Chemistry 4000/5000/6000
Part 3 - Introduction to X-rays, their generation and interaction
with matter
Recap: Additional consequences of the atomic
scattering factors
1. For electromagnetic radiation, the square of the amplitude is the intensity.
2. The most intense diffracted X-ray beams will occur from those atoms with the most
electrons, i.e. those with the largest atomic number!
3. Crystals composed only of "light" atoms, i.e. up to atomic number 10, tend to diffract weakly
4. Light" atoms such as carbon can be very hard to locate in the proximity of very heavy atoms,
such as platinum or uranium, because of the very strong scattering of the nearby heavy
atoms.
5. Alternative that solves some of the difficulties with the location of hydrogen atoms or light
atoms close to heavy atoms is to do a complementary experiment where the same crystal is
p
to neutron diffraction.
also exposed
6. The atomic scattering factor diagram shows that the intensity of all diffracted beams will fall
off with increasing sin(2θ)/λ
Chemical Crystallography
The structure factor (amplitude)
o Massa, p. 37-40.
o For a geometric derivation of the structure factor formula, see West, pp. 157 – 159.
o The structure factor is the main geometric factor affecting the intensity of the
reflections
o It provides the essential information relating the diffraction pattern to chemical
structure
o Hence it is the most important equation in crystallography
o The structure factor does not specify the location in space of the reflections (i.e. the
d-spacing)
o That qquantityy is indeed always
y obtained from the Bragg
gg equation
q
o Hence the combination of Bragg equation and the structure factor will allow us to
completely specify the peaks in a powder pattern
o The structure factor is a general concept, and will be used extensively in our
discussion of single-crystal X-ray diffraction
What is the structure factor?
o The intensity of the diffracted beam from any crystal plane (hkl) is a geometric sum
of all the atomic scattering factors
o It is represented by the symbol Fhkl
o It depends on where the atoms are located with respect to this lattice plane, and
which atom type is located there
¾ For example, if a heavy atom is located directly in a certain lattice plane, this
plane will have a very intense reflection
¾ If a plane misses most atoms in the crystal (or glances through their periphery),
that plane will have a very weak reflection.
o There is a structure factor for each and every Miller index (h k l).
1
It is a phase rule
o There are two major contributors to the total amplitude of the reflection of any given
Miller index
¾ The atomic structure factors of the atoms involved
¾ The relative phase of the scattered X-ray from each atom involved
o Thus the structure factor has the form of a phase rule
N
Fhkl = ∑ f j e
N
Fhkl = ∑ f j e
Since we usually measure intensity Ihkl, but work out the amplitudes Fhkl, we note that the
square root of the intensity gives only the absolute value of the structure factors
Note their phase relationships.
Fhkl ∝ I hkl
This is called the modulus of the structure factor
The fact that phases are not directly available from the experimental data constitutes
the greatest challenge for X-ray crystallography.
Corrections also need to be undertaken to the intensity data.
This process is known as data reduction
1. Lorentz correction, L, which depends on the geometry of data collection and is
provided by the manufacturer of the instrument you are using
2. Polarization correction, p.
3. Scale factor K
Fhkl =
[2π i ( hx j + ky j + lx j )]
j =1
o
N
= ∑ f j [cos 2π ( hx j + ky j + lz j )] + i[sin 2π (hx j + ky j + lz j )]
j =1
Since the intensity of any diffracted X-ray beam depends on the square of the amplitude, we
can obtain the intensity of the reflection as follows:
N
Obtaining |SF| from Intensities
o
o
The equation can be expressed as the complex-number cos+sin expansion of an
exponential
*
I hkl ∝ Fhkl × Fhkl
= ∑ f j {[cos 2π (hx j + ky j + lz j )] + i[sin 2π (hx j + ky j + lz j )]}
o In this equation, we are instructed to take a sum over
¾ all the different atom positions in the unit cell,
cell x,
x yy, z
¾ for each Miller index (h k l)
¾ using the atomic scattering factor (form factor, fj) that applies at the specific
value of sinθ/λ from the Bragg equation for that reflection
o The e2πi term is the phase-sensitive portion of the equation.
