With the help of a labelled diagram show that there are four octahedral voids per unit cell in a cubic close-packed structure.

In cubic close packing, since atoms are present at corners and face centers, total atoms are:

Total atoms = atoms at corners + atoms at face centre

= $(8\times \frac{1}{8})+(6\times \frac{1}{2})=1+3=4$

So, the total number of atoms in cubic close packing = $4$.

The octahedral voids are present at the edge centres and the body centre. All octahedral voids are shown in the below diagram:



Number of octahedral voids per unit cell in cubic close packing =$12\times \frac{1}{4}$ ( i.e., edge center voids ) $+1$ ( i.e., body centre void )
= $3+1=4$

Thus, the number of atoms in unit cell = total number of octahedral void.

Hence, there are four octahedral voids per unit cell in a cubic close-packed structure.