Ferrous oxide has a cubie structure and edge length of the uint cell is ` 5.0Å` . Assuming the density o ferrous oxide to be `3.84 g//cm^(3)` , the no. Of `Fe^(2+)` and `O^(2-)` ions present in each unit cell be : (use `N_(A)= 6xx 10^(23)`):

Options

A. `4 Fe^(2+) and 4 O^(2-)`
B. `2 Fe^(2+) and 2 O^(2-)`
C. `1 Fe^(2+) and 1 O^(2-)`
D. `3 Fe^(2+) and 4 O^(2-)`

Answer

To solve the problem of finding the number of \( \text{Fe}^{2+} \) and \( \text{O}^{2-} \) ions in the unit cell of ferrous oxide (FeO), we will follow these steps:

### Step 1: Understand the given data
- **Edge length of the unit cell (a)**: \( 5.0 \, \text{Å} = 5.0 \times 10^{-8} \, \text{cm} \)
- **Density (\( \rho \))**: \( 3.84 \, \text{g/cm}^3 \)
- **Avogadro's number (\( N_A \))**: \( 6 \times 10^{23} \, \text{mol}^{-1} \)
- **Molecular weight of FeO**: \( 72 \, \text{g/mol} \)

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
Substituting the value of \( a \):
\[ V = (5.0 \times 10^{-8} \, \text{cm})^3 = 1.25 \times 10^{-22} \, \text{cm}^3 \]

### Step 3: Use the density formula to find the number of formula units per unit cell (Z)
The formula for density is given by:
\[ \rho = \frac{Z \cdot m}{N_A \cdot V} \]
Where:
- \( Z \) = number of formula units per unit cell
- \( m \) = mass of one formula unit (FeO) in grams

First, we calculate the mass of one formula unit:
\[ m = \frac{72 \, \text{g/mol}}{N_A} = \frac{72 \, \text{g/mol}}{6 \times 10^{23} \, \text{mol}^{-1}} = 1.2 \times 10^{-25} \, \text{g} \]

Now, substituting the values into the density formula:
\[ 3.84 = \frac{Z \cdot 1.2 \times 10^{-25}}{6 \times 10^{23} \cdot 1.25 \times 10^{-22}} \]

### Step 4: Solve for Z
Rearranging the equation to solve for \( Z \):
\[ Z = \frac{3.84 \cdot 6 \times 10^{23} \cdot 1.25 \times 10^{-22}}{1.2 \times 10^{-25}} \]

Calculating this gives:
\[ Z = \frac{3.84 \cdot 7.5 \times 10^{1}}{1.2 \times 10^{-25}} = \frac{28.8}{1.2} = 24 \]

### Step 5: Determine the number of ions in the unit cell
Since the formula of ferrous oxide is \( \text{FeO} \), for every formula unit of FeO, there is one \( \text{Fe}^{2+} \) ion and one \( \text{O}^{2-} \) ion. Therefore, if \( Z = 4 \):
- Number of \( \text{Fe}^{2+} \) ions = 4
- Number of \( \text{O}^{2-} \) ions = 4

### Final Answer
The number of \( \text{Fe}^{2+} \) and \( \text{O}^{2-} \) ions present in each unit cell is:
- \( 4 \, \text{Fe}^{2+} \)
- \( 4 \, \text{O}^{2-} \)

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