In a compound $X Y_{2} O_{4}$ oxide ions are arranged in ccp structure. Cations X are present in octahedral voids and cation Y are equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is x .Find the value 10x.

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$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{6}$

answer is E.

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Detailed Solution

A CCP lattice has a face centred cubic unit cell. So the number of particles present in the unit cell are 4 or the number of oxide ions in the given crystal packing is n=4
The number of tetrahedral voids are given by $= 2 \times n = 8$
The number of octahedral voids are given by $n = 4$
The formula of the compound says, $X Y_{2} O_{4}$ , the number of cations X in one unit cell $= 1$ and the number of cations Y in one unit cell $= 2$
It is also given in the question, that the cation Y is distributed equally in octahedral and tetrahedral voids, thus, we can say that one Y is in octahedral void and one in tetrahedral void. Also, X occupies the octahedral voids. So the number of octahedral voids occupied $= 1 + 1 = 2$
Hence, the fraction of octahedral voids occupied is $= 2 / 4 = \frac{1}{2}$