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What is the total number of voids (tetrahedral + octahedral) present in of a compound forming a stabSolid State Chemistry Question

Question

What is the total number of voids (tetrahedral + octahedral) present in $0.5 \text{ moles}$ of a compound forming a stable hexagonal close-packed (HCP) structure?

Answer: C

💡 Solution & Explanation

In close packing, for $N$ host particles, tetrahedral voids $=2N$ and octahedral voids $=N$. Total voids $=3N$. For $0.5$ mol host particles, total voids $=3\times0.5N_A=1.5N_A=9.033\times10^{23}$.

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