Solid State Test - 5

The lattice energy depends on the charge on anion and the type of crystal structure. As both NaCl and CaO have the same crystal structure but different product value of charges, i.e. for NaCl, the charges are +1 and -1, and the product value is 1; and for CaO, the charges are +2 and -2, and the product value is 4.
Therefore, the lattice energy of CaO is 4 times that of NaCl with the same crystal structure.

Before removing face centred atoms:
Number of 'A' atoms = (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 [There are 8 'A' atoms at the corners and 6 'A' atoms at the face centres]
Number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4 [There are 12 'B' atoms at the edges and 1 'B' atom at the centre of the unit cell]
After removing two face centred 'A' atoms along one of the axes:
Number of 'A' atoms = (8 x 1/8) + (4 x 1/2) = 1 + 2 = 3 [There are now only 4 'A' atoms at the face centres]
Number of 'B' atoms = (12 x 1/4) + (1 x 1) = 3 + 1 = 4
Therefore, the stoichiometry after removing these atoms is A₃B₄.

Calcium titanate is CaTiO₃. Since Ti⁴⁺ is at the body centre, it represents octahedral holes. Total octahedral holes are four, out of which only one is occupied by Ti⁴⁺ ions. This makes them 25% of the holes.

When sodium chloride is doped with strontium chloride, the type of defect is called an impurity effect. To maintain electrical neutrality, each Sr²⁺ ion replaces 2Na⁺ ions. Thus, the density of the crystal decreases due to cationic vacancies produced.

An octahedral hole at the centre of a unit cell, occupied by titanium ions, has 6 nearest neighbour oxide ions. The other octahedral holes, located at the centres of the edges of unit cells, have 6 nearest neighbour each, as the case with any octahedral hole, but 2 of the 6 neighbours are barium ions (at unit cell corners termination the given edge) and 4 oxide ions. The proximity of two cations Ba²⁺ and Ti⁴⁺ would be electrostatically unfavourable. So, titanium cannot occupy all of the octahedral holes with restrictions as stated above.