## Locating tetrahedral and octahedral voids Chemistry Notes

→ Both tetrahedral and octahedral voids are present in close packed structures. The determination of these voids in cubic close packing (ccp) or face centred cubic (fcc) or hexagonal close packing (hep) is done as follows

→ Locating Tetrahedral Void : Unit cell of ccp or hcp or foc lattice can be divided into eight small cubes. Atoms are present at each corner in each cube. So, four atoms will be present in each small cube which on combination form tetrahedron. Hence, a tetrahedral void is present in each small cube. Therefore, the total number of tetrahedral voids will be eight times which will be twice the number of atoms. The formation of tetrahedral void is represented in figure 1.40.

→ Locating Octahedral Void : If we see unit cell of ccp or foc lattice carefully then it will be known that here cube does not contain body centre but its face is surrounded by six atoms present at centres. If these face centres are connected with each other then a octahedron is formed. So, here an octahedral void is present in unit cell that is present at body centre of cube.

→ Similarly, an octahedral void is also present at middle point of each edge because this point is also surrounded by six atoms in which four atoms are of that unit cell while two atoms are of neighbouring unit cell. Figure 1.41 clears that each corner of cube is surrounded by four unit cells. So, octahedral void will also be surrounded by four unit cells. Similarly, an octahedral void will contribute only 1/4th part in its unit cell.

→ So, the number of octahedral void at body centre of cube in cubic close packing is 1 and 12 octahedral voids are present at each corners which are shared between four unit cells Hence, Number of octahedral voida = 12 × $$\frac{1}{4}$$ = 3

→ So, Total number of octahedral voids = 1 + 3 = 4 Since we known that 4 atoms are present in each unit cell in cop structure. So, the number of octahedral voids is equal to this number.

Note:

• Number of tetrahedral voids = 2 × Number of atoms in unit cell
• Number of octahedral voids = Number of atoms in unit cell