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In a normal spinel type structure the oxide ion are arranged in ccp whereas 1/8 tetrahedral holes are occupied by zn +2 ions and 50% of octa hedral holes are occupied by Fe +3 ions the formula of the compund is??? In a normal spinel type structure the oxide ion are arranged in ccp whereas 1/8 tetrahedral holes are occupied by zn+2ions and 50% of octa hedral holes are occupied by Fe+3 ions the formula of the compund is???
Since Zn2+ atoms occupy one-eighth of all the tetrahedral voids and the Fe3+ atoms occupy half of all the octahedral voids in the oxygen lattice. Thus the formula becomes "ZnFe2O4
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