Based on solute-solvent interactions, arrange the following in order…

The correct answer is Option (2) → B, C, D, A

Solubility follows the rule “like dissolves like”. A solute dissolves well only when its attraction to the solvent is similar to the attractions it must break, both within the solute and within the solvent. n-Octane is non-polar, so its molecules attract each other and a solute only through weak London (dispersion) forces. On this basis:

• Ionic solute: strong ion–ion attractions, which a non-polar solvent cannot replace. Least soluble.

• Polar solute with hydrogen bonding: its molecules hold each other strongly, and that energy is not returned by the solvent. Poorly soluble.

• Polar solute without hydrogen bonding: only dipole–dipole attractions to overcome. Somewhat more soluble.

• Non-polar solute: London forces on both sides, just like the solvent. Most soluble.

So the order of increasing solubility in n-octane is:

    \( \displaystyle \text{ionic} \lt \text{polar, H-bonded} \lt \text{polar, not H-bonded} \lt \text{non-polar} \)

    \( \displaystyle \ce{KCl} \lt \ce{CH3OH} \lt \ce{CH3CN} \lt \ce{C6H12} \)

Matching the given compounds:

• B. Potassium chloride (KCl): ionic

• C. Methanol, \( \ce{CH3OH} \): polar, strongly hydrogen bonded

• D. Acetonitrile, \( \ce{CH3CN} \): polar, no hydrogen bonding

• A. Cyclohexane: non-polar hydrocarbon

Shortcut: KCl (B) is the least soluble and cyclohexane (A) the most, so only Options (2) and (4) remain. Hydrogen-bonded methanol is less soluble than acetonitrile, so C comes before D, which is Option (2).

NCERT (2026-27 edition):
Solubility of a Solid in a Liquid – “… It is observed that polar solutes dissolve in polar solvents and non polar solutes in non-polar solvents. In general, a solute dissolves in a solvent if the intermolecular interactions are similar in the two or we may say like dissolves like.”

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