Match List-I with List-II with their boiling points. List-I Compound…

The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

The boiling point of a liquid depends on how strongly its molecules attract one another. Molecules with an O–H group form intermolecular hydrogen bonds, which need extra energy to break, so alcohols and water boil far higher than ethers of similar size. Ethers have no H on oxygen, so their molecules cannot hydrogen bond with one another, and they boil like alkanes of comparable molecular mass.

• (A) \( \ce{CH3CH2OH} \) — This matches (III) 351K. Ethanol molecules are hydrogen bonded through their –OH groups, so ethanol boils much higher than its isomer dimethyl ether.

• (B) \( \ce{CH3OCH3} \) — This matches (IV) 248K. Dimethyl ether has no O–H bond and a low molecular mass; only weak dipole and van der Waals forces act, so it is a gas at room temperature.

• (C) \( \ce{H2O} \) — This matches (I) 373K. Each water molecule has two H atoms and two lone pairs, so it forms an extensive network of hydrogen bonds; water boils at 100 °C.

• (D) \( \ce{CH3CH2OCH2CH3} \) — This matches (II) 348K, by elimination. Diethyl ether (ethoxyethane) cannot hydrogen bond with itself; its real boiling point is 307.6 K, close to that of n-pentane (309.1 K), so the printed 348 K does not fit it. It is the only value left once (A), (B) and (C) are matched.

Shortcut: Water boils at 373 K, so (C)-(I); this leaves Options (1) and (2). Ethanol boils at 351 K (78 °C), so (A)-(III), which is Option (1).

Note 1: It seems that there is a mistake in the original NTA paper. The boiling point of diethyl ether (ethoxyethane) is printed as 348K; its actual boiling point is 307.6 K.

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: