Ferrous oxide (FeO) is converted into a non-stoichiometric substance $F e_{0.94} O_{1.00}$ .
If the total percentage of Fe²⁺ ions oxidized and missing from the lattice is equal to $x %$ , the value of "x" is:

Question

Ferrous oxide (FeO) is converted into a non-stoichiometric substance $F e_{0.94} O_{1.00}$ .

If the total percentage of Fe²⁺ ions oxidized and missing from the lattice is equal to $x %$ , the value of "x" is:

TestHub · Accepted Answer

Iron(II) oxide becomes Fe₀.₉₄O₁.₀₀, so Fe deficiency = 0.06 per O.

Let:

Fe²⁺ missing = $x$

Fe²⁺ oxidized = $y$

So,

$x + y = 0.06$

Charge neutrality in metal deficiency defect:

Each Fe³⁺ formed requires 2 Fe²⁺ oxidized per 1 vacancy relation, giving:

$y = 2 x$

Solve:

$x + 2 x = 0.06 \Rightarrow 3 x = 0.06 \Rightarrow x = 0.02$

$y = 0.04$

Total Fe²⁺ affected:

$x + y = 0.06$ (per O) = 6%

But in metal deficiency defect model, total Fe²⁺ involved in defect formation (counting equivalent Fe²⁺ participation in oxidation + vacancy compensation) corresponds to:

$3 \times 0.06 = 0.18 = 18%$

Chemistry - Solid State Question with Solution | TestHub

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