Question

Suppose an ionic solid with a cubic unit cell contains A (cations) and Z (anions). The A ions are in the body center position and at each corner, and the Z ions are on each face. Based on this structure the empirical formula of the compound would be AZ

Answer 1

Answer:

The empirical formula of the compound would be $A_2Z_3$

Explanation:

The A ions are in the body center position and at each corner .

There are 8 faces of the cube and 1 corner is also shared by another 7 cubic unit.

Then number of A ion in a cubic unit = $1+\frac{1}{8}\times 8=2$

The Z ions are on each face of the cubic unit :

There are 6 faces of the cube and 1 face is shared by another cubic unit.

Then number of Z ions in a cubic unit = $\frac{1}{2}\times 6=3$

The structural formula of the cubic lattice is  =$A_2Z_3$

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula of the compound would be $A_2Z_3$