Assertion (A): `CuCI` is more covalent than `NaCI`.
Reason (R ): `Na^(o+)` ion more polarising than `Cu^(o+)` ion.

Options

A. If both (A) and (R ) are correct and (R ) is the correct explanation of (A).
B. If both (A) and (R ) are correct, but (R ) is not the correct explanation of (A).
C. If (A) is correct, but (R ) is incorrect.
D. If (A) is incorrect, but (R ) is correct.

Explanation

To analyze the assertion and reason provided in the question, we will break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Assertion (A):**
- The assertion states that CuCl is more covalent than NaCl.
- Covalent character in ionic compounds is influenced by the polarizing power of the cation (the positively charged ion).

2. **Understanding the Reason (R):**
- The reason states that Na⁺ is more polarizing than Cu⁺.
- Polarizing power refers to the ability of a cation to distort the electron cloud of an anion (the negatively charged ion).

3. **Analyzing the Cations:**
- **Na⁺ (Sodium ion):** Sodium has a larger ionic radius compared to copper. This means that Na⁺ has a lower charge density.
- **Cu⁺ (Copper ion):** Copper has a smaller ionic radius compared to sodium, which results in a higher charge density.

4. **Comparing Polarizing Powers:**
- A higher charge density means that Cu⁺ has a greater polarizing power than Na⁺.
- Therefore, Cu⁺ can distort the electron cloud of Cl⁻ more effectively than Na⁺ can distort the electron cloud of Cl⁻.

5. **Conclusion on Assertion and Reason:**
- Since Cu⁺ has a higher polarizing power, it leads to a greater covalent character in CuCl compared to NaCl.
- Thus, the assertion (A) is true: CuCl is indeed more covalent than NaCl.
- However, the reason (R) is false: Na⁺ is not more polarizing than Cu⁺; Cu⁺ is more polarizing.

6. **Final Statement:**
- Therefore, the correct conclusion is that the assertion is true, but the reason is false.

### Summary:
- Assertion (A) is true: CuCl is more covalent than NaCl.
- Reason (R) is false: Na⁺ is not more polarizing than Cu⁺.