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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.What percentage of the tetrahedral voids is occupied by A and B?

$\begin{array}{1 1}13.5\%,60\%\\12.5\%,50\%\\14.5\%,70\%\\17.5\%,80\%\end{array} $

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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 eight tetrahedral and four octahedral voids.
Out of the eight tetrahedral voids,one is occupied by A and out of the four octahedral voids two are occupied by B.
Percentage of tetrahedral voids occupied by A=$\large\frac{1}{8}$$\times 100$
$\Rightarrow 12.5\%$
Percentage of tetrahedral voids occupied by B=$\large\frac{2}{4}$$\times 100$
$\Rightarrow 50\%$
answered Aug 1, 2014 by sreemathi.v

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