COMEDK · Chemistry · 16. Solid State
In a normal spinel type structure, the oxide ions are arranged in ccp, whereas \(1 / 8\) tetrahedral holes are occupied by \(\mathrm{Zn}^{2+}\) ions and \(50 \%\) of octahedral holes are occupied by \(\mathrm{Fe}^{3+}\) ions. The formula of compound is
- A \(\mathrm{ZnFe}_2 \mathrm{O}_4\)
- B \(\mathrm{ZnFe}_2 \mathrm{O}_3\)
- C \(\mathrm{ZnFe}_2 \mathrm{O}_2\)
- D \(\mathrm{Zn}_2 \mathrm{Fe}_2 \mathrm{O}_4\)
Answer & Solution
Correct Answer
(A) \(\mathrm{ZnFe}_2 \mathrm{O}_4\)
Step-by-step Solution
Detailed explanation
In a ccp arrangement of oxide ions, let the number of oxide ions (\(O^{2-}\)) be \(N\). The number of tetrahedral voids is \(2N\) and the number of octahedral voids is \(N\). According to the problem, \(\mathrm{Zn}^{2+}\) ions occupy \(1/8\) of the tetrahedral voids. Therefore,…
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