Question

A binary solid $\left(A^{+}{B}^{-}\right)$ has a zinc blende structure with $B^{-}$ ions constituting the lattice and $A^{+}$ ions occupying $25\%$ tetrahedral voids. The formula of solid is

• A. $AB$
• B. $A_{2}B$
• C. $A{B}_2$
• D. $A{B}_4$

Answer 1

The correct option is C $A{B}_2$

Since $A^{+}{B}^{-}$ has a zinc blende structure, so it has a cubic close packed structure.
Since $B^{-}$ ions constituting the lattice, thus number of $B^{-}$ ions $=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
Again, number of tetrahedral voids would be $8$
So, number of $A^{-}$ ions would be $=\frac{1}{4}\times 8=2$
SO the formula of the solid is $A_2{B}_4\text{or}A{B}_2$