In a A_x B_y crystal structure,A^{+y} ions oc | vedclass

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Question

(A) In a $BCC$ unit cell,the number of $B^{-x}$ ions per unit cell is $2$. The number of tetrahedral voids in a crystal is twice the number of atoms p

Vedclass · Accepted Answer

$A_4 B_2$

Vedclass · Answer

(A) In a $BCC$ unit cell,the number of $B^{-x}$ ions per unit cell is $2$. The number of tetrahedral voids in a crystal is twice the number of atoms per unit cell. Since $B^{-x}$ ions form the $BCC$ lattice,the number of tetrahedral voids $= 2 	imes 2 = 4$. $A^{+y}$ ions occupy all the tetrahedral voids,so the number of $A^{+y}$ ions $= 4$. The ratio of $A:B = 4:2$,which simplifies to $2:1$. Therefore,the formula of the compound is $A_2 B$.

In a $A_x B_y$ crystal structure,$A^{+y}$ ions occupy all the tetrahedral voids and $B^{-x}$ ions form a $BCC$ unit cell. What is the formula of the compound?

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