Question

A forms $hcp$ lattice and B are occupying $1/3^{rd}$ of tetrahedral voids, then the formula of compound is

• A. $AB$
• B. $A_3B_2$
• C. $A_2B_3$
• D. $AB_4$

Answer 1

Answer: (B)

$A_3B_2$

Answer 2

Let no. of atoms of A used in close packing = n Number of tetrahedral voids = 2n Number of atoms of $B=\frac{1}{3}\times2n=\frac{2}{3}n$ $A : B=n : \frac{2}{3}n=3 : 2$ Formula of the compound =$A_{3}B_{2}$