Example 4. A real-life Example:
This is the structure of a cadmium compound from Prof. Kim Dunbar's group.
Their code for this project is "SUN33".
Initial information:
- molecule (chemical formula) contains: C54H66B2Cd1F8O18P2
- space group: C2/c (#15, b axis unique, cell choice 1)
- unit cell contains eight molecules (Z = 8)
Notes:
- molecule is expected to contain a P-Cd-P fragment with
180° P-Cd-P angle and 2.4Å Cd-P distances.
- the height of the Patterson map origin peak appears to have been suppressed
by some artifact of the Fourier program (by a factor of ~2?).
[The effects of this "error" carry through this example.]
- the list of observed Patterson map peaks also includes the distances from
the peak coordinates to the origin (the Patterson vector length). [This
information can be useful in matching the observed peaks with the expected
peaks.]
The Patterson map peaks (coordinates, multiplicities, distances) are:
Observed: Expected for C2/c:
dist
1. 0.0000 0.0149 0.0000 5824 0.17 a. 0, 0, 0 8.
2. 0.5000 0.2149 0.5000 2011 18.13 b. 2X, 0, 1/2+2Z 4.
3. -0.2248 0.0166 0.4314 2009 20.14 c. 0, 2Y, 1/2 4.
4. 0.0014 0.0137 0.0768 1172 2.46 d. 2X, 2Y, 2Z 2.
5. -0.2756 0.2198 0.0680 1115 12.54
6. 0.2191 0.0121 0.4896 726 13.51
etc.
SOLUTION:
- Calculate:
- origin peak height;
- scale factor relating the calculated origin peak height to the observed
origin peak height; and,
- expected scaled Cd-Cd and Cd-P peak heights.
- Origin peak:
zi zi ni Z
Cd1 48 * 48 * 1 * 8 = 18432.
P2 15 * 15 * 2 * 8 = 3600.
F8 9 * 9 * 8 * 8 = 5184.
O18 8 * 8 * 18 * 8 = 9216.
C54 6 * 6 * 54 * 8 = 15552.
B2 5 * 5 * 2 * 8 = 400.
H66 1 * 1 * 66 * 8 = 528.
------
HO = 52912.
- Scale factor:
SF = 5824. / 52912. = 0.110
- Expected peak heights:
zi zj
Cd - Cd 48 * 48 * SF = 254 (multiplicity = 1)
507 (multiplicity = 2)
1014 (multiplicity = 4)
2029 (multiplicity = 8)
Cd - P 15 * 48 * SF = 79 (multiplicity = 1)
159 (multiplicity = 2)
317 (multiplicity = 4)
634 (multiplicity = 8)
1268 (multiplicity =16)
(etc.)
- Match observed peaks with expected peaks (note that the expected scaled
2X and 4X Cd-Cd peak heights are about one half of the observed peak heights):
assignment---------------
2. 0.5000 0.2149 0.5000 2011. | c. 0, 2Y, 1/2 1013.
3. -0.2248 0.0166 0.4314 2009. | b. 2X, 0, 1/2+2Z 1013.
5. -0.2756 0.2198 0.0680 1115. | d. 2X, 2Y, 2Z 506.
Apply Patterson map space group (C2/m) symmetry operations to calculate
the positions of equivalent Patterson map peaks:
use -1/2+X, 1/2+Y, Z with peak 2
use -X, -Y, 1-Z with peak 3
use 1/2+X, 1/2+Y, Z with peak 5
resulting in:
assignment---------------
2' 0.0000 0.7149 0.5000 2011. | c. 0, 2Y, 1/2 1013.
3' 0.2248 -.0166 0.5686 2009. | b. 2X, 0, 1/2+2Z 1013.
5' 0.2244 0.7198 0.0680 1115. | d. 2X, 2Y, 2Z 506.
- Calculate coordinates for the Cd atom:
The above gives two estimates of 2X (0.2248 and 0.2244)
- which average to 2X = 0.2246 and lead to X = 0.1123;
two estimates of 2Y (0.7149 and 0.7198)
- which average to 2Y = 0.7174 and lead to Y = 0.3587; and
two estimates of 2Z (0.0680 and 0.0686)
- which average to 2Z = 0.0683 and lead to Z = 0.0342.
Together these give the fractional coordinates for the Cd atom:
Cd 0.1123 0.3587 0.0342
- Calculate coordinates for two P atoms:
Assign peak 4 as the Cd-P vector (because of the Patterson vector
length and the observed peak height):
4. 0.0014 0.0137 0.0768 1172.
As we expect a linear P-Cd-P fragment, we first add peak 4 coordinates to
the Cd coordinates to get one P position:
Cd 0.1123 0.3587 0.0342
4. 0.0014 0.0137 0.0768
------ ------ ------
P1 0.1136 0.3724 0.1110
Then we subtract peak 4 coordinates from the Cd coordinates to get the other
P position:
Cd 0.1123 0.3587 0.0342
4. -.0014 -.0137 -.0768
------ ------ ------
P2 0.1109 0.3450 -.0426
This results in a P-Cd-P fragment with 180° P-Cd-P angle and 2.46Å
Cd-P distances. The above is consistent with, and correctly assigns, the
strongest four non-origin peaks in the Patterson map.
- Starting with the three atoms
Cd 0.1123 0.3587 0.0342
P1 0.1136 0.3724 0.1110
P2 0.1109 0.3450 -.0426
a few itterations of structure factor calculations and difference Fourier maps
revealed the complete structure.
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