Iron (II) oxide has a cubic structure and each unit cell has side 5 Å . If the density of the oxide is 4 g ` cm^(-3)` Calculate the number of ` Fe^(2+) and O^(2+)` ions presnent in each unit cell ( Molar mass of FeO = 72 ` " g mol"^(-1)`
`N_(A) = 6.02 xx 10^(23) "mol"^(-1)` )

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- The molar mass of Iron (II) oxide (FeO) = 72 g/mol
- Density of FeO = 4 g/cm³
- Side length of the unit cell (a) = 5 Å = 5 x 10⁻¹⁰ m

### Step 2: Convert the side length to cm
Since the density is given in g/cm³, we need to convert the side length from Ångstroms to centimeters:
\[ a = 5 \, \text{Å} = 5 \times 10^{-10} \, \text{m} = 5 \times 10^{-8} \, \text{cm} \]

### Step 3: 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:
\[ V = (5 \times 10^{-8} \, \text{cm})^3 = 1.25 \times 10^{-22} \, \text{cm}^3 \]

### Step 4: Use the density formula to find Z (number of formula units per unit cell)
The formula for density (D) is given by:
\[ D = \frac{Z \times \text{Molar mass}}{N_A \times V} \]
Where:
- \(D\) = density (4 g/cm³)
- \(Z\) = number of formula units per unit cell (what we need to find)
- Molar mass = 72 g/mol
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)
- \(V\) = volume of the unit cell

Rearranging the formula to solve for Z:
\[ Z = \frac{D \times N_A \times V}{\text{Molar mass}} \]

### Step 5: Substitute the values into the equation
Substituting the known values:
\[ Z = \frac{4 \, \text{g/cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1} \times 1.25 \times 10^{-22} \, \text{cm}^3}{72 \, \text{g/mol}} \]

### Step 6: Calculate Z
Calculating the numerator:
\[ 4 \times 6.022 \times 10^{23} \times 1.25 \times 10^{-22} = 3.0075 \times 10^{2} \]
Now, dividing by the molar mass:
\[ Z = \frac{3.0075 \times 10^{2}}{72} \approx 4.17 \approx 4 \]

### Step 7: Interpret the value of Z
Since \(Z = 4\), this indicates that there are 4 formula units of FeO in each unit cell.

### Step 8: Determine the number of ions in the unit cell
In the case of Iron (II) oxide (FeO):
- Each unit cell contains 4 Fe²⁺ ions and 4 O²⁻ ions.

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

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