Body Centered Cubic (bcc) Structure
A more efficiently packed cubic structure is the 'body-centered cubic' (bcc). The first layer of a square array is expanded slightly in all directions. Then, the second layer is shifted so its spheres nestle in the spaces of the first layer (Figures 5a, b). This repeating order of the layers is often symbolized as 'ABA...'. Like Figure 3b, the considerable space shown between the spheres in Figure 5b is misleading: spheres are closely packed in bcc solids and touch along the body diagonal. The packing efficiency of the bcc structure is about 68%. The coordination number for an atom in the bcc structure is eight. How many total atoms are there in the unit cell for a bcc structure? Draw a diagonal line connecting the three atoms marked with an 'x' in Figure 5b. Assuming the atoms marked 'x' are the same size, tightly packed and touching, what is the value of this body diagonal as a function of r, the radius? Find the edge and volume of the cell as a function of r.


Cubic Closest Packed (ccp)
A cubic closest packed (ccp) structure is created by layering close packed arrays. The spheres of the second layer nestle in half of the spaces of the first layer. The spheres of the third layer directly overlay the other half of the first layer spaces while nestling in half the spaces of the second layer. The repeating order of the layers is 'ABC...' (Figures 6 & 7). The coordination number of an atom in the ccp structure is twelve (six nearest neighbors plus three atoms in layers above and below) and the packing efficiency is 74%.


Figure 6: Close packed Array Layering. The 1st and 3rd layers are represented by light
spheres; the 2nd layer, dark spheres. The 2nd layer spheres nestle in the spaces of the 1st
layer marked with an “x”. The 3rd layer spheres nestle in the spaces of the 2nd layer that
directly overlay the spaces marked with a “·” in the 1st layer.


Figure 7a & 7b: Two views of the Cubic Close Packed Structure

If the cubic close packed structure is rotated by 45° the face centered cube (fcc) unit cell can be viewed (Figure 8). The fcc unit cell contains 8 corner atoms and an atom in each face. The face atoms are shared with an adjacent unit cell so each unit cell contains ½ a face atom. Atoms of the face centered cubic (fcc) unit cell touch across the face diagonal (Figure 9). What is the edge, face diagonal, body diagonal, and volume of a face centered cubic unit cell as a function of the radius?


Figure 8: The face centered cubic unit cell is drawn by cutting a diagonal plane through
an ABCA packing arrangement of the ccp structure. The unit cell has 4 atoms (1/8 of
each corner atom and ½ of each face atom).


Figure 9a:Space filling model of fcc. Figure 9b: The face of fcc. Face diagonal = 4r.

Ionic Solids
In ionic compounds, the larger ions become the lattice point “spheres” that are the framework of the unit cell. The smaller ions nestle into the depressions (the “holes”) between the larger ions. There are three types of holes: 'cubic', 'octahedral', and 'tetrahedral'. Cubic and octahedral holes occur in square array structures; tetrahedral and octahedral holes appear in close-packed array structures (Figure 10). Which is usually the larger ion – the cation or the anion? How can the periodic table be used to predict ion size? What is the coordination number of an ion in a tetrahedral hole? an octahedral hole? a cubic hole?


Figure 10. Holes in ionic crystals are more like 'dimples' or 'depressions' between the
closely packed ions. Small ions can fit into these holes and are surrounded by larger ions
of opposite charge.

The type of hole formed in an ionic solid largely depends on the ratio of the smaller ion’s radius the larger ion’s radius (rsmaller/rlarger). (Table 1).


Empirical Formula of an Ionic Solid
Two ways to determine the empirical formula of an ionic solid are: 1) from the number of each ion contained within 1 unit cell 2) from the ratio of the coordination numbers of the cations and anions in the solid.


Example: Find the empirical formula for the ionic compound shown in Figures 11 & 12.
First Method: When using the first method, remember most atoms in a unit cell are shared with other cells. Table 2 lists types of atoms and the fraction contained in the unit cell. The number of each ion in the unit cell is determined: 1/8 of each of the 8 corner X ions and 1/4 of each of the 12 edge Y ions are found within a single unit cell. Therefore, the cell contains 1 X ion (8/8 = 1) for every 3 Y ions (12/4 = 3) giving an empirical formula of XY3. Which is the cation? anion? When writing the formula of ionic solids, which comes first?


Second Method: The second method is less reliable and requires the examination of the crystal structure to determine the number of cations surrounding an anion and vice versa. The structure must be expanded to include more unit cells. Figure 12 shows the same solid in Figure 11 expanded to four adjacent unit cells. Examination of the structure shows that there are 2 X ions coordinated to every Y ion and 6 Y ions surrounding every X ion. (An additional unit cell must be projected in front of the page to see the sixth Y ion ). A 2 to 6 ratio gives the same empirical formula, XY3.

Summary:

Simple Cubic: 1 total atom per cell (1/8 each corner)
Body Centered Cubic: 2 atoms per cell (1 in center and 1/8 for each corner)
Face Centered Cubic: 4 atoms per cell
(1/2 per face and 1/8 for each corner)