`FeO` crystallizes in `NaCl`-type of crystal lattice. The crystals of `FeO` are deficient in iron and are always non-stoichiometric. Some cationic sites are vacant and some contain `Fe^(3+)` ions but the combination is such that the structure is elctrically neutral. The formula approximates to `Fe_(0.95)O`.
a. What is the ratio of `Fe^(2+)` to `Fe^(3+)` ions in the solid?
b. What percentage of cation sites are vacant?

Answer

a. `8.5`, b. `5%`

Explanation

The compound is `Fe_(0.95)O -= Fe_(95)O(100)`.
Let the compound has `x%` of `Fe^(2+)` and `(95 - x)%` of `Fe_(3+)` ion.
For electrical neutrality:
Total potitive charge `=` Total negative charge
`2x + 3(95 - x) = 2 xx 100 implies x = 85%`
Thus, `Fe^(2+) = 85%` and `Fe^(3+) = (95 - 85) = 10%`
Hence, `(Fe^(2+))/(Fe^(2+)) = (85)/(10) = 8.5`, b. Number of vacant sites `= 100- 95 = 5%`