



To determine which carbocation is the most stable, we should consider the factors that stabilize carbocations. Carbocation stability is generally enhanced by:
Let us analyze each option one by one:
Therefore, the second option is the correct answer as it represents a tertiary carbocation, offering maximum stability among the given options.
Given:
- The carbocation intermediate is stabilized by the following effects: - **Inductive Effect (+I)**: Electron-withdrawing groups stabilize the carbocation. - **Resonance Effect (+M)**: Groups with lone pairs can donate electron density through resonance to stabilize the carbocation. - **Hyperconjugation**: Delocalization of electrons from adjacent C-H bonds to the empty p-orbital of the carbocation center stabilizes the carbocation.
In **Option 2**, the carbocation is conjugated with a stronger +M group \( -OCH_3 \), which provides resonance stabilization to the positive charge on the carbocation. \[ \text{Resonance Effect:} \quad -OCH_3 \text{ provides electron density through resonance}. \] This leads to increased stabilization of the carbocation in **Option 2** compared to others.
The carbocation in **Option 2** is the most stable due to the resonance stabilization provided by the \( -OCH_3 \) group.
A molecule with the formula \( AX_4Y \) has all its elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among A, X, and Y. Elements X and Y have first and second highest electronegativity values respectively among all the known elements.
The shape of the molecule is:
Statement-1: \( \text{ClF}_3 \) has 3 possible structures.
Statement-2: \( \text{III} \) is the most stable structure due to least lone pair-bond pair (lp-bp) repulsion.

Which of the following options is correct?