Assertion: `sp^(3)` hybrid carbon atom is more electronegative than sp hybrid carbon atom.
Reason: `sp^(3)` hybrid orbitals are more closer to the nucleus.

Options

A. If both assertion and reason are true and reason is the correct explanation of assertion
B. If both assertion and reason are true but reason is not the correct explanation of assertion
C. If assertion is true but reason is false.
D. If both assertion and reason are false.

Explanation

To solve the question, we need to analyze both the assertion and the reason provided.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states that an `sp^3` hybrid carbon atom is more electronegative than an `sp` hybrid carbon atom.
- Electronegativity is a measure of an atom's ability to attract and hold onto electrons.

2. **Understanding Hybridization**:
- In `sp^3` hybridization, there is 25% `s` character and 75% `p` character.
- In `sp` hybridization, there is 50% `s` character and 50% `p` character.

3. **Electronegativity and `s` Character**:
- The electronegativity of an atom is directly related to the percentage of `s` character in its hybrid orbitals.
- The more `s` character an orbital has, the closer it is to the nucleus, and thus the more electronegative the atom tends to be.

4. **Comparing Electronegativity**:
- For `sp^3` hybridized carbon (25% `s` character), it is less electronegative compared to `sp` hybridized carbon (50% `s` character).
- Therefore, the assertion is **false** because `sp` hybridized carbon is more electronegative than `sp^3` hybridized carbon.

5. **Understanding the Reason**:
- The reason states that `sp^3` hybrid orbitals are closer to the nucleus.
- However, since `sp` hybrid orbitals have more `s` character (50% vs. 25%), they are actually closer to the nucleus than `sp^3` hybrid orbitals.

6. **Conclusion**:
- Both the assertion and the reason are false. The correct relationship is that `sp` hybrid carbon is more electronegative than `sp^3` hybrid carbon, and `sp` hybrid orbitals are indeed closer to the nucleus than `sp^3` hybrid orbitals.

### Final Answer:
Both the assertion and the reason are false.