Question

Calculate the total number of tetrahedral and octahedral voids formed in 0.6 mol of a compound if it forms hcp structure.

• A. $3.613 \times 10^{24}$
• B. $7.226 \times 10^{24}$
• C. $1.084 \times 10^{24}$
• D. $2.913 \times 10^{24}$

Hint:

Total voids = 3 $\times$ Total particles. Octahedral = 1 particle; Tetrahedral = 2 per particle.

Answer 1

The Correct Option is C

Step 1: Concept
In a close-packed structure (hcp or ccp), if there are $N$ atoms, the number of octahedral voids is $N$ and the number of tetrahedral voids is $2N$. Total voids = $3N$.

Step 2: Meaning
$N$ is the total number of particles (atoms/molecules) in the given moles.

Step 3: Analysis
Number of particles $N = \text{moles} \times N_A = 0.6 \times 6.022 \times 10^{23} = 3.6132 \times 10^{23}$. Total voids = $3 \times N = 3 \times 3.6132 \times 10^{23} = 10.8396 \times 10^{23} = 1.084 \times 10^{24}$.

Step 4: Conclusion
The total number of voids is $1.084 \times 10^{24}$. Final Answer: (C)