A) \[Z{{n}_{2}}F{{e}_{2}}{{O}_{4}}\]
B) \[ZnF{{e}_{2}}{{O}_{3}}\]
C) \[ZnF{{e}_{2}}{{O}_{4}}\]
D) \[ZnF{{e}_{2}}{{O}_{2}}\]
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: \[ZnF{{e}_{2}}{{O}_{4}}\]
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