Question

Atom A forms FCC. B occupies all td. voids and C occupies all oct. voids. If all atoms along a body diagonal of this cube are removed what is the new possible formula?

Answer 1

Hint: Here, FCC stands for face centered cubic lattice, td. voids for tetrahedral voids and Oct. voids for octahedral voids. The different types of lattice structures are fcc, bcc and hcp which means Face centered cubic, body centered cubic and hexagonal closest packing.

Complete answer: Like all other lattices, face centered cubic (fcc) lattice has lattice points at all eight corners of the unit plus additional points at the center of each face of the unit cell.
In fcc, unit cell number of atoms present = $4$
Number of tetrahedral voids = $2 \times $number of octahedral voids and, number of octahedral voids = number of atoms present in cubic unit cell
So,
A=$4$, B=$4$, C= $8$
On the body diagonal, there is one octahedral void (at body center)
Two tetrahedral voids (at one fourth distance from each corners) After the removal of atoms at body diagonal we will be left with:
$A = 6 \times \dfrac{1}{2} + \dfrac{{(8 - 2)}}{2} = 3 + \dfrac{3}{4} = \dfrac{{15}}{4}$
 (face) (corners)
$B = 4 - 1 = 3$
and $C = 8 - 2 = 6$
Therefore,

$A$	$B$	$C$
$\dfrac{{15}}{4}$	$3$	$6$

Multiply by 4

$A$	$B$	$C$
$\dfrac{{15}}{4} \times 4$	$3 \times 4$	$6 \times 4$

On simplification we have

$A$	$B$	$C$
$15$	$12$	$24$

Now we can reduce this in simplest form by dividing it by 3.
Then on dividing by 3 we have

$A$	$B$	$C$
$5$	$4$	$8$

So our required new formula would be, ${A_5}{B_4}{C_8}$.

Note:
The above formula can also be written as ${A_{15}}{B_{12}}{C_8}$ but we always write it in the lowest possible form therefore, the correct answer would be ${A_5}{B_4}{C_8}$. The two lattices with highest densities are fcc that is face centered cubic and hcp that is hexagonal close packing. Make sure to use the correct number of octahedral and tetrahedral voids in each case.