When an atom or an ion is missing from its normal lattice site, a lattice vacancy (Schottky detect) is created In stoichiometric ionic crystals, a vacancy of one ion has to be accompanied by the vacancy of the oppositely charged ion in order to maintain electrical neutrality.
In a Frenkel defect an ion leaves its position in the lattice and occupies an interstitial void.This is the Frenkel defect commonly found along with the Schottky defects and interstitials.In pure alkali halides, Frendel defect are not found since the ions cannot get into the intenstitial sites.Frenkel defects are found in silver halides because of the small size of the `Ag^+` ion.Unlike Schottky defects. Frenkel defects do not change the density of the solids.In certain ionic solids (e.g. AgBr) both Schottky and Frenkel defects occur.
The defects discussed above do not disturb the stoichiometry of the crystalline meterial.There is large Such solids showing deviations from the ideal stoichiometric composition form an important group of solids For example in the vanadium oxide, `VO_x`, x can be anywhere between 0.6 and 1.3.There are solids which are difficult to prepare in the stoichiometric composition.Thus, the ideal composition in compounds such as FeO is difficult to obtain (normally we get a composition of `Fe_(0.85) O` but it may range from `Fe_(0.93)` O to `Fe_(0.96)O`).Non-stoichiometric behaviour is most commonly found for transition metal compounds through is also known for some lanthanoids and actinodes.
Zinc oxide loses oxygen reversibly at high temperature and turns yellow in colour.The excess metal is accomdated interstitially, giving rise to electrons trapped in the neighbourhood.the enhanced electrical conductivity of the non-stoichiometric ZnO arises from these electrons .
Anion vacancies in alkali halides are produced by heating the alkali halide crystals in an atmosphere of the alkali metal vapour.When the metal atoms deposite on the surface they diffuse into the crystal and after ionisation the alkali metal ion occupies cationic vacancy whereas electron occupies anionic vacancy.Electrons trapped in anion vacancies are referred to as F-centres (from Farbe the German word for colour) that gives rise to interesting colour in alkali halides.Thus, the excess of potassium in KCl makes the crystal appear violet and the excess of lithium in LiCl makes it pink.
In the crystal of `Fe_(0.93)O`, the percentage of Fe (II) will be

Options

A. `15%`
B. `85%`
C. `30%`
D. `78%`

Explanation

To find the percentage of Fe²⁺ ions in the crystal of Fe₀.93O, we can follow these steps:

### Step 1: Understand the Composition
The formula Fe₀.93O indicates that there are 0.93 moles of Fe for every mole of O. In an ideal FeO compound, the ratio of Fe to O is 1:1, meaning that for every Fe²⁺ ion, there is one O²⁻ ion.

### Step 2: Calculate Missing Fe²⁺ Ions
Since the formula is Fe₀.93O, we can determine how many Fe²⁺ ions are missing:
- Ideal FeO would have 1 mole of Fe²⁺ for 1 mole of O.
- In Fe₀.93O, the number of Fe²⁺ ions is 0.93.
- Therefore, the number of missing Fe²⁺ ions is:
\[ 1 - 0.93 = 0.07 \text{ moles of Fe²⁺} \]

### Step 3: Determine Charge Balance
Each Fe²⁺ ion has a charge of +2. The total positive charge missing due to the absence of 0.07 moles of Fe²⁺ is:
\[ 0.07 \text{ moles} \times 2 \text{ (charge of Fe²⁺)} = 0.14 \text{ positive charge} \]

### Step 4: Compensating Charge with Fe³⁺ Ions
To maintain electrical neutrality, the missing positive charge of +0.14 must be compensated by Fe³⁺ ions. Each Fe³⁺ ion has a charge of +3. The number of moles of Fe³⁺ ions required to balance the charge is:
\[ \text{Let } x \text{ be the moles of Fe³⁺. Then, } 3x = 0.14 \Rightarrow x = \frac{0.14}{3} \approx 0.04667 \text{ moles of Fe³⁺} \]

### Step 5: Calculate the Total Moles of Iron
The total moles of iron (Fe) in the crystal is the sum of Fe²⁺ and Fe³⁺ ions:
\[ \text{Total moles of Fe} = \text{moles of Fe²⁺} + \text{moles of Fe³⁺ = } 0.93 + 0.04667 \approx 0.97667 \]

### Step 6: Calculate the Percentage of Fe²⁺
Now, we can calculate the percentage of Fe²⁺ ions in the crystal:
\[ \text{Percentage of Fe²⁺} = \left(\frac{0.93}{0.97667}\right) \times 100 \approx 95.3\% \]

### Step 7: Final Calculation for Fe²⁺ Percentage
Since we need the percentage of Fe²⁺ ions, we can also calculate it directly from the total amount:
\[ \text{Percentage of Fe²⁺} = 100\% - \text{Percentage of Fe³⁺} \]
Where the percentage of Fe³⁺ can be calculated as:
\[ \text{Percentage of Fe³⁺} = \left(\frac{0.04667}{0.97667}\right) \times 100 \approx 4.8\% \]
Thus, the percentage of Fe²⁺ would be:
\[ 100\% - 4.8\% \approx 95.2\% \]

### Conclusion
The percentage of Fe²⁺ in the crystal of Fe₀.93O is approximately **95.2%**.

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