Consider Na2O with a fcc unit cell, what fraction of tetrahedralholes are occupied by Na+ if O2- occupy lattice points?

To determine the fraction of tetrahedral holes occupied by Na+ in Na2O with a fcc unit cell, we need to consider the arrangement of the O2- ions on the lattice points.

In a fcc unit cell, there are 4 lattice points per unit cell. Each lattice point is shared by 8 adjacent unit cells, resulting in a total of 8 tetrahedral holes per unit cell.

Since O2- ions occupy the lattice points, there are 4 O2- ions per unit cell. Each O2- ion is located at the center of a tetrahedral hole. Therefore, there are also 4 tetrahedral holes occupied by O2- ions per unit cell.

Now, let's calculate the fraction of tetrahedral holes occupied by Na+ ions. Since Na2O has a 1:2 ratio of Na+ to O2- ions, there are 2 Na+ ions per unit cell.

Out of the 8 tetrahedral holes per unit cell, 4 are already occupied by O2- ions. Therefore, there are 8 - 4 = 4 tetrahedral holes available for Na+ ions.

Since there are 2 Na+ ions per unit cell, the fraction of tetrahedral holes occupied by Na+ ions is given by:

Fraction of tetrahedral holes occupied by Na+ = (Number of tetrahedral holes occupied by Na+ ions) / (Total number of tetrahedral holes)

Fraction of tetrahedral holes occupied by Na+ = 4 / 8

Simplifying, we find that the fraction of tetrahedral holes occupied by Na+ ions in Na2O with a fcc unit cell is 1/2.