Assignment 6 – Rob Whittock.
1. The 2 theta values of peaks used for indexing the well-known organic substance are shown below:
|
2
theta (degrees) |
2
theta (degrees) |
|
7.5 |
18.07 |
|
8.07 |
18.27 |
|
10.25 |
18.81 |
|
11.77 |
19.36 |
|
14.82 |
20.63 |
|
15.05 |
21.64 |
|
16.09 |
22.67 |
|
16.64 |
23.45 |
|
16.90 |
24.27 |
|
17.15 |
24.41 |
Of
the first twenty three peaks, twenty peaks were chosen which did not have
noticeable tails or of weak intensity.
The peaks with noticeable tails were rejected since peak position would
be inaccurate, which could lead to difficulties in indexing the pattern. The very weak peak at 20.84 theta was
rejected since it was very weak and there were better peaks to choose.
The
output data from the indexing program 210 was :
Wavelength = 1.54056 Å Zero Error = 0.000°
The 10 Most Probable Solutions
|
N |
Q(A) |
Q(B) |
Q(C) |
Q(D) |
Q(E) |
Q(F) |
Lines |
Figure |
Lattice |
|
1 |
18.08 |
262.19 |
83.70 |
0.00 |
10.67 |
0.00 |
20 |
72.1 |
C |
|
2 |
18.08 |
262.19 |
83.70 |
0.00 |
10.67 |
0.00 |
20 |
72.1 |
C |
|
3 |
83.68 |
18.08 |
153.82 |
11.08 |
71.35 |
10.81 |
20 |
10.4 |
P |
|
4 |
54.98 |
18.09 |
83.72 |
10.65 |
22.20 |
10.58 |
20 |
7.8 |
P |
|
5 |
62.77 |
47.30 |
82.79 |
46.34 |
10.45 |
37.78 |
20 |
7.3 |
P |
|
6 |
83.68 |
18.09 |
157.08 |
10.90 |
72.88 |
10.75 |
19 |
11.9 |
P |
|
7 |
59.40 |
18.08 |
83.66 |
10.64 |
24.05 |
10.68 |
19 |
9.0 |
P |
|
8 |
13.40 |
70.22 |
64.09 |
0.00 |
5.15 |
0.00 |
19 |
7.4 |
P |
|
9 |
53.19 |
18.10 |
83.20 |
10.54 |
19.93 |
10.36 |
19 |
4.8 |
P |
|
10 |
56.27 |
18.08 |
82.71 |
10.35 |
20.47 |
10.22 |
19 |
4.3 |
P |
The Equivalent Real-Space Unit Cells
|
N |
a (Å) |
b (Å) |
c (Å) |
a (°) |
b (°) |
g (°) |
V (Å3) |
|
1 |
23.742 |
6.176 |
11.035 |
90.00 |
97.88 |
90.00 |
1602.7 |
|
2 |
23.742 |
6.176 |
11.035 |
90.00 |
97.88 |
90.00 |
1602.7 |
|
3 |
11.589 |
23.798 |
8.512 |
93.74 |
107.73 |
96.44 |
2209.6 |
|
4 |
13.824 |
24.000 |
11.149 |
96.45 |
98.28 |
98.56 |
3585.6 |
|
5 |
13.484 |
16.678 |
11.856 |
111.65 |
86.32 |
110.19 |
2319.8 |
|
6 |
11.602 |
23.787 |
8.432 |
93.56 |
107.95 |
96.44 |
2188.2 |
|
7 |
13.304 |
23.986 |
11.166 |
96.44 |
98.72 |
98.22 |
3453.0 |
|
8 |
27.423 |
11.933 |
12.540 |
90.00 |
95.04 |
90.00 |
4087.7 |
|
9 |
14.026 |
23.995 |
11.161 |
96.53 |
97.48 |
98.60 |
3648.4 |
|
10 |
13.624 |
23.981 |
11.193 |
96.46 |
97.55 |
98.22 |
3555.5 |
The
best solution was solution 1 and 2 (highlighted in blue),
thus the unit cell parameters for the best solution are shown in the table
below:
|
a |
b |
c |
|
23.742 |
6.176 |
11.035 |
|
alpha |
beta |
gamma |
|
90 |
97.88 |
90 |
Cell parameters
|
a (Å) |
b (Å) |
c (Å) |
a (°) |
b (°) |
g (°) |
V (Å3) |
|
23.74200 |
6.17600 |
11.03500 |
90.000 |
97.880 |
90.000 |
1602.8 |
Wavelength = 1.54056 Å 2q Maximum = 60.000° Zero Error = 0.000° Minimum d-spacing = 1.54056 Å
Space group is C2/m Lattice = C 1st symbol = 1 2nd symbol = 2/m 3rd symbol = 1 Crystal class is 1 2/m 1
Reflection conditions are hkl: h + k = 2n
The
Bravais lattice is monoclinic, it has
an axis of order 2 imposed as a symmetry constraint forcing two of the unit
cell angles to 90° and 4 independent
lattice parameters a, b, c,
and b.
The
space group is C2/m and the crystal class is 1/2m 1.
Comment: You need to be very careful here and check all the peaks
since the system is pseudo C-centred due the reflection condition
h0l: h=2n. If you check the h1l reflections, some break the C-centring condition.
2 Marks
hko: h+k=2n
(glide plane)
hol: h=2n (glide plane)
okl: k=2n
(glide plane)
oko: k=2n (screw axes)
hoo: h=2n (screw axes).
Three glide planes and two screw axes are present in
this crystal system. The presence of
the glide planes and screw axes was determined by establishing the systematic
absences, which did not satisify the reflection condtions shown above.
Comment: You should not consider hk0 and 0kl for glide plane conditions when
the system is monoclinic.
2 Marks
4. The most likely space group is C2/m.
Comment: You need to justify your choice. You get one mark given
previous errors.
1 Mark
|
2q |
h |
k |
l |
|
2q |
h |
k |
l |
|
2q |
h |
k |
l |
|
7.50 |
2 |
0 |
0 |
|
14.82 |
1 |
1 |
0 |
|
16.90 |
-2 |
0 |
2 |
|
8.07 |
0 |
0 |
1 |
|
15.05 |
4 |
0 |
0 |
|
17.15 |
1 |
1 |
1 |
|
10.25 |
-2 |
0 |
1 |
|
16.09 |
-4 |
0 |
1 |
|
18.07 |
4 |
0 |
1 |
|
11.77 |
2 |
0 |
1 |
|
16.64 |
-1 |
1 |
1 |
|
18.27 |
3 |
1 |
0 |
|
|
a |
b |
c |
alpha |
beta |
gamma |
|
Value |
23.72716 |
6.17063 |
11.02685 |
90 |
97.9052 |
90 |
|
Error |
0.00855 |
0.00171 |
0.00369 |
0 |
0.0145 |
0 |
Comment: Excellent tables spoilt only by giving too many d.p. for the cell parameters given the magnitude of the esd values.
2
thetazero = -0.0161 ± 0.0046
The 2 theta zero refined
value is small which suggests the X-ray diffractometer is well aligned.
High angle data is difficult
to include in this particular data set due to weak or broad peaks at high 2
theta values.
Good attempt at this part of the assignment.
8 Marks