Solid State Question 56
Question: In a normal spinel type structure, the oxide ions are arranged in ccp whereas 1/8 tetrahedral holes are occupied by $ Z{{n}^{2+}} $ ions and 50% of octahedral holes are occupied by $ F{{e}^{3+}} $ ions .The formula of the compound is
Options:
A) $ Zn_2Fe_2O_4 $
B) $ ZnFe_2O_3 $
C) $ ZnFe_2O_4 $
D) $ ZnFe_2O_2 $
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Answer:
Correct Answer: C
Solution:
[c] Number of O-atoms per unit cell $ =\frac{1}{8}\times 8+\frac{1}{2}\times 6=4 $
Number of octahedral holes per unit cell $ =1\times 4=4 $
Number of $ F{{e}^{3+}} $ ions per unit cell $ =\frac{50\times 4}{100}=2 $
Number of tetrahedral voids per unit cell $ =2\times 4=8. $
Number of $ Z{{n}^{2+}} $ ions per unit cell $ =\frac{1}{8}\times 8=1 $ Hence, formula: $ ZnFe_2O_4 $
BETA
