Question

A solid $A^{+}{B}^{-}$ has the $B^{-}$ ions arranged in a bcc lattice. If the $A^{+}$ ions occupy half of the tetrahedral sites in the structure. The formula of the solid is ?

• A. $AB$
• B. $A{B}_2$
• C. $A_{2}B$
• D. $A_3{B}_4$

Answer 1

The correct option is A $AB$

In a closed packed structure, the number of tetrahedral voids per atom of the crystal is $4$.
No. of $B^{-}$ ions per unit cell = 2
Since, half of the tetrahedral voids are occupied by $A^{+}$, the number of $A^{+}$ is same as that of $B^{-}$ in the crystal.

Thus, the formula is AB.
Option A is correct.