Which of these symmetry operations cannot be found in the crystal structure of an enantiomorphic molecule, e.g. a naturally-occurring sugar such as D-glucose? Give brief reasons for your answer.
[5 Marks]
The simplest crystallographic point groups are 1,2,3,4 and 6 all of which
posses a single rotation axis only. These are referred to as proper rotations.
Improper rotations combine a rotation with an inversion through a point on
the axis of rotation and are the basis for the point groups -1, m(i.e. -2),
-3, -4 and -6. The latter relates an object to its inverse, known as an
enantiomorph. An object and its enantiomorph are, in general, physically distinct
entities (e.g. right and left hands).
As a consequence of the above, enantiomorphic molecules cannot contain symmetry
operations that change the handedness of the molecule.They possess no improper
rotations. Thus,
Lattice translations. These will be translations along a plane and so do
not change handeness. They can be found in enantiomorphic molecules.
Rotations Straight forward rotations will not change handneness (a twist of
a left hand will not change it into a right hand!).
Rotary inversions These will produce an enantiomorph and will, therefore, not be
found in enantiomorphic molecules.
Screw-rotations. Screw rotations contain a rotation and a translation along the axis
of rotation. Both these operations do not produce an enantiomorph and will, therefore, not be
found in enantiomorphic molecules. Helical screw axes can change handedness along
one axis, but do not change handedness along the other 2 axes and, therefore can also be
found in enantiomorphic molecules.
Mirror planesMirror planes behave the same as a two-fold rotation
axis combined with an inversion on the axis of rotation. Mirror planes produce enantiomorphs
and are not found in enantiomorphic molecules.
Glide planesThese consist of a translation along a mirror plane.
They are not found in enatiomorphic molecules for the same reason given
in Mirror planes.
If the coordinates of the central metal atom, Co, in this compound are
0.271,0.250,0.458, determine the coordinates of any other Co atoms
within the unit cell, showing your workings.
Is the Co atom sited on a special position in this space group? Give
brief reasons for your answer.
[5 Marks]
From the notes, the P21/c space group has 8 points
of inversion, 4 two-one screw axes and 2 c-glide planes. The symmetry operators
that relate positions in the unit to a point at co-ordinates x, y, z are as
follows
Substitution for the known values of x,y,z in the above symmetry operators
gives the following co-ordinates for symmetry equivalent Co atoms in the unit
cell.
Comment: The last part is incorrect, though
the initial statement is correct. Helical screw axes do not change
handedness. Also not confuse symmetry in molecules with symmetry
in crystals.
Comment: Excellent clear answer, except for
screw axis part.
5 Marks
Comment: There are only 4 symmetry operators per
unit cell for this space group - see model answer
For the Co atom to be on a 'special position' it would need to be positioned
on one of the two-one screw axes or on a point of inversion or on a c-glide
plane. When an atom lies on a special position,the symmetry operators for that space group
generate fewer unique positions within the unit cell.
Comment: See logic is correct, though not the
application of it (due to the previous errors)
The Co atom is not on a special position
because there are 6 symmetry operators and they have
produced 6 unique positions for the Co atom.
Comment: Your logic is correct, though not the
application of it (due to the previous errors)
For example, the four two-one screw axes are at (0,y,0.25), (0,y,0.75), (0.5,y,0.25) and
(0.5,y,0.75). These do not correspond to the positions of the 6 Co atoms
identified above. In a similar manner, and by a study of the space group
views in the notes, the Co atoms are also not on C-glide planes nor at the positions
of centres of inversion.
2 Marks