A crystalline solid is made of X,Y and Z elem | vedclass

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(A) In an $FCC$ unit cell:Number of $X$ atoms (at corners and face-centres) $= 8 	imes \frac{1}{8} + 6 	imes \frac{1}{2} = 4$.Number of $Y$ atoms (i

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$X_5Y_4Z_8$

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(A) In an $FCC$ unit cell:Number of $X$ atoms (at corners and face-centres) $= 8 	imes \frac{1}{8} + 6 	imes \frac{1}{2} = 4$.Number of $Y$ atoms (in octahedral voids) $= 4$.Number of $Z$ atoms (in tetrahedral voids) $= 8$.Along one body diagonal,there are two corner $X$ atoms,one octahedral void $(Y)$ at the body centre,and two tetrahedral voids $(Z)$.Removing these atoms:Number of $X$ atoms $= 4 - (2 	imes \frac{1}{8}) = 4 - 0.25 = 3.75 = \frac{15}{4}$.Number of $Y$ atoms $= 4 - 1 = 3$.Number of $Z$ atoms $= 8 - 2 = 6$.The ratio $X : Y : Z = \frac{15}{4} : 3 : 6$.Multiplying by $4$,we get $15 : 12 : 24$.Dividing by $3$,we get $5 : 4 : 8$.Therefore,the simplest formula is $X_5Y_4Z_8$.

$A$ crystalline solid is made of $X$,$Y$ and $Z$ elements. Atoms of $X$ form $FCC$ packing; atoms of $Y$ occupy octahedral voids while atoms of $Z$ occupy tetrahedral voids. What will be the simplest formula of the solid if atoms along one body diagonal are removed?

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