Advanced Certificate in Powder Diffraction On the Web - Assignment 3 - Kennedy Miller

1. a) The Unit Cell is chosen by the crystallographer and it is determined by the repeatability of the atoms within a given space. It generally consists of a section of the crystal lattice with the fewest number of lattice points. The packing of the atoms/molecules is important but ultimately it is the crystallographer who will pick the unit cell to aid their work. Different crystallographer can pick different unit cells from the same structure.

b) The crystal lattice is determined by the packing of the atoms/molecules in a crystal and can be described as an infinite number of equally spaced points. The lattice is produced by packing molecules/atoms into a three dimensional structure. The lattice can be dissected into numerous unit cells connected together by translations in various directions.

c) The Origin in reciprocal space is determined by the apparatus used for the experiment, so it is neither the crystallographer nor the packing of the atoms/molecules.

d) The Origin in the crystal will be determined by the crystallographer and it is imortant that a point is chosen to help the crystallographer solve the structure and can help make calculations simpler. It often corresponds to a point of high symmetry.

e) The space-group symmetry is a method of minimising the amount of positions of atoms/molecules required to descibe a unit cell. The packing of the atoms/molecules will determine the structural symmetry of the atoms/molecules in the lattice since this leads to symmetry elements being present. It is still up to the crystallographer to assign the symmetry and this is a tool used to help the crystallographer.

2. The Laue classes are a selection of point groups, that are related to each other by being centrosymmetric. This common factor in cenrosymmetric groups is a point of inversion, i.e a symmetric centre point. The Laue group symmetry is the maximum symmetry that a particular crystal can exhibit.

The easiest example to explain is the Monoclinic, 2/m symmetry point group, where a two-fold rotation in one plane followed by a mirror symmetry in the perpedicular plane results in the the generation of two points that can be connected via a straight line through the origin. As the symmetry of the crystal systems get higher, there are laue classes for each crystal system.

The tetragonal system has 4/m and 4/mmm point groups that feature a point of inversion. To get from one atom to another in the 4/m system, a rotation around one axis of 90^o followed by a mirror reflection inthe perpendicular axis will result in the position of another atom/molecule. From this position, all the other atoms/molecules positions can be determined.

3. Enantiomorphic point groups all have right and left handedness within the symmetry of the molecules.

Lattice Translation is a pure translation and doesn't involve a change of handedness

Rotations cannot change the handedness of the atom/molecule since it only changes the angle at which the atom/molecule is viewed.

Rotary Inversions will change the handedness since they invert the molecule along with the rotation.

Screw rotations are a combination of rotational and translational operations, neither of which lead to a change of handedness.

A glide plane invlves a reflection followed by a translation, therefore will change the handedness.

Since D-Glucose is a chiral molecule, neither glide planes, nor rotary inversions will be found since they will change the handedness of the molecule.

4. The P2₁/c space group is indicative of a monoclinic system with 2/m symmetry

There are four equivalent posoitions related by the points are

1) x,y,z (0.271, 0.250, 0.458)

2) -x, 1/2+y, 1/2-z (-0.271, 0.75, 0.042)

3) -x, -y, -z (-0.271, -0.250, -0.458)

4) x, 1/2-y, 1/2+z (0.271, -0.25, 0.958)

The Co atom is sited on a special position since there is a c-glide plane situated at (x,1/4,z). Since these match the co-ordinates of the Co atom, this is indicative of where an atom lying on a centre of symmetry. Therefore, this molecule must show point group symmetry that can coexist with space group symmetry.

5. The symmetry is low-order,

There is one mirror symmetry present through the Cl-C-C-Cl plane

There is a twofold axis perperdicular to the mirror plane half-way between the C-C

There is a point of inversion mid-way between the C-C

This would suggest that the molecular point symmetry for 1,2-dichloroethylene is 2/m

6. i) Lens shaped symbol is a two-fold axis normal to the plane of paper and at the point (0,0,z)

Dark solid line is a 'Mirror plane perpendicular to a'

Dotted line is an 'a Glide plane perperdicular to b'

ii) and iii) Only J (left handedness), K (right handedness) and Z (left handedness) are related by elements within the unit cell,

Please find my workings below

A Not related

B Translation of (0,1,0) - Same handedness

C Mirror followed by (0,1,0) translation - Change of handedness

D Not related

E Not related

F Translation by 1,1,0 - Same handedness

G Two-fold rotation around (0,1/2,0) - Same handedness

H Not related

I Not related

J Glide followed by translation of (0,1,0) OR a two-fold rotation around (1/2,1/2,z) followed by mirror in (3/4,y,z) - Change to left handedness

K Related 2-fold rotation around (1/2,1/2,z) - Same right handedness

L Not related

M to X Not related due to non allowance of half-cell translations

Y Not related

Z Mirror reflection at (1/4,y,z) - Change to left handedness

a Not related

b Not related

c Translation of (1,0,0) - Alternatively, 2 glides - Same handedness (due to two changes of handedness)

d Two-fold rotation around (0,0,z) - Same handedness

e Not related

f Not related

g Glide plane at (x,0,z) - Change of handedness

h Two-fold rotation around (1/2,0,z) - Same handedness

i Not related

iv.) Orthorhombic because it has a single 2 axes plus two mutually perpendicular axes.

There are also symmetry elements on every axis, where all angles are equal to 90 degrees.