Ferrous oxide has cubes structure and each edge of the unit cell is `5.0 Å` .Assuming of the oxide as `4.0g//cm^(3)` then the number of `Fe^(2+) and O^(2)` inos present in each unit cell will be

Options

A. two `Fe^(2+)` and four `O^(2-)`
B. three `Fe^(2+)` and three `O^(2-)`
C. four `Fe^(2+)` and two `O^(2-)`
D. four `Fe^(2+)` and four `O^(2-)`

Explanation

To solve the problem, we need to determine the number of Fe²⁺ and O²⁻ ions present in each unit cell of ferrous oxide (FeO) given the unit cell edge length and the density of the compound.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Edge length of the unit cell (a) = 5.0 Å = \(5.0 \times 10^{-8}\) cm
- Density (D) = 4.0 g/cm³
- Molar mass of FeO = Molar mass of Fe (56 g/mol) + Molar mass of O (16 g/mol) = 72 g/mol

2. **Calculate the Volume of the Unit Cell:**
\[ \text{Volume} = a^3 = (5.0 \times 10^{-8} \text{ cm})^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

3. **Use the Density Formula:**
The density formula relates the density (D), the number of formula units per unit cell (Z), the molar mass (M), and the volume of the unit cell (V):
\[ D = \frac{Z \cdot M}{V \cdot N_A} \]
where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \text{ mol}^{-1}\)).

4. **Rearranging the Formula to Solve for Z:**
\[ Z = \frac{D \cdot V \cdot N_A}{M} \]

5. **Substituting the Values:**
\[ Z = \frac{4.0 \text{ g/cm}^3 \cdot 1.25 \times 10^{-22} \text{ cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1}}{72 \text{ g/mol}} \]

6. **Calculating Z:**
\[ Z = \frac{4.0 \cdot 1.25 \cdot 6.022}{72} = \frac{30.1}{72} \approx 4.0 \]

7. **Conclusion:**
Since Z = 4, this means there are 4 formula units of FeO in each unit cell. Therefore, the number of Fe²⁺ ions is 4 and the number of O²⁻ ions is also 4.

### Final Answer:
- Number of Fe²⁺ ions = 4
- Number of O²⁻ ions = 4