A compound is formed by elements X, Y, and Z. | vedclass

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(A) In an $FCC$ lattice, the number of atoms per unit cell is $4$. Since $Z$ forms the $FCC$ lattice, the number of $Z$ atoms $= 4$. The number of oct

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$X_3Y_2Z_3$

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(A) In an $FCC$ lattice, the number of atoms per unit cell is $4$. Since $Z$ forms the $FCC$ lattice, the number of $Z$ atoms $= 4$. The number of octahedral voids is equal to the number of atoms in the lattice, so the number of $X$ atoms $= 4$. The number of tetrahedral voids is twice the number of atoms in the lattice, so the number of tetrahedral voids $= 2 	imes 4 = 8$. Given that $Y$ occupies $\frac{1}{3}$ of the tetrahedral voids, the number of $Y$ atoms $= \frac{1}{3} 	imes 8 = \frac{8}{3}$. The ratio of atoms $X:Y:Z = 4 : \frac{8}{3} : 4$. Multiplying by $3$ to get the simplest whole number ratio: $12 : 8 : 12$, which simplifies to $3 : 2 : 3$. Therefore, the formula of the compound is $X_3Y_2Z_3$.

$A$ 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:

Similar Questions

$A$ compound is formed by atoms of $A$,$B$ and $C$. Atoms of $C$ form $hcp$ lattice. Atoms of $A$ occupy $50\%$ of octahedral voids and atoms of $B$ occupy $\frac{2}{3}$ of tetrahedral voids. What is the molecular formula of the solid?

The molecular formula of a compound is $AB_2O_4$. Atoms of $O$ form a $ccp$ lattice. Atoms of $A$ (cation) occupy $\frac{1}{8}$ of the tetrahedral voids. Atoms of $B$ (cation) occupy a fraction of the octahedral voids. What is the fraction of vacant octahedral voids?

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

In a face-centred cubic $(FCC)$ unit cell,what is the total volume occupied by the atoms?

In a metal oxide,$O^{2-}$ ions are arranged in $hcp$ lattice while metal ions occupy $2/3$ of the octahedral voids. The simplest formula of the oxide is:

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