KCl crystallizes int the same type of lattic as done NaCl . Given that `r_(Na^(+))//r_(Cl^(-)) = 0.50` and `r_(Na^(+))//r_(K^(+)) = 0.70` , Calcualte the ratio of the side of the unit cell for KCl to that for NaCl:

Options

A. `1.143`
B. `1.224`
C. `1.414`
D. `0.875`

Answer

To solve the problem, we need to calculate the ratio of the side of the unit cell for KCl to that for NaCl using the given ratios of ionic radii. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Crystal Structure
KCl crystallizes in the same type of lattice as NaCl, which is the rock salt structure. In this structure, the anions (Cl⁻) form a face-centered cubic (FCC) lattice, and the cations (K⁺ or Na⁺) occupy the octahedral voids.

### Step 2: Define the Ionic Radii
Given:
- \( r_{Na^+} / r_{Cl^-} = 0.50 \)
- \( r_{Na^+} / r_{K^+} = 0.70 \)

Let:
- \( r_{Na^+} = r \)
- \( r_{Cl^-} = 2r \) (from the first ratio)
- \( r_{K^+} = \frac{10}{7} r \) (from the second ratio)

### Step 3: Write the Edge Length Equations
For the rock salt structure, the edge length \( A \) of the unit cell can be expressed in terms of the ionic radii:
- For NaCl:
\[ A_{NaCl} = 2r_{Cl^-} + r_{Na^+} = 2(2r) + r = 4r + r = 5r \]

- For KCl:
\[ A_{KCl} = 2r_{Cl^-} + r_{K^+} = 2(2r) + \frac{10}{7}r = 4r + \frac{10}{7}r \]

### Step 4: Simplify the Expression for \( A_{KCl} \)
To combine the terms for \( A_{KCl} \):
\[ A_{KCl} = 4r + \frac{10}{7}r = \left(4 + \frac{10}{7}\right)r = \left(\frac{28}{7} + \frac{10}{7}\right)r = \frac{38}{7}r \]

### Step 5: Calculate the Ratio of the Edge Lengths
Now we can find the ratio of the edge lengths:
\[ \frac{A_{KCl}}{A_{NaCl}} = \frac{\frac{38}{7}r}{5r} = \frac{38}{7 \cdot 5} = \frac{38}{35} \]

### Step 6: Simplify the Ratio
The ratio simplifies to:
\[ \frac{A_{KCl}}{A_{NaCl}} = \frac{38}{35} \approx 1.086 \]

### Final Answer
Thus, the ratio of the side of the unit cell for KCl to that for NaCl is approximately \( 1.086 \).

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