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In an ionic compound the anion $(N^-)$ from cubic close type of packing,while the cation $(M^+)$ ions occupy one third of the tetrahedral voids.Deduce the empirical formula of the compound and the coordination number of $(M^+)$ ions.

$\begin{array}{1 1}M_2N_3,4\\M_3N_3,3\\MN_3,2\\M_2N_2,5\end{array} $

1 Answer

Number of N ion in each fcc unit cell = 4
Number of tetrahedral voids =$4\times 2=8$
Fraction occupied by $(M^+)$ ions =$8\times \large\frac{1}{3}=\frac{8}{3}$
Ratio of N : M = 4 : $\large\frac{8}{3}$
$\Rightarrow 4\times \large\frac{3}{8}$
$N : M = 3 : 2= M_2N_3$
Coordination number of $M^+$ ion = 4
answered Aug 4, 2014 by sreemathi.v
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