In the distance between `Na^(+)` and `CI^(-)` ions in sodium
chloride crystal is y pm , the length of the edge of the unit cell is

Options

A. 4y pm
B. `y//4` pm
C. `y//2`pm
D. 2y pm

Answer

To solve the problem, we need to determine the length of the edge of the unit cell of sodium chloride (NaCl) given the distance between the sodium ion (Na⁺) and the chloride ion (Cl⁻) is y picometers.

### Step-by-Step Solution:

1. **Understanding the Structure of NaCl**:
Sodium chloride crystallizes in a face-centered cubic (FCC) structure. In this structure, the Na⁺ ions occupy the face centers, and the Cl⁻ ions occupy the corners of the cube.

2. **Identifying Ionic Radii**:
In an ionic crystal, the distance between the cation (Na⁺) and anion (Cl⁻) is equal to the sum of their ionic radii. Therefore, if we denote the ionic radius of Na⁺ as R_Na and that of Cl⁻ as R_Cl, the distance between them can be expressed as:
\[ \text{Distance (y)} = R_{Na} + R_{Cl} \]

3. **Relating Edge Length to Ionic Radii**:
For an FCC lattice, the relationship between the edge length (a) of the unit cell and the ionic radii is given by:
\[ a = 2(R_{Na} + R_{Cl}) \]
This is because in the FCC structure, the face diagonal contains one Na⁺ ion and two Cl⁻ ions.

4. **Substituting the Given Distance**:
From the problem, we know that the distance between Na⁺ and Cl⁻ is given as y pm. Therefore, we can substitute this into the equation:
\[ a = 2y \]

5. **Final Result**:
Thus, the length of the edge of the unit cell (a) in sodium chloride is:
\[ a = 2y \text{ pm} \]

### Conclusion:
The length of the edge of the unit cell of sodium chloride is \(2y\) picometers.