There are eleven classes of crystal groups, descernable from 2D Laue photographs (giving, hence, only 2D crystallographic groups). The diffraction pattern unavoidably "adds" a center of symmetry to the actual point group, so that the observed Laue class can be regarded as a "hint" for that group.

Derivation of Laue class from a point group: one should remove translational components from operators. Here are two examples, for groups C2/c and P2₁c, respectively:

1. C2/c
   

   	Symmetry		After removal of	Symbol
   	operator		translational components
   	x, y, z	=>	x, y, z	1
   	-x+1/2, -y, z	=>	-x, -y, z	2
   	-x, -y, -z	=>	-x, -y, -z	-1
   	x+1/2, y, -z	=>	x, y, -z	m

   
   => group C2/c correponds to Laue class 2/m.



2. P2₁c
   

   	Symmetry		After removal of	Symbol
   	operator		translational components
   	x, y, z	=>	x, y, z	1
   	-x+1/2, y+1/2, -z+1/2	=>	-x, y, -z	2
   	-x, -y, -z	=>	-x, -y, -z	-1
   	x+1/2, -y+1/2, z+1/2	=>	x, -y, z	m

   
   => group P2₁c correponds to Laue class 2/m either.

Corrected HTML file just here: Added </DD></DL>
Comment: Full marks can only be achieved when two different Laue classes are given, despite the fact that you clearly understand how the Laue class relates to the real-space symmetry of the crystal.
Minor comment: Safest to state explicitly that the Laue class has the symmetry of the reciprocal lattice.
4 Marks
3. Crystallographic symmetry operations may be grouped into 5 distinct groups, namely, lattice translations, rotations, rotary-inversions, screw-rotations, and glide planes.

   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.

   ---

   Answer:

   All the symmetry operations which change handedness are out of question in the case of an enantiomorphic molecule. Practically, this means: symmetry operations which contain inversion. From given operations such are rotary-inversion, and glide planes (mirroring is combination of rotation by 180 degrees, plus inversion).
   Comment: Correct symmetry operators identified, but no explanation of the property of an enantiomorphic molecule which makes the rotary-inversion and planes forbidden symmetry operations. (It is tautological to simply state that an enenatiomorphic molecule is one to which these symmetry operations cannot be applied.) Answer is just too brief.
   3 Marks

4. An organometallic cobalt compound crystallises in space group P2₁c.

   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.

   ---

   Answer:

   This group has four symmetry operators, and their action upon coordinates of given Co atom is shown in the following table:

   Symmetry operation		New x	New y	New z		Final x	Final y	Final z
   x, y, z	:	0.271	0.250	0.458
   -x, 1/2+y, 1/2-z	:	-0.271	0.750	0.042	=>	0.729	0.750	0.042
   -x, -y, -z	:	-0.271	-0.250	-0.458	=>	0.729	0.750	0.542
   x, 1/2-y, 1/2+z	:	0.271	0.250	0.958

   [The second and the third operation have given atom positions outside the unit cell, and had to be corrected by an appropriate translation.]
   Model answer beautifully tabulated!
   3 Marks

   Apparently, there are four different Co atoms in the unit cell. Not a one of them is on a special position, i.e. at a point of inversion, or on a two-one screw axis, or a glide plane.
   Comment: Glide planes and two-one axes are not special positions. Also the Co atoms do lie on the glide plane, even though this not a special position.
   1 Mark

5. Three different views are shown below of 1,2-dichloroethylene in its most-likely (minimum-energy) conformation. What symmetry elements are present (if any)? Hence suggest the molecular point symmetry for this particular conformation.

   View A	View A, but then rotated »35° about Y	View A, but then rotated 90° about Y

   ---

   Answer:

   Only one symmetry element is present: point of inversion, situated in the middle of the bond between two central atoms.
   Comment: See model answer
   1 Mark

   Corresponding molecular point symmetry, therefore, would be -1.
   Minor comment: Correct deduction if previous answer were correct
   2 Marks

6. You are supplied with the diagram of the primitive space group shown below:
   
   1. Identify the three types of symmetry element present in the diagram.
   2. Given the position of the right-handed object shown in red, specify (using the letters A-Z, a-i) the position of symmetry-related objects within the unit cell, stating for each the symmetry element which relates the red and magenta objects.
   3. For each symmetry-related object specified in (ii) above, state its handedness giving a brief reason for your answer.
   4. Specify the crystal system to which this space group belongs giving a brief reason for your answer.

   ---

   Answer:
   1. The three types of symmetry element present in the diagram are mirror planes, twofold rotation axes, and glide planes (perpendicular to the plane of the screen, with glide along the dashed line).
   
   Excellent: 3 Marks
   2. They transform the right-handed object shown in red into objects marked with K (symmetry operation: -x, -y, z -- twofold rotation axes), J (1/2+x, -y, z -- glide planes), and Z (1/2-x, y, z -- mirror planes).
   
   Minor comment: You are in danger of mixing up the generic space-group operators with the actual ones for a specific operation, e.g. the symmetry operator to reach position K is actually given by 1-x,1-y,z
   3 Marks
   3. Only K does not change initial handedness; J and Z do, since they are obtained through transformations based on inversion.
   
   2 Marks
   4. It is immediately clear that we are not dealing with a triclinic crystal system. Not a 120 degrees gamma angle excludes trigonal and hexagonal, and lack of four-fold axes tetragonal crystal system. Also, this can not be a cubic system -- a and b are not of the same length. Finally, a different from 90 degrees beta angle is ruled out due to the presence of mirror planes normal to ab-plane, which excludes monoclinic crystal system. Therefore, the diagram describes an orthogonal [Typo? I assume you mean orthorhombic!] crystal system.
   
   Excellent logic: 2 Marks