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. 241.5 pm
B. 165.7 pm
C. 275.1 pm
D. 322.5 pm

Answer

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

### Step 1: Understand the NaCl Structure
In the NaCl structure, the cations (in this case, \( X^+ \)) occupy octahedral voids in a face-centered cubic (FCC) lattice formed by the anions (in this case, \( Y^- \)).

### Step 2: Use the Relationship Between Cation and Anion Radii
For a compound with an NaCl structure, the ratio of the radius of the cation \( r_{X^+} \) to the radius of the anion \( r_{Y^-} \) is given by the inequality:
\[ 0.414 \leq \frac{r_{X^+}}{r_{Y^-}} \leq 0.732 \]
This means that we can express the radius of the anion in terms of the radius of the cation.

### Step 3: Calculate the Radius of the Anion
Given that the radius of the cation \( r_{X^+} = 100 \) pm, we can use the lower limit of the ratio to find the radius of the anion:
\[ \frac{r_{X^+}}{r_{Y^-}} = 0.414 \]
Rearranging this gives:
\[ r_{Y^-} = \frac{r_{X^+}}{0.414} \]

### Step 4: Substitute the Value of the Cation Radius
Now, substituting the value of \( r_{X^+} \):
\[ r_{Y^-} = \frac{100 \text{ pm}}{0.414} \approx 241.5 \text{ pm} \]

### Conclusion
Thus, the radius of the anion \( Y^- \) is approximately \( 241.5 \) pm.

### Final Answer
The radius of the anion \( Y^- \) is \( 241.5 \) pm.

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