Question

A compound with formula $A{B}_2{O}_4$ adopts a crystal structure in which $O^{2-}$ ions forms ccp arrangement. $A^{2+}$ ions occupy octahedral voids while $B^{3+}$ ions are distributed equally between tetrahedral and octahedral voids. The fraction of octahedral and tetrahedral voids occupied respectively are:

• A. $12.5\%,50\%$
• B. $50\%,12.5\%$
• C. $25\%,50\%$
• D. $50\%,25\%$

Answer 1

The correct option is B $50\%,12.5\%$

The chemical formula of the given compound is $A{B}_2{O}_4$
$O^{2-}$ ions forms ccp arrangement
$\therefore$ Effective Number of atoms = 4
Number of octahedral voids = 4
Number of tetrahedral voids = 8
According to the question $A^{2+}$ ion occupies octahedral voids.
$1B^{3+}$ ion occupies octahedral void and the other $B^{3+}$ ion occupies tetrahedral void.
$\therefore$ Fraction of octahedral voids occupied =$\frac{2}{4}\times 100=50\%$
Fraction of tetrahedral void occupied =$\frac{1}{8}\times 100=12.5\%$