Question

A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 many of these are tetrahedral voids?

Answer 1

Number of atoms in close packaging = 0.5 mol

1 has $6.022\times {10}^{23}$ particles

So that

Number of close-packed particles

$=0.5\times 6.022\times {10}^{23}=3.011\times {10}^{23}$

Number of tetrahedral voids = 2 × number of atoms in close packaging

Plug the values we get

Number of tetrahedral voids $=2\times 3.011\times {10}^{23}$

$=6.022\times {10}^{23}$

Number of octahedral voids = number of atoms in close packaging

So that

Number of octahedral voids $=3.011\times {10}^{23}$

Total number of voids = Tetrahedral void + octahedral void

$=6.022\times {10}^{23}+3.011\times {10}^{23}$

$=9.03\times {10}^{23}$