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

Explanation

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

### Step-by-Step Solution:

1. **Understand the Structure of NaCl**: In the NaCl crystal structure, the chloride ions (Cl⁻) occupy the corners of the cube, and the sodium ions (Na⁺) occupy the octahedral voids. The edge length of the unit cell is given as 564 pm.

2. **Identify the Given Values**:
- Edge length of NaCl unit cell (A) = 564 pm
- Radius of Cl⁻ ion (Rₗ) = 181 pm

3. **Relationship Between Edge Length and Ionic Radii**:
The relationship between the edge length (A) of the NaCl unit cell and the ionic radii can be expressed as:
\[ A = R_{Cl^-} + 2R_{Na^+} + R_{Cl^-} \]
This simplifies to:
\[ A = 2R_{Cl^-} + 2R_{Na^+} \]
or
\[ A = 2(R_{Cl^-} + R_{Na^+}) \]

4. **Substituting the Known Values**:
Substitute the known values into the equation:
\[ 564 = 2(181 + R_{Na^+}) \]

5. **Solving for Rₙₐ⁺**:
Divide both sides by 2:
\[ 282 = 181 + R_{Na^+} \]
Now, isolate Rₙₐ⁺:
\[ R_{Na^+} = 282 - 181 \]
\[ R_{Na^+} = 101 \text{ pm} \]

6. **Conclusion**:
The size of the sodium ion (Na⁺) is 101 pm.

### Final Answer:
The size of the Na⁺ ion is **101 pm**.

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