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A solid has three types of atoms $$X,\ Y$$ and $$Z$$. $$X$$ forms an $$FCC$$ lattice with $$Y$$ atoms occupying all the tetrahedral voids and $$Z$$ atoms occupying half of the octahedral voids. The formula of the solid is


A
X2Y4Z
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B
XY2Z4
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C
X4Y2Z
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D
X4Y2Z3
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Solution

The correct option is A $$X_2Y_4Z$$
Given that $$X$$ forms an fcc lattice. Thus, the number of $$X$$ atoms $$ = 4$$ (number of atoms per unit cell).
Now, since $$Y$$ occupies all the tetrahedral voids, thus, the number of atoms of $$Y = 4\times 2 = 8$$ (number of tetrahedral voids $$=$$ twice the number of octahedral voids).
Finally, we have $$Z$$ atoms occupying half of the number of octahedral voids. Hence, $$Z = \dfrac 12 \times 4 = 2$$
Thus, the formula of the solid is $$X_4Y_8Z_2$$ or $$X_2Y_4Z$$.

Chemistry

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