Question

The anions $\left(A\right)$ form hexagonal close packing and the cations $\left(C\right)$ occupy only $\frac{2}{3}$ of the octahedral voids formed. The simplest formula of the ionic compound is:

• A. $CA$
• B. $C_3{A}_2$
  
• C. $C_4{A}_3$
  
• D. $C_2{A}_3$
  

Answer 1

The correct option is D $C_2{A}_3$


For hexagonal close packing, the number of atoms per unit cell is $6$.
Number of octahedral voids is also $6$.
Hence,
number of atoms of $A=6$
Number of atoms of $C=6\times \frac{2}{3}=4$
So, the simplest formula of the ionic compound is $C_2{A}_3$.