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Question

Oxide ion forming a fcc lattice in which Fe2+ ions are occupying 1/8th of tetrahedral void and Fe3+ ions are occupying 12 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.
Fe2+ occupying 18th tetrahedral voids, i.e., 18×8=1 void=1 atom of Fe2+
There are 4 octahedral voids in FCC.
Thus, Fe3+ occupy 12 of octahedral voids, i.e., 12×4=2 voids2 ions of Fe3+
Oxide ions are present in FCC lattice. Thus, no. of oxide ion is 4. Thus, the formula of compound is Fe(II)Fe2(III)O4 or FeFe2O4.

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