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:

  1. molecule (chemical formula) contains: C54H66B2Cd1F8O18P2
  2. space group: C2/c (#15, b axis unique, cell choice 1)
  3. unit cell contains eight molecules (Z = 8)

Notes:

  1. molecule is expected to contain a P-Cd-P fragment with 180° P-Cd-P angle and 2.4Å Cd-P distances.
  2. 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.]
  3. 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:

  1. Calculate:
    1. origin peak height;
    2. scale factor relating the calculated origin peak height to the observed origin peak height; and,
    3. expected scaled Cd-Cd and Cd-P peak heights.
       
    1. Origin peak:
    2.         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.
      
    3. Scale factor:
    4.   SF = 5824. / 52912. = 0.110
      
    5. Expected peak heights:
    6.            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.)
      
  2. 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.
    
  3. 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
    
  4. 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.
     
  5. 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.