The total number of voids (octahedral + tetra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1.8 	imes 10^{24}$

Vedclass · Answer

(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:

Similar Questions

Which of the following lattices has the highest packing efficiency: $(i)$ simple cubic,$(ii)$ body-centred cubic,and $(iii)$ hexagonal close-packed lattice?

In a close-packed lattice,how does the size of an octahedral void compare to that of a tetrahedral void?

$A$ compound forms a hexagonal close-packed $(HCP)$ structure. How many octahedral voids are present in $0.3 \, mol$ of it?

What is the number of octahedral and tetrahedral voids present respectively in $0.25 \ mol$ of a substance having $hcp$ structure?

In a $ccp$ structure of a mixed oxide,$1/8$ of the tetrahedral voids are occupied by divalent $X^{+2}$ ions and $50\%$ of the octahedral voids are occupied by trivalent $Y^{+3}$ ions. The formula of the oxide is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module