Close packing and the holes between the atoms

How spheres stack in hcp and ccp, how to count the atoms and holes in a unit cell, and how the holes get filled to make ionic structures.

Describe hcp and ccp stacking and their 12 nearest neighbours.

Count atoms in the ccp cell by sharing atoms between cells.

Count octahedral and tetrahedral holes per cell and per ccp atom.

Read the table of which holes are filled in common structures.

Spheres stack in layers, and there are two ways to add the third

Intuition

Stack oranges in a crate. The first layer sits in a flat honeycomb pattern. Each next layer rests in the dips of the one below. After two layers, the third layer has two choices, and the choice gives two different crystals.

Many elemental metals are close-packed, and each atom has 12 nearest neighbours. The two stacking patterns are hexagonal close-packed (hcp) with layers ABAB, and cubic close-packed (ccp, also called face-centred cubic, fcc) with layers ABCABC. The packing efficiency of fcc is 74%.

Intuition

An atom sitting on a corner is also sitting in seven other boxes. So each box gets only an eighth of it. Add up the shares, not the atoms you can see.

Try it: worked example

How many atoms are in one ccp unit cell?

Show answer and steps

Answer4 atoms: 1 from the vertices and 3 from the faces.

8 vertex atoms, each shared by 8 cells: $8 \times \tfrac{1}{8} = 1$ .
6 face atoms, each shared by 2 cells: $6 \times \tfrac{1}{2} = 3$ .
Add: $1 + 3 = 4$ atoms per ccp unit cell.

The holes between close-packed atoms hold the other ions

Between the packed atoms there are gaps of two shapes. An octahedral hole has six neighbours around it. A tetrahedral hole has four. The lecture says the holes have sizes $r_h = 0.414\,r$ for octahedral and $r_h = 0.255\,r$ for tetrahedral, where $r$ is the radius of the packed atoms.

Optional aside

Lecture 7 gives the tetrahedral ratio as $0.225$ , computed from $1/\sin(109.47/2) - 1$ . Lecture 5 gives $0.255$ . The two slides disagree. Lecture 7 uses $0.225$ as the boundary in its radius-ratio table, so $0.225$ is the value consistent with that table. Check the lecture recording before quoting the number in an answer.

Try it: worked example

How many octahedral and tetrahedral holes are there per ccp atom?

Show answer and steps

Answer1 octahedral and 2 tetrahedral per atom.

Octahedral hole at the cube centre: 1, fully inside the cell.
Holes at the 12 cube edges, each shared by 4 cells: $12 \times \tfrac{1}{4} = 3$ .
Total octahedral holes: $3 + 1 = 4$ per cell. With 4 atoms per cell, that is 1 per atom.
Tetrahedral holes: 8 per cell (one per octahedron face, or per cube vertex).
With 4 atoms per cell: $8/4 = 2$ per atom.

Which holes are filled decides the structure

Exam detail

Test 1 asks for structural descriptions of ionic lattices. The table from the lecture (anion packing, which holes are filled, structure and examples):

Anion packing	Filling (Td / Oh)	Structure	Examples
ccp (fcc)	none / all Oh	rock-salt	NaCl, MgO, LaN
ccp	none / half Oh	cadmium chloride	CdCl $_2$ , NiCl $_2$
ccp	half Td / none	zinc blende (sphalerite)	ZnS, CuCl, BN
ccp	all Td / none	anti-fluorite	Na $_2$ O, Li $_2$ S
ccp	1/8 Td / half Oh	spinel	MgAl $_2$ O $_4$
hcp	none / all Oh	nickel arsenide	NiAs, FeS
hcp	none / half Oh	cadmium iodide	CdI $_2$ , Mg(OH) $_2$
hcp	half Td / none	wurtzite	ZnS, BeO
hcp	none / half Oh	rutile	TiO $_2$ , CrO $_2$

The ccp rows were checked against the slide image. The hcp rows come from the raw slide text and were not checked against the image. Rutile appears with the same filling as cadmium iodide in the table. Lecture 5 describes rutile as a distorted hcp anion array, so check the hcp rows against the slide.

Check yourself

Select all that apply

Spheres stack in layers, and there are two ways to add the third

Counting atoms means counting the share that belongs to the cell

The holes between close-packed atoms hold the other ions

Which holes are filled decides the structure

Check yourself