With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close packed structure.
Cubic close packod structure contains one atom at each uf eight corners of a unit cell and one atom at each of six laces which can be represented below
As we know any atom surrounded by six atoms (hard sphere) creates an octahedral void. In case of fcc body centre is surrounded by six identical atoms present at face centre hence, there is a octahedral void at body centre of each unit cell.
Beside the body centre there is one octahedral void at centre of each of 12 edge as shown below
Since, each void is shared by 4 unit cell. Therefore, contribution of octahedral void to each edge of a unit cell is
Number of octahedral void at centre of 12 edge = x 12 = 3
Number of octahedral void at body centre = 1
Therefore, total number of octahedral void at each ccp lattice = 3 + 1 = 4
