Calculate the lattice parameter of the rocksalt structure in terms of the cat- ion and anion radii rc and r . Do the same for the wurtzite and cubic fluorite structures. Using the ionic radii in Table 1.2, calculate the lattice parameters of MgO, ZnO, and Ceo).2 Discuss using Pauling’s rules why some compounds of ABO, formula are of the ilmenite or lithium niobate structures (FeTio,. LiNbo,) while others form the perovskite structure (LaGaO, PbTiO,). Note that the cation va- lences vary. Determine the Madelung constant for a two-dimensional crystal consisting of two interpenetrating square lattices. Lithium oxide forms in the antifluorite structure. (a) Compute the value of the lattice constant. (b) Calculate the density of Lio, and the density of an 0.01 mole% SrO 11. 7. 8. 5. solid solution in Li,o (c) What is the maximum radius of a cation which can be accommodated in the vacant interstice of the anion array?
The Correct Answer and Explanation is :
Lattice Parameters of Crystalline Structures
- Rocksalt Structure (NaCl-type): In the rocksalt structure, each cation (M) and anion (X) are octahedrally coordinated, forming a face-centered cubic (FCC) lattice. The lattice parameter ( a ) is related to the ionic radii ( r_{\text{M}} ) and ( r_{\text{X}} ) by: [ a = 2 \times (r_{\text{M}} + r_{\text{X}}) ] This equation assumes that the cations and anions touch along the edge of the unit cell.
- Wurtzite Structure: The wurtzite structure is a hexagonal close-packed (hcp) arrangement where each cation is tetrahedrally coordinated to four anions. The lattice parameters ( a ) and ( c ) are related to the ionic radii by: [ a = 2 \times (r_{\text{M}} + r_{\text{X}}) ] [ c = \frac{8}{\sqrt{3}} \times (r_{\text{M}} + r_{\text{X}}) ] These relations are derived from the geometry of the hcp lattice.
- Cubic Fluorite Structure (CaF₂-type): In the fluorite structure, the cations form a face-centered cubic lattice, and the anions occupy all tetrahedral interstitial sites. The lattice parameter ( a ) is related to the ionic radii by: [ a = 2 \times (r_{\text{M}} + r_{\text{X}}) ] This equation assumes that the cations and anions touch along the edge of the unit cell.
Calculations for MgO, ZnO, and CeO₂:
Using the ionic radii from Table 1.2:
- MgO (Rocksalt Structure):
- ( r_{\text{Mg}} = 0.72 \, \text{Å} )
- ( r_{\text{O}} = 1.40 \, \text{Å} ) [ a = 2 \times (0.72 + 1.40) = 4.24 \, \text{Å} ]
- ZnO (Wurtzite Structure):
- ( r_{\text{Zn}} = 0.74 \, \text{Å} )
- ( r_{\text{O}} = 1.40 \, \text{Å} ) [ a = 2 \times (0.74 + 1.40) = 4.28 \, \text{Å} ]
[ c = \frac{8}{\sqrt{3}} \times (0.74 + 1.40) \approx 10.44 \, \text{Å} ] - CeO₂ (Cubic Fluorite Structure):
- ( r_{\text{Ce}} = 1.14 \, \text{Å} )
- ( r_{\text{O}} = 1.40 \, \text{Å} ) [ a = 2 \times (1.14 + 1.40) = 4.08 \, \text{Å} ]
Discussion on ABO₃ Structures:
Pauling’s rules provide insights into the stability of crystal structures based on ionic radii and charge distribution. Compounds like FeTiO₃ (ilmenite structure) and LiNbO₃ (lithium niobate structure) adopt these structures due to favorable ionic radii ratios and charge distributions. In contrast, compounds like LaGaO₃ and PbTiO₃ form the perovskite structure, which is stabilized by different ionic radii and charge considerations.
Madelung Constant for a Two-Dimensional Crystal:
The Madelung constant quantifies the electrostatic potential energy per ion in a crystal lattice. For a two-dimensional square lattice with two interpenetrating square lattices, the Madelung constant ( M ) can be calculated by summing the Coulombic interactions between all pairs of ions in the lattice. This involves considering the distances between ions and their respective charges. The exact value depends on the specific arrangement and charge distribution within the lattice.
Antifluorite Structure of Lithium Oxide (Li₂O):
(a) Lattice Constant:
In the antifluorite structure, the anions occupy the FCC lattice, and the cations occupy the tetrahedral interstitial sites. The lattice constant ( a ) is related to the ionic radii by:
[ a = 2 \times (r_{\text{Li}} + r_{\text{O}}) ]
Using ( r_{\text{Li}} = 0.76 \, \text{Å} ) and ( r_{\text{O}} = 1.40 \, \text{Å} ):
[ a = 2 \times (0.76 + 1.40) = 4.32 \, \text{Å} ]
(b) Density of Li₂O:
The density ( \rho ) is given by:
[ \rho = \frac{m}{V} ]
where ( m ) is the mass of the unit cell and ( V ) is its volume. The volume ( V ) is ( a^3 ).