2 2 2 2 2 2 2 sin θ = (λ 4)[1 d ] sin θ = A(h + k ) + Cl ; A = λ 4a ; C = λ 4c h h h or [1 d 2] = (h2 + k 2 ) a2 + l2 c2 cosγ cosβ 1 cosβ 1 1 a a a

h k k k k k Hexagonal 1 cosα + cosγ cosα + cosα 1 2 2 2 2 2 2 2 2 a b b b b b sin θ = A(h + hk + k ) + Cl ; A = λ 3a ; C = λ 4c l l l or [1 d 2] = 4(h2 + k 2 ) 3a2 + l2 c2 cosα 1 cosβ 1 cosβ cosα 1 c c c Orthorhombic = d 2 1 cosα cosβ sin2 θ = Ah2 + Bk 2 + Cl2; A = λ 2 4a2 ; B = λ2 4b2 ; C = λ 2 4c2 cosα 1 cosα or [1 d 2] = h2 a2 + k 2 b2 + l2 c2 cosβ cosα 1

43 44 Example Results

• The distances between lattice planes are on If a hexagonal crystal has a (110) reflection the order of lengths of unit cells (~10 Å)

at 2θ = 8.5˚, at what 2θ angle would you • 50 Å ≥ dhkl ≥ 0.5 Å for most “small molecule expect to observe the (110) reflection? structures” • Two most common sources:

λCu = 1.540562 Å (Kα1)

λMo = 0.71073 Å (Kα1)

45

45 46

Neutron Diffraction Electron Diffraction • In neutron diffraction, so-called thermal neutrons are • In high-energy electron usually used. diffraction, electrons with 100 • Thermal neutrons have deBroglie wavelengths that are keV kinetic energies are comparable to typical X-ray wavelengths: commonly used • e–’s have short deBroglie λCu = 1.5418 Å, wavelengths:

λMo = 0.71069 Å, λCu = 1.5418 Å, -1/2 λn = h/pn = h[3mnkBT] = 1.45 Å λMo = 0.71069 Å, λ = h/p = h[2m E]-1/2 = 0.0039 Å • A major advantage of neutron diffraction is in the very e,100keV e different form factors, which have magnitudes that do • Ewald sphere is very large in not scale with atomic number (good for many light comparison with reciprocal lattice spacings. Therefore, a atoms, especially deuterium). large section of any plane tangent • Neutron spin (magnetic moment) can be exploited to to the Ewald sphere is effectively probe magnetic ordering. “on the sphere” at the same time.

47 48 Summary Multiple Atom Structures • The geometrical aspects of a crystal (cell • Within translationally related sets of atoms, the conditions for dimensions, translational symmetry, etc.) constructive interference will be determine the direct lattice. B B B satisfied simultaneously. However, different sets of atoms will generally • The direct lattice, in turn, determines the A A A diffract the X-ray with different reciprocal lattice. d d phases. • Example: For the “crystal” shown • There is a one-to-one correspondence B B here, when the set of A atoms all A A A scatter the X-ray constructively between the RLVs and the vectors through with respect to each other, the set which the incident radiation is diffracted. d d d of B atoms will also scatter the X- B B B ray constructively. But the the sets • The geometry of the diffraction pattern is {Ai} and {Bi} will generally scatter determined by the cell dimensions and A A A with different phases. d d d • The resultant diffraction intensities symmetry - no specific structural details will be determined by the sum of beyond symmetry and dimensions of the unit B B B the diffraction amplitudes due to A A A {Ai} and {Bi} - including the effect of cell affect the “positions” of diffraction peaks. different phases!

