Question:
Atoms of element B form hcp lattice and atoms of element A occupy \(\frac{2}{3}\) of tetrahedral voids. The formula of the compound formed by the elements A and B is
Atoms of element B form hcp lattice and atoms of element A occupy \(\frac{2}{3}\) of tetrahedral voids. The formula of the compound formed by the elements A and B is
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For close packed structures:
\[
\text{Tetrahedral voids}=2N
\]
\[
\text{Octahedral voids}=N
\]
These formulas solve almost every CUET Solid State void problem.
Updated On: Jun 17, 2026
- \(A_3B_4\)
- \(A_4B_3\)
- \(AB_4\)
- \(A_4B\)
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The Correct Option is B
Solution and Explanation
Concept:
Questions involving tetrahedral and octahedral voids are among the most important numerical concepts from the Solid State chapter.
The key facts are:
\[
\text{Number of tetrahedral voids}
=
2N
\]
\[
\text{Number of octahedral voids}
=
N
\]
where \(N\) is the number of close-packed particles.
Step 1: Assume number of B atoms. Let the hcp lattice contain: \[ N \] atoms of B. Thus: \[ B=N \]
Step 2: Calculate tetrahedral voids. Total tetrahedral voids: \[ 2N \]
Step 3: Determine number of A atoms. Given: \[ \frac{2}{3} \] of tetrahedral voids are occupied. Therefore: \[ A = \frac{2}{3}(2N) = \frac{4N}{3} \]
Step 4: Determine ratio. \[ A:B = \frac{4N}{3}:N \] Dividing by \(N\): \[ = \frac{4}{3}:1 \] Multiplying throughout by 3: \[ = 4:3 \] Thus: \[ \boxed{A_4B_3} \]
Step 5: Final conclusion. \[ \boxed{\text{Option (B)}} \]
Step 1: Assume number of B atoms. Let the hcp lattice contain: \[ N \] atoms of B. Thus: \[ B=N \]
Step 2: Calculate tetrahedral voids. Total tetrahedral voids: \[ 2N \]
Step 3: Determine number of A atoms. Given: \[ \frac{2}{3} \] of tetrahedral voids are occupied. Therefore: \[ A = \frac{2}{3}(2N) = \frac{4N}{3} \]
Step 4: Determine ratio. \[ A:B = \frac{4N}{3}:N \] Dividing by \(N\): \[ = \frac{4}{3}:1 \] Multiplying throughout by 3: \[ = 4:3 \] Thus: \[ \boxed{A_4B_3} \]
Step 5: Final conclusion. \[ \boxed{\text{Option (B)}} \]
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