`CdO` has `NaCl` like structure with density `8.27g//c c`. If the ionic radius of `O^(2-)` is `1.24A^(@)` determine the ionic radius of `Cd^(2+)`:

Options

A. `1.5 Å`
B. `1.1 Å`
C. `1.9 Å`
D. `1.5 Å`

Answer

To determine the ionic radius of \( \text{Cd}^{2+} \) in cadmium oxide \( \text{CdO} \) with a NaCl-like structure, we can follow these steps:

### Step 1: Understand the Structure
Cadmium oxide has a NaCl-like structure, which means it has a face-centered cubic (FCC) arrangement. In this structure:
- The oxide ions \( \text{O}^{2-} \) occupy the FCC lattice points.
- The cadmium ions \( \text{Cd}^{2+} \) occupy the octahedral sites.

### Step 2: Identify the Number of Atoms in the Unit Cell
In the FCC structure:
- There are 4 formula units of \( \text{CdO} \) per unit cell (4 \( \text{O}^{2-} \) and 4 \( \text{Cd}^{2+} \)).

### Step 3: Use the Density Formula
The density \( \rho \) of a crystalline solid is given by the formula:
\[ \rho = \frac{Z \cdot M}{N_A \cdot A^3} \]
Where:
- \( Z \) = number of formula units per unit cell = 4
- \( M \) = molar mass of \( \text{CdO} \) = 128 g/mol
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \) mol\(^{-1}\)
- \( A \) = edge length of the unit cell (in cm)

### Step 4: Rearrange the Density Formula to Find Edge Length
Rearranging the formula to solve for \( A^3 \):
\[ A^3 = \frac{Z \cdot M}{\rho \cdot N_A} \]
Substituting the known values:
\[ A^3 = \frac{4 \cdot 128 \, \text{g/mol}}{8.27 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 5: Calculate the Edge Length
Calculating \( A^3 \):
\[ A^3 = \frac{512}{8.27 \cdot 6.022 \times 10^{23}} \approx 1.032 \times 10^{-22} \, \text{cm}^3 \]
Taking the cube root to find \( A \):
\[ A \approx (1.032 \times 10^{-22})^{1/3} \approx 4.68 \times 10^{-8} \, \text{cm} = 4.68 \, \text{Å} \]

### Step 6: Relate the Edge Length to Ionic Radii
In the NaCl structure, the relationship between the edge length \( A \) and the ionic radii is given by:
\[ A = 2R_{O^{2-}} + 2R_{Cd^{2+}} \]
Where:
- \( R_{O^{2-}} = 1.24 \, \text{Å} \)

Substituting the values:
\[ 4.68 = 2(1.24) + 2R_{Cd^{2+}} \]
\[ 4.68 = 2.48 + 2R_{Cd^{2+}} \]
\[ 2R_{Cd^{2+}} = 4.68 - 2.48 = 2.20 \]
\[ R_{Cd^{2+}} = \frac{2.20}{2} = 1.10 \, \text{Å} \]

### Final Answer
The ionic radius of \( \text{Cd}^{2+} \) is \( 1.10 \, \text{Å} \).

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