Given below are two statements :
Statement I : Hyperconjugation is a permanent effect.
Statement II: Hyperconjgation is ethyl cation `(CH _(3) -overset(+) (C)H _(2))` involves the overlapping of `C _( sp ^(2)) - H _(1s)` bond with empty 2p orbial of other carbon.
Choose the correct option :

Options

A. Both statement I and statement II are false
B. Statement I is incorrect but statement II is true
C. Statement I is correct but statement II is false
D. Both Statement I and statement II are true.

Answer

To solve the question, we need to analyze the two statements given about hyperconjugation and the ethyl cation.

### Step-by-step Solution:

1. **Understanding Hyperconjugation**:
- Hyperconjugation is an effect that involves the delocalization of electrons in sigma bonds (C-H or C-C) with adjacent empty p-orbitals or π-bonds.
- It is considered a permanent effect because it contributes to the stability of carbocations and alkenes.

2. **Evaluating Statement I**:
- **Statement I**: "Hyperconjugation is a permanent effect."
- This statement is **True**. Hyperconjugation is indeed a permanent effect that stabilizes carbocations and other species through electron delocalization.

3. **Evaluating Statement II**:
- **Statement II**: "Hyperconjugation in ethyl cation `(CH3-CH2+)` involves the overlapping of `C_(sp^2) - H_(1s)` bond with the empty 2p orbital of the other carbon."
- In the ethyl cation, the carbon atom with the positive charge is `sp^2` hybridized. The adjacent carbon atom is `sp^3` hybridized.
- The overlap in hyperconjugation occurs between the `C-H` sigma bond of the `sp^3` carbon and the empty `p` orbital of the `sp^2` carbon, not the other way around as stated.
- Therefore, the statement is **False**. The overlap is not between `C_(sp^2) - H_(1s)` and the empty p-orbital of the other carbon, but rather between `C_(sp^3) - H_(1s)` and the empty p-orbital of the `sp^2` carbon.

4. **Conclusion**:
- Statement I is True.
- Statement II is False.
- Therefore, the correct option is that Statement I is true and Statement II is false.

### Final Answer:
- Statement I is True.
- Statement II is False.