A compound XY crystallizes in BCC lattice with unit cell - edge length of 480 pm , if the radius of Y = is 225 pm , then the radius of X is

Options

A. `95.34` pm
B. `225` pm
C. `127.5` pm
D. `190.7` pm

Answer

To find the radius of X in the compound XY that crystallizes in a BCC lattice, we can follow these steps:

### Step 1: Understand the BCC Structure
In a Body-Centered Cubic (BCC) lattice, there are two types of atoms: one at the center of the cube and one at the corners. The relationship between the radii of the two different atoms (X and Y) can be expressed in terms of the unit cell edge length (a).

### Step 2: Use the BCC Relationship
The relationship for the radii of the atoms in a BCC lattice is given by:
\[ R_X + R_Y = \frac{\sqrt{3}}{2} a \]
where:
- \( R_X \) is the radius of atom X,
- \( R_Y \) is the radius of atom Y,
- \( a \) is the edge length of the unit cell.

### Step 3: Substitute the Known Values
Given:
- Edge length \( a = 480 \, \text{pm} \)
- Radius of Y \( R_Y = 225 \, \text{pm} \)

Substituting these values into the equation:
\[ R_X + 225 = \frac{\sqrt{3}}{2} \times 480 \]

### Step 4: Calculate the Right Side
Calculate \( \frac{\sqrt{3}}{2} \times 480 \):
\[ \frac{\sqrt{3}}{2} \approx 0.866 \]
\[ \frac{\sqrt{3}}{2} \times 480 \approx 0.866 \times 480 \approx 415.68 \, \text{pm} \]

### Step 5: Solve for \( R_X \)
Now, we can solve for \( R_X \):
\[ R_X + 225 = 415.68 \]
\[ R_X = 415.68 - 225 \]
\[ R_X = 190.68 \, \text{pm} \]

### Step 6: Round the Answer
Rounding \( R_X \) to three significant figures gives:
\[ R_X \approx 190.7 \, \text{pm} \]

### Final Answer
The radius of X is approximately **190.7 pm**.

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