Match List I with List II.
              List I                                                         List II.
A. Propanamine and N-Methylethanamine          I. Metamers
B. Hexan-2-one and Hexan-3-one                        II. Positional isomers
C. Ethanamide and Hydroxyethanimine             III. Functional isomers
D. o-nitrophenol and pnitrophenol                     IV. Tautomers
Choose the correct answer from the options given below :-

To identify the correct matches, examine the type of isomerism exhibited by each pair.

A. Propanamine and N-Methylethanamine

Propanamine: $$CH_3CH_2CH_2NH_2$$

N-Methylethanamine: $$CH_3CH_2NHCH_3$$

Both compounds have the same molecular formula but differ in the distribution of alkyl groups on either side of the polyvalent nitrogen atom. Therefore, they are metamers.

$$A \rightarrow I$$

B. Hexan-2-one and Hexan-3-one

Hexan-2-one: $$CH_3COCH_2CH_2CH_2CH_3$$

Hexan-3-one: $$CH_3CH_2COCH_2CH_2CH_3$$

The functional group remains the same, but its position changes from carbon-2 to carbon-3. Therefore, they are positional isomers.

$$B \rightarrow II$$

C. Ethanamide and Hydroxyethanimine

Ethanamide: $$CH_3CONH_2$$

Hydroxyethanimine: $$CH_3C(OH)=NH$$

These structures differ by migration of a proton and shifting of a double bond. Therefore, they are tautomers.

$$C \rightarrow IV$$

D. o-Nitrophenol and p-Nitrophenol

In these compounds, the nitro group occupies different positions on the benzene ring relative to the hydroxyl group.

The change is from ortho-position to para-position, so they are positional isomers.

$$D \rightarrow II$$

Hence, the correct matching is:

$$A \rightarrow I,\quad B \rightarrow II,\quad C \rightarrow IV,\quad D \rightarrow II$$

Therefore, the correct answer is

$$\boxed{A-I,;B-II,;C-IV,;D-II}$$