The density of an ionic compounds `(Mw = 58.5)` is `2.165 kg m^(-3)` and the edge length of unit cell is `562` pm, then the closest distance between `A^(o+)B^(ɵ)` and `Z_(eff)` of unit cell is

Options

A. `281` pm, `4`
B. `562` pm, `2`
C. `562` pm, `4`
D. `281` pm, `2`

Explanation

To solve the problem, we need to find the closest distance between the ions \( A^{+}B^{-} \) and the effective nuclear charge \( Z_{\text{eff}} \) in the unit cell of the ionic compound. We have the following information:

- Molecular weight \( Mw = 58.5 \, \text{g/mol} \)
- Density \( \rho = 2.165 \, \text{kg/m}^3 \)
- Edge length of the unit cell \( a = 562 \, \text{pm} = 562 \times 10^{-12} \, \text{m} \)

### Step 1: Convert the molecular weight to kg/mol
Since the density is given in kg/m³, we should convert the molecular weight from grams to kilograms:
\[ Mw = 58.5 \, \text{g/mol} = 0.0585 \, \text{kg/mol} \]

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the unit cell can be calculated using the edge length \( a \):
\[ V = a^3 = (562 \times 10^{-12} \, \text{m})^3 \]
Calculating this gives:
\[ V = 1.778 \times 10^{-28} \, \text{m}^3 \]

### Step 3: Use the density formula to find \( Z_{\text{eff}} \)
The density formula relates density, molecular weight, volume, and Avogadro's number:
\[ \rho = \frac{Z_{\text{eff}} \times Mw}{V \times N_A} \]
Rearranging this formula to solve for \( Z_{\text{eff}} \):
\[ Z_{\text{eff}} = \frac{\rho \times V \times N_A}{Mw} \]
Substituting the known values:
- \( \rho = 2.165 \, \text{kg/m}^3 \)
- \( V = 1.778 \times 10^{-28} \, \text{m}^3 \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)
- \( Mw = 0.0585 \, \text{kg/mol} \)

Calculating \( Z_{\text{eff}} \):
\[ Z_{\text{eff}} = \frac{2.165 \times 1.778 \times 10^{-28} \times 6.022 \times 10^{23}}{0.0585} \]
Calculating this gives:
\[ Z_{\text{eff}} \approx 4 \]

### Step 4: Determine the closest distance between ions
For an FCC (Face-Centered Cubic) structure, the closest distance between the cation \( A^{+} \) and anion \( B^{-} \) is given by:
\[ \text{Distance} = \frac{a}{2} \]
Substituting the edge length:
\[ \text{Distance} = \frac{562 \, \text{pm}}{2} = 281 \, \text{pm} \]

### Final Answer
The closest distance between \( A^{+}B^{-} \) and \( Z_{\text{eff}} \) of the unit cell is:
\[ \text{Distance} = 281 \, \text{pm}, \quad Z_{\text{eff}} = 4 \]