`r_(Na^(+))= 195 `pm and `r_(Cl^(-)) =281 `pm in NaCl ( rock salt ) structure. What is the shortest distance between `Na^(+)` ions ?

Options

A. 778.3 pm
B. 673 .06 pm
C. 195.7pm
D. 390.3 pm

Answer

To find the shortest distance between \( \text{Na}^+ \) ions in NaCl (rock salt) structure, we can follow these steps:

### Step 1: Understand the Structure
NaCl crystallizes in a face-centered cubic (FCC) lattice structure where \( \text{Cl}^- \) ions occupy the corners and face centers of the cube, while \( \text{Na}^+ \) ions occupy the octahedral voids, which are located at the body center and edge centers of the cube.

### Step 2: Define the Radii
Given:
- Radius of \( \text{Na}^+ \) ion, \( r_{\text{Na}^+} = 195 \, \text{pm} \)
- Radius of \( \text{Cl}^- \) ion, \( r_{\text{Cl}^-} = 281 \, \text{pm} \)

### Step 3: Calculate the Edge Length of the Unit Cell
In the NaCl structure, the edge length \( a \) of the unit cell can be calculated using the relationship between the radii of the ions. The edge length \( a \) is given by:

\[ a = 2 \times r_{\text{Na}^+} + 2 \times r_{\text{Cl}^-} \]

Substituting the values:

\[ a = 2 \times 195 \, \text{pm} + 2 \times 281 \, \text{pm} \]
\[ a = 390 \, \text{pm} + 562 \, \text{pm} = 952 \, \text{pm} \]

### Step 4: Calculate the Shortest Distance Between \( \text{Na}^+ \) Ions
The shortest distance between two \( \text{Na}^+ \) ions can be found by considering the geometry of the unit cell. The shortest distance between two \( \text{Na}^+ \) ions located at the edge centers is given by:

\[ d = \frac{a}{\sqrt{2}} \]

Substituting the value of \( a \):

\[ d = \frac{952 \, \text{pm}}{\sqrt{2}} \approx \frac{952 \, \text{pm}}{1.414} \approx 673.26 \, \text{pm} \]

### Conclusion
The shortest distance between \( \text{Na}^+ \) ions in NaCl is approximately \( 673.26 \, \text{pm} \).

---