In a crystal lattice,C anions form a cubic cl | vedclass

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Question

(A) Let the number of $C$ anions in the $CCP$ structure be $N = 4$.Number of octahedral voids = $N = 4$.Number of tetrahedral voids = $2N = 8$.Cations

Vedclass · Accepted Answer

$A_2BC_2$

Vedclass · Answer

(A) Let the number of $C$ anions in the $CCP$ structure be $N = 4$.Number of octahedral voids = $N = 4$.Number of tetrahedral voids = $2N = 8$.Cations $(A)$ occupy $50\%$ of tetrahedral voids: $A = 0.50 	imes 8 = 4$.Cations $(B)$ occupy $50\%$ of octahedral voids: $B = 0.50 	imes 4 = 2$.Ratio $A:B:C = 4:2:4 = 2:1:2$.Therefore,the molecular formula is $A_2BC_2$.

In a crystal lattice,$C$ anions form a cubic close-packed $(CCP)$ structure. Cations $(A)$ occupy $50\%$ of the tetrahedral voids and cations $(B)$ occupy $50\%$ of the octahedral voids. Determine the molecular formula of the compound.

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