A compound is formed by atoms of A,B and C. A | vedclass

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(A) Let the number of atoms of $C$ in the $hcp$ lattice be $n = 4$. Number of octahedral voids $= n = 4$. Number of tetrahedral voids $= 2n = 8$. Atom

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$A_3 B_8 C_6$

Vedclass · Answer

(A) Let the number of atoms of $C$ in the $hcp$ lattice be $n = 4$. Number of octahedral voids $= n = 4$. Number of tetrahedral voids $= 2n = 8$. Atoms of $A$ occupy $50\%$ of octahedral voids,so $A = 4 	imes 0.5 = 2$. Atoms of $B$ occupy $\frac{2}{3}$ of tetrahedral voids,so $B = 8 	imes \frac{2}{3} = \frac{16}{3}$. The ratio of $A:B:C$ is $2 : \frac{16}{3} : 4$. Multiplying by $3$ to get whole numbers,we get $6 : 16 : 12$. Dividing by $2$,the simplest ratio is $3 : 8 : 6$. Thus,the molecular formula is $A_3 B_8 C_6$.

$A$ 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?

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