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In a crystalline solid, having formula $AB_2O_4$, oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids.(ii) What percentage of the octahedral voids is occupied by B?

$\begin{array}{1 1}(A)\;50\%&(B)\;50.5\%\\(C)\;45\%&(D)\;60\%\end{array} $

1 Answer

In a cubic close packed lattice of oxide ions there would be two tetrahedral voids and one octahedral void for each oxide ion.
$\therefore$ For four oxide ions there would be 8 tetrahedral and four octahedral voids two are occupied by B.
Percentage of tetrahedral voids occupied by B=$\large\frac{2}{4}$$\times 100=50\%$
Hence (A) is the correct answer.
answered May 28, 2014 by sreemathi.v