Structure solution
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Answers part 2
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The Patterson syntheses calculated in the last part were:
Why is the Patterson density a periodic function in real space ?
 A Patterson shows the interatomic vectors of the structure.
Since the structure itself is periodic, the distribution of
interatomic vectors is periodic as well. You can deduce this
straight from the formula:
The complex exponential is identical for the argument mod 2 PI, or
for hu mod 1. Thus, if an integer vector u is added,
the result remains invariant. Note that this holds only if
h is restricted to integer values.
Explain the differences between the Xray and neutron Patterson.
Why are the neutron Patterson peaks at 1/2 1/2 0 and at 0.2 0 0 negative ?
 Each peak in the Patterson represents an interatomic vector. Its
height is determined by the number of identical vectors and by
the product of the scattering factors of the two atoms.
 The Xray intensities are dominated by the metals. The oxygen
contributes only a minor fraction to the intensities. Accordingly
the Xray Patterson is dominated by the metalmetal vectors.
 The neutron scattering lengths of the atoms do not increase with
increasing atomic number. The scattering lengths of the elements
in this example are:
Zirconium 0.69
Titanium 0.33
Oxygen 0.575

 The peak at 1/2 1/2 0 represents the metalmetal vector. Since
the scattering length of Titanium is negative, this peak is
negative. Accordingly the peak at 0.2 0 0 represents the
Titanium  Oxygen vector. Note that this distinction between
the TiO and ZrO could not be made as unambiguously from the
Xray Patterson.
Solve the structure, at least for the metal atoms.
 Using the Patterson peaks and the space group information the
structure can easily be solved:
Zirconium 0.5 0.5 0.0
Titanium 0.0 0.0 0.0
Oxygen 0.2 0.0 0.0

For the final structure determination go to the next section ..
