In which of the following second ion is more stable than the first:

Options

A. `underset("I")(CH_(2)=Coverset(Theta)H) and underset("II")(CH-=overset(Theta)C)`
B. `underset("I")(C_(6)H_(5)-Coverset(Theta)H_(2)) and underset("II")(CH_(2)=CH-Coverset(Theta)H_(2))`
C. `CH_(3)-Coverset(Theta)H-CH=CH_(2) and Coverset(Theta)H_(2)-CH=CH_(2)`
D. None

Answer

To determine in which of the given options the second ion is more stable than the first, we will analyze each option step by step.

### Step 1: Analyze the First Option
The first option is:
- **First Ion:** CH2=CH− (vinyl anion)
- **Second Ion:** CH≡C− (alkynyl anion)

**Hybridization:**
- The carbon in the vinyl anion (CH2=CH−) is sp² hybridized.
- The carbon in the alkynyl anion (CH≡C−) is sp hybridized.

**Electronegativity:**
- The sp hybridized carbon (alkynyl anion) has more s character (50% s character) compared to the sp² hybridized carbon (33% s character).
- Therefore, the sp hybridized carbon can stabilize the negative charge better due to its higher electronegativity.

**Conclusion for First Option:**
The second ion (CH≡C−) is more stable than the first ion (CH2=CH−).

### Step 2: Analyze the Second Option
The second option is:
- **First Ion:** C6H5CH2− (benzyl anion)
- **Second Ion:** CH=CH− (allylic anion)

**Resonance Stabilization:**
- The benzyl anion (C6H5CH2−) is highly resonance stabilized due to the ability of the phenyl group to delocalize the negative charge.
- The allylic anion (CH=CH−) has resonance as well, but it is less stable than the benzyl anion.

**Conclusion for Second Option:**
The first ion (C6H5CH2−) is more stable than the second ion (CH=CH−), so this option does not satisfy the condition.

### Step 3: Analyze the Third Option
The third option is:
- **First Ion:** CH3−CH−CH2= (allylic anion)
- **Second Ion:** CH2=CH− (another allylic anion)

**Resonance Stabilization:**
- The first ion (CH3−CH−CH2=) can participate in resonance, and the negative charge can be delocalized.
- The second ion (CH2=CH−) also has resonance but is less stable compared to the first ion.

**Conclusion for Third Option:**
The first ion (CH3−CH−CH2=) is more stable than the second ion (CH2=CH−), so this option does not satisfy the condition.

### Final Conclusion
From the analysis:
- The first option (CH2=CH− vs. CH≡C−) shows that the second ion is more stable than the first.
- The second option (C6H5CH2− vs. CH=CH−) shows that the first ion is more stable than the second.
- The third option (CH3−CH−CH2= vs. CH2=CH−) shows that the first ion is more stable than the second.

Thus, the only option where the second ion is more stable than the first is the **first option**.