Question:
In ionic solid, anions are arranged in ccp array and cations occupy $1/3$ tetrahedral voids. What is the formula of ionic compound?
[Consider $\text{A} = \text{cation}; \text{B} = \text{anion}$]
In ionic solid, anions are arranged in ccp array and cations occupy $1/3$ tetrahedral voids. What is the formula of ionic compound?
[Consider $\text{A} = \text{cation}; \text{B} = \text{anion}$]
[Consider $\text{A} = \text{cation}; \text{B} = \text{anion}$]
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Tet voids = $2 \times$ Atoms; Oct voids = Atoms.
Updated On: May 14, 2026
- $\text{AB}_3$
- $\text{A}_2\text{B}_3$
- $\text{A}_3\text{B}_2$
- $\text{AB}_4$
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The Correct Option is B
Solution and Explanation
Step 1: Concept In a ccp (or fcc) lattice, if the number of atoms is $N$, the number of tetrahedral voids is $2N$.
Step 2: Meaning Let the number of anions (B) be $N$. Then number of tetrahedral voids is $2N$.
Step 3: Analysis Cations (A) occupy $1/3$ of tetrahedral voids.
Number of A = $\frac{1}{3} \times 2N = \frac{2N}{3}$.
Ratio A : B = $\frac{2N}{3} : N = 2 : 3$.
Step 4: Conclusion The chemical formula is $\text{A}_2\text{B}_3$. Final Answer: (B)
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