The increasing order of \(\mathrm{pK}_{\mathrm{a}}\) values of the following alkylammonium ions would be \(\begin{array}{lll}\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+} & \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}{ }^{+} & \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{3}{ }^{+}\end{array}\) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}\) (A) (B) (C) (D) (a) \((\mathrm{D})<(\mathrm{A})<(\mathrm{C})<(\mathrm{B})\) (b) \((B)<(C)<(A)<(D)\) (c) \((A)<(B)<(D)<(C)\) (d) \((A)<(C)<(B)<(D)\)

The conjugate bases are formed by removing a proton (H+) from each of the given alkylammonium ions. The conjugate bases are: 1. CH3NH2 for CH3NH3(+) 2. (CH3)2NH for (CH3)2NH2(+) 3. CH3CH2NH2 for CH3CH2NH3(+) 4. (CH3)3N for (CH3)3NH(+)

We will now analyze the stability of the conjugate bases by considering the inductive effect of the alkyl groups. More stable conjugate bases correspond to stronger acids, so we are looking for a more stable conjugate base. The inductive effect of an alkyl group stabilizes the negative charge on the nitrogen atom. The more alkyl groups we have surrounding the nitrogen atom, the more stable the conjugate base will be. Considering this, we can see that (CH3)3N is the most stable conjugate base due to the presence of three alkyl groups. CH3CH2NH2 is more stable than CH3NH2 because CH3NH2 has only one alkyl group whereas CH3CH2NH2 has two alkyl groups at different positions. (CH3)2NH has two alkyl groups but at the same position, which reduces its stabilizing effect compared to CH3CH2NH2.

From the stability of the conjugate bases, we can derive the order of the pKa values because stronger acids (lower pKa) have more stable conjugate bases. The order of pKa values will be: (CH3)3NH(+) < CH3CH2NH3(+) < (CH3)2NH2(+) < CH3NH3(+)

Using the derived pKa order, we can find which option matches the correct order: (a) (D) < (A) < (C) < (B) (b) (B) < (C) < (A) < (D) (c) (A) < (B) < (D) < (C) (d) (A) < (C) < (B) < (D) Comparing the orders, we can see that the correct order is given by option (a). So, the answer to this question is \((\mathrm{D})<(\mathrm{A})<(\mathrm{C})<(\mathrm{B})\).