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Problem Set 4 - Wolfgang Seidl

1 a) unit cell

ii, the crystallographer defines a suitable unit cell. Mostly the defined unit cell is not the smallest possible one. For certain purposes the choice of a larger unit cell allows to make different symmetry elements in the cell more obvious.

1 Mark

b) crystal lattice

i, the packing of atoms/molecules determines the crystal lattice, because the thermodynamic stable crystal seeks the lowest possible energy state in the system. Hence the crystal tries to build up a dense packing of atoms/molecules. Dense packing is an intrinsic element which is most important!

1 Mark

c) origin in reciprocal space

iii, the origin in reciprocal space is always a point of inversion, even if the crystal displayed has no inversion symmetry! The origin is created by building up the reciprocal space. Hence neither the packing nor the crystallographer are responsible for it.

1 Mark

d) origin in the crystal

ii, it is up to the crystallographer which origin in a crystal he chooses. To prevent chaos there are some rules to set the origin to high points of symmetry or a point of inversion. But one could also choose a completely different origin.

1 Mark

e) space group symmetry of the crystal

i, the packing of atoms/molecules determines certain symmetry elements in the crystal, and the combined symmetry elements for a crystal give the space group symmetry.

1 Mark

Excellent answer

2. What is Laue Class?

A member of the 11 point groups of the Laue Classes includes a point of inversion, thus the symmetry operator -x, -y, -z is present. Examples would be the linear ethylene which has its inversion point in the center of the C-C double bond (mmm), or Ni(CO)4, which is planar and has its point of inversion at the center of the Ni atom. Point group would be 4/mmm.

Comment: You need to point out (and briefly discuss) the fact that Laue class is concerned with diffraction symmetry (i.e. reciprocal space); your examples are correct in terms of point group, but then you give real space examples of these point groups.
1 + 2 Marks

3. Neither mirror/glide planes nor rotary inversions can be found. If the enantiomorphic molecule is mirrored somewhere in the crystal it will change into the opposite
form! This allows no symmetry operators which include a mirror symmerty in any way. This is obvious for mirror/gilde planes and also for the rotary inversion,
because an inversion is a twofold rotation combined with a mirror plane perpendicular to it.

Comment: basically correct, but you should mention handedness/chirality explicitly.
2 + 2 Marks

4. The four symmetry operators for P2₁/c create other Co atoms at the follwoing positions::

x, y, z = 0.271, 0.250, 0.458, the given atom

-x, 1/2+y, 1/2-z = -0.271, 0.5+0.250, 0.5-0.452 = -0.271, 0.750, 0.042 (rotation)

-x, -y, -z = -0.271, -0.250, -0.458 (inversion)

x, 1/2-y, 1/2+z = 0.271, 0.5-0.250, 0.5+0.458 = 0.271, 0.250, 0.948 (glide plane)
Comment: You need to translate final coordinates so that atoms are within unit cell (i.e. so that coordinate values lie in the range 0 to 0.99999999
2 Marks

Yes the Co atom lies on a special position. If I am right, it is sited on the glide plane, which is not a general position. The effect is, that the Co atom seems to move to point (0.271, 0.250, 0.948) by a translation and not a glide plane. The atom is positioned on the plane itself, so it is mirrored to itself and finally translated.
Comment: See model answer. A mark given for spotting that the atom is on the glide plane, but note that atoms on glide planes are not in special positions, in contrast to those on mirror planes.
1 Mark

5. The symmetry elements are:

point of inversion in the center of the C-C bond
a mirror plane, with Cl-C-C-Cl lying in the plane
a twofold rotation axis perpendicular to the mirror plane mentioned above

This should give the molecular point group symmetry 2/m.
Excellent
3 + 2 Marks

6. i) symmetry elements:

twofold rotation axis perpendicular to a/b plane
mirror plane
glide plane

Comment: OK, though a little more detail as in model answer should be given
3 Marks

ii) The red object can move to:

    Z by the mirror plane at 0.25, y, z
    K by the twofold rotation axis in the center of the unit cell
    Z is rotated to J by the twofold rotation in the center of the unit cell

The glide plane operator mirrors the red object to H and then translates it parallel to the glide plane to ... who knows?
Because of the missing value for the glide I assume that it is 0.5. This would bring the red object to J even with the right handness!
(Perfectly correct logic and quite a sensible way to attempt this problem - note that the translation will always be 1/2 except for d-glide planes, which come in pairs.)
3 Marks

    iii) Z opposite handness due to the mirror plane
        J opposite handness due to the rotation of Z which doesn´t change the handness or due to the glide plane
        K has the same handness as the red object after the twofold rotation which has no effect on the handness
3 Marks

iv) The crystal system is orthorhombic due to three occuring 2-fold symmetry characterstics: twofold rotation, mirror plane and glide plane.
Minor comment: better to say symmetry elements of order 2 rather than use the expression "2-fold" which refers to proper rotation axes strictly speaking.
1 Mark