We know that close packed structures have both tetrahedral and octahedral voids. Let us take 𝐜𝐜𝐩 (or 𝐟𝐜𝐜) structure and locate these voids in it.
(𝐚) 𝐋𝐨𝐜𝐚𝐭𝐢𝐧𝐠 𝐓𝐞𝐭𝐫𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : Let us consider a unit cell of 𝐜𝐜𝐩 or 𝐟𝐜𝐜 lattice [Fig. 1(a)]. The unit cell is divided
into eight small cubes. Each small cube has atoms at alternate corners [Fig. 1(a)]. In all, each small cube has 4 atoms. When joined to each other, they make a regular tetrahedron. Thus, there is one tetrahedral void in each small cube and eight tetrahedral voids in total. Each of the eight small cubes have one void in one unit cell of 𝐜𝐜𝐩 structure. We know that 𝐜𝐜𝐩 structure has 4 atoms per unit cell. Thus, the number of tetrahedral voids is twice the number of atoms.
(𝐛) 𝐋𝐨𝐜𝐚𝐭𝐢𝐧𝐠 𝐎𝐜𝐭𝐚𝐡𝐞𝐝𝐫𝐚𝐥 𝐕𝐨𝐢𝐝𝐬 : Let us again consider a unit cell of 𝐜𝐜𝐩 or 𝐟𝐜𝐜 lattice [Fig. 2(a)]. The body centre of the cube, C is not occupied but it is surrounded by six atoms on face centres. If these face centres are joined, an octahedron is generated. Thus, this unit cell has one octahedral void at the body centre of the cube.
Besides the body centre, there is one octahedral void at the centre of each of the 12 edges. [Fig. 2(b)]. It is surrounded by six atoms, four belonging to the same unit cell (2 on the corners and 2 on face centre) and two belonging to two adjacent unit cells. Since each edge of the cube is shared between four adjacent
unit cells, so is the octahedral void located on it. Only `1/4` th of each void belongs
to a particular unit cell.
Thus in cubic close packed structure:
Octahedral void at the body-centre of the cube = 1
12 octahedral voids located at each edge and shared between four unit cells ` = 12 xx 1/4 = 3`
`therefore` Total number of octahedral voids `= 4`
We know that in 𝐜𝐜𝐩 structure, each unit cell has `4` atoms. Thus, the number of octahedral voids is equal to this number.