If radius of a metal atom (A) is 5pm and radius of an electronegative atom (B) is 20 pm, then in the unit cell

Options

A. A in octahedral voids, B in FCC unit
B. A in FCC unit, B in tetrahedral void
C. A in BCC unit, B in cubic void
D. A in tetrahedral void, B in FCC unit

Answer

To solve the problem, we need to analyze the given information about the radii of the metal atom (A) and the electronegative atom (B) and determine their positions in the unit cell based on the radius ratio and coordination numbers.

### Step-by-Step Solution:

1. **Identify the Radii of Atoms**:
- The radius of metal atom (A) = 5 pm
- The radius of electronegative atom (B) = 20 pm

2. **Determine the Type of Atoms**:
- Metal atom (A) is typically a cation (positively charged).
- Electronegative atom (B) is typically an anion (negatively charged).

3. **Calculate the Radius Ratio**:
- The radius ratio (r+) = radius of cation (A) / radius of anion (B) = 5 pm / 20 pm = 0.25.

4. **Apply the Radius Ratio Rule**:
- According to the radius ratio rule, we can determine the coordination number based on the calculated ratio:
- 0.155 < r+ < 0.225 → Coordination number = 3
- 0.225 < r+ < 0.414 → Coordination number = 4
- 0.414 < r+ < 0.732 → Coordination number = 6
- 0.732 < r+ < 1 → Coordination number = 8
- Since our radius ratio (0.25) falls in the range of 0.225 to 0.414, the coordination number is 4.

5. **Determine the Type of Void**:
- A coordination number of 4 corresponds to a tetrahedral void in the crystal structure.

6. **Identify Positions in the Unit Cell**:
- The smaller atom (A, radius = 5 pm) will occupy the tetrahedral void.
- The larger atom (B, radius = 20 pm) will be present in the face-centered cubic (FCC) unit cell.

7. **Final Conclusion**:
- Atom A (metal) is in the tetrahedral void, and atom B (electronegative) is in the FCC unit cell.

### Final Answer:
Atom A is in the tetrahedral void, and atom B is in the FCC unit cell.