Question
A compound forms hexagonal close packed (hcp) structure. What is the number of (i) Octahedral voids (il) Tetrahedral voids (iii) Total voids formed in 0.4 mol of it?

Answer

  • In the close packing of spheres, certain hollows left vacant, these holes in the crystal s is called interstitial voids.
  • There are two types of voids : octahedral voids and tetrahedral voids. Therefore, the total number of voids is equal to sum of octahedral voids and tetrahedral voids.
  • Number of octahedral voids is equal to number of atoms (lattice points) present in the crystal.
  • Number of tetrahedral voids is equal to twice of number of number of atoms.
We have given 0.4 moles of compound.
We know that number of atoms in 1 mole = 6.023×10²³ atoms
Then number of atoms in 0.4 moles = 0.4×6.023×10²³
= 2.41 × 10²³ atoms
(a) Number of octahedral voids = Number of atoms in 0.4mol of compound
Number of octahedral voids = 2.41 × 10²³
(b) Number of tetrahedral voids = 2× Number of atoms in 0.4mol crystal
= 2× 2.41 × 10²³
= 4.82 × 10²³
(c) Total number of voids = octahedral voids + tetrahedral voids
Total voids = 2.41 × 10²³ + 4.82 × 10²³
Total voids = 7.23 × 10²³ Voids.

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