Oxide ion forming a fcc lattice in which $F e^{2 +}$ ions are occupying $1 / 8^{t h}$ of tetrahedral void and $F e^{3 +}$ ions are occupying $\frac{1}{2}$ of the octahedral void. What will be the formula of the compound?

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Solution

In FCC lattice, 8 tetrahedral voids are present.
At each body diagonal two tetrahedral voids are present.
$F e^{2 +}$ occupying $\frac{1}{8}$ th tetrahedral voids, i.e., $\frac{1}{8} \times 8 = 1$ void $= 1$ atom of $F e^{2 +}$
There are 4 octahedral voids in FCC.
Thus, $F e^{3 +}$ occupy $\frac{1}{2}$ of octahedral voids, i.e., $\frac{1}{2} \times 4 = 2$ voids $⟹ 2$ ions of $F e^{3 +}$
Oxide ions are present in FCC lattice. Thus, no. of oxide ion is 4. Thus, the formula of compound is $F e (I I) F e_{2} (I I I) O_{4}$ or $F e F e_{2} O_{4}$ .

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