A solid has CCP arrangement having atoms A, B | vedclass

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Question

(A) In a $CCP$ (cubic close packing) arrangement:$1$. Number of atoms at corners $(B)$ = $8 	imes (1/8) = 1$.$2$. Number of atoms at face centres $(A

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$A_3 B C_4$

Vedclass · Answer

(A) In a $CCP$ (cubic close packing) arrangement:$1$. Number of atoms at corners $(B)$ = $8 	imes (1/8) = 1$.$2$. Number of atoms at face centres $(A)$ = $6 	imes (1/2) = 3$.$3$. Total number of tetrahedral voids = $2 	imes (	ext{number of atoms in } CCP) = 2 	imes 4 = 8$.$4$. Number of $C$ atoms occupying $50\%$ of tetrahedral voids = $0.50 	imes 8 = 4$.Therefore,the ratio of atoms is $A:B:C = 3:1:4$.The molecular formula is $A_3 B C_4$.

$A$ solid has $CCP$ arrangement having atoms $A, B, C$. If $A$ atoms are present at face centres,$B$ at corners and $C$ atoms occupy $50\%$ tetrahedral voids,then the molecular formula of the solid will be:

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