In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by Zn2+ ions and 50% of octahedral holes are occupied by Fe3+ ions. The formula of the compound is –
A
Zn2Fe2O4
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B
ZnFe2O3
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C
ZnFe2O4
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D
ZnFe2O2
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Solution
The correct option is CZnFe2O4 In one unit cell no. of O2−=4
The no. of Zn2+=18×8=1
The no. of Fe3+=12×4=2
∴ molecular formula of given spinel structure is ZnFe2O4