Hexagonal close-packed (HCP) Structure

HEXAGONAL CLOSE-PACKED (HCP) STRUCTURE

A unit cell of a close-packed hexagonal structure is shown in fig 1.24 (a).

In this type of crystal structure, the unit cell has one atom at each of the 12 corners of the hexagonal face, one atom each at the centre of the two hexagonal faces. 3 atoms symmetrically arranged inside the unit cell.

There are 3 layers of atoms in it. At the bottom layer (A), the central atom has '6' nearest neighbouring atoms in the same plane. The second layer (B) which is at a distance c/2 from the bottom layer has 3 atoms as shown in fig. 1.24 (b).

The third layer ie., the layer similar to the bottom layer which is at a distance 'c' from the bottom layer. Thus, HCP structure has the stacking sequence of AB, AB ...

1. Number of atoms per unit cell

Each corner atom is shared by 6 surrounding unit cells, i.e., each corner atom gives 1/6 th of its share to unit cell.

Number of atoms due to corner atoms in upper hexagonal plane of unit cell = 1/6 × 6 = 1

Number of atoms due to corner atoms in lower hexagonal plane of unit cell = 1/6 × 6 = 1

Each central atom in the upper and lower planes (base central atom) is shared by two unit cells. It means that the upper and lower planes of the given unit cell contain 1/2 atom each.

⸫ Total number of central atoms in both upper and lower planes = 1/2 × 2 = 1

There are '3' atoms completely inside the unit cell which are not shared by any other adjacent unit cells.

⸫ Total number of atoms in a hcp unit cell = 1+1+1+3 = 6

2. Coordination number

Consider the bottom layer, the central atom has 6 nearest neighbouring atoms in the same plane.

At a distance ‘c’/2 from the bottom layer, there are two layers, one above and the other below the bottom layer containing 3 atoms in each layer are also neighbouring atoms.

In total, there are 12 (3 + 6 + 3) nearest neighbouring atoms. (Fig. 1.24 (c))

⸫ Coordination number = 12

3. Atomic radius (r)

Atoms touch each other along the edges of the hexagon. Thus, from the figure 1.24 (d), the nearest neighbouring distance a = 2 r.

4. Calculation of c/a ratio

Let 'c' be the height of the unit cell of HCP structure and 'a' be the distance between two neighbouring atoms.

Consider a triangle ABO in the bottom layer of unit cell of HCP structure (fig 1.25). Here A, B and O are the lattice points and exactly above these atoms at a perpendicular distance 'c'/2, the next layer atom lies at C.

Δ ABO is an equilateral triangle. A line AY is drawn ⊥r to BO (Fig. 1.25 (b)). Then,

cos 30° = AY / AB

X is a centre for Δ ABO

Substituting for AY, we get

In the ΔAXC, plane AXC is perpendicular to the plane ABO

Taking square root on both sides, we have

5. Packing factor

Volume of all the atoms in a unit cell (v)

Number of atoms per unit cell n = 6

Atomic radius r = a/2

Volume of all 6 atoms in the unit cell

v = 6 × 4/3 π r3

Substituting for r, we have

Volume of the unit cell (V)

Area of the base of the hexagon = 6 × Area of triangle AOB

Substituting v and V, we have

Therefore, packing factor is 74% ie., 74% of the volume of unit cell is occupied by the atoms and the remaining 26% volume is vacant.

Note: Packing factors of FCC and HCP structures are same and they are known as closely packed structures

Common examples of HCP structure

Magnesium, zinc, titanium, zirconium, beryllium and cadmium.