Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App