The total number of voids (octahedral + tetra | vedclass

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(A) In a hexagonal close-packed $(HCP)$ structure, the number of atoms per unit cell $(N)$ is $6$. For $1$ mole of the compound, the number of atoms i

Vedclass · Accepted Answer

$1.8 	imes 10^{24}$

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(A) In a hexagonal close-packed $(HCP)$ structure, the number of atoms per unit cell $(N)$ is $6$. For $1$ mole of the compound, the number of atoms is $N_A = 6.022 	imes 10^{23}$. In any close-packed structure, the number of octahedral voids is equal to the number of atoms $(N)$, and the number of tetrahedral voids is equal to $2N$. Total number of voids = (Number of octahedral voids) + (Number of tetrahedral voids) = $N + 2N = 3N$. For $1$ mole, total voids = $3 	imes N_A = 3 	imes 6.022 	imes 10^{23} = 1.8066 	imes 10^{24}$. Rounding to two significant figures, we get $1.8 	imes 10^{24}$.

The total number of voids (octahedral + tetrahedral) in $1$ mole of a compound forming a hexagonal close-packed $(HCP)$ structure is:

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