Page 1.
Q1.
(a) The Unit Cell is chosen by the crystallographer. This is an arbitrary
choice as any unit cell may be described as primitive triclinic, but it is usually
chosen so as to include all the symmetry elements relating the molecules within
the crystal lattice, but not including the lattice translations.
1 Mark
(b) The Crystal Lattice is chosen by the crystallograhper. Again this may
be described as a primitive triclinic lattice, but this will not always describe
the full symmetry present such as face-centring, so higher order lattices would be
chosen.
0 Marks
(c) The Origin in Reciprocal Space is neither chosen by the crystallographer
or due to the arrangement of the atoms/molceules in the crystal. It is derived
from the geometry of the Ewald sphere and the theory of diffraction.
1 Mark
(d) The Origin in the Crystal is chosen by the crystallographer. Certain
space groups may be described with different origins, but usually an origin at a
centre of symmetry is chosen.
1 Mark
(e) The Space-Group Symmetry of the Crystal is due to the packing of the
atoms/molecules within the crystal. Although the unit cell may be chosen arbitrarily,
the symmetry relating various moieties remains the same (hence the choice of unit
cell should be chosen to reflect this symmetry).
1 Mark
Q2.
The term Laue Class refers to a set of space groups possessing a centre of symmetry.
Comment: Not space groups, but point
groups
All translational elements removed from a space group will leave the Laue class of which it is
a subgroup.
Comment: See model answer
1 Mark
Two examples are 4/m and 4/mmm.
2 Marks
Q3.
Any symmetry operation within a unit cell must by definition act on every molecule
within that cell. Enantiomorphic molecules possess "handedness", ie. they are
mirror images of the opposite enanitomorphs. This means any symmetry operation
involving reflection cannot exist (except in a racaemic mixture). Lattice
translations are inherent in a crystal so must exist. Rotations and
screw-rotations do not involve reflection and so may exist within the crystal, but
rotary inversions and glide planes do involve reflection and so cannot exist in
the crystal structure of a pure enantiomeric molecule.
Comment: Explain why the molecules possess "handedness"
4 Marks
Q4.
The coordinates (0.271, 0.250, 0.458) refer to the position (x, y, z) within the crystal. Along with the position (x, y, z), the symmetry elements within the space group P21/c produce equivalent positions at (-x, -y, -z) , (-x, 0.5+y, 0.5-z), and (x, 0.5-y, 0.5+z). These positions produce coordinates at (-0.271, -0.25, -0.458), (-0.271, 0.75, 0.042) and (0.271, 0.250, 0.958).
These coordinates do not lie within the unit cell. Any translation of 1 along any
unit cell axis produces the same position in the adjacent unit cell. Therefore,
the coordinates may be translated to (0.729, 0.750, 0.542), (0.729, 0.750, 0.042)
and (0.271, 0.250, 0.958).
3 Marks
The Co is situated on the c-glide plane at (x, 0.25, z).
Comment: True, but you should state explicitly
that this is not a special position and why it is not
0 Marks
Q5.
There is a centre of symmetry present lying between the two carbon (black) atoms.
Comment: See model answer
1 Mark
The molecular point symmetry is -1.
Minor comment: Correct deduction if previous
answer were correct
2 Marks
Q6. (i) The three types of symmetry element in the diagram are mirror planes, a-glide planes and two-fold rotation axes.
(ii) The symmetry related positions within the cell are at positions labeled
J : a-glide plane at (x, 0.5, z)
K : two-fold rotation axis at (0.25, 0.25, z)
Minor comment: Correct, but the axis is actually at
(0.5,0.5,z)
Z : mirror plane at (0.25, y, z)
(iii) Position J is left-handed due to the reflection across the glide plane.
Position K is right-handed as a rotation axis does not involve reflection.
Position Z is left-handed due to the reflection in the mirror plane.
(iv) The space group is Pma2 with equivalent positions (x, y, z), (0.5+x, -y, z),
(-x, -y, z) and (0.5-x, y, z). The crystal system is orthorhombic as all angles
must be 90 but no axes have to be the same length.
Excellent answer: 10 Marks
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