A solid `XY` has `NaCl` structure. If radius of `X^(+)` is `100 p m`. What is the radius of `Y^(-)` ion ?

Options

A. 120 pm
B. (`136.6` to `241.6`) pm
C. 280 pm
D. Unpredictable

Answer

To solve the problem step by step, we will use the relationship between the radii of cations and anions in a solid with NaCl structure.

### Step-by-step Solution:

1. **Understanding the NaCl Structure**:
- In the NaCl structure, the ratio of the radius of the cation (R₊) to the radius of the anion (R₋) is given by the formula:
\[ \frac{R_{X^+}}{R_{Y^-}} = 0.414 \]
- Here, \( R_{X^+} \) is the radius of the cation \( X^+ \) and \( R_{Y^-} \) is the radius of the anion \( Y^- \).

2. **Given Data**:
- The radius of the cation \( X^+ \) is given as \( R_{X^+} = 100 \) picometers (pm).

3. **Setting Up the Equation**:
- We need to find the radius of the anion \( Y^- \) (denoted as \( R_{Y^-} \)).
- Using the ratio from the NaCl structure:
\[ \frac{R_{X^+}}{R_{Y^-}} = 0.414 \]
- Rearranging this gives us:
\[ R_{Y^-} = \frac{R_{X^+}}{0.414} \]

4. **Substituting the Values**:
- Substitute \( R_{X^+} = 100 \) pm into the equation:
\[ R_{Y^-} = \frac{100 \text{ pm}}{0.414} \]

5. **Calculating the Radius of \( Y^- \)**:
- Now, perform the calculation:
\[ R_{Y^-} \approx \frac{100}{0.414} \approx 241.5 \text{ pm} \]

6. **Conclusion**:
- The radius of the anion \( Y^- \) is approximately \( 241.5 \) picometers.

### Final Answer:
The radius of the \( Y^- \) ion is approximately \( 241.5 \) picometers.

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