In a face-centered cubic (FCC) arrangement,at | vedclass

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(B) In an $FCC$ unit cell,there are $8$ corners and $6$ face centers. Number of atoms of $A$ initially at corners $= 8 	imes \frac{1}{8} = 1$. After

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$AB_6$

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(B) In an $FCC$ unit cell,there are $8$ corners and $6$ face centers. Number of atoms of $A$ initially at corners $= 8 	imes \frac{1}{8} = 1$. After removing $4$ atoms from corners,remaining atoms of $A = (8 - 4) 	imes \frac{1}{8} = 4 	imes \frac{1}{8} = \frac{1}{2}$. Number of atoms of $B$ at face centers $= 6 	imes \frac{1}{2} = 3$. The ratio of $A:B = \frac{1}{2} : 3 = 1 : 6$. Therefore,the empirical formula is $AB_6$.

In a face-centered cubic $(FCC)$ arrangement,atoms $A$ are present at the corners and atoms $B$ are present at the face centers. If $4$ atoms of $A$ are removed from the corners,what is the empirical formula of the compound?

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