Question:
A compound consists of atoms A and B. The atoms of B form hcp lattice. The atoms of A occupy \(\frac{1}{3}\) rd of octahedral voids and \(\frac{1}{3}\) rd of tetrahedral voids. What is the molecular formula of the compound?
A compound consists of atoms A and B. The atoms of B form hcp lattice. The atoms of A occupy \(\frac{1}{3}\) rd of octahedral voids and \(\frac{1}{3}\) rd of tetrahedral voids. What is the molecular formula of the compound?
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Calculate molecular formula by considering atoms in lattice and occupancy of voids.
Updated On: Jun 3, 2025
- \({A}_2{B}\)
- \({AB}_2\)
- \({AB}_3\)
- \({AB}\)
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The Correct Option is D
Solution and Explanation
- Number of atoms of B in hcp lattice per unit cell = 6.
- Number of octahedral voids per atom = 1.
- Number of tetrahedral voids per atom = 2.
Atoms of A occupy:
\[ \frac{1}{3} \times 6 = 2 \quad \text{octahedral voids} \]
\[ \frac{1}{3} \times (2 \times 6) = 4 \quad \text{tetrahedral voids} \]
Total atoms of A = \(2 + 4 = 6\). So ratio of A to B is \(6:6 = 1:1\).
Hence, molecular formula is \(\mathrm{AB}\).
- Number of octahedral voids per atom = 1.
- Number of tetrahedral voids per atom = 2.
Atoms of A occupy:
\[ \frac{1}{3} \times 6 = 2 \quad \text{octahedral voids} \]
\[ \frac{1}{3} \times (2 \times 6) = 4 \quad \text{tetrahedral voids} \]
Total atoms of A = \(2 + 4 = 6\). So ratio of A to B is \(6:6 = 1:1\).
Hence, molecular formula is \(\mathrm{AB}\).
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