o
o
o
o
[2π i ( hx j + ky j + lz j )]
j =1
o
Manipulation of the SF equation
j =1
N
{[ 2π (hx
h
∑ f {[cos
×
j =1
j
j
+ ky
k j + lz
l j )] − i[[sin
i 2π ( hx
h j + ky
k j + lz
l j )]}
N
= ∑ f j [cos 2π (hx j + ky j + lz j )]2 + [sin 2π ( hx j + ky j + lz j )]2
j =1
o
Because:
i 2 = −1
Building models
o In most forms of X-ray crystallography (protein crystallography is a bit of an
exception), we construct an atomic model (biocrystallography often builds a model
of the electron density, which is an alternative approach.)
o From that model we calculate the expected structure factors for the diffraction
pattern
o The correctness of the crystal structure is then obtained by comparing the similarity
or difference between the calculated and the measured structure factors for the
diffracted beams corresponding to each Miller index of the crystal
o The NMR analogy
gy
o NaCl by Braggs and DNA by Watson and Crick
KI hkl
Lp
2
Our example is CaF2, the fluorite structure
o
o
Structure factors for fluorite
The menu hkl/uvw can be used in CaRIne Crystallography to show planes of atoms that
belong to various Miller indices or to vectors perpendicular to those planes
Two different, parallel, sets of atoms in (1 1 1) planes in the Fluorite unit cell are shown
o
No.
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
Ca 2+
Caa2+
Ca 2+
a
Ca 2+
Ca 2+
Ca 2+
Ca 2+
Ca 2+
F 1-
Ca 2+
F 1-
Cac2+
c
F 1F 1-
F 1F 1-
F 1F 1-
Ca 2+
Ca 2+
x
Ca 2+
Ca 2+
x
Ca 2+
z
z
C 2+
Ca
F 1-
F 1-
F 1-
Ca 2+
F 1-
y Ca 2+
y
Ca 2+
F 1-
F 1- Ca 2+
F 1-
Ca 2+
F 1-
Ca 2+
Ca 2+
Ca 2+
Ca 2+
Cab2+
Coordinate positions for fluorite in Fm3m , No. 225
b
Ion
Ca
Ca
Ca
Ca
F
F
F
F
F
F
F
F
x
0
1/2
1/2
0
1/4
1/4
1/4
3/4
3/4
3/4
1/4
3/4
y
0
1/2
0
1/2
1/4
1/4
3/4
1/4
3/4
1/4
3/4
3/4
z
0
0
1/2
1/2
1/4
3/4
1/4
1/4
1/4
3/4
3/4
3/4
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.00
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
1.0000
NonEq.
Eq.
Eq.
Eq.
NonEq.
Eq.
Eq.
Eq.
Eq.
Eq.
Eq.
Eq.
Ca 2+
Ca 2+
Generating the SF Master Equation
N
Fhkl = ∑ f j e
o
[2π i ( hx j + ky j + lz j )]
j =1
Now fill in all twelve atom positions into this equation, and simply terms as much as possible:
N
Fhkl = ∑ f j e
[[2π i ( hx j + kyy j + lx j )]
j =1
Evaluating for (2 0 2)
o
o
Fhkl = f Ca [[cos 0 + cos 2π + cos 4π + cos 2π ]
= f Ca [cos 2π (0) + cos π ( h + k ) + cos π ( h + l ) + cos π ( k + l )]
+if Ca [sin 0 + sin 2π + sin 4π + sin 2π ]
+if Ca [sin 2π (0) + sin π ( h + k ) + sin π ( h + l ) + sin π ( k + l )]
+ f F [cos
+ cos
π
2
+ sin
o
o
2
2
π
2
π
( h + k + l ) + cos
(3h + 3k + l ) + cos
+if F [sin
π
π
( h + k + l ) + sin
(3h + 3k + l ) + sin
2
π
2
π
π
2
2
( h + k + 3l ) + cos
(3h + k + 3l ) + cos
( h + k + 3l ) + sin
(3h + k + 3l ) + sin
π
2
π
π
2
2
π
2
( h + 3k + l ) + cos
( h + 3k + l ) + sin
2
π
2
( h + 3k + 3l ) + sin
+ f F [cos 2π + cos 4π + cos 2π + cos 4π
π
( h + 3k + 3l ) + cos
To be observed, a reflection must have hkl either all odd, or all even.