49 50

How do data determine “Structures” ? Structure Factors • The positions of diffraction peaks tell us only about the lattice parameters. • With n atoms distributed throughout the unit cell, • The intensities of the peaks tell us about the with fixed positions with respect to each other, nature and positions of the atoms within the scatter X-rays that produce interference patterns. unit cell. Amplitude of scattered wave = n • The central problem of crystallography is in A S f ′(K) exp 2 iK d ∝ K = ∑ j { π i j } working backwards from the peak intensities j=1 !"# !##"##$ to locations and identities of atoms in the unit geometric factor cell. atomic form factor Easier: How do “Structures” determine data? (determined by charge distribution of the jth atom)

51 52 The Structure Factor expression Recall: BCC treated as Primitive & Systematic Absences (a/2)(–x^ + y^ + ^z) a ^ a = xˆ + yˆ − zˆ az • Symmetry operations place restrictions on 2 ( ) a how atoms in the cell must be distributed. b = (−xˆ + yˆ + zˆ) a 2 c b (Physically, perhaps it is the other way a c = (xˆ − yˆ + zˆ) around!) 2 (a/2)(x^ – y^ + z^) n ay^ A ∝ S = f ′(2θ) × exp 2πiK i d K ∑ j { j } Use to find Recip. Lattice Basis: ax^ j=1 !#"#$ !##"##$ form factor geometric factor (a/2)(x^ + y^ – z^) b × c c × a a × b a* = ; b* = ; c* = a • This means that there are symmetry V V V relationships between the list of fj’s and dj’s. V = (a × b)• c

53 54

BCC treated with conventional cell (2 atoms/cell) • All recip. Lattice vectors are of the form K = ha∗ + kb∗ + lc∗ a = axˆ ⎫ hkl ⎪ xˆ yˆ zˆ ⎡ ⎧ ⎛ h k l ⎞ ⎫⎤ b = ayˆ ⎬ ⇒ a* = ; b* = ; c* = S = f × ⎢exp 2πi(h i 0 + k i 0 + l i 0 + exp⎨2πi + + ⎬⎥ a a a hkl Mo { } ⎝⎜ 2 2 2⎠⎟ ˆ ⎪ (a/2)(–x^ + y^ + ^z) ⎣⎢ ⎩ ⎭⎦⎥ c = az ⎭ S = f × ⎡1+ exp πi h + k + l ⎤ ^ hkl Mo ⎣ { ( )}⎦ Atomic Positions: az (a/2)(x^ + y^ + ^z) If h k l d1 = 0 + + = 1 d = (a + b + c) b a even no. : 2 2 c exp{iπ (h + k + l)} = 1 ⇒ Shkl = 2 fMo Form Factors: (a/2)(x^ – y^ + ^z) odd no. : exp iπ (h + k + l) = −1 ⇒ Shkl = 0 ′ ′ ay^ { } f1 (2θ) = f2 (2θ) = fM′ o ax^ ∴ if h + k + l = odd, reflection will not be observed. (a/2)(x^ + y^ – ^z) a

55 56 Recall: BCC treated as Primitive ^ ^ (1/a)(y + z) Example: NaCl ⎧dNa1 = 0 ⎪ d = 1 (a + b) 1 , 1 ,0 1 ⎪ Na2 2 ( 2 2 ) a* xˆ yˆ Na: ⎨ = ( + ) d = 1 (a + c) 1 ,0, 1 a (1/a)z^ ⎪ Na3 2 ( 2 2 ) 1 ⎪d = 1 (b + c) 0, 1 , 1 b* = (yˆ + zˆ) ⎩ Na4 2 ( 2 2 ) a b* 1 (1/a)(x^ + z^) c* c* = xˆ + zˆ 1/a a ( ) (1/a)y^ (1/a)x^ ⎧Cl1 1 ,0,0 a* ⎪ ( 2 ) Cl2 0, 1 ,0 ⎪ ( 2 ) Cl: ⎨ (1/a)(x^ + y^) Cl3 0,0, 1 ⎪ ( 2 ) If we use the conventional cell (which leaves out half the lattice ⎪Cl4 1 , 1 , 1 ⎩ ( 2 2 2 ) points, the reciprocal lattice includes the ‘open’ points too: xˆ yˆ zˆ a = axˆ ; b = ayˆ ; c = azˆ ⇒ a* = ; b* = ; c* = a a a 57 58

