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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

  1. A \(\mathrm{ZnFe}_2 \mathrm{O}_4\)
  2. B \(\mathrm{ZnFe}_2 \mathrm{O}_3\)
  3. C \(\mathrm{ZnFe}_2 \mathrm{O}_2\)
  4. D \(\mathrm{Zn}_2 \mathrm{Fe}_2 \mathrm{O}_4\)
Verified Solution

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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