Sponsor Area

The Solid State

Question
CBSEENCH12005514

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

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 This means that x M2+ have been replaced by M3+ ions.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

∴     % of M3+ ions  = 0.080.96×100 = 8.33%

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 13rd of these are occupied by the element M. Hence the ratio of number of atoms of M and is 2×13:1  or  2:3.  So, the formula of the compound is M2N3.