Q.The ratio of close packed atoms to tetrahedral hole in cubic packing is

Building close-packed layers in the ABAB... pattern (Section 6.6.2, each sphere touching 6 neighbours) and stacking a third layer over them in two different ways gives the two most efficient 3D packings possible:

• aba stacking -> hexagonal close-packed (hcp): the third layer sits directly over the first.
• abc stacking -> cubic close-packed (ccp): the third layer is offset from BOTH earlier layers (built on the fcc unit cell).

Both reach coordination number 12 (6 in-layer + 3 above + 3 below) and a packing efficiency of ~74% -- the highest of any arrangement in this unit (derived from the fcc relation 4r = root-2*a, giving r = (root-2/4)a).

Voids. Stacking two close-packed layers creates two kinds of gap: a tetrahedral void (4 touching spheres, 3 from the lower layer + 1 from the upper, centres forming a tetrahedron) and an octahedral void (6 touching spheres, 3+3, centres forming an octahedron). If n is the number of close-packed spheres, the number of octahedral voids = n and the number of tetrahedral voids = 2n -- so the ratio of close-packed atoms to tetrahedral holes is always 1:2. This directly gives the formula of a compound where a second atom occupies some fraction of the voids of a ccp/fcc lattice of a first atom (e.g. atom M in ccp with atom N filling 1/3 of the tetrahedral voids: M:N = 4 : (1/3 x 8) = 4 : 8/3 = 3:2, formula $M_3N_2$).