Question

A= HCP, B= 2/3 of tetrahedral voids, C= 1/2 of octahedral voids, then what is the formula?​

Answer 1

In Hexagonal Close packing,

effective number atoms = (6x1/6) + (12x1/12) + 2 = 6

2 Tetrahedral voids are present on each body diagonal. There are 4 body diagonals which means there are 8 Tetrahedral voids in a lattice structure.

Given 2/3 of the Tetrahedral voids are filled with B = (2 x 8)/3 = 16/3

Octahedral voids are present at edge centres and body centres,

So effective number of Octahedral voids in lattice structure = (12x1/4) + 1 = 4

Given 1/2 of the Octahedral voids are occupied by C = 4/2 = 2

∴The ratio of A : B : C = 6 : 16/3 : 2

which can also be written as 18 : 16 : 6 = 9 : 8 : 3

The formula thus becomes A₉B₈C₃

Hope it helps!