IB Mineral Sciences Module B: Reciprocal Space, Symmetry and Crystallography BD3

Solving the crystal structure of massicot (PbO)

Massicot is a naturally-occurring form of lead(II) oxide produced by the oxidation of lead ores. Unfortunately, it is not one of the most visually spectacular minerals known to mankind. But it is one of the most dense: a cubic centimetre of the material weighs in at just under 10 g, which is only slightly less than metallic lead itself. Together with its dimorphic cousin litharge (the more common and more colourful form of PbO), massicot has been used for centuries in the manufacture of paints, ceramics, glasses, and – more recently – in batteries.

In this demonstration we are going to attempt to solve the crystal structure of massicot us- ing x-ray diffraction data. The idea of the exercise is to bring together the different aspects of the course: we are going to need to use our knowledge of symmetry and reciprocal space and our understanding of diffraction experiments themselves. Hopefully, by the end of the exercise you will have an appreciation for how all the techniques we have learned come together to enable us to determine the arrangement of atoms in minerals. This demonstration uses a java applet, which is located at: http://rock.esc.cam.ac.uk/∼alg44/teaching ib bd3.html

Unit cell

(i) Diffraction photographs taken parallel to the three crystal axes are shown overleaf. The axis labels are arbitrary, but consistent across the three images. To what point group does the diffraction pattern belong, and hence in which crystal system does massicot crystallise?

(ii) It is known from independent physical measurements that the crystal structure of massicot must be centrosymmetric. To what point group must its structure belong? IB Mineral Sciences Module B: Reciprocal Space, Symmetry and Crystallography BD3 IB Mineral Sciences Module B: Reciprocal Space, Symmetry and Crystallography BD3

(iii) The central squares drawn on each diffraction photograph are scaled such that they cut the axes at points in the diffraction pattern exactly 1 A˚ −1 away from the central (000) peak. Use this to calculate the (real-space) lattice parameters of massicot.

(iv) Recalling that the density of PbO is close to 10 g cm−3, calculate how many formula units must be contained within each unit cell. (Atomic masses: Pb = 207.2 g mol−1; O = 15.999 g mol−1)

(v) Illustrate on the diagrams which are the (020) and (102)¯ reflections.

Space group

(vi) List the reflection conditions for each of the three planes shown in the diffraction photographs, and decide the minimum number of rules needed to account for all systematic absences.

(vii) Hence determine the space group, taking into account the fact that massicot is cen- trosymmetric.

(viii) Draw the corresponding space group diagram, using a view that includes the a and b axes within the plane of the paper. Include on your diagram the rotation or screw axes that are generated by other symmetry elements, and also indicate the locations of the centres of symmetry.

(ix) How many general positions are there in this space group? What general implication does this have for the positions of the Pb and O atoms?

Atom positions

(x) The atomic number of Pb is 82 while that of O is 8. What does this tell us about the sensitivity of x-ray diffraction patterns to the positions of the Pb and O atoms?

(xi) We know that the Pb atoms must lie on a symmetry element. Using your space group diagram, see if you can determine the four different sets of positions possible for the Pb atoms. As a hint, two of these can be obtained by placing Pb atoms on centres of symmetry, and the other two will involve other symmetry elements.

In order to distinguish between these four possibilities, we are going to use the Patterson method. This involves calculating the Fourier transform of the observed intensities, which gives us a map of the different interatomic vectors (the Patterson map). Instead of calcu- lating the entire three-dimensional Fourier transform, we are going to restrict ourselves to the (hk0) reflections, which will give us information about the structure as if it had been projected onto the (a,b) plane.

(xii) For each of the four possible sets of Pb atom positions, sketch the distribution of the first few interatomic vectors projected onto the (a,b) plane. Use these to suggest what the corresponding Patterson map would be expected to look like in each case. IB Mineral Sciences Module B: Reciprocal Space, Symmetry and Crystallography BD3

Now open the web page http://rock.esc.cam.ac.uk/∼alg44/teaching ib bd3.html to access the java applet we are going to use:

The way this applet works is that we can enter the intensities of (hk0) reflections and it will automatically calculate the Patterson map as we go. To enter intensities, you have to select the appropriate reflection first. This can be done by clicking the individual reflection in the top right-hand panel, or by entering the h and k values in the top left-hand panel. We then enter the intensity value in the “|F|” box (making sure the phase is set to zero), and select “Set SF”. You should notice that the relevant square on the reflection diagram becomes coloured red (the intensity being proportional to the value just entred), and the Patterson map in the bottom right-hand corner should change. The program already knows Friedel’s law, so entering an intensity for (hk0) will add the same value for (h¯k0);¯ however, you will need to apply any additional Laue symmetry yourself.

(xiii) The strongest measured intensities for |h,k| ≤ 4 are listed overleaf. Enter these into the applet and sketch the Patterson map it calculates. Does this correspond to any of the maps you were expecting? Can you now deduce the positions of the Pb atoms?

(xiv) What additional information would you need to be able to determine the positions of the O atoms?

(xv) Draw a quick sketch of the crystal structure of massicot, illustrating just the Pb atoms. IB Mineral Sciences Module B: Reciprocal Space, Symmetry and Crystallography BD3

Reflection Intensity (arb. units) Reflection Intensity (arb. units) (000) 1000 (230) 60 (010) 88 (400) 152 (200) 938 (040) 151 (020) 901 (420) 152 (220) 481 (240) 144 (030) 132 (440) 103 (310) 115