



The stability of carbocations depends on the resonance stabilization and inductive effects from substituents on the benzene ring. In this question, we are asked to identify the most stable carbocation among the following options.
- Option 1: \(\text{C}_6\text{H}_5\text{CH}_3^+ \): The methyl group \( \text{CH}_3 \) is an electron-donating group through inductive effects, which tends to destabilize carbocations.
- Option 2: \(\text{C}_6\text{H}_5\text{CH}_2^+ \): The benzylic carbocation is stabilized by resonance, as the positive charge can delocalize into the aromatic ring, making it more stable than an alkyl carbocation.
- Option 3: \(\text{C}_6\text{H}_4\text{CH}_3^+ \): Similar to option 1, the methyl group donates electrons inductively, which destabilizes the carbocation.
- Option 4: \(\text{C}_6\text{H}_5\text{OCH}_3^+ \): The oxygen atom in the methoxy group \( \text{OCH}_3 \) is highly electronegative and donates electron density via resonance into the ring, stabilizing the carbocation formed at the para-position. This makes this carbocation the most stable among the given options. Therefore, the most stable carbocation is \( \boxed{\text{C}_6\text{H}_5\text{OCH}_3^+} \).
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are


What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)