Problem Set 4 - Wolfgang Seidl


1 a) unit cell

    b) crystal lattice     c) origin in reciprocal space     d) origin in the crystal     e) space group symmetry of the crystal



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.


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.


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)

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.


5. The symmetry elements are:

This should give the molecular point group symmetry 2/m.


6. i) symmetry elements:

    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!

    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

   iv) The crystal system is orthorhombic due to three occuring 2-fold symmetry characterstics: twofold rotation, mirror plane and glide plane.