Assertion: On mixing `500 mL` of `10^-6 M Ca^(2+)` ion and `500 mL` of `10^-6 MF^-` ion, no precipitate of `CaF_2` will be obtained. `K_(sp) (CaF_2 = 10^-18)`.
Reason: If `K_(sp)` is greater than ionic product, a precipitate
will develop.

Options

A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.
C. If Assertion is True but the Reason is False.
D. If both Assertion & Reason are False

Explanation

To solve the problem, we need to analyze the assertion and reason provided regarding the mixing of calcium ions (Ca²⁺) and fluoride ions (F⁻) and their potential to form a precipitate of calcium fluoride (CaF₂).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two solutions:
- 500 mL of \(10^{-6} \, M \, Ca^{2+}\)
- 500 mL of \(10^{-6} \, M \, F^{-}\)
- The solubility product (\(K_{sp}\)) of \(CaF_2\) is given as \(10^{-18}\).

2. **Calculating the Concentration After Mixing:**
- When we mix equal volumes (500 mL each), the total volume becomes:
\[ V_{total} = 500 \, mL + 500 \, mL = 1000 \, mL = 1 \, L \]
- The concentration of \(Ca^{2+}\) after mixing:
\[ [Ca^{2+}] = \frac{10^{-6} \, M \times 500 \, mL}{1000 \, mL} = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, M \]
- The concentration of \(F^{-}\) after mixing:
\[ [F^{-}] = \frac{10^{-6} \, M \times 500 \, mL}{1000 \, mL} = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, M \]

3. **Calculating the Ionic Product (IP):**
- The ionic product for the precipitation of \(CaF_2\) is given by:
\[ IP = [Ca^{2+}] \times [F^{-}]^2 \]
- Substituting the concentrations:
\[ IP = (5 \times 10^{-7}) \times (5 \times 10^{-7})^2 = (5 \times 10^{-7}) \times (25 \times 10^{-14}) = 125 \times 10^{-21} = 1.25 \times 10^{-19} \]

4. **Comparing Ionic Product with Solubility Product:**
- Now we compare the ionic product with the \(K_{sp}\):
\[ K_{sp} (CaF_2) = 10^{-18} \]
- Since \(IP = 1.25 \times 10^{-19} < K_{sp} = 10^{-18}\), no precipitate will form.

5. **Conclusion:**
- The assertion is true: no precipitate of \(CaF_2\) will be obtained.
- The reason is false: the correct statement should be that if the ionic product is less than the \(K_{sp}\), no precipitate will form.

### Final Answer:
- Assertion: True
- Reason: False