A compound is formed by two elements M and N. | vedclass

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(B) In a hexagonal closed pack $(HCP)$ structure,the number of atoms per unit cell is $6$. Since the number of octahedral voids is equal to the number

Vedclass · Accepted Answer

$M_2 N_3$

Vedclass · Answer

(B) In a hexagonal closed pack $(HCP)$ structure,the number of atoms per unit cell is $6$. Since the number of octahedral voids is equal to the number of atoms in the close-packed structure,the number of octahedral voids is $6$. Given that $M$ occupies $2/3$ of the octahedral voids,the number of $M$ atoms per unit cell is $\frac{2}{3} 	imes 6 = 4$. The ratio of $M$ atoms to $N$ atoms is $4:6$,which simplifies to $2:3$. Therefore,the formula of the compound is $M_2 N_3$.

$A$ compound is formed by two elements $M$ and $N$. Element $N$ forms a hexagonal closed pack $(HCP)$ array with $2/3$ of the octahedral holes occupied by $M$. The formula of the compound is $....$

Similar Questions

In the closest packing of atoms,

Calculate the total volume occupied by all particles in an $fcc$ unit cell if the volume of the unit cell is $6.4 \times 10^{-23} \ cm^3$.

$A$ compound is formed by two elements $X$ and $Y$. Atoms of the element $Y$ (anions) form an $hcp$ structure and atoms of the element $X$ (cations) occupy all the tetrahedral voids. Determine the formula of the compound.

If the anion $(A)$ forms a hexagonal closest packing and the cation $(C)$ occupies only $2/3$ of the octahedral voids in it,then the general formula of the compound is :-

In a close-packed lattice,the number of tetrahedral voids compared to the number of octahedral voids is:

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