A solid compound `XY` has `NaCl` structure. If the radius of the cation is `100` pm, the radius of the anion `(Y^(-))` will be

Options

A. `275.1 "pm"`
B. `322.5 "pm"`
C. `241.5 "pm"`
D. `165.7 "pm"`

Explanation

To find the radius of the anion \( Y^{-} \) in the solid compound \( XY \) with a NaCl structure, we can follow these steps:

### Step 1: Understand the NaCl Structure
In the NaCl structure, the cations and anions are arranged in a face-centered cubic (FCC) lattice. The cations occupy the octahedral voids, while the anions form the FCC structure.

### Step 2: Use the Relationship Between Cation and Anion Radii
In an ionic solid with NaCl structure, the radius of the octahedral void (where the cation resides) is related to the radius of the anion. The formula for the radius of the octahedral void \( R_{oct} \) is given by:
\[ R_{oct} = 0.414 \times R_{anion} \]
where \( R_{anion} \) is the radius of the anion.

### Step 3: Given Data
We are given:
- Radius of the cation \( R_{cation} = 100 \) pm

### Step 4: Calculate the Radius of the Anion
Since the cation occupies the octahedral void, we can set the radius of the cation equal to the radius of the octahedral void:
\[ R_{cation} = R_{oct} \]
Thus, we can write:
\[ 100 \, \text{pm} = 0.414 \times R_{anion} \]

### Step 5: Solve for the Radius of the Anion
Rearranging the equation to find \( R_{anion} \):
\[ R_{anion} = \frac{100 \, \text{pm}}{0.414} \]
Calculating this gives:
\[ R_{anion} \approx 241.5 \, \text{pm} \]

### Conclusion
The radius of the anion \( Y^{-} \) is approximately \( 241.5 \) pm.

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