K • 0,0,0 K • 1 , 1 ,0 K • 1 ,0, 1 K • 0, 1 , 1 hkl ( ) hkl ( 2 2 ) hkl ( 2 2 ) hkl ( 2 2 ) Systematic extinctions due to face centering ⎡ ⎧ ⎛ h k ⎞ ⎫ ⎧ ⎛ h l ⎞ ⎫ ⎧ ⎛ k l ⎞ ⎫⎤ h k l S = f ⎢1+ exp⎨2πi + ⎬ + exp⎨2πi + ⎬ + exp⎨2πi + ⎬⎥ hkl Na ⎝⎜ 2 2⎠⎟ ⎝⎜ 2 2⎠⎟ ⎝⎜ 2 2⎠⎟ ⎣⎢ ⎩ ⎭ ⎩ ⎭ ⎩ ⎭⎦⎥ even even even Shkl = fNa i 4 + fCl i 4 = 4( fNa + fCl )

⎡ ⎧ ⎛ h⎞ ⎫ ⎧ ⎛ k ⎞ ⎫ ⎧ ⎛ l ⎞ ⎫ Cl4 ⎤ odd odd odd Shkl = fNa i 4 + fCl i (−4) = 4( fNa − fCl ) ⎢exp⎨2πi⎜ ⎟ ⎬ + exp⎨2πi⎜ ⎟ ⎬ + exp⎨2πi⎜ ⎟ ⎬ ⎥ ⎩ ⎝ 2⎠ ⎭ ⎩ ⎝ 2⎠ ⎭ ⎩ ⎝ 2⎠ ⎭ + f ⎢ ⎥ odd even even Shkl = fNa i 0 + fCl i 0 = 0 Cl ⎢ ⎧ ⎛ h k l ⎞ ⎫⎥ Cl1 Cl2 Cl3 + exp 2πi + + ⎢ ⎨ ⎜ ⎟ ⎬⎥ even odd even S = 0 (0,0,2) (0,2,2) ⎣⎢ ⎩ ⎝ 2 2 2⎠ ⎭⎦⎥ hkl 1 ,0,0 0, 1 ,0 0,0, 1 ( 2 ) ( 2 ) ( 2 ) 1 , 1 . 1 even even odd S = 0 ( 2 2 2 ) hkl (2,2,2)

⎡ ⎤ odd odd even Shkl = 0 (1,1,1) Shkl = fNa 1+ exp πi(h + k) + exp πi(h + l) + exp πi(k + l) ⎣ { } { } { }⎦ c* ⎡ ⎤ odd even odd S = 0 + fCl exp πi(h) + exp πi(k) + exp πi(l) + exp πi(h + k + l) hkl a* b* ⎣ { } { } { } { }⎦ (0,2,0) = f ⎡1+ (−1)h+k + (−1)h+l + (−1)k+l ⎤ + f ⎡(−1)h + (−1)k + (−1)l + (−1)h+k+l ⎤ S 0 Na ⎣ ⎦ Cl ⎣ ⎦ even odd odd hkl = (2,0,0) (2,2,0)

