If the radius of a Chloride ion is 0.154 nm, then what is the maximum size of a cation that can fit in each of its octahedral voids?

Options

A. 1.15 x `10^(-1)` nm
B. 1.21 x `10^(-1)` nm
C. 1.18 x `10^(-1)` nm
D. 1.13 x `10^(-1)` nm

Answer

To find the maximum size of a cation that can fit in the octahedral voids created by chloride ions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given data:**
- The radius of the chloride ion (R⁻) is given as 0.154 nm.

2. **Understand the relationship between the radii:**
- For octahedral voids, the relationship between the radius of the cation (R⁺) and the radius of the anion (R⁻) is given by the formula:
\[ \frac{R^+}{R^-} = 0.732 \]

3. **Rearrange the formula to find R⁺:**
- We can express the radius of the cation (R⁺) in terms of the radius of the anion (R⁻):
\[ R^+ = 0.732 \times R^- \]

4. **Substitute the value of R⁻ into the equation:**
- Now, substitute R⁻ = 0.154 nm into the equation:
\[ R^+ = 0.732 \times 0.154 \]

5. **Calculate R⁺:**
- Performing the multiplication:
\[ R^+ = 0.732 \times 0.154 = 0.112488 \text{ nm} \]

6. **Convert the result to a more standard form:**
- This can also be expressed in scientific notation:
\[ R^+ = 1.12488 \times 10^{-1} \text{ nm} \approx 0.1125 \text{ nm} \]

7. **Final answer:**
- The maximum size of the cation that can fit in each of the octahedral voids is approximately **0.1125 nm**.