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 element X in the compound XY that crystallizes in a Body-Centered Cubic (BCC) lattice, we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC lattice, there are two types of atoms: one at the center and one at the corners of the cube. The relationship between the atomic radii and the edge length (a) of the unit cell is given by the formula:
\[ R = \frac{\sqrt{3}}{4} a \]
where R is the effective radius of the atoms in the BCC structure.

### Step 2: Set Up the Equation
For the compound XY, we can express the total radius in terms of the radii of X and Y:
\[ R_X + R_Y = \frac{\sqrt{3}}{2} a \]
where \( R_X \) is the radius of X and \( R_Y \) is the radius of Y.

### Step 3: Substitute Known Values
Given:
- Edge length \( a = 480 \) pm
- Radius of Y \( R_Y = 225 \) 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 rearrange the equation to solve for \( R_X \):
\[ R_X = 415.68 - 225 \]
\[ R_X \approx 190.68 \text{ pm} \]

### Step 6: Round to Significant Figures
Rounding \( 190.68 \) pm gives us approximately \( 190.7 \) pm.

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

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