In an ionic solid anions are arranged in hcp array and cations occupy $\frac{2}{3}$ of octahedral voids. What is the formula of ionic compound? [consider A = cation; B = anion]}
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Number of Octahedral Voids = $N$; Number of Tetrahedral Voids = $2N$.
Step 1: Understanding the Question:
This is a stoichiometry problem in solid-state chemistry. We determine the ratio of atoms based on their lattice positions and void occupancy. Step 2: Key Formula or Approach:
In a closed-packed lattice with \( n \) atoms:
Number of Octahedral Voids = \( n \)
Number of Tetrahedral Voids = \( 2n \) Step 3: Detailed Explanation:
1. Let the number of anions (\( B \)) in the hcp lattice be \( n \).
2. The total number of octahedral voids available is also \( n \).
3. Cations (\( A \)) occupy \( \frac{2}{3} \) of these octahedral voids.
4. Therefore, the number of cations (\( A \)) = \( \frac{2}{3} \times n \).
5. Ratio of \( A : B \) is \( \frac{2}{3}n : n \).
6. Ratio \( A : B = 2 : 3 \). Step 4: Final Answer:
The formula of the compound is \( A_2B_3 \).