59 60 Other systematic absences Consider h = k = 0 • Example: Systematic absences expected for … and l = odd a crystal with a 4 screw axis. ⎛ ⎞ 1 S ∝ f exp 2πilz ⎡1+ eπi 2 + eπi + e3πi 2 ⎤ = 0 hkl ⎜ ∑ p ⎟ { ( )} i ⎣ ⎦ 41 screw axis along c ⎝ p ⎠ 1st nd 3rd 4th 41 1 41 1 41 3 2 x, y,z ⎯⎯→ y,x,z + 4 ⎯⎯→ x, y,z + 2 ⎯⎯→ y,x,z + 4 term term term term (Every atom belongs to set of 4 symmetry-related atoms.) … and l = 2, 6, 10 … ⎧exp 2πi hx + ky + lz ⎫ ⎛ ⎞ { ( )} πi πi ⎪ ⎪ S ∝ f exp 2πilz i ⎡1+ e +1+ e ⎤ = 0 (πi 2)l hkl ⎜ ∑ p ⎟ { ( )} ⎣ ⎦ ⎪+exp{2πi(−hy + kx + lz)}i e ⎪ ⎝ p ⎠ (-1) (-1) S ∝ f hkl ∑ p ⎨ (πi)l ⎬ p ⎪+exp{2πi(−hx − ky + lz)}i e ⎪ … and l = 4, 8, 12 … ⎪ 3πi 2 l ⎪ +exp 2πi hy − kx + lz i e( ) S ∝ ∼ ⎡1+1+1+1⎤ (non zero) ⎩⎪ { ( )} ⎭⎪ hkl ⎣ ⎦ (p refers to summation over the sets of symmetry-related atoms.) For h = k = 0, only reflections with l = 4, 8, 12 … will be observed.

61 62

Other systematic absences Symmetry of the Diffraction Pattern; • Example: crystal with an a-glide ⊥ to a 41 axis. Equivalent Reflections - Laue Symmetry

(41 screw axis along c, ∴ a glide || to the ab-plane) The pattern of single crystal diffraction “spots” is often a−glide 1 a−glide x, y,z ⎯⎯⎯⎯→ x + 2 , y,z ⎯⎯⎯⎯→ x +1, y,z referred to as the intensity-weighted reciprocal space. ⎫ Including the intensities, what symmetries should the ⎪⎧exp 2πi(hx + ky + lz) ⎪ S f { } intensity-weighted reciprocal space exhibit? hkl ∝ ∑ p ⎨ ⎬ p ⎪+exp 2πi(h(x +1 2) + ky − lz) ⎪ – In the absence of absorption and/or “anomalous scattering”, ⎩ { }⎭ reflections related to each other by inversion in reciprocal space, (h,k,l) and (-h,-k,-l), should be of equal intensity. This is called This will be zero when the term in brackets is zero: Friedel’s Law and it applies even for noncentrosymmetric space groups (no inversion center). exp{2πi(hx + ky + lz)} = −exp 2πi(h(x +1 2) + ky − lz) { } The Patterson symmetry adds an inversion center if the exp 2 i lz exp 2 i h 2 lz eπih exp 2 i lz { π ( )} = − { π ( − )} = − { π (− )} space group is acentric. Therefore, in centrosymmetric exp{2πilz} = −(−1)h exp{−2πilz} cases, the Patterson symmetry is the same as the symmetry of the diffraction pattern. If l = 0,then absences occur where h = odd. For b-glide ⊥ to c-axis, find absences where k = odd (l = 0).

63 64 Symmetry of the Diffraction Pattern; Tl(1+x)BaSrCa(1–x)Cu2O(7-δ) Superconductor Equivalent Reflections - Laue Symmetry Translation of a crystal leaves reciprocal space • Within the disorder model (x = 0.22), the unaffected. Therefore, the translational parts of space group is symmetry operations can be ignored in finding symmorphic: P4/mmm. reflections that are expected to be equivalent. • The reciprocal lattice – Screw operations have same symmetry effect as intensities reflect each simple rotational point group operations symmetry operation – Glide operations have the same symmetry effect as • No systematic simple mirror planes absences

See: Table 11.4, p. 382 in Cotton. a = 3.8234 Å; c = 12.384 Å

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Tl(1+x)BaSrCa(1-x)Cu2O(7-δ) Superconductor Tl(1+x)BaSrCa(1-x)Cu2O(7-δ) Superconductor

• Simulated electron • Simulated electron diffraction pattern diffraction • 100 plane. • 001 plane. • The reciprocal • The reciprocal lattice lattice intensities intensities reflect reflect each each symmetry symmetry operation operation • No systematic • No systematic absences absences

