Two statements are given below :
Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series.
Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass.
Choose the most appropriate option :

Solution

Chemistry ma’am · 4 steps

Check each statement against real data: melting points zig-zag, and solubility falls as the chain grows.

1. Even beats odd. In an acid with an even number of carbons, the -COOH group and the terminal -CH₃ sit on opposite sides of the zig-zag chain. These molecules pack more tightly in the crystal, so more energy is needed to melt them.

   Acid	Carbon atoms	Melting point
   Acetic acid	2 (even)	17 °C
   Propanoic acid	3 (odd)	-21 °C
   Butanoic acid	4 (even)	-5 °C
   Pentanoic acid	5 (odd)	-34 °C
   Hexanoic acid	6 (even)	-3 °C

2. Statement I. Each even acid melts higher than the odd acids on either side of it. For example, butanoic acid (-5 °C) is above both propanoic acid (-21 °C) and pentanoic acid (-34 °C). So Statement I is correct.
3. Statement II. The -COOH group forms hydrogen bonds with water, but the alkyl chain is hydrophobic. The acids from C₁ to C₄ mix fully with water. As the chain gets longer, solubility falls, and higher acids are almost insoluble. So Statement II is correct.
4. So. Both statements are correct, so the answer is (a).

The idea to keep. Even-carbon acids pack better in the crystal, so they melt higher than the odd-carbon acids next to them. A longer alkyl chain means a less polar molecule, so it dissolves less in water.

Answer from the NTA final answer key. Solution by PrepWiser.