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(C) In a face-centered cubic $(FCC)$ lattice,there are $8$ corners and $6$ face centers. Number of $X$ atoms at corners = $8 - 2 = 6$. Contribution of

Vedclass · Accepted Answer

$XY_4$

Vedclass · Answer

(C) In a face-centered cubic $(FCC)$ lattice,there are $8$ corners and $6$ face centers. Number of $X$ atoms at corners = $8 - 2 = 6$. Contribution of each corner atom = $\frac{1}{8}$. Total $X$ atoms = $6 	imes \frac{1}{8} = \frac{6}{8} = \frac{3}{4}$. Number of $Y$ atoms at face centers = $6$. Contribution of each face-centered atom = $\frac{1}{2}$. Total $Y$ atoms = $6 	imes \frac{1}{2} = 3$. Ratio of $X:Y = \frac{3}{4} : 3 = 3 : 12 = 1 : 4$. Therefore,the formula of the compound is $XY_4$.

In a face-centered cubic lattice of $X$ and $Y$ atoms,$X$ atoms are present at the corners while $Y$ atoms are at the face centers. If two atoms of $X$ are missing from the corners,what is the formula of the compound?

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