67 68 Tl BaSrCa Cu O Superconductor (1+x) (1-x) 2 (7-δ) Cs-Leucite, low symmetry (298K) • Simulated electron diffraction pattern • At high temp (473K), • 110 plane. Cs-leucite • The reciprocal (CsAlSi2O6) adopts a lattice intensities cubic structure reflect each (space group: I a 3 d ), symmetry a = 13.7062Å. operation • At room temperature, • No systematic the symmetry is absences lower 41 2 • What is it? I a 3d

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Leucite (298K), l = 0 plane Leucite (298K), h = 0 plane

• Simulated electron • What symmetries are diffraction pattern suggested? • What symmetries are possible?

71 72 Leucite (298K), l = 0 plane Leucite (298K), l = 0 plane

• Adjacent spots have • Intense peaks are a separation of indexed. (How was 0.153 Å–1. this done?) 1 • /(0.153 Å–1) = 6.536 Å. What does this number mean?

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Powder Diffraction NaCl: h = 0 and h = k planes • In principle, powder diffraction is nothing more than a ‘spatially averaged’ version of single-crystal diffraction. • Much of the rich spatial information one gathers in single-crystal data is lost in powder diffraction, only the 2θ information remains. • However, once (if) the cell can be inferred, • The circles correspond to reflections on the powder diffraction data is still quite same “cone”, i.e., with the same 2θ values. powerful.

75 76 • Any RLV is written as K = ha∗ + kb∗ + lc∗ Laue Equations hkl K i K = h2a∗ i a∗ + k 2b∗ i b∗ + l2c∗ i c∗ • These are relations connecting observed 2θ hkl hkl 4 values and indices: 2hka∗ b∗ 2hla∗ c∗ 2klb∗ c∗ sin2 + i + i + i = 2 θ Recall: a∗ i a = b∗ i b = c∗ i c = 1 λ a∗ i b = a∗ i c = b∗ i a = b∗ i c = c∗ i a = c∗ i b = 0 earlier we found • Special cases: Orthorhombic 2 4 ∗ 1 ∗ 1 ∗ 1 K sin K K sin2 a = xˆ ; b = yˆ ; c = zˆ hkl = θ hkl i hkl = 2 θ a b c λ λ 2 h2 k 2 l2 ⎛ λ ⎞ ⎛ h2 k 2 l2 ⎞ or K 2 sin2 = 2 + 2 + 2 ⇒ θ = ⎜ ⎟ ⎜ 2 + 2 + 2 ⎟ 2 a b c ⎝ 2 ⎠ ⎝ a b c ⎠ λ 2 ⎛ λ ⎞ sin2 K 2 θ = hkl = ⎜ ⎟ 4 ⎝ 2dhkl ⎠

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• Reciprocal lattice • Cubic and Tetragonal - special cases of

orthorhombic: 2 ∗ b × c acxˆ 2 2 ⎛ λ ⎞ 2 2 2 a = = = xˆ sin θ = (h + k + l ) cubic V 2 ⎝⎜ 2a⎠⎟ ( 3 2)a c 3a 2 2 2 2 2 ⎛ h + k l ⎞ sin tetragonal ∗ 1 ⎛ xˆ ⎞ ∗ 1 θ = λ ⎜ 2 + 2 ⎟ b yˆ ; c zˆ ⎝ 4a 4c ⎠ = ⎜ + ⎟ = • Hexagonal a ⎝ 3 ⎠ c c out of plane of paper • Laue equation 2 2 2 2 2 α = β = 90° γ = 120° 3 1 2 λ λ ⎡ 2 ∗ 2 ∗ 2 ∗ ∗ ∗ ⎤ a = a( xˆ − yˆ) sin θ = K i K = ⎢h a + k b + l c + 2hka i b ⎥ 2 2 4 4 ⎣ ⎦ b = ayˆ λ 2 λ 2 sin2 θ = h2 + k 2 + hk + l2 c = czˆ 3a2 ( ) 4c2 3 V = a i (b × c) = a2c 2 79 80