Sodium (Na =23) crystallizen in bcc arrangement with the interfacial separation between the atoms at the edge `53.6 "pm"`. The density of sodium crystal is:

Options

A. `2.07 g//"cc"`
B. `2.46 g//"cc"`
C. `1.19 g // "cc"`
D. None of these

Answer

To find the density of sodium (Na) crystallizing in a body-centered cubic (BCC) arrangement, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Atomic weight of sodium (Na) = 23 g/mol
- Interfacial separation (distance between atoms) = 53.6 pm (picometers)
- The BCC structure has 2 atoms per unit cell (Z = 2).

2. **Convert Interfacial Separation to Meters:**
- 1 picometer (pm) = \(10^{-12}\) meters.
- Therefore, \(53.6 \, \text{pm} = 53.6 \times 10^{-12} \, \text{m}\).

3. **Relate Edge Length (a) and Atomic Radius (r):**
- In a BCC structure, the interfacial separation is given by:
\[ a - 2r = 53.6 \, \text{pm} \]
- The body diagonal in a BCC unit cell is given by:
\[ \sqrt{3}a = 4r \]
- From this, we can express \(r\) in terms of \(a\):
\[ r = \frac{\sqrt{3}}{4}a \]

4. **Substitute \(r\) into the Interfacial Separation Equation:**
- Substitute \(r\) into the interfacial separation equation:
\[ a - 2\left(\frac{\sqrt{3}}{4}a\right) = 53.6 \, \text{pm} \]
- Simplifying gives:
\[ a - \frac{\sqrt{3}}{2}a = 53.6 \, \text{pm} \]
- This can be rearranged to:
\[ a\left(1 - \frac{\sqrt{3}}{2}\right) = 53.6 \, \text{pm} \]

5. **Calculate Edge Length (a):**
- Calculate the numerical value:
\[ a = \frac{53.6 \, \text{pm}}{1 - \frac{\sqrt{3}}{2}} \approx 400 \, \text{pm} \]

6. **Convert Edge Length to Centimeters:**
- Convert \(a\) from picometers to centimeters:
\[ a = 400 \, \text{pm} = 400 \times 10^{-12} \, \text{m} = 4 \times 10^{-8} \, \text{cm} \]

7. **Calculate Density Using the Density Formula:**
- The formula for density (\(d\)) is:
\[ d = \frac{Z \cdot m}{N_a \cdot a^3} \]
- Where:
- \(Z = 2\) (number of atoms per unit cell for BCC)
- \(m = 23 \, \text{g/mol}\)
- \(N_a = 6.022 \times 10^{23} \, \text{mol}^{-1}\) (Avogadro's number)
- \(a = 4 \times 10^{-8} \, \text{cm}\)

8. **Calculate Volume of the Unit Cell:**
- The volume \(V\) of the unit cell is:
\[ V = a^3 = (4 \times 10^{-8} \, \text{cm})^3 = 64 \times 10^{-24} \, \text{cm}^3 \]

9. **Substitute Values into the Density Formula:**
\[ d = \frac{2 \cdot 23 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 64 \times 10^{-24} \, \text{cm}^3} \]
- Calculate the density:
\[ d \approx 1.19 \, \text{g/cm}^3 \]

### Final Answer:
The density of sodium crystal is approximately **1.19 g/cm³**.