In a close-packed arrangement,A type of atoms | vedclass

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(D) In a $fcc$ unit cell: Number of $A$ atoms (corners) = $8 	imes \frac{1}{8} = 1$. Number of $B$ atoms (face centers) = $6 	imes \frac{1}{2} = 3$.

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(D) In a $fcc$ unit cell: Number of $A$ atoms (corners) = $8 	imes \frac{1}{8} = 1$. Number of $B$ atoms (face centers) = $6 	imes \frac{1}{2} = 3$. Number of $C$ atoms (tetrahedral voids) = $8$. Number of $D$ atoms (octahedral voids) = $4$. Along one body diagonal,there are $2$ corners and $2$ tetrahedral voids. Removing these: $A = 1 - (2 	imes \frac{1}{8}) = 1 - \frac{1}{4} = \frac{3}{4}$. $B = 3$ (no change). $C = 8 - 2 = 6$. $D = 4$ (no change). Formula: $A_{3/4}B_3C_6D_4$. Multiplying by $4$: $A_3B_{12}C_{24}D_{16}$.

In a close-packed arrangement,$A$ type of atoms are at corners,$B$ type of atoms are at face centers,$C$ type of atoms are at tetrahedral voids,and $D$ type of atoms are at octahedral voids. If all the atoms along any one body diagonal are removed,what will be the formula of the compound?

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