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

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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 $X_4Y_3$

Hence answer is (a)

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