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Question

In a normal spinel types structure, the oxide ions are arranged in ccp whereas 18 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 C

ZnFe2O4


In one unit cell no. of O2=4
The no. of Zn2+=18×8=1


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