A metal M forms a hexagonal close-packed stru | vedclass

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(C) In a hexagonal close-packed $(hcp)$ structure,the number of octahedral voids is equal to the number of atoms $(N)$,and the number of tetrahedral v

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) In a hexagonal close-packed $(hcp)$ 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 per atom $= N + 2N = 3N$.Given amount of metal $= 0.02 \ mol$.Number of atoms $= 0.02 	imes 6.02 	imes 10^{23} = 1.204 	imes 10^{22}$.Total number of voids $= 3 	imes (1.204 	imes 10^{22}) = 3.612 	imes 10^{22} = 36.12 	imes 10^{21}$.The nearest integer is $36$.

$A$ metal $M$ forms a hexagonal close-packed structure. The total number of voids in $0.02 \ mol$ of it is $......... \times 10^{21}$ (Nearest integer) (Given $N_{A} = 6.02 \times 10^{23}$)

Similar Questions

$A$ compound is formed by elements $X$, $Y$, and $Z$. Atoms of $Z$ (anions) make $FCC$ lattice. Atoms of $X$ (cations) occupy all the octahedral voids. Atoms of $Y$ (cations) occupy $\frac{1}{3}$ rd of the tetrahedral voids. The formula of the compound is:

How many tetrahedral voids are present in $0.4 \ mol$ of a compound that forms an $hcp$ structure?

What is the packing efficiency of $fcc$ crystal structure (in $\%$)?

Give two examples of metals that crystallize in the $hcp$ (hexagonal close-packed) structure.

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