1. Statement I: Tropolone is an aromatic compound with a conjugated ring system containing 8 \(\pi\) electrons: - 6 \(\pi\) electrons are endocyclic (within the ring). - 2 \(\pi\) electrons from the \(>\text{C}=\text{O}\) group are exocyclic (outside the ring). Hence, tropolone satisfies the conditions of aromaticity. Therefore, Statement I is correct.
2. Statement II: The \(\pi\) electrons of the \(>\text{C}=\text{O}\) group are not involved in the aromaticity of the compound, as they are exocyclic and do not contribute to the conjugated system of the ring. Therefore, Statement II is incorrect. ### Diagram: \[ \text{Tropolone: Aromatic compound with 6 \(\pi\) endocyclic electrons} \] \[ \text{Structure:} \] \[ \text{OH} \quad \text{C}=\text{O} \]
Conclusion: Statement I is true, but Statement II is false.
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: