Account for the following observations :
(i) `pK_(b)` for aniline is more than that for methylamine.
(ii) Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide.
(iii) Aniline does not undergo Friedel-Crafts reaction.

Explanation

(i) In aniline, the lone pair of electrons on the N-atom are delocalized over the benzene ring. As a result, electron density on the nitrogen decreases. In contrast, in `CH_(3)NH_(2)^(+)` I-effect of `CH_(3)` increases the electron density on the N-atom. Therefore, aniline is a weaker base than methylamine and hence its `pK_(b)` value is higher than that of methylamine.
(ii) Methylamine being more basic than water, accepts a proton from water liberating `OH^(-)` ions.

These `OH^(-)` ions combine with `Fe^(3+)` ions present in `H_(2)O` to form brown ppt. of hydrated ferric oxide.
`FeCl_(3) rarrFe^(3+) 3Cl^(-) 2Fe^(3+)+6OH^(-)rarrundersetunderset(("Brown ppt."))("Hydrated ferric oxide")(2Fe (OH)_(3))`.
(iii) Aniline being a Lewis base reacts with Lewis acid `AlCl_(3)` to form a salt
`underset("Lewis base")(C_(6)H_(5)NH_(2))+underset("Lewis acid")(AlCl_(3))rarrC_(6)H_(5)N^(+)H_(2)AlCl_(3)^(-)
`
As a result, N of aniline acquires +ve charge and hence it acts a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel-Crafts reactions.