Crystallography- Hexagonal-based Structures and Ternary Compounds

Question 1

Wurtzite

Answer 1

ZnS. Based on hcp (where S is). T+ site has Zn, T- empty, Oct empty. Forms corner sharing tetrahedra. CN=4. Covalent and ionic bonding. ABAB stacking of tetrahedra for ZnS. Has single polarity axis [0001].

Question 2

Nickel arsenide

Answer 2

NiAs. As forms hcp. Ni in oct sites. T+ and T- empty. Anion is close packed hexagonal so ABAB. Metal is primitive hexagonal so AAAA (hexagonal directly above each other). Bonding is metallic/heterovalent. As has CN=6. Polyhedral are edge shared prisms or corner shared octahedra. Tolerant to significant non-stoichiometry towards AB2 or A2B

Question 3

Rutile

Answer 3

TiO2. O forms pseudo-hcp. Ti fills half oct sites. T+ and T- empty. Edge and corner sharing octahedra. Bonding is purely ionic, Ti has CN=6. Close packed oxygen layers not planar as hex-to-tetragonal distortion. ABA sequence in x and y.

Question 4

Corundum

Answer 4

Al2O3. Oxygen forms hcp. Al fills 2/3 of oct sites. T+ and T- empty. Get distorted OAl4 tetrahedra and AlO6 octahedra. Bonding is ionic/covalent and CN=4. Packing ACBCACBC. If was fcc then some oxygens would have more Al neighbours than others. Second hardest structure after diamond as distortion leads to difficult slip

Question 5

Spinel

Answer 5

MgAl2O4. Oxygen forms fcc. Al half fills oct sites. Mg 1/8 fills T+/T-. Get MgO4 tetrahedra, AlO6 octahedra, OMgAl3 tetrahedra. Form of AB2O4. Inverse spinel when A atoms on octahedral sites instead of tet, half of B atoms on tet instead of oct while other half stay as with standard spinel.

Question 6

Why can ternary structures be more complex?

Answer 6

If the two cations forming a joint oxide have two differing preferred oxidation states and therefore CN and bond numbers

Question 7

Perovskite

Answer 7

ABO3. MeTiO3 where Me is Ca, Sr or Ba. Get pseudo-fcc of Ca-O layers. Ti fills 1/4 of oct sites. Get corner sharing TiO6 octahedra. Bonding ionic/metallic with CN 12 and 6. With decreasing temperature goes cubic, tetragonal, orthorhombic, rhombohedral.

Question 8

Perovskite derivatives

Answer 8

ABO3. ReO3-NaxWO3. Like perovskite but omit all or part of metal atoms in pseudo-fcc layers. Get corner sharing ReO6/WO6 octahedra. Bonding is ionic/metallic with CN 12 and 6. W must have flexible oxidation states to account for Na vacancy in its neighbourhood. x<1