In a crystalline solid anion $B$ are arranged in a $c c p$ lattice. Cation $A$ are equally distributed between octahedral and tetrahedral voids. If all octahedral voids are occupied. What is the formula of the solid?

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Solution

Number of atoms in CCP =4
Therefore, number of atoms in close packing, $B = 4$
We know octahedral voids are equal to the number of atoms in close packing
Therefore octahedral voids =4
and tetrahedral voids $= 2 \times (n u m b e r o f a t o m s i n c l o s e p a c k i n g) = 8$
Given, A atoms are equally distributed between both voids and A atoms occupies all octahedral voids. Therefore, total $A a t o m s = 4 \times (O c t a h e d r a l a t o m s) + 4 \times (t e t r a h e d r a l a t o m s) = 8 a t o m s$
Therefore formula of $A B = A_{8} B_{4}$ OR $A_{2} B$

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In a crystalline solid anion B are arranged in a ccp lattice. Cation A are equally distributed between octahedral and tetrahedral voids. If all octahedral voids are occupied. What is the formula of the solid?

In a crystalline solid anion $B$ are arranged in a $c c p$ lattice. Cation $A$ are equally distributed between octahedral and tetrahedral voids. If all octahedral voids are occupied. What is the formula of the solid?