If the radius of `Cl^(1) ion 181"pm, and the radius of" `Na^+` ion is 101pm then the edge length of unit cel l is:

Options

A. 282pm
B. 285.71pm
C. 512pm
D. 564pm

Answer

To find the edge length of the unit cell for the NaCl structure given the radii of the chloride ion (Cl⁻) and sodium ion (Na⁺), we can follow these steps:

### Step 1: Understand the structure
NaCl crystallizes in a face-centered cubic (FCC) structure. In this structure, the chloride ions (Cl⁻) occupy the face centers and corners of the cube, while the sodium ions (Na⁺) occupy the octahedral sites, which are located at the center of the cube and at the edge centers.

### Step 2: Identify the radii
We are given:
- Radius of Cl⁻ ion (R⁻) = 181 pm
- Radius of Na⁺ ion (R⁺) = 101 pm

### Step 3: Determine the relationship between the ions and the edge length
In the NaCl structure, the chloride ions and sodium ions touch each other along the edge of the cube. The relationship can be expressed as:
\[ a = R⁻ + R⁺ + R⁻ + R⁺ \]
Where:
- \( a \) = edge length of the unit cell
- \( R⁻ \) = radius of Cl⁻ ion
- \( R⁺ \) = radius of Na⁺ ion

This simplifies to:
\[ a = 2R⁻ + 2R⁺ \]

### Step 4: Substitute the values
Now, substituting the values of the radii into the equation:
\[ a = 2(181 \, \text{pm}) + 2(101 \, \text{pm}) \]

### Step 5: Calculate the edge length
Calculating this gives:
\[ a = 2(181) + 2(101) \]
\[ a = 362 + 202 \]
\[ a = 564 \, \text{pm} \]

### Conclusion
Thus, the edge length of the unit cell is:
\[ \boxed{564 \, \text{pm}} \]

---