In a crystalline solid,anion B are arranged i | vedclass

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(B) In a $ccp$ lattice,the number of effective atoms per unit cell is $4$. Therefore,the number of $B$ atoms = $4$. The number of octahedral voids = $

Vedclass · Accepted Answer

$A_3B_2$

Vedclass · Answer

(B) In a $ccp$ lattice,the number of effective atoms per unit cell is $4$. Therefore,the number of $B$ atoms = $4$. The number of octahedral voids = $4$. The number of tetrahedral voids = $2 	imes 4 = 8$. Cation $A$ occupies $50\%$ of octahedral voids and $50\%$ of tetrahedral voids. Number of $A$ atoms = $(0.50 	imes 4) + (0.50 	imes 8) = 2 + 4 = 6$. The ratio of $A:B = 6:4 = 3:2$. Thus,the formula of the crystalline solid is $A_3B_2$.

In a crystalline solid,anion $B$ are arranged in $ccp$ lattice and cation $A$ occupy $50\%$ of the octahedral voids and $50\%$ of tetrahedral voids. What will be the formula of the crystalline solid?

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