Question

A crystal is made of particles A, B and C. A forms fcc packing, B occupies all octahedral voids of fcc and C occupies all tetrahedral voids of fcc.
If all the particles along one body diagonal are removed, then what will be the formula of the crystal?

• A. $A_5{B}_4{C}_8$
• B. $A_{5}B{C}_8$
• C. $A_8{B}_4{C}_5$
• D. $A_5{B}_2{C}_8$

Answer 1

The correct option is A $A_5{B}_4{C}_8$

In crystal of ABC, atom A forms the fcc packing. Thus, the atom A occupies the corner of the cube and face centres of the cuble.
In fcc lattice, the octahedral voids are present at 12 edge centre and at the body centre of the unit cell.
Here, two tetrahedral voids on the every body diagonal of fcc. Thus, totally 8 tetrahedral voids present inside the unit cell.

On the body diagonal, there is one octahedral void (at body center) and two tetrahedral voids.

Thus, removing one body diagonal will result in removal of two A particles from corners, one B particles from octahedral void and two C particles from tetrahedral voids.
Thus, the no. of particles are as follows
$\text{No. of A- particles}=6\times 1/8+6\times \frac{1}{2}=\frac{15}{4}$
$\text{No. of B - particles}=\frac{12}{4}=3$
$\text{No. of C - particles}=\frac{6}{1}=6$
So, the formula is $A_{\frac{15}{4}}{B}_3{C}_{6}or{A}_{15}{B}_{12}{C}_{24}$
Option a) is the correct answer
i.e. $A_5{B}_4{C}_8$