The edge length of unit cell of a metal having molecular weight 75 g/mol is 5Å which crystallises in cubic lattice. If the density is 2 g/c.c., then the radius of the metal atom in pm is

Options

A. 216
B. 320
C. 432
D. 108

Explanation

To find the radius of the metal atom in picometers (pm), we can follow these steps:

### Step 1: Calculate the number of atoms per unit cell (Z)
Since the metal crystallizes in a cubic lattice, we need to determine the type of cubic lattice. The most common types are:
- Simple Cubic (SC): Z = 1
- Body-Centered Cubic (BCC): Z = 2
- Face-Centered Cubic (FCC): Z = 4

For this problem, we will assume it is BCC, where Z = 2.

### Step 2: Use the formula for density
The density (D) of a substance can be calculated using the formula:
\[ D = \frac{Z \cdot M}{N_a \cdot V} \]
Where:
- \( D \) = density (g/cm³)
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass (g/mol)
- \( N_a \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) atoms/mol)
- \( V \) = volume of the unit cell (cm³)

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the cubic unit cell can be calculated using the edge length \( a \):
\[ V = a^3 \]
Given that the edge length \( a = 5 \) Å, we need to convert this to cm:
\[ 5 \text{ Å} = 5 \times 10^{-8} \text{ cm} \]
Now, calculate the volume:
\[ V = (5 \times 10^{-8} \text{ cm})^3 = 1.25 \times 10^{-22} \text{ cm}^3 \]

### Step 4: Rearrange the density formula to find Z
Substituting the known values into the density formula:
\[ 2 = \frac{2 \cdot 75}{6.022 \times 10^{23} \cdot 1.25 \times 10^{-22}} \]
Now, we can solve for Z, but since we already assumed Z = 2, we can proceed to the next step.

### Step 5: Relate the radius to the edge length
For a BCC lattice, the relationship between the radius \( r \) and the edge length \( a \) is given by:
\[ 4r = \sqrt{3}a \]
Rearranging gives:
\[ r = \frac{\sqrt{3}a}{4} \]

### Step 6: Substitute the edge length into the radius formula
Now substitute \( a = 5 \times 10^{-8} \text{ cm} \):
\[ r = \frac{\sqrt{3} \cdot (5 \times 10^{-8})}{4} \]
Calculating this gives:
\[ r = \frac{1.732 \cdot 5 \times 10^{-8}}{4} \]
\[ r \approx \frac{8.66 \times 10^{-8}}{4} \]
\[ r \approx 2.165 \times 10^{-8} \text{ cm} \]

### Step 7: Convert the radius to picometers
To convert centimeters to picometers:
\[ 1 \text{ cm} = 10^{10} \text{ pm} \]
Thus:
\[ r \approx 2.165 \times 10^{-8} \text{ cm} \times 10^{10} \text{ pm/cm} \]
\[ r \approx 216.5 \text{ pm} \]

### Final Answer
The radius of the metal atom is approximately **216.5 pm**.