Question

Consider the ionic solid A_xB_y which has the unit cell below. The B ions are packed in a cubic arrangement, where each face has this structure:

There is one B in the center of the cube. The structure can also be described in terms of three parallel planes of B’s of the type shown above. The resulting structure thus contains eight intersecting cubes of B’s. The A ions are found in the centers of alternate intersecting cubes (that is, four of every eight cubes have A’s in the center). What is the formula of ? In the extended structure, how many B’s surround each A? What structure do the B’s form?

Short answer

The formula of $A_x{B}_y=A{B}_2$ and B’s atoms form the cubic packed unit cell structure.

Step-by-step solution

The values of x and y in the ionic bond AXBy.

From the structure given, we have 8 atoms of B available in the corners of the cube, 6 atoms of B available at the face of the cube, 12atoms of available on the edgesof the cube, and 1 atom of B is available in the center,

So, the atoms of B:

$$\begin{gathered} y=8\times \frac{1}{8}+6\times \frac{1}{2}+12\times \frac{1}{4}+1 \\ =8 \end{gathered}$$

It is given that A is available in the centers of alternate intersecting cubes, that is, four of every cube has A's in the center.

So, there are 4 atoms of A are present.

Now, $A:B=4:8\text{or}1:2$

Hence, $A_x{B}_y=A{B}_2$where, x = 1 and y = 2.

The structure B

From the structure given, we can see that each atom ofA is surrounded by 8 atoms of B , and at the center, corners, faces, and edges of the structure, atoms of B are present.

So, the B’s atoms form the cubic packed unit cell structure.