The reduction of permanganate anion (MnO$_4^-$) in alkaline medium involves the following process:
\[ \text{MnO}_4^- \, (\text{oxidation state of Mn} = +7) \rightarrow \text{Mn}^{4+} \, (\text{oxidation state of Mn} = +4) \] The reduction from Mn$^{7+}$ to Mn$^{4+}$ involves the gain of 3 electrons, as the change in oxidation number is from +7 to +4.
Therefore, the reduction of permanganate anion involves the gain of 3 electrons.
Given below are two statements:
Statement I: The conversion proceeds well in a less polar medium. \[ {CH}_3{CH}_2{CH}_2{CH}_2{Cl} \xrightarrow{{HO}^-} {CH}_3{CH}_2{CH}_2{CH}_2{OH} + {Cl}^- \] Statement II: The conversion proceeds well in a more polar medium. \[ {CH}_3{CH}_2{CH}_2{CH}_2{Cl} \xrightarrow{{R}_3{N}} {CH}_3{CH}_2{CH}_2{CH}_2{NH}_2 + {Cl}^- \] In the light of the above statements, choose the correct answer from the options given below:
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:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: