UPSC NDA I 2023 General Ability Solved Paper

Concept:Boiling point of straight-chain alkanes increases with the number of carbon atoms due to stronger van der Waals forces.Explanation:Longer carbon chains have more electrons, leading to stronger London dispersion forces between molecules. This raises the boiling point as molecular size and surface area increase. List the hydrocarbons with their carbon count: Methane ($\mathrm{CH}_4$) has 1 carbon atom. Ethane ($\mathrm{C}_2\mathrm{H}_6$) has 2 carbon atoms. Propane ($\mathrm{C}_3\mathrm{H}_8$) has 3 carbon atoms. Butane ($\mathrm{C}_4\mathrm{H}_{10}$) has 4 carbon atoms. Pentane ($\mathrm{C}_5\mathrm{H}_{12}$) has 5 carbon atoms. Heptane ($\mathrm{C}_7\mathrm{H}_{16}$) has 7 carbon atoms. Octane ($\mathrm{C}_8\mathrm{H}_{18}$) has 8 carbon atoms. Now check each option for increasing carbon number: Option A: Methane (1), Butane (4), Propane (3), Heptane (7) — not in increasing order. Option B: Propane (3), Butane (4), Pentane (5), Octane (8) — carbon count steadily increases, so boiling points increase. Option C: Propane (3), Butane (4), Heptane (7), Methane (1) — Methane at the end breaks the order. Option D: Octane (8), Ethane (2), Methane (1), Propane (3) — decreasing order. Thus only option B shows the correct sequence.Answer:Option B: Propane, Butane, Pentane, Octane.