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In a compound atoms of element Y form ccp lattice and those of element X occupy ${\large\frac{2}{3}}^{rd}$ of tetrahedral voids. The formula of the compound will be

$(a)\;X_4Y_3\qquad(b)\;X_2Y_3\qquad(c)\;X_2Y\qquad(d)\;X_3Y_4$

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• Let, the number of close packed spheres be N,
then:
The number of octahedral voids generated = N
The number of tetrahedral voids generated = 2N
Number of atoms of Y = N = 4
Number of atoms of X = $\large\frac{2}{3}\times 2N = \large\frac{2}{3}\times 8= \large\frac{16}{3}$
$\therefore$ Formula of the compound will be $X_4Y_3$
edited Apr 2, 2014