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In a compound, nitrogen atoms $(N)$ make cubic close packed lattice and metal atoms $(M)$ occupy one-third of the tetrahedral voids present. Determine the formula of the compound formed by $M$ and $N$ ?
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Let the number of $N$ atoms in ecp is $x$.
$\therefore$ Number of tetrahedral voids $=2 x$.
As $1 / 3$ rd of tetrahedral voids are occupied by atoms of metal $M$, therefore number of atoms of $M$ present $=\frac{1}{3} \times 2 x$
$\frac{\text { Number of } N \text { atoms }}{\text { Number of } M \text { atoms }}=\frac{3 x}{2 x}=\frac{3}{2}$
So, the formula of the compound is $M_2 N_3$.
$\therefore$ Number of tetrahedral voids $=2 x$.
As $1 / 3$ rd of tetrahedral voids are occupied by atoms of metal $M$, therefore number of atoms of $M$ present $=\frac{1}{3} \times 2 x$
$\frac{\text { Number of } N \text { atoms }}{\text { Number of } M \text { atoms }}=\frac{3 x}{2 x}=\frac{3}{2}$
So, the formula of the compound is $M_2 N_3$.
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