Question

A crystal is made of particles A and B. A forms fcc packing and B occupies all the octahedral voids. If all the particles along the plane as shown in figure are removed, then the formula of the crystal will be

• A. $AB$
• B. $A_5{B}_7$
• C. $A_7{B}_5$
• D. none of the above

Answer 1

Answer: (A)

$AB$

In fcc, a total of $4$ octahedral voids are present.

So, it contains $4A$ and $4B$ atoms but all the particles along the plane as shown in the figure given in the question are removed.
This plane contains $\frac{3}{2}$ octahedral void atoms ($1$ at body center and $2$ at edge center) and $\frac{3}{2}$ lattice atoms ($2$ face centers and $4$ corners).

So, the formula will be $A_{3/2}{B}_{3/2}=AB$.