What will be value of `r_(Na^(+))+r_(CI^(-))` in NaCI crystal having edge length 'a'

Options

A. `(asqrt(3))/(2)`
B. `sqrt(2a)`
C. `(a)/(2)`
D. `(a)/(2sqrt2a)`

Answer

To find the value of \( r_{Na^+} + r_{Cl^-} \) in a NaCl crystal with edge length \( a \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of NaCl**:
- NaCl crystallizes in a face-centered cubic (FCC) lattice structure. In this structure, the chloride ions (Cl^-) are located at the corners and the face centers of the cube, while the sodium ions (Na^+) are located at the edge centers and the body center of the cube.

2. **Understanding the Edge Length**:
- The edge length of the cube is given as \( a \). In an FCC lattice, the ions touch each other along the face diagonal.

3. **Determine the Relationship Between Ion Radii and Edge Length**:
- In the FCC structure, the distance between the centers of the Na^+ and Cl^- ions along the edge of the cube can be expressed as:
\[ 2r_{Na^+} + 2r_{Cl^-} = a \]
- This equation arises because there are two Na^+ ions and two Cl^- ions along the edge length \( a \).

4. **Simplify the Equation**:
- Dividing the entire equation by 2 gives:
\[ r_{Na^+} + r_{Cl^-} = \frac{a}{2} \]

5. **Final Result**:
- Therefore, the value of \( r_{Na^+} + r_{Cl^-} \) in the NaCl crystal is:
\[ r_{Na^+} + r_{Cl^-} = \frac{a}{2} \]

### Conclusion:
The answer to the question is:
\[ r_{Na^+} + r_{Cl^-} = \frac{a}{2} \]