﻿ Analysis shows that a metal oxide has the empirical formula M0.98 O1.00. Calculate the percentage of M2+ and M3+ ions in this crystal.OrIn 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. from Chemistry The Solid State Class 12 Nagaland Board

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Analysis shows that a metal oxide has the empirical formula M0.98 O1.00. 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:

M0.98 O1.00 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 This means that x M2+ have been replaced by M3+ ions.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

∴

Thus, M2+ is 92% and M3+ is 8% in given sample having formula, M0.96 O1.00.

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   So, the formula of the compound is M2N3.

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