166. Analysis shows that a metal oxide has the empirical formula M_0.98 O_1.00. Calculate the percentage of M²⁺ and M³⁺ ions in this crystal.

Or

In an ionic compound the anion (N) form cubic close type of packing. While the cation (M⁺) ions occupy one third of the tetrahedal voids. Deduce the empirical formula of the compound and the coordination number of (M) ions.

Solution:

M_0.98 O_1.00 is non-stoichiometric compound and is a mixture of M²⁺ and M³⁺ ions.
Let x atoms of M³⁺ are present in the compound. This means that x M²⁺ have been replaced by M³⁺ ions.
∴  Number of M²⁺ ions = 0.96 – x.
For electrical neutrality, positive charge on compound

= negative charge on compound.
∴ 2(0.96 – x) + 3x = 2
1.92 – 2x + 3x = 2
or x = 2 – 1.92 = 0.08

∴     $\% \mathrm{of} {\mathrm{M}}^{3+} \mathrm{ions} = \frac{0.08}{0.96}\times 100 = 8.33\%$

Thus, M²⁺ is 92% and M³⁺ is 8% in given sample having formula, M_0.96 O_1.00.

Or

The number of tetrahedral voids formed is equal

to twice the number of atoms of element N and only $\frac{1}{3}\mathrm{rd}$ of these are occupied by the element M. Hence the ratio of number of atoms of M and is $2\times \left(\frac{1}{3}\right):1 \mathrm{or} 2:3.$  So, the formula of the compound is M₂N₃.