Chemistry · Ch 11 — The Solid State

Packing Efficiency in hcp and ccp Structures

11.7.1

Packing Efficiency in hcp and ccp Structures

Both close-packed arrangements — hcp and ccp — are equally efficient, so it is enough to work out the efficiency for the ccp (fcc) structure. Let the edge length of the unit cell be $a$ and the face diagonal $AC = b$ (Fig. 1.24).

In the right triangle $ABC$ :

AC^2 = b^2 = BC^2 + AB^2 = a^2 + a^2 = 2a^2 \quad\Rightarrow\quad b = \sqrt{2}\,a

Along the face diagonal the spheres touch, so if $r$ is the radius of a sphere:

b = 4r = \sqrt{2}\,a \quad\Rightarrow\quad a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}\,r \qquad\left(\text{equivalently } r = \frac{a}{2\sqrt{2}}\right)

A ccp unit cell contains 4 spheres, so the volume occupied is $4 \times \tfrac{4}{3}\pi r^3$ , while the volume of the cube is $a^3 = (2\sqrt{2}\,r)^3$ . Hence: …

Figure 1.24Cubic close packing; other sides are not provided with spheres for sake of clarity.

What this figure shows. An fcc unit cell drawn for the packing-efficiency derivation: a cube of edge length 'a' with spheres shown along one face; the face diagonal AC = b is marked, with triangle ABC used to derive b = sqrt(2) a and 4r = sqrt(2) a. …

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