The edge length of sodium chloride unit cell is 564 pm. If the size of `Cl^(-)` ion is 181 pm. The size of `Na^(+)` ion will be

Options

A. 101 pm
B. 181 pm
C. 410 pm
D. 202 pm

Answer

To find the size of the sodium ion (Na⁺) in a sodium chloride (NaCl) unit cell, we can follow these steps:

### Step 1: Understand the relationship between the edge length and ionic radii
In a face-centered cubic (FCC) structure like NaCl, the relationship between the edge length (A) of the unit cell and the ionic radii of the cation (Na⁺) and anion (Cl⁻) is given by:
\[ R_{Na^+} + R_{Cl^-} = \frac{A}{2} \]
where \( R_{Na^+} \) is the radius of the sodium ion and \( R_{Cl^-} \) is the radius of the chloride ion.

### Step 2: Substitute the known values
We know:
- The edge length of the NaCl unit cell, \( A = 564 \) pm
- The radius of the chloride ion, \( R_{Cl^-} = 181 \) pm

Now, we can substitute these values into the equation:
\[ R_{Na^+} + 181 \, \text{pm} = \frac{564 \, \text{pm}}{2} \]

### Step 3: Calculate \( \frac{A}{2} \)
Calculate \( \frac{A}{2} \):
\[ \frac{564 \, \text{pm}}{2} = 282 \, \text{pm} \]

### Step 4: Rearrange the equation to find \( R_{Na^+} \)
Rearranging the equation gives:
\[ R_{Na^+} = 282 \, \text{pm} - 181 \, \text{pm} \]

### Step 5: Perform the subtraction
Now, perform the subtraction:
\[ R_{Na^+} = 282 \, \text{pm} - 181 \, \text{pm} = 101 \, \text{pm} \]

### Final Answer
The size of the sodium ion \( R_{Na^+} \) is **101 pm**.

---