Question

A compound forms hexagonal close packed (hcp) structure. What is the number of (i) Octahedral voids, (ii) Tetrahedral voids, (iii) Total voids formed in 0.7 mol of it.

Answer 1

Number of atoms in 0.7 mol = 0.7 x 6.023 x 10²³

= 4.2161 x 10²³ atoms

1. Octahedral voids
   = Number of atoms
   = 4.2161 x 10²³ voids
2. Tetrahedral voids
   = 2 x Number of atoms
   = 2 x 4.2161 x 10²³
   = 8.4322 x 10²³ voids
3. Total voids in 0. 7 mol of it:
   = Octahedral voids + Tetrahedral voids
   = 4.2161 x 10²³ + 8.4322 x 10²³
   = 12.6483 x 10²³ voids