Indexing cubic powder patterns systematic absences for Bravais lattices (Simple cubic, body centered cubic and face centered cubic) to index a simple cubic powder pattern and identify the lattice type (Use the program given to you. Input can be either two theta or d values).

Bragg’s Law 2dhkl sin = n (n is generally ignored. Why??) d-spacing equation for orthogonal crystals – cubic, tetragonal or orthorhombic)

for cubic this simplifies as, Hence, i.e.,

How many lines? Bragg peak at lowest angle means, the indices will be the lowest i.e., (h2 + k2 + l2) will be minimum. Remember, h,k,l are all integers, so the lowest value is 1. For a cubic material, the largest d-spacing may be assigned the reflection, 100 or 010 or 001 (Multiplicity or equivalent reflections). This is true only for a primitive lattice (P). For BCC and FCC, the first Bragg peak will be 110 and 111.

How many lines? Note: 7 and 15 are not possible. Note: we start with the largest d-spacing and work down. Largest d-spacing = smallest 2 This is for PRIMITIVE only.

Remember ….. Not all reflections are present in every substance. What are the limiting (h2 + k2 + l2) values or where do you expect the last reflection? or sin2 has a limiting value of 1, so for this limit:

Wavelength of the X-ray source  = 1.54 Å  = 1.22 Å The number of observable reflections are wavelength dependent (Relative intensities remain the same but the position changes) A smaller wavelength will access higher hkl values

Indexing Powder Patterns Indexing a powder pattern means correctly assigning the Miller index (hkl) to the peak in the pattern, i.e., to estimate the unit cell dimensions. If we know the unit cell parameters (known compound or known structure), You may generate ‘d’ values by using the program HKLGEN.

Indexing Powder Patterns Note finding the unit cell from the powder pattern, is not trivial even for cubic systems. The unit cell of copper is 3.613 Å. What is the Bragg angle for the lowest observable reflection with CuK radiation ( = 1.5418 Å)?

Question  = 12.32o, so 2 = 24.64o BUT….

Systematic Absences Due to symmetry, certain reflections cancel each other out (out of phase). These are non-random – hence “systematic absences” For each Bravais lattice, there are thus rules for allowed reflections: P (primitive): no restrictions (all allowed) I (Body centered): h+k+l =2n allowed F (face centered): h,k,l all are odd or all even

Reflection Conditions (Extinctions) For each Cubic Bravais lattice:   PRIMITIVE B.C.C F.C.C. h2 + k2 + l2 All possible h+k+l=2n h,k,l all odd/even 1 1 0 0 2 1 1 0 3 1 1 1 4 2 0 0 5 2 1 0 6 2 1 1 8 2 2 0 9 2 2 1, 3 0 0 10 3 1 0 11 3 1 1 12 2 2 2 13 3 2 0 14 3 2 1 16 4 0 0

General rule Characteristic of every cubic pattern is that all 1/d2 values have a common factor. The highest common factor is equivalent to 1/d2 when (hkl) = (100) and hence = 1/a2. The multiple (m) of the hcf = (h2 + k2 + l2) We can see how this works with an example

Indexing example 2 d (Å) 1/d2 m 21.76 4.08 0.06 25.20 3.53 0.08 35.88 h k l 21.76 4.08 0.06 25.20 3.53 0.08 35.88 2.50 0.16 42.38 2.13 0.22 44.35 2.04 0.24 51.57 1.77 0.32  = 1.5418 Å 3 4 8 11 12 16 1 1 1 2 0 0 2 2 0 3 1 1 2 2 2 4 0 0 Highest common factor = 0.02 So 0.02 = 1/a2 a = 7.07Å Lattice type? (h k l) all odd or all even  F-centred

In real life, the numbers are rarely so “nice”! Try another… d (Å) 1/d2 m h k l 3.892 2.752 2.247 1.946 1.741 1.589 1.376 1.297 Highest common factor = So a = Å Lattice type? In real life, the numbers are rarely so “nice”!

Watch out! You may have to revise your hcf… …and another Highest common factor = So a = Å d (Å) 1/d2 m h k l 3.953 2.795 2.282 1.976 1.768 1.614 1.494 1.398 Lattice type? Watch out! You may have to revise your hcf…

Remember, the error is more when you calculate cell dimension with larger ‘d’ values. Hence, unit cell dimensionas should be obtained only by curve fitting. So a plot of (h2 +k2 + l2) against sin  has slope 2a/

KCl has rock salt (NaCl) structure KCl has rock salt (NaCl) structure. Explain why the patterns are different. Notice the reflections (111) and (131) are barely visible in the case of KCl.

Shape of the crystals do not change the position of the peaks.