A solid contains elements A and B. Anions of | vedclass

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(B) Let the number of atoms of $B$ in the $ccp$ lattice be $n$. Since $B$ forms a $ccp$ lattice, the number of octahedral voids is $n$ and the number

Vedclass · Accepted Answer

$A_3B_2$

Vedclass · Answer

(B) Let the number of atoms of $B$ in the $ccp$ lattice be $n$. Since $B$ forms a $ccp$ lattice, the number of octahedral voids is $n$ and the number of tetrahedral voids is $2n$. Cations of $A$ occupy $50 \%$ of octahedral voids, so number of $A$ atoms in octahedral voids $= 0.5 	imes n = 0.5n$. Cations of $A$ occupy $50 \%$ of tetrahedral voids, so number of $A$ atoms in tetrahedral voids $= 0.5 	imes 2n = n$. Total number of $A$ atoms $= 0.5n + n = 1.5n$. The ratio of $A:B = 1.5n : n = 1.5 : 1 = 3 : 2$. Therefore, the molecular formula of the solid is $A_3B_2$.

$A$ solid contains elements $A$ and $B$. Anions of $B$ form $ccp$ lattice. Cations of $A$ occupy $50 \%$ of octahedral voids and $50 \%$ of tetrahedral voids. What is the molecular formula of the solid?

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