Q.A compound forms hcp structure. Number of octahedral and tetrhedral voids in 0.5 mole of substance is respectively a. $3.011\times10^{23}$, $6.022\times10^{23}$
b. $6.022\times10^{23}$, $3.011\times10^{23}$
c. $4.011\times10^{23}$, $2.011\times10^{23}$
d. $6.011\times10^{23}$, $12.022\times10^{23}$

[!NOTE]
The textbook numbers this question vii. — and then prints vii. again for the next question of the same exercise (the book's own misnumbering: its eighth item is never labelled viii.). We keep the book's printed sequence.

Constituent particles of a crystal pack as tightly as possible, treated as hard touching spheres; the coordination number (number of touching neighbours) measures how compact that packing is. Building up from one dimension (a simple touching row) to two dimensions gives two packings: square close packing (spheres aligned in rows and columns, 2D coordination number 4, 'AAAA...' stacking) and hexagonal close packing (staggered rows, 2D coordination number 6, more efficient than square packing). Stacking these 2D layers in three dimensions gives three structures: stacking square layers directly above one another ('AAAA...') gives the simple cubic structure, coordination number 6 (polonium is the only metal that adopts it); stacking hexagonal layers so a third layer realigns with the first ('ABAB...') gives the hexagonal close-packed (hcp) structure; stacking hexagonal layers so a third, distinct layer covers the octahedral voids instead ('ABCABC...') gives the cubic close-packed (ccp, equivalent to fcc) structure -- both hcp and ccp reach the maximum coordination number of 12 (6 in-layer + 3 above + 3 below). Stacking two close-pack …