The correct order of increasing basicity in aqueous medium for

(I) \(NH_3\)   (II) \(CH_3CH_2NH_2\)     (III) \((CH_3CH_2)_2NH\)   (IV) \((CH_3CH_2)_3N\)

(A) (II) < (I) < (IV) < (III)

(B) (II) < (I) < (III) < (IV)

(C) (I) < (II) < (IV) < (III)

(D) (I) > (II) > (IV) > (III)

Choose the correct answer from the options given below:

(C)

The correct answer is option 1. (C) i.e.,  (I) < (II) < (IV) < (III):

(I) \(NH_3\) has no alkyl groups attached to the nitrogen atom, so it has the highest basicity among the given compounds because it can readily donate its lone pair of electrons to a proton (H+) in the aqueous medium.

(II) \(CH_3CH_2NH_2\) has one alkyl group (ethyl group, \(CH_3CH_2-\)) attached to the nitrogen atom. The presence of one alkyl group reduces the availability of lone pairs on nitrogen for proton donation, making it less basic than \(NH_3\) but more basic than the compounds with more alkyl groups.

(IV) \((CH_3CH_2)_3N\) has three alkyl groups attached to the nitrogen atom. With three alkyl groups, it has the least availability of lone pairs on nitrogen for proton donation, making it the least basic among the given compounds.

(III) \((CH_3CH_2)_2NH\) has two alkyl groups attached to the nitrogen atom, which is in between \(CH_3CH_2NH_2\) and \((CH_3CH_2)_3N\) in terms of basicity.

So, the order of increasing basicity is (I) < (II) < (IV) < (III), which matches option (C).