Work out the allowed (2 0 2) structure factor:
(3h + k + l )
π
2
(3h + 3k + 3l )]
(3h + k + l )
π
2
+ cos 4π + cos 6π + cos 4π + cos 6π ]
+if F [sin 2π + sin 4π + sin 2π + sin 4π
(3h + 3k + 3l )]
This is the structure factor master equation for CaF2
It is more economical to first deal with all the x, y, z, and then calculate F for each reflection,
i.e. for the {h k l}.
+ sin 4π + sin 6π + sin 4π + sin 6π ]
o
Evaluate each expression, noting that cos0 = 1, cos2π = 1, cosπ = –1, sin0=0, sinπ=0,
sin2π=0, etc
F202 = f Ca [1 + 1 + 1 + 1] + if Ca [0 + 0 + 0 + 0]
+ f F [1 + 1 + 1 + 1 + 1 + 1 + 1 + 1] + if F [0 + 0 + 0 + 0 + 0 + 0 + 0 + 0]
= 4 f Ca + 8 f F
3
Numerical value requires form factors
o
λ
2
=
1 2 2 + 02 + 2 2
8
sin θ
=
=
;
=
d2
5.462 2
5.462 2
λ
8
= 0.259
4 × 5.462 2
hkl
(1 1 1)
(2 0 2)
(3 1 1)
(4 0 0)
(3 3 1)
I
100
57
16
5
4
Mult.
8
12
24
6
24
I/(mult,Lp)
0.409
0.476
0.098
0 193
0.193
0.047
Fobs
0.640
0.690
0.313
0 439
0.439
0.217
Fcalc
67
97
47
75
39
Fobs scaled
90
97
44
62
31
|Fobs-Fcalc|
23
0
3
12
8
o In solving unknown crystal structures, the object is always to obtain a model
structure for which the calculated structure factors, Fcalc(h k l), are in good
agreement with those obtained from the experimental intensities, i.e. Fobs(h k l)
o For the first five lines of CaF2, the values for Fcalc(h k l) are given in column 7
o The experimental
p
intensities are ggiven in column 4
o The intensities must be corrected for the multiplicity of each diffraction (which is
not taken into account in the structure factor master formula), as well as the Lorentz
and polarization factors, to correct the measured intensity data
o Observed structure factor, Fobs(h k l) related to the intensity by Fobs(h k l) = √Icorrected
o Multiplication of each Fobs(h k l) value by 141 gives the scaled values in column 8.
Looking up the intercepts on the graph:
fCa (2 0 2) = 12.65 and
o
d(Å)
3.143
1.929
1.647
1 366
1.366
1.254
We now use the Bragg law to obtain values for the atomic scattering factors:
4 sin 2 θ
o
A fit of the calculation to data in CaF2
fF (2 0 2) = 5.8
Hence, by filling in the values we finally
get:
F202 = 97
The R factor (residual factor)
o
The measure of agreement between the calculated and observed structure factors is given
by the residual factor, usually shortened to the R-factor, given by the equation:
∑ F −F
∑F
obs
R=
o
obs
For the fluorite structure analyzed in the table, this becomes:
∑F
o
o
o
o
calc
obs
scaled = 324;;
∑F
obs
− F calc = ∑ ΔF = 47;; R =
∑ ΔF
∑F
obs
= 0.15
A 15% “disagreement” seems high, but fit is only to 5 lines
Major disagreement with (1 1 1) may be an experimental error
10% residual is usually the correct structure
A final refined model should have 2 – 6% residual – better agreement is not possible
4