KCl crystallises in the same type of lattice as does NaCl Given that `r_(Na^(+))//r_(Cl^(-))=0.55` and `r_(K^(+))//r_(Cl^(-))=0.74`, the ratio of the side of unit cell for KCl to that of NaCl is

Options

A. `1.122`
B. `0.891`
C. `1.414`
D. `0.414`

Explanation

To solve the problem, we need to find the ratio of the side of the unit cell for KCl to that of NaCl based on the given radius ratios of the ions.

### Step-by-Step Solution:

1. **Identify the given ratios**:
- Radius ratio for NaCl: \( \frac{r_{Na^+}}{r_{Cl^-}} = 0.55 \)
- Radius ratio for KCl: \( \frac{r_{K^+}}{r_{Cl^-}} = 0.74 \)

2. **Express the radius ratios in terms of \( r_{Cl^-} \)**:
- From the NaCl ratio:
\[ r_{Na^+} = 0.55 \cdot r_{Cl^-} \]
- From the KCl ratio:
\[ r_{K^+} = 0.74 \cdot r_{Cl^-} \]

3. **Calculate the effective ionic radii for both salts**:
- For NaCl:
\[ r_{Na^+} + r_{Cl^-} = 0.55 \cdot r_{Cl^-} + r_{Cl^-} = (0.55 + 1) \cdot r_{Cl^-} = 1.55 \cdot r_{Cl^-} \]
- For KCl:
\[ r_{K^+} + r_{Cl^-} = 0.74 \cdot r_{Cl^-} + r_{Cl^-} = (0.74 + 1) \cdot r_{Cl^-} = 1.74 \cdot r_{Cl^-} \]

4. **Set up the ratio of the effective ionic radii**:
- The ratio of the effective ionic radii for KCl to NaCl is:
\[ \frac{r_{K^+} + r_{Cl^-}}{r_{Na^+} + r_{Cl^-}} = \frac{1.74 \cdot r_{Cl^-}}{1.55 \cdot r_{Cl^-}} = \frac{1.74}{1.55} \]

5. **Calculate the ratio**:
- Performing the division:
\[ \frac{1.74}{1.55} \approx 1.122 \]

6. **Conclusion**:
- The ratio of the side of the unit cell for KCl to that of NaCl is approximately \( 1.122 \).

### Final Answer:
The ratio of the side of the unit cell for KCl to that of NaCl is \( 1